Willis Eschenbach and I have been defending ourselves on Facebook against Joe Postma’s claims we have “flat Earth” beliefs about the radiative energy budget of the Earth. The guy is obviously passionate, as our discussion ended with expletive-laced insults hurled my way (I suspect Willis decided the discussion wasn’t worth the effort, and withdrew before the fireworks began).
Joe advertises himself as an astrophysicist who works at the University of Calgary. I don’t know his level of education, but his claims have considerable influence on others, which is why I am addressing them here. He has numerous writings and Youtube videos on the subject of Earth’s energy budget and greenhouse effect, and the supposed errors the climate research community has made. I get emails and comments on my blog from others who invoke his claims, and so he is difficult to ignore.
Here I want to address just one of his claims (repeated by others, and the basis of his accusation I am a flat-Earther), recently described here, regarding the value of solar flux at the top of the atmosphere that is found in many simplified diagrams of the Earth’s energy budget. I will use the same two graphics used in that article, one from Harvard and one from Penn State:
Joe’s claim (as far as I can tell) is that that the solar flux value (often quoted to be around 342 W/m2) is unrealistic because it is for a flat Earth. But as an astrophysicist, he should recognize the division by 4 (“Fs(1-A)/4” and “S/4”) in the upper-left portion of both figures, which takes the solar constant at the distance of the Earth from the sun (about 1,370 W/m2) and spreads it over the spherical shape of the Earth. Thus, the 342 W/m2 value represents a spherical (not flat) Earth.
Just because someone then draws a diagram using a flat surface representing the Earth doesn’t mean the calculation is for a “flat Earth”.
Next in that article, Joe’s (mistaken) value for the solar constant is then used to compute the resulting Earth-Sun distance implied by us silly climate scientists who believe the solar constant is 342.5 W/m2 (rather than the true value of 1,370 W/m2). He gets twice the true, known value of the Earth-Sun distance, simply because he used a solar flux that was off by a factor of 4.
Now, I find it hard to believe an actual astrophysicist could make such an elementary error. I can ignore Joe’s profane personal insults, but he ends up influencing many people, and then I have to deal with their questions individually. Sometimes it’s better if I can just point them to a blog post, which is why I wrote this.
UPDATE: (June 6, 2019): Joe Postma has posted a YouTube video rebutting my article. If you listen to him from 2:30 to 2:45, Joe refuses to accept that the S=1,370 W/m2 “solar constant” energy that is intercepted by the cross-sectional area of the Earth must then get spread out, over time, over the whole (top-of-atmosphere) surface area of Earth. [This why S gets divided by 4 in global average energy budget diagrams, it’s the difference between the area of a circle and the area of a sphere with the same radius.] I am at a loss for words how he can refuse to accept something that is so obviously true — it’s simple geometry. I stand by everything I have written here.
Thanks. It gets bloody aggravating
The main error of the so called energy budget is that it is a radiation calculation for only 1 second (!) in Watt/m^2 = Joule/sm^2. Joule is the unit for energy, and in one second the radiation from the sun hits only half of the earth, and not evenly all over, but 1362 W/m^2 at equator and 0 at the poles (during equinox), so the factor should rather be 2, instead og 4. The outgoing radiation will be from the whole sphere though. The rest of the calculation will carry this error, and so the energy budget will be erroneous. It is simply unphysical.
In addition to this error, the analytical US Standard Atmosphere principles and calculations supported by the empirical evidence of Nikolov & Zeller explains the global temperature, which is modulated by clouds as prof. Dr. Henrik Svensmark has proved.
The surface area of a hemisphere is 2πr2, which is twice the surface area of disc of the same radius, so the average insolation of the sunlit half of Earth’s surface is half of the 1362 W/m2 and the average over the entire surface is half of that or 340.5 W/m2. Of this, about 100 W/m2 is reflected by the atmosphere, clouds and Earth’s surface, and another 77 W/m2 is absorbed by the atmosphere, so only 163.5 W/m2 is absorbed by the surface. Even if you double this (= 327 W/m2 for the sunlit surface only), the average surface temperature of the sunlit half could not exceed 2.6⁰C even if the Earth stopped rotating. Even if all the solar radiation absorbed by the atmosphere (77 W/m2) was radiated to the surface of that hemisphere, the combined 404 W/m2 could not raise the average surface temperature beyond about 17.5⁰C. The average daytime temperature for the sunlit hemisphere is about 20.7⁰C, which means it would be emitting about 422 W/m2 assuming an emissivity of 1.0. So it has to be receiving more energy from somewhere, especially as it is not stationary but rotating into night, and that somewhere is the radiating atmosphere and clouds.
Averaging insolation across the entire globe in the simple ‘flat earth’ model does distort reality, because the sunlit surface is at least 12⁰C warmer on average than the shaded hemisphere and therefore radiates more energy (proportional to T4). Calculating each hemisphere separately adds 1.1 W/m2 to the total energy lost from the surface in the model, and dividing the surface into small grids to do the maths results in an extra 5-6 W/m2 being lost to space. In other words, doing the calculations correctly results in the need for even more atmospheric radiation to compensate for energy lost from a warmer surface by day. The simple ‘flat-earth’ model thus underestimates the greenhouse effect which Postma denies!
D. Weston Allen says:
“The simple flat-earth model thus underestimates the greenhouse effect which Postma denies!”
And that is why insolation and atmospheric pressure explain the global temperature of space objects with atmosphere in a better way, see Nikolov & Zeller
( https://www.omicsonline.org/open-access/New-Insights-on-the-Physical-Nature-of-the-Atmospheric-Greenhouse-Effect-Deduced-from-an-Empirical-Planetary-Temperature-Model.pdf )
Petter,
You’ve just explained the factor of 4 right there in your own words:
” not evenly all over, but 1362 W/m^2 at equator and 0 at the poles (during equinox), so the factor should rather be 2″
So that is one factor of 2, from the uneven distribution on the lit side.
The average square meter on the lit side is receiving 681 W.
while the incoming “radiation from the sun hits only half of the earth”
That is ANOTHER factor of 2 from only hitting half the Earth at a time.
Together, this makes 4.
Averaged over a 24 h period, the average square meter on Earth is receiving 340.5 W.
The radiation FLUX calculation is energy per second per m^2(J/sm^2 = W/m^2), and only half the earth is hit by solar radiation, averaged to
1/2 x 1360 J/sm^2 = 680 J/sm^2 as a projected disk.
The incoming ENERGY in 1 second is 680 J/m^2.
For an ENERGY budget, it is this incoming ENERGY at half of the earth that shall be compared to the outgoing ENERGY from the whole earth in this second.
It doesn’t matter if we multiplate both sides of the equation with 24x60x60 s (24 h).
Just to be clear then, the outgoing flux must be 340 W/m^2 for full Earth, to balance that 690 W/m^2 for 1/2 Earth?
arghh, 680 W/m^2
Dividing the solar constant by four gives a value for the average OUTPUT from Earth’s systems assuming the hypothesis of radiative balance is correct – it is NOT a suitable value for insolation in my humble opinion.
10 hours of 340 W/m2 (BB temp of ~5°C) input does not have the same thermodynamic effect as 1 hour of 3400 W/m2 (BB temp of ~257°C).
A sun capable of inducing only minus 18°C, and that is what all of the models explicitly claim, is not only an unrealistic starting point for any hypothesis it also denies reality. Even NASA claim the Moon, with no gases at all let alone greenhouse ones, has a “Blackbody” temperature 15°C higher than Earth’s as listed in their Planetary Fact Sheets.
And don’t cite albedo for this difference because the majority of Earth’s albedo is due to atmospheric effects – primarily clouds.
Roy comments “He gets twice the true, known value of the Earth-Sun distance, simply because he used a solar flux that was off by a factor of 4.”
I am afraid this is simply not true at all – it is the people who use the divide by four calculation who are effectively moving Earth from its current orbit to an orbit twice the distance from the Sun – not Postma.
Using the values from NASA’s Planetary Fact Sheets you can easily calculate the solar constant of 1361 W/m2 as listed on their site using the average orbit and the inverse square law.
To arrive at 340 W/m2 requires an orbit of twice the distance and Postma has repeatedly stated this – there is no error on his part – treating insolation as 340 W/m2 is simply incorrect in my humble opinion.
The factor of 4 is for the average flux hitting the average square meter of the Earth.
If you want to calculate the suns distance, thats a different calculation altogether, that doesnt require the solar constant to be divided by 4.
If Joe Postma wants to divide by 4 anyway, well, that’s just plain dumb.
Rosco, no that is not correct. The 340 W/m^2 is the average flux over the entire Earth during one full orbital cycle. Postma’s calculation is incorrect because he conflates this value with the solar constant.
In my opinion, yes. Energy in = energy out
Roy Spencer
I think it is good you challenge his education. I am not sure he has any science background. He allowed me to post on his blog a couple posts before he went lunatic with his cursing. He is not a rational person and you will not be able to rationally discuss ideas with him.
He is more like a cult leader of a few brainwashed followers who also lack the least bit of reasoning ability.
I think he loves this cult mentality and leading his flock around. He is also one who will not do an experiment. I think one of his cult followers posts often on your blog.
I did attempt to explain to him about his mistakes but you will not be able to reach him. I wonder if he will have a blog post devoted to you with a lot of highly negative words about you. He told me “God hates me”. Not a rational human.
Norm,
Oh come on, Norm. Postma has a MSc in Astrophyics. This is not some weak field like astronomy. He is employed by the University of Calgary, AB as a research scientist. He is also under contract with the Canadian Space Agency. He has numerous published scientific papers related to the astrophysics field.
You trying to explain his “mistakes” is laughable.
Postma is a really smart guy but his idea that one dimensional models don’t work is totally false.
Like so many academics he goes bonkers when anyone points out the errors in his analysis. Is he a genius or a lunatic? Some people are both.
Maybe I should mention that he banned me from his web site when ho could not defend his false equations.
Sorry, but all the credentials in the world can’t help him on this particular subject. He’s simply wrong and stubborn and abusive about it.
SketpicGoneWild
He demonstrates a complete and total lack of any understanding of heat transfer knowledge. I gave him verbatim textbook physics and he said I was wrong and banned me from his Looney Tunes blog. He is a deranged human with zero ability to tolerate anyone telling him he is wrong. I would suggest you steer clear of this person when seeking understanding of anything related to heat transfer. Open a textbook instead and read actual science. It is what I did. You will see how wrong Joe Postma actually is and if you tell him, he will wig out on you and ban you as well.
Best to let the ignorant sycophants praise him. I would not want to be a member of this club. His followers understand as much physics of heat transfer as he does.
He could be intelligent if he did not have this total inability to grasp he could be wrong about something, learn and grow. He blows up when you tell him he might not be getting something right.
“You trying to explain his ‘mistakes’ is laughable.”
Pfft.
There are in fact tens of thousands of scientists, like Dr. Roy, with better credentials than him, who totally disagree with him.
You want to take his word for it, go ahead, but you’d be better off learning the facts from an unbiased textbook.
Smart people make mistakes all of the time; even those with degrees in astrophysics. The arrogance of the mistake may be concerning, but I’m still willing to cut the guy some slack because I know I make mistakes from time to time as well.
How very humble of you.
For making this comment, Norman, I challenge your education, since a little research on your part would have revealed Postmas’s education and science background.
Robert,
I have also experienced JP saying completely wrong things about heat transfer, and then mercilessly insulting and banning people, such as me, who disagree. Even those who are professional, respected scientists, like Roy Spencer.
Seriously, why do you admire someone who feels the need to bully people? That only indicates a deep insecurity.
Nate, please stop trolling.
Doesn’t “joe schmoe” ever wonder why it doesn’t stay as hot at the north pole than it does at the equator? Willis was right, Dr S, it weren’t worth the time (especially yours)…
Yes, Joe by himself isn’t worth the time, but as my post says, Joe has influence on many others who then pepper me with questions, assert silly claims, and so it’s easier for me if I can just point them to this post than to endlessly repeat myself.
i would think that the “pepperers” aren’t worth the time of day either. (anybody who can’t figure it out all on their lonesome should be left to themselves, too)…
Fonzie .
Reasonable, patient, respectful comment or rebuttal is the way.
Moving to name calling or labels ( say…’pepperers’ or ‘denier’ or ‘basket of deplorables’) is revealing of ‘the pot calling the kettle black’. It doesn’t help matters and should be avoided.
“Reasonable, patient, respectful comment or rebuttal is the way.”
Yeah, but i’m THE FONZ (i gotta reputation to uphold… ☺️)
Roy, you wrote:
“Just because someone then draws a diagram using a flat surface representing the Earth doesn’t mean the calculation is for a ‘flat Earth’.”
Roy, you’re not getting it. It’s not just the appearance of a flat line that Joe is talking about — it’s the solar flux divided by four, placed onto this flat line.
One fourth of the solar flux is attributed to the visual appearance of a flat line that represents a flat plane. What else could this line/plane represent but a flat Earth? — it only has one fourth of the solar flux projected upon it too?
Look at the actual numerical figure associated with the graphic that leads to the only conclusion possible — the physics involved here with this picture representation is for a flat Earth.
What is the surface upon which the one-fourth-solar-power falls? First it is the disc intercept with Earth’s radius, which is a flat disc, which DOES properly represent the solar flux on HALF of a sphere. But then, this flat half of a sphere’s power, correctly represented by the flat disc intercept, is magically wrapped around the whole sphere — that’s where the absurdity comes in.
You either have to accept that solar flux exists at one fourth power on the dark side of Earth where zero solar power actually exists, or you have to accept that the intercept disc for the half sphere remains flat AND its power divided by four, which places it twice the distance from the sun. There is no other rational way to interpret this combination of numbers and pictures.
In my reality, solar power falls on a hemisphere, in real time, at the given power for the hemisphere. Solar power does NOT fall on the entire sphere, all the time, at one fourth power.
The division-by-four is how we figure the Earth’s output. The division by four is not the solar input.
Robert Kernodle
YOU: “The division-by-four is how we figure the Earth’s output. The division by four is not the solar input.”
The output must equal the input if you are to remain at a steady state temperature. If the input is greater than the output, the temperature goes up. If the input is less than the output the temperature goes down.
The division by 4 for solar output is okay if you want to average the solar input over all the Earth’s surface.
In reality you have much different amounts of solar energy reaching different regions (poles vs equator). When all the energy is added that reaches the Earth it is the same leaving the Earth even if you have great regional differences.
If you had a ball the size of Earth in space, uniformally heated with 70% of the total energy that reaches the Earth from the Sun, it would reach an equilibrium temperature of 255 K.
Even if we agree that output equals input, the input does not come in at the average, nor equally all over the sphere instantaneously. That’s the mistake in this division-by-four move.
The input comes in on an area that is 2pi(r)^2 and goes out on an area of 4pi(r)^2. The areas over which these two figures occur are different.
Even more, there is never a moment in time when all areas of the Earth are receiving an average flux to drive the many processes that determine temperature.
Temperature is not just a function of surface area and flux.
“Even more, there is never a moment in time when all areas of the Earth are receiving an average flux to drive the many processes that determine temperature.”
No one sensible is saying that it is.
“Temperature is not just a function of surface area and flux.”
Of course, no one thinks it is.
Averaging is clearly a simplification.
But it is still useful for studying energy balance over a longer time scales.
It is ordinary in science to make simplifications, if they are useful.
Nate, please stop trolling.
“Joes claim (as far as I can tell)…”
Thanks, Robert, for your clarification. It might be a good idea for us to figure out what exactly Joe’s claim is. (there’s no point in leading him into expletive lace sin… ☺️)
BTW, i always enjoyed that jaworowski piece that you did.
Fonzie- I didnt see the Jaworowski piece you mention. I was always interested in his views on ice coring because it struck a chord with my post grad work on marine cores and not to be believed analysis of the carbonate system in the interstial pore waters. Can you point me towrds this comment by Robert on Jaworoski plse? Thanks
https://hubpages.com/education/ICE-Core-CO2-Records-Ancient-Atmospheres-Or-Geophysical-Artifacts
Lauchie, i think the original date of this piece is 2012 (but it was updated in 2017). It’s a very good easy to read piece for the layman as i recall. Not too heady, just fills you in on the basics of Jaworowski’s work…
Many thanks Fonzie for that link. Some of the information I had read (all of the Jarowoski papers + the Beck 2007 work) and some I hadnt. Really appreciate you posting this link.
“Expletive lace sin” — (^_^) …
I think Joe has stated exactly what his claim is. I think it might be a good idea for us to figure out why PhDs can’t get it.
How does an average flux tell us what energy is being delivered at each latitude, and how the collective differences of energy delivered at each latitude collectively produce the circulation patterns and other processes that determine real-world, real-time temperatures?
‘How does an average flux tell us what energy is being delivered at each latitude, and how the collective differences of energy delivered at each latitude collectively produce the circulation patterns and other processes that determine real-world, real-time temperatures?’
It is not meant to tell us all that.
Thats what General Circulation models are for, and require big computers.
Nate, please stop trolling.
Remember, we are talking about the *average* flux received over the entire surface area for one full orbital cycle. Everybody already agrees that sunlight only falls on 1/2 of the Earth at any given time. Everybody also agrees (at least they should) that the flux on the lit side of Earth is NOT the same at every point. In fact, the only places that can even receive the true solar constant is between 23.5N and 23.5S latitude and the exact latitude where the flux is perpendicular changes throughout the year.
Also, in terms of the energy budget the input and output are generally balanced. If the in-out > 0 then there is a net positive energy imbalance and the Earth will gain heat. If in-out < 0 then there is a net negative energy imbalance and the Earth will lose heat.
bdgwx says, June 4, 2019 at 12:07 PM:
No, bdgwx. You got it completely turned on its head. To get it straight, there won’t simply be an ENERGY balance, but a HEAT imbalance [Q_in – Q_out], leading to a positive or negative NET HEAT [Q_net]. And the Earth will not, as a result, gain/lose HEAT [Q]. It will gain/lose INTERNAL ENERGY [U]. This topic apparently still confuses you. Heat [Q] is a net transfer of energy from a cooler to a warmer place due to the temperature difference between those two places. An increase/decrease in internal energy [U] is what results from such a transfer. It’s that simple … Yet you keep confusing the two.
In terms of the Earth (or the Earth’s surface) only HEAT [Q] heats or cools. More thermodynamically precise, only the NET HEAT [Q_net = Q_in – Q_out] is capable of heating/cooling the system in question (the Earth system as a whole, and/or one of its ‘subsystems’, like the surface), that is, raising or lowering its temperature [T]. “Back radiation”, for instance, does NOT (!!!) constitute net heat (or incoming heat, for that matter), and therefore by itself does absolutely NOTHING to change the temperature of Earth’s global surface. It’s only part of a (much) greater whole. And the whole is what matters. Looking at (and obsessing over) single variables like this in isolation tells you nothing about the total situation. It only reveals a POTENTIAL effect, not of an ACTUAL (as in ‘realised’) effect.
Once you gain a thorough understanding of this simple principle, you will realise how straightforward it is to debunk the idea of “an anthropogenically enhanced GHE” as a cause of ‘global warming’ (“AGW”) using just the relevant available observational data from the real Earth system.
I agree Kristian. I should have used “energy” in place of “heat”.
bdgwx: ‘If the in-out > 0 then there is a net positive energy imbalance and the Earth will gain heat.’
K: ‘To get it straight, there wont simply be an ENERGY balance…’
Everybody can understand what bdgwx is saying here. The net flow of heat is inward.
The rambling semantics lecture adds nothing new to it, except vague and confusing statements like:
“It only reveals a POTENTIAL effect, not of an ACTUAL (as in ‘realised’) effect.”
Somebody needs to give Kristian a swirly.
To be fair…I’m kind of stickler on using terminology as appropriately as possible and I didn’t really do that in my post. That’s my bad. But, yeah, most people should have been able to understand what I meant.
Nate says, June 4, 2019 at 2:21 PM:
Yes, so why not write that. Physics is also very much about precision in language.
It’s essential to get the semantics right, Nate, otherwise people end up speaking past each other. Like they do on most all discussions on this particular subject here on this blog. If you don’t get HEAT straight, then you end up believing and promoting notions that are physically meaningless, like “atmospheric back radiation” adding energy to the surface to make it warmer.
Only people who want to perpetuate such confusion will find my “rambling semantics lecture” annoying. The rest – apparently including bdgwx here – will get the point, appreciate it, and change their terminology usage.
You, on the other hand, are just a lost case not worth the time and effort.
‘Only people who want to perpetuate such confusion will find my ‘rambling semantics lecture’ annoying.’
Makes my point, K.
You believe is is always the reader’s fault if they don’t appreciate, agree with, or understand your ‘brilliant’ posts.
Nate, please stop trolling.
“But then, this flat half of a spheres power, correctly represented by the flat disc intercept, is magically wrapped around the whole sphere thats where the absurdity comes in.”
No magic at all, Robert. The Earth is spinning, and the dark side will be lit in 12 h.
The average each side recieves over a day is therefore HALF of what the sun delivers to your disk.
No different the if my home’s furnace, is on 50% of the time. Average power=1/2 of maximum
Robert Kernodle says:
The surface area of the sphere is four times the area of its shadow.
You get the same number by three dimensional integration of the projected area over half the sphere.
No, just the input, the output is over the surface area.
Yes, day/night must be considered for an exact calculation because of T^4.
Clearly, Svante you simply say “no” and add nothing further to justify it — like in a child argument where one says, “is not”, to which the other replies, “is too”.
My only response, then, is … “is too”. (^_^)
Sorry, get a better explanation from a professor of physics:
https://tinyurl.com/y3pnvmmb
Svante, please stop Svante-ing.
There is absolutely nothing “flat Earth” involved in the determination of the AVERAGE amount of irradiation per square meter per solar day. Look at the link I provide and go down to 2.6. The page carefully explains how they estimate the AVERAGE and how that estimate in turn can be used to predict REAL irradiance, and then see how the REAL measured irradiance at different locations on the globe was very close to the estimate. That estimate used SPHERICAL geometry not “flat Earth” geometry. https://www.itacanet.org/the-sun-as-a-source-of-energy/part-2-solar-energy-reaching-the-earths-surface/#2.1.-The-Solar-Constant
It’s a big problem. People like Al Gore who don’t really know what they are talking about but have a large following of devoted admirers who hang on their every misinformed word. It makes them feel important, raises their self esteem.
I’m pretty sure Al Gore is not dependent upon your approval or fandom.
David, please stop trolling.
You can not divide by 4, why would you, please explain ???
You say the averge is 1360 W/m^2. So lets say I dropped a bomb on the Earth where the Sun was shining for that day, I would get a hit or dent in the ground of 1360 W/m^2 in that area, So how and why could you divide by 4 ??? It does not make any sense ??? Or make it simple, you drop a 1 pound brick down and calculate the force hitting the ground, you dont divide by 4 ???
Yes I know its not observed everywhere, but where its observed, its putting out a power of 1360 W/m^2, so you cant divide this ???
I have a car traveling quarter a way around the Earth, my averge speed is 80MPH, you would not divide this by 4 ??? Or would you ??? It seems you would, but does that make sense to you this car speed being devude by 4 ???
Wayne
Roy. You’re talking past one another. Joe is saying the location and time specific solar isolation is critical. That it generates the climate. He understands where the 342 comes from. He’s saying it makes no sense for climate science to not fully explore the ‘real time’ (and he should add ‘real place’) paradigm. From the latter frame, the average emergy per unit area does seem like a massive abstraction. It is definitely weird that your next post’s really very interesting foray into exploring the real time idea is so novel, because its novelty suggests hardly anyone has actually tried to seriously explore the ‘real time and place’ physics, including surface composition (i.e. land types, ocean) and atmosphere (with and without the ‘bad’ ghg’s for example).
It seems fairly intuitive to think that the angle of solar incidence has a big influence on temperature (at our given location) throughout the day, and that earth surface and atmospheric features will store or slow down the loss of energy, raising temp as the day wears on until the solar incidence can no longer support rising temps and they start to fall. Yes, convection and all sorts of other neat things influence this as well. What I think Joe is not saying (but implying) is the idea that the ‘real time and place’ amount of absorbed energy of the sun can trigger other events, events that can’t be taken into consideration if the average value of 342 is treated as a constant.
When the solar constant is spread out evenly across latitudes, what impact does that have on energy budget calculations when cumulus cloud decreases between 20N 20S? Over the period of a year, latitudes between 20N 20S receive a higher value of that otherwise evenly spread out across latitudes solar constant than what would latitudes between 20S 40S for example. How does the energy budget cope with such a scenario?
The solar constant isn’t “spread out evenly”. The incidence angle between the sunlight and the Earth’s surface is taken into account, which includes variations with latitude, Earth’s tilt, the obliquity of our orbit around the sun, and time-of-day. Also, the regional variations in cloud cover and thus albedo are included. All of those are taken into account when computing the global- and yearly-average energy budget numbers for our climate system, which are then put on simplified diagrams that some people mistakenly believe represent a flat Earth.
Roy, you wrote:
“The solar constant isnt ‘spread out evenly’.”
Dividing the solar constant by four, then, does what? What, then, does this division physically mean, if not “spread out” over the entire Earth sphere all at once, all the time?
The incidence angle is quite completely accounted for in the calculus that simplifies how this power falls onto a hemisphere, which is the intercept disc with Earth’s radius.
The error occurs in spreading the amount that this intercept disc represents over a whole sphere, where some of it does not exist at all to begin with, and where some of it is missing from where it should be.
The solar constant is the flux perpendicular to the surface.
The cross sectional area of a sphere is 1/4 of its total area.
To get the average flux at TOA you can do it two different ways. You can either integrate the flux over the entire surface and for one full orbital period or you can simply take the solar constant and project it onto the cross sectional area of Earth. The later is the simplest since all you have to do is divide it by 4.
Hi bdgwx,
You seem to be doing it here as well, I mean dividing by 4 ??? As the cross sectional area of a sphere is 1/4 of its total area, this is the area you are taking for the solar constant. Then as the Sun is heating each one of these cross sectional areas, you must x by 4, not divided by 4 ???
Let’s go for one day 48 hour. The Suns heat goes all around the Earth. So let’s just say in one second, one quarter is heated up 1, in 6 hours this Sun heats this quarter up 21,600, this Sun goes onto the next quarter, and heats it up 21,600, and for the next 2 quarters, totaling 86,400. The next time around there is going to be heat left in the seas and earth, thus the total will be more.
“HOWEVER” do you get that you need to multiplayer not divide ??? Your way of thinking would be a static Earth that does not rotate, and just one quarter of the Earth is heated up, thus to get the averge temp. on Earth, we have to divide, by 4, but with a rotating Earth, we have to multiply by 4.
Wayne
It’s an average energy/heat budget, Robert. The Earth is a rotating sphere in space with the solar flux coming in from one side only. And so, to get the average input from the Sun (over the full globe, and over the full day/year), you have to divide the solar constant by four. Why is this so hard to grasp?
It is only in pretending to be able to derive actual temperatures (even average ones) directly from this average input/output budget that “Modern Climate Science” goes wrong …
https://okulaer.files.wordpress.com/2014/10/drivhuseffekten.png
(Flux values from Stephens et al., 2012.)
You seem to be doing it here as well, I mean dividing by 4 ??? As the cross sectional area of a sphere is 1/4 of its total area, this is the area you are taking for the solar constant. Then as the Sun is heating each one of these cross sectional areas, you must x by 4, not divided by 4 ???
Let’s go for one day 48 hour. The Suns heat goes all around the Earth. So let’s just say in one second, one quarter is heated up 1, in 6 hours this Sun heats this quarter up 21,600, this Sun goes onto the next quarter, and heats it up 21,600, and for the next 2 quarters, totaling 86,400. The next time around there is going to be heat left in the seas and earth, thus the total will be more.
“HOWEVER” do you get that you need to multiplayer not divide ??? Your way of thinking would be a static Earth that does not rotate, and just one quarter of the Earth is heated up, thus to get the averge temp. on Earth, we have to divide, by 4, but with a rotating Earth, we have to multiply by 4.
Wayne
Robert Kernodle says:
Dividing the solar constant by four, then, does what?
Averages.
It averages.
You can do all the integrals if you want, but the canonical geometrical argument is a lot simpler to understand.
Yes, I know, David A, it averages, and THAT’s the problem.
“Canonical geometric argument” is a nice phrase, but it does not address the problem. The problem is the misuse of the result of the “canonical geometric argument”.
David, please stop trolling.
GC is not me (gallopingcamel). I support Dr. Roy’s refutation of “GC”‘s statement.
Yep, I’ve also had Postma go Postal on me. Apparently for having the audacity to politely question his calculations and logic.
Not sure why his followers think his Dear Leader persona is an admirable quality.
Glad to see him exposed as a fraud. Thanks, Roy.
Nate, please stop trolling.
You seem to be doing it here as well, I mean dividing by 4 ??? As the cross sectional area of a sphere is 1/4 of its total area, this is the area you are taking for the solar constant. Then as the Sun is heating each one of these cross sectional areas, you must x by 4, not divided by 4 ???
Let’s go for one day 48 hour. The Suns heat goes all around the Earth. So let’s just say in one second, one quarter is heated up 1, in 6 hours this Sun heats this quarter up 21,600, this Sun goes onto the next quarter, and heats it up 21,600, and for the next 2 quarters, totaling 86,400. The next time around there is going to be heat left in the seas and earth, thus the total will be more.
“HOWEVER” do you get that you need to multiplayer not divide ??? Your way of thinking would be a static Earth that does not rotate, and just one quarter of the Earth is heated up, thus to get the averge temp. on Earth, we have to divide, by 4, but with a rotating Earth, we have to multiply by 4.
Wayne
Dr. Spencer,
The “342 W/m2 value represents a spherical (not flat) Earth” only if you flatten out the global surface area and average the solar input as if it were striking the that area constantly 24/7. Joe’s point is that doesn’t happen and therefore all the calculations based on dividing the 1,370 W/m2 by four lead to gross and most likely misleading conclusions.
I don’t approve of bad manners and inappropriate language, but I would rule in Joe’s favor on the value of his model over the flat earth model you regurgitated in your post above.
As always, love your work and this blog, from which today’s post will initiate some fervent discussions, I’m sure.
I would also have thought that flattening of the earth argument alludes to having a 4*pi*r^2 area being illuminated at 342W uniformly, thereby simplifying the dynamics too much.
But to think it be a good idea to use the symplified model to calculate a sun-earth distance is ridiculous and shows nothing.
Chic,
Yeah I’m not convinced that’s an accurate model either. I would agree that the Sun is so far away that at any given instant in time 50% of the planet is receiving sunlight and that at any given instant in time 50% of the planet is receiving almost no sunlight. Also, at any given instant in time one small point on the planet is receiving 100% direct sunlight and every place else is receiving less than 100% all the way to zero percent. Not an easy problem.
NO. No one assumes that the solar flux is spread out evenly across the Earth, or that the sun shines day and night! Why must people infer such silliness? If I measure 24 hours of solar flux at the surface of the Earth, and do a 24 hour average of that, it is the actual average amount of sunlight that my location received in 24 hours. It does NOT imply that the sun shines at night!
Are you saying that if you shine constant light on a given area 24/7 that the energy absorbed or other effects observed would be the same as the actual input simulating the diurnal situation?
I would like to see that data.
The Arctic and Antarctic vary seasonally continuous daylight through diurnal illumination to continuous darkness and back again. What does this do to local conditions?
Well, you can read all night long without artificial light in summer and sleep all day long in winter, I suppose, although I’ve never been to either place.
Yeah, now I want to look at this
Sorry, Roy, but I think that you might be confusing amount of sunlight with amount of daylight. Otherwise, your suggested average makes no sense to me.
Solar flux is watts per meter squared, which is joules per second per meter squared. It is a rate of flow. It is not an accumulation of anything.
The flux, or rate of power flow, would be the same over the twenty-four hours.
If I drove 60 miles per hour (a rate) for 24 hours, would dividing 60 by 24 make any sense?
‘If I drove 60 miles per hour (a rate) for 24 hours, would dividing 60 by 24 make any sense?’
Good analogy, Robert.
He is saying driving 60 mph for 12 h = 720 miles, then resting for 12 h, gives you the same result as driving 30 mph for 24 h straight.
How did he find the average 24 h rate of 30 mph?
He divided 720 mi by 24 h.
Nate, please stop trolling.
Chic,
‘the value of his model over the flat earth model.’
As Roy made abundantly clear: ‘Just because someone then draws a diagram using a flat surface representing the Earth doesn’t mean the calculation is for a ‘flat Earth’.’
‘dividing the 1,370 W/m2 by four lead to gross and most likely misleading conclusions.’
Such as? This is simply the spatial average over a day. Climate change happens over decades.
“He gets twice the true, known value of the Earth-Sun distance, simply because he used a solar flux that was off by a factor of 4.”
Now that is misleading, and intentionally so.
Nate,
Do I have to spell it out for you? There are many questions about what is causing climate change over whatever period of time you want to discuss. When first entering the debate over the degree to which humans influence the change, the simple flat earth diagrams were helpful for me to investigate the factors involved. I moved beyond the (IMO, misleading) simplified diagrams precisely because they ignore the factors Dr. Spencer outlined and obscure an in-depth understanding of the effects of CO2 on temperature.
If anything is misleading, it is the last Dr. Roy statement you quoted.
“He gets twice the true, known value of the Earth-Sun distance, simply because he used a solar flux that was off by a factor of 4.
Whats wrong with this quote?
Solar flux is subject to interpretation. Is it the absolute amount received at a specific distance on a defined surface exposed to the flux or is it the averaged amount per surface area taking into consideration the geometry and rotation of a sphere at that distance? I think there would be no argument about the calculation if all parties were constrained to the same definitions.
‘I think there would be no argument about the calculation if all parties were constrained to the same definitions.’
Yes that’s what science generally does, uses common definitions. The factor of 4 is generally used, and why it is used has been thoroughly explained many times in many papers and books (as you know).
The fact that Postma misunderstood it is entirely on him, and likely intentional, for trolling climate scientists.
Nate, please stop trolling.
You said “all the calculations based on dividing the 1370 W/m2 by four lead to gross and most likely misleading conclusions.” That is incorrect. Go to paragraph 2.6 in the link below. Look at the estimate the derive, and then look at the real world observations. They are a match, and that is science.
https://www.itacanet.org/the-sun-as-a-source-of-energy/part-2-solar-energy-reaching-the-earths-surface/#2.1.-The-Solar-Constant
Had a problem with him with regards to transfer of energy from a cool surface to a hot surface. What it boiled down to was how can he explain how a ROOM TEMPERATURE thermal imaging camera works.
Without the use of negative focussed dark rays!(got banned)
RT thermography relies on all temps above absolute zero adding to the energy hitting the sense element of the sensor (sitting at perhaps 30C. (the -20C imaging limit is set by the sensor noise compared to the low energy of the focused thermal radiation frpm the -20C object)
You are really stretching the intent of the division by 4. The incident solar flux at the top of atmosphere that hits the earth’s cross sectional area really does get reduced by a factor of 4 when spread over the area of the Earth. It doesnt get spread evenly, of course, but the factor of reduction is still 4. That’s simple geometry. Quit trying to turn it into something that was never intended. The fact Joe then used it to compute the earth-sun distance proves he thinks it’s the solar constant, which it is not.
Not sure to who you are replying, but I am not objecting to or promoting a factor of 4 calculation. I’m simply arguing that the effect of solar insolation at full sun averaged with no sun will not result in a realistic estimate of the amount of energy absorbed or the consequences thereof.
The *average* solar flux (when combined with albedo) does indeed determine the amount of total absorbed solar energy over time. It doesn’t matter how the instantaneous flux at some location (or even over the whole earth) varies — day versus night, with latitude, with the obliquity of the orbit, or tilt of the Earth’s rotational axis — all of those time-variations averaged together over 1 year into a single time-averaged absorbed solar flux (W/m2) at the surface, will result in the total solar energy absorbed when multiplied by the number of seconds in a year.
Yes, if you measure the absorbed energy and average over time as you describe, then you get the average absorbed energy ipso facto. However, if you apply the equivalent amount of radiant energy uniformly to an identically defined surface, you won’t necessarily get the same energy absorbed as from a cyclically applied source. The absorp-tion and dissipation rates will not average out the same. Think of the ocean as an exemplary surface.
I could be wrong, as I’m relying on intuition rather actual data.
[“The incident solar flux at the top of atmosphere that hits the earths cross sectional area really does get reduced by a factor of 4 when spread over the area of the Earth. It doesn’t get spread evenly, of course, but the factor of reduction is still 4. “]
“Spread over the area of the Earth” — What does this phrase mean, then, if not “spread evenly”? Is there some way that dividing by four could make such a spread UNeven?
[“Thats simple geometry.”]
In simple geometry, a division by four for a continuous sphere means a continuous, even resulting spread. Dividing by four is not a discontinuous operation. If I divide 4 square feet by 4, then I get 1 square foot, four times — no discontinuities, no differences from one square to the other, no different definitions of what a square foot is. A square foot is a square foot.
If I calculate the area of a sphere, using 4 pi r^2, then the resulting area is a continuous surface. Now if I project a plane onto this continuous surface, then this projected plane is also a continuous surface fused to the continuous spherical surface. Dividing the intercept disc (of Earth’s radius) by 4, then, is like projecting a continuous planar surface onto the continuous spherical surface, which is the same as saying “spread evenly”. A meter-squared is a meter squared.
[“Quit trying to turn it into something that was never intended.”]
What exactly, then, was intended? If science endorses the value of exactness, and this is supposed to be science, then why does this science exactly show what Joe has tried to explain.
[“The fact Joe then used it to compute the earth-sun distance proves he thinks its the solar constant, which it is not.”]
But that’s what the diagram represents — again, one fourth of the solar constant falling on the entire spherical surface area of the Earth, all the time.
There is no other solar constant in the diagram. If so, then where is it? Where is the sun in that diagram? How much power does the sun have in that diagram? ANSWER: The only sun in the diagram is the sun that has one fourth of the solar constant shining continuously, everywhere on Earth, all the time.
This is not the input of the earth. This is the divide-by-four OUTPUT of the Earth, amazingly positioned and represented visually as the real solar input.
Robert,
Equating input and output doesn’t make for a strong case, because the surfaces we are dealing with are not mirrors or the like. A variety of specific heats and heat transfer variables are involved.
First take input alone. Is the solar energy received at a specific angle equivalent to 342 W/m2 24 hours/day the same as 684 W/m2 in 12 hours?
Assuming it is, what is the likely output from a given material receiving the same input under either condition? My guess is it will not be the same and I believe this is at the heart of Joe’s argument.
Chick B,
You are getting at something that I am trying to get at — solar energy received at a specific angle (latitude), for a given number of hours has a certain real-time, real-impact effect. It would seem quite a complex problem to figure out all these specific latitudinal energies and exact real-time effects caused by the limiting energies of the specific latitudes.
There are no averages causing those effects. What causes those effects is the flux at that specific angle and duration of time for the flux to operate.
Ultimately, it’s about so much more than simple surface area and “average flux”, input-output arguments.
But baby steps. (^_^)
Well done Robert. I was going to debate here, but you have it all to hand. You wrote “But thats what the diagram represents again, one fourth of the solar constant falling on the entire spherical surface area of the Earth, all the time.”
Roy and Willis failed to answer several of my questions. What you have done is what Robert has told you, this is not the way physics works sorry. You can NOT hold a flame up that warms one quarter of the Earth, then dilute it and says its hitting the whole Earth, but that’s what your doing to say without the atmosphere the Earth would be just -18c ??? So your saying its just -18c, but cos of the atmosphere its 20% warmer, close to 30c is that right ??? So you claim the Sun is not warming but just the GHG’s are giving us all this heat ??? Ok answer me this please;
So on a summers day in England to Dubai, I am not feeling different temp. from the Sun, but it’s the GHG’s that are making for temp. from 25c to 45c ??? You know that’s not right. Its that area pointing to the Sun, and the Sun is the only one putting those temps out. Its like when the Sun Hits your nose, it gets the most heat. Your way or thinking makes the GHG’s heat up Dubia far more than England, how come this then ???
Wayne
“You can NOT hold a flame up that warms one quarter of the Earth, then dilute it and says its hitting the whole Earth”
You can do a fun experiment in your home. Take a flashlight and hold it perpendicular to the floor. It works better on a tile floor so that you can more easily estimate the surface area the light is covering. Now tilt the flashlight so that the light is hitting the floor at angle. The same amount of light now spreads out over a large surface area. So the average luminosity per square meter decreases.
The flame (Sun) shines on 1/2 of the Earthly sphere at any given time. The shadow cast by Earth has an area that is 1/2 the hemispherical area facing the flame (Sun). This is because solar flux has to spread out over 2x the area. And when considering the other 1/2 of Earth is unlit the total reduction factor is 4x.
“So on a summers day in England to Dubai, I am not feeling different temp. from the Sun”
Well, yes, you are actually “feeling” a different temperature from the Sun. At least you are measuring a different amount of solar radiation. This is because of the difference in latitude, but also because of the opaqueness of the atmosphere which can be (and usually) is different in different locations.
“Its that area pointing to the Sun, and the Sun is the only one putting those temps out. Its like when the Sun Hits your nose, it gets the most heat. Your way or thinking makes the GHGs heat up Dubia far more than England, how come this then ???”
There are lot of other factors in play here. The temperature at any given location is the net effect of all agents acting at that location. This includes but is not limited to warm air advection, cold air advection, solar flux, aerosol optical depths, upwelling longwave radiation, downwelling radiation, latent heat, sensible heat, cloudiness, convection, and the list goes on and on. All of these factors are highly variable both spatially and temporally. The temperature at any given location is a product of everything going on at that location. GHGs are not the reason why Dubai is warmer than England.
Yes I was making another point. If as some say without the GHG’s the Earth would be -18c, then the Sun would not be say 25c in Uk and 45c in Dubai, it would be the same in each Country ??? Yes. BUt, its not the GHG’s heating us up at all, but keeping us cooler, as we feel the different heats of different Countries its the Sun that is hearing us up.
Wayne
Without an atmosphere the number of factors affecting the temperature at any given location are reduced substantially. In fact, sans an atmosphere the temperature would be almost entirely modulated by solar radiation. But, solar radiation is different at different latitudes because the Earth is tilted on it’s axis. This is a bit counter-intuitive, but not only would there still be a difference in temperature between England and Dubai, but the difference could actually be even greater. The reason is that the atmosphere works to homogenize temperatures by advective processes. Consider the range of temperatures on the Moon which is essentially receives the same amount of energy per square meter as the Earth.
wayne rowley says:’s
If as some say without the GHGs the Earth would be -18c, then the Sun would not be say 25c in Uk and 45c in Dubai, it would be the same in each Country ???
Not at all in any way whatsoever.
Jeez.
Nor does anyone think the Earth’s average non-GHG temperature is -18 C.
David, please stop trolling.
‘There is no other solar constant in the diagram. If so, then where is it? Where is the sun in that diagram? How much power does the sun have in that diagram? ANSWER: The only sun in the diagram is the sun that has one fourth of the solar constant shining continuously, everywhere on Earth, all the time.’
Its a 1-D diagram Robert, based on a 1-D energy balance model. Everyone knows the world is 3D. Thats why the model is abstract, not meant to be taken as a literal picture of the Earth, but only capture the energy flows in and out.
The fact that you and Joe want to take it literally, when that is not its intent at all, is all your doing.
The fact that Postma, misunderstanding the factor of 4, then misuses it (likely on purpose) in a bogus calculation of the sun’s distance, is his doing. Is that sophistry?
‘but only capture the energy flows in and out.’
More precisely, the global average, daily energy flows in and out.
Nate, please stop trolling.
Roy is right, Robert.
This is Basic Science 101.
It may seem hard to believe, Robert, but most of the “regulars” on this blog are still in denial that even the Green Plate Effect is debunked! Let alone the real thing…
☺️
DREMT makes another sad attempt at baiting and trolling. That’s all he can offer humanity.
DREMT, I want to hear you way in on this. Dr. Spencer is trying to explain energy budgets. Please post something besides a troll.
As if these discussions haven’t all been had several times before.
I’m just laughing at all the creative ways to miss such an obvious and simple point.
‘As if these discussions haven’t all been had several times before.’
OMG!
If only this DREMT could explain that to the other DREMT who keeps bringing up the same two topics!
I’m just laughing at all the creative ways to miss such an obvious and simple point.
Roy, are you going to get back to your El Nino claims?
You have the situation exactly backwards — the MEIv2 trend is DOWN, not up.
This destroys your entire argument.
http://www.drroyspencer.com/2019/06/uah-global-temperature-update-for-may-2019-0-32-deg-c/#comment-355666
yes, I just saw the comments today, and admitted — in 3 places to make sure people noticed — that I was wrong in that assertion.
The point that Joe trys to make is that if a flat disk of twice the radius of Earth (exposed side area equal to
Earth’s surface area) were located in a fixed position, normal to theSun at 2 AU, it would receive a uniform 340
W/m^2, which, with and albedo of 0.3, is equivalent to the starting point for the surface temperature calculations you use to determine the claimed greenhouse effect of 33K.
If you don’t think a uniform 340 W/m^2 to the top of the atmosphere is like a “flat Earth”, what is it like?
Starting with the average fails to capture the power of the Sun, which is capable of heating the exposed half of Earth’s surface to an average temperature of about 303K and is capable of local surface heating to greater than 400K.
If you consider the storage of energy in matter of Earth’s oceans, atmosphere and land, heated by the Sun, the 33K “greenhouse effect” isn’t necessary.
Nobody is saying that the 340 W/m^2 figure is uniform. All that is being said is that it is the average. It’s no different than the 15C global mean temperature. Nobody is saying that 15C is uniform. What they say is that it is the average. And these simple 3 layer models are not meant to be used to analyze diurnal cycles or specific effects at specific locations at specific times. They are used to more easily communicate the different energy transfer process and the energy budget as a whole on a global scale and over 1 sidereal year.
And, no, Postma’s point as articulated in 3 different videos is that these simple 3 layer models are wrong because they look flat in the diagram and because he conflates the 1360 W/m^2 figure with 340 W/m^2 and uses the later to compute an orbital distance 2x further away than it really is to show how wrong the diagram is. The problem is that he totally misrepresents the diagram. The diagram only appears flat because they extend it left and right to make room for labeling otherwise there is no need for width and thus an appearance of flatness or curvature either way. And we know the model is for a spherical Earth because of the labels Fs(1-A)/4 and S/4. There are various other strawman arguments and inaccuracies in his videos which have been discussed in this thread already so there’s no need to rehash them all here.
The problem with starting with the 340 W/m^2 average is that no part of the surface ever gets heated to the temperature which it is actually heated to, by the Sun; leading people to incorrectly conclude the imaginary radiative greenhouse effect exists and is necessary to explain why the average surface temperature is ~33K higher than the average equilibrium temperature of a sphere with a 0.3 albedo at 1 AU.
Dr. Roy Spencer, PhD, did the calculation per hour and latitude here:
https://tinyurl.com/y5km5an4
The outcome was the same,
there is a greenhouse effect,
I’ll be darned.
Dr. Roy’s calculation does not include the massive atmosphere heated on the day side with its mass maintaining the lapse rate and keeping the oceans from releasing heat by increasing saturation temperature due to increased saturation pressure.
bdgwx, please stop trolling.
No, I’m not saying that. I didnt even imply it. Dont change the subject by erecting a strawman.
Thank you Dr. Spencer for trying to explain this to the posters. They prefer fighting strawmen.
Craig T, please stop trolling.
Dr. Roy, thanks for this. I keep getting the same nonsense. You hit the nail on the head when you said:
People have this bizarre idea that an average of a variable, say sunshine, somehow implies that the variable is constant and equal everywhere. But that’s not the case.
It’s just an average, folks.
Go outside as Dr. Roy describes. Measure the amount of sunlight every five minutes over a 24 hour period. You’ll get a number that goes from hundreds of watts per square meter (W/m2) during the day to zero at night.
Now, average them. I think we can all agree that what you get is the average sunlight that has fallen on that spot over 24 hours. AVERAGE.
But that average does NOT mean that the sun shines at night. It’s just an average, which is very useful for a variety of calculations.
The same is true for a global average. Let’s repeat Dr. Roy’s thought experiment around the planet. Suppose we could get thousands and thousands of people all over the globe to take the measurements just described. Everyone everywhere measures the sunlight every five minutes for 24 hours.
We take all of those measurements and we average them.
I think we can all agree that what you get is the average sunlight that has fallen on the globe over 24 hours. It will be on the order of 170 W/m2 or so.
But again, that does NOT imply that either the sun shines at night, nor that the earth is flat.
As before,it’s just an average. It doesn’t imply anything at all about the flatness of the planet.
There are lots of valid arguments that climate models and global energy balances have problems.
The false claim that climate scientists and climate models deal with the world as if it were flat is not one of those valid arguments …
Regards to all, and Dr. Roy, thanks for your blog. It’s always interesting and worthwhile.
w.
Willis, what exactly was Joe claiming? That the number should be 1370 W/M2 or less than 340 W/M2? (or something else?) Not getting exactly what his position was from Dr Roy’s post…
(thanks)…
He is conflating the average solar flux at Earth’s TOA with the solar constant. The former is about 340 W/m^2. The later is about 1360 W/m^2.
He is claiming that climate scientists think the Earth is 2x further away than it really is. He makes this strawman argument by computing the distance using the 340 W/m^2 value because he doesn’t understand that there is a difference between the two values and because he uses the wrong value.
And instead of considering that the mistakes were all on his part he arrogantly thinks he’s figured something out that no one else has.
Arrogance, yes. All climate deniers are arrogant, thinking they’ve hit upon something that has somehow escaped all the world’s experts over the last 100 years.
bdgwx, certain clowns have NOT figured it out.
You probably don’t want names, do you?
Youll get a number that goes from hundreds of watts per square meter (W/m2) during the day to zero at night.
Of course the Sun shines at night, it shines throughout the 24 hour day.
AT NIGHT ??? why would you mention night ??? The Sun does NOT have night, the Suns hits the Earth all over in 24 hours all the time. THIS is another area your going wrong. We are NOT measuring the area hit by the Sun when it get dark, you cant do this. We are measuring the total flex of the Sun in its 24 hour period. Night, why mention night ??? We are NOT measuring the average temp. of the Earth over one full day, then you would add in night. We are adding up the total flex hitting Earth in 24 hours.
But we don’t want averge do we ???Why are you looking at the average ??? Lets say I used 4000cal. as in energy, in 24 hours, running around the Earth. My averge cal. for 6 hours was 1000cal. BUT, more important, my overall, or total cal. = 4000.
AGAIN, why are you few here working the total flux into the averge flux ??? The Sun is constantly hitting the Earth with energy/heat. So its to total we need, not the average.
Wayne
This situation depends only on very basic geometry, Wayne.
Did you take 10th grade geometry?
Actually DA, it involves physics.
We know that leaves you out.
JD, which side of this argument are you on? You need to take a position and not just troll.
“Curious fact. If you light a candle outside in the daytime, the sun gets warmer …”
Willis Eschenbach
😂
Dr. Spencer is correct. 340 W/m^2 is the average flux received at TOA as measured by integrating the flux over the entire surface for one full orbital cycle. This provides a very convenient and simple way for estimating the constituent pieces of Earth’s energy budget. One convenient aspect of this is that you can easily switch between flux and energy quickly. For example, at 340 W/m^2 the total energy received in one year is thus 340 W-years/m^2. The reason why this is 1/4 the solar constant is because the solar constant is measured perpendicular to the surface so you have to compensate for incident angle or curvature of Earth. The easiest way to do this is to take the cross sectional area of Earth. The cross sectional area of a sphere is 1/4 of its total area.
340 W/m^2 is the average flux received at TOA as measured by integrating the flux over the entire surface. BUT that’s the point, you cant average one quarter of the Earth into an averge for the entire surface, we need the full flux received by just the one quarter, x that by 4, not divided by 4.
Look, if heat coming in to one quarter it 200, the averge for 4 quarters is 50. BUT, if 200 hit one quarter, 200 hits the second quarter, 200 hits the third quarter, and 200 hits the fourth quarter, GET IT NOW ???
Wayne
Yes you can. It’s easy. Here’s how it is done. The Earth receives 5.472e24 joules of energy at TOA over one full orbital cycle.
(5.472e24 j) / (365.24 * 24 * 3600 s) = 173e15 j/s
(173e15 j/s) / (510,000,000,000,000 m^2) = 340 W/m^2
also
(5.472e24 j) / (510,000,000,000,000 m^2) = 340 W-years/m^2
This matches up exactly with solar constant of 1360 W/m^2. Since the solar constant is perpendicular to the surface you must compensate for the curvature of Earth to get the average amount falling on a hemisphere. The area of a sphere is 4x the cross sectional area so the solar constant flux is spreading out over 4x the area. To get the average flux over the entire surface area of Earth you must divide the solar constant by 4. One final point…the intersection of the Earthly sphere with the plane of solar radiation forms the great circle or cross section of the Earthly sphere. But, solar radiation isn’t a 2D plane it is actually a 3D sphere. The intersection of two spheres does not technically etch out another cirlce. It etches out a curved 3D circle. This means the true cross sectional area of Earth is actually slightly higher than 4, but the difference is negligible so we usually just simplify the geometry as a 3D sphere intersecting with a 2D plane and use 4 as divisor and call it day.
To get the average flux over the entire surface area of Earth you must divide the solar constant by 4 ??? WHY, do you divide this by 4 ??? If the quarter cross section area the flux is measured on, = say 100 over 6 hours. You would then x this by 4, = 400, not divide by 4 25 ??? Why are you dividing ???
The flex is NOT spread out over the entire surface area of Earth at any one time, its only hitting one quarter, so why think or even say its hitting the entire surface area of Earth when its not ??? This is the part where people don’t get or agree with you.
How much flux hits one quarter of the Earth in 6 hours ??? To get the total, you x by 4. YES ???
Wayne
One way to explain this is that the quarter cross section is the same thing as the area of the shadow cast by Earth. This area is visualized by intersecting a plane through the center of Earth to form a circle. The area of this circle is 2*pi*r^2 = 3.14 * (6.37e6 m)^2 = 125e12 m^2. But the area of Earth itself is 4*pi*r^2 = 4 * 3.14 * (6.37e6 m)^2 = 500e12 m^2. Notice that the area of the shadow is 1/4 the area of Earth.
One important thing to mention is that the solar constant is quantity measured perpendicular to the surface. Or using or shadow visualization it is the flux being blocked by Earth and thus hitting Earth. So the power received by Earth is 1360 W/m^2 * 125e12 m^2 = 170e15 W. But the Earth is not a flat disk. It is spherical. So if you want to know the average flux it would be 170e15 W / 500e12 m^2 = 340 W/m^2!
Your last question is a bit confusing and difficult to answer. It’s confusing because sunlight falls on 1/2 of Earth at any given time; not 1/4. You’d have to identify which 1/4 you are talking about because the answer won’t be 1/2 of the 1/2. It’ll actually be some component of the sine of the latitude we are considering. It is difficult to answer because you have to know what time of year it is. This is important because the solar constant is just the average over one full orbital cycle. The actual flux varies by 8% because Earth’s orbit elliptical.
Wayne Rowley says:
To get the average flux over the entire surface area of Earth you must divide the solar constant by 4 ??? WHY, do you divide this by 4 ???
Have you ever read the 1st chapter of any textbook on climate science?
No?
Then get off your rear end and go read one, and learn something for a change. All you’re doing is coming here ignorant and making a mess of things.
David, please stop trolling.
Why are you looking at the average ??? Lets say I used 4000cal. as in energy, in 24 hours, running around the Earth. My averge cal. for 6 hours was 1000cal. BUT, more important, my overall, or total cal. = 4000.
AGAIN, why are you few here working the total flux into the averge flux ??? The Sun is constantly hitting the Earth with energy/heat. So its to total w need, not the average.
Wayne
We are interested in the average for a few reasons. First, the actual solar flux changes throughout the year. 1360 W/m^2 is itself an average computed over one full orbital cycle. The actual flux changes because Earth’s orbit is elliptical. Second, this allows us to mentally compute the total energy received by Earth over a period of time. Third, most of the other energy budget constituents are themselves yearly averages.
Knowing that 340 W/m^2 is the average flux received at TOA we can immediately conclude the energy received per square meter over one year is…wait for it…340 W-years/m^2! Over a 10 year period it is 3400 W-years/m^2. And, of course, you can multiple these number by 500e12 m^2 to get the total over the entire Earth.
Wayne Rowley says:
Lets say I used 4000cal. as in energy, in 24 hours, running around the Earth.
Is there some reason you can’t use the ACTUAL numbers that pertain to the Earth?
(No, there isn’t.)
David, please stop trolling.
I don’t think this is about the average radiative flux from the sun as such really. It is about calculating equilibrium temps under the assumtion of average radiative flux. Basically it is about the T^4. During noon your part of the planet gets to radiate at higher temps, whereas your midnight counterpart has to radiate at lower temps. Now is it justified to average it all out?
Lets look at an extreme case, a planet rotating just such that it alwyas faces the sun in the same way. Would you be confident to assign an equilibrium average temperature by assuming uniform flux and calculating equilibrium?
Clueless.
Yuri, in the case of a non-rotating plant the division by 4 would not even be used, because the night side of the planet never receives solar energy. But your point is worth discussing, because the rotation rate of the planet does matter if you want to approximate average radiating temperature from average solar flux. For the moon, the approximation is poor because the moon rotates so slowly that daytime temperatures go very high on the non-linear Stefan-Boltzmann curve. On the Earth, however, the approximation of getting an average emitting temperature from an average absorbed solar flux is pretty good: http://www.drroyspencer.com/2016/09/errors-in-estimating-earths-no-atmosphere-average-temperature/
Thanks for digging up that post, Dr. Spencer.
Naaaah!
You can use S over the cross section of Earth, which is a circle with the Earth radius.
You can use S over the hemissphere to account for the curvature and to get a better picture of potential temperature if you wanted to.
You cannot use S any other way because you exhausted the real physical area which is sun lit, with the second step.
More so in both applicable cases you can measure the solar spectrum and you will find little differnce of the incoming spectrum.
You must not use S over a larger aread because no more is exposed to sun light at any second.
Spreading out the soalr constant over the full area of the Earth is a flat Earth argument take it or leave it.
Climate science can do better than to debate the obvious, which is what we call night and day. Averaging is part of the climate science because statistic is all they have and mathematically there is no night and day, but when it come to physics it exists.
Spreading out the solar constant over the full area of the spherical Earth cannot possibly be a flat Earth argument take it, leave it, or see the top post again.
AP said…”Spreading out the soalr constant over the full area of the Earth is a flat Earth argument take it or leave it.”
Umm…no. It is the exact opposite of a flat Earth argument. It is an argument that treats the Earth as a sphere which is what it really is. Not spreading the solar constant out over a spherical shape is the flat Earth argument.
By the way, the solar constant S is itself an average over one sidereal year which includes 365 rotations of Earth (relative to the Sun). Therefore it is a quantity that can be used (if done the right way) to represent the energy received over the entire Earth during one year.
Ball4,
are you saying there is no night and day and you are not familiar what a Watt is?
The sun lit area of Earth does not extend to the whole spherical area at any second.
The day has 24 hours, but the sun hits Earth only for 12 hours continuously on the day side.
If you say that you can just use Power by half over a larger area then you omit the most important part that you need to get to an amount of Energy and this is time.
While there is sunshine at any second, there is only sunshine for 12 hours (averaged of course) on any location of Earth.
Closing the eyes to the obvious instead of working with reality and admitting day and night cycle into the equation is weird.
Its weird why we are having this discussion.
Daily observation is on Joes and my side!
bdgwx says:
June 4, 2019 at 6:00 PM
It is the exact opposite of a flat Earth argument. It is an argument that treats the Earth as a sphere which is what it really is.
No you mean to say it realises that Earth rotates and we know that. And it is a sphere. But at any second there is only half of that sphere lit up by the sun.
You argue against night and day due to an misunderstanding of what a Watt is.
This does not help Dr. Spencer nor anyone to defend this position.
You can maybe calculate the little Joules that remain on any part of Earth due to the solar energy and rotational changes of the amount hitting it, but the Power at which it comes is double then what you are calculating. Power double, area half and then calculate Joule by multiplying with time.
Mathematically it is the same. Physically the impact off power is different.
“are you saying there is no night and day and you are not familiar what a Watt is?”
I agree the GHCN thermometer field covers the whole ~sphere (more or less) with continuous min/max readings night and day; take the mean of their output so have to take the mean of their input for comparison. It works, is physically meaningful. You get ~288K from the thermometers, ~288K from using the ~340 net of albedo (and that was done BEFORE satellites confirmed it). I am familiar with the SI derived unit Watt.
Closing one’s eyes to the obvious apples to apples comparison isn’t helpful.
Ball4, please stop trolling.
Ball4,
we are not talking about a thermometer net work that covers the whole world or at least the surface areas and sparingly so if you look a little bit closer.
We are talking about an amount of Watt’s that you claim you know what it is, per square meter. This is power over area, which is any single second a Joule on that area.
You are should just honestly agree that what you defend here is physically the wrong position to describe incoming solar radiation.
The way you try to distract with your last comment lets me think you understand this and you are throwing a smoke bomb.
This is another way of saying that you have a none-defend-able position.
Thanks
The position is defended by both 1LOT theory and measurement so Another Joe needs to find a fundamental flaw in the 1LOT AND the instrumentation to defend his position. Another Joe is not going to be successful.
Ball4,
which measurement would support your claim?
Any measurement of an incoming spectra of sun light will show the solar constant. This is why it is called a solar constant. It is constant for the whole Earth.
This is what you can measure.
All else is assumption and the divide by four is wrong since the impact area of the sun is always just a hemisphere.
If any you can assume that the curvature of Earth requires to divide by two to estimate the amount of energy by square meter per second.
I wonder how you want to apply the first law of Thermodynamics to Power. Show your workings there. I am very interested.
The measurements are here:
http://www.drroyspencer.com/2019/06/uah-global-temperature-update-for-may-2019-0-32-deg-c/
The global historical climate network thermometers are also used and result in the same answer.
Numerous authors take the system input power and have annualized it over 4-15 periods to get energy in/out which is conserved.
Another Joe,
Maybe this can help.
There are many devices in our daily lives that use the same principle.
Ovens, furnaces, microwave ovens, air conditioners.
Like the sun, many of these devices have only one power output, call it 100 %.
If I want to use a lower power, eg. on the microwave to defrost meat, I can tell it to use 30% power. In fact, that means it still turns the power on to 100%, but it is only on 30% of the time.
It gives the same result as a 30% power level on all the time.
In my house, on a medium cold day, the furnace may turn on 50% of the time. Same result as being @ 50% power for 100% of the time.
This is the same principle with the sun and the rotating Earth.
Ball4,
so you are using the measured temperatures as a proof for the theory tgat tries to proof that the temperatures are like this?
It is sort of a circular reasoning and on top of this completly off topic since we were simply talking about spectral analyses of the arriving solar radiation.
Nate,
you are using the microwave as an analogy.
Can you tell me if you think that you can bring water to the boil with the microwave when you run it with less power – 30% but continously?
Would this have the same effect?
“It is sort of a circular reasoning..”
Experimental proof of the theory is never circular reasoning.
Sorry Ball4,
I was not aware that the data of the “global historical climate network thermometers” you refer to is experimental.
My wrong, I assumed you wanted to proof the measured temperatures by means of bringing up measured temperatures.’
I would be obliged if you could give me a link to the experiment. I feel that the link you send above might have been misplaced.
Thanks
“I was not aware that the data of the “global historical climate network thermometers” you refer to is experimental.”
Now you are.
#2
Ball4, please stop trolling.
Postma doesn’t deserve an epsilons worth of attention.
David, please stop trolling.
Gentlemen, I have to say that I am very honoured to be here with you. Reading your expert discussions makes me feel like I am on board of the Starship Enterprise. I understand nothing and I am just waiting for someone to say: “engage!”.
There, I hope I made your day 🙂
{rolls eyes}
To put it simply, someone offers to employ you at $10/hour.
Then an astrophysicist with nothing better to do, says: “But that’s not $30/hour because you won’t be working all the time” 24/7. Then someone else says: “but they are only you paying you once a week … so you’ll only be earning in that hour when you’re paid”. Then someone else says: “but you won’t be spending it all the time”, etc.
And then after you’re fed up with all the waffle on meaningless things & are sitting with a beer … you work out that that they will be deducting $25/hour for accommodation, clothing, tools, etc. (in this scenario = clouds).
Roy, Thank you for engaging in debate and for that we respect you, even while disagreeing. We now have your post at Principia Scientific International: https://principia-scientific.org/on-the-flat-earth-rants-of-joe-postma/
Truly,
John O’Sullivan
No serious person should be engaging Postma or Sullivan in anything.
Roy, you really should know better.
How about getting back to your wrong sign for the El Nino trend?
yes, I responded to that today. I have other things to do than monitor comments on my blog all day long.
We certainly have no wish to engage with you, Mr Appell
John,
Why not?
Because he’s a troll.
Surely it is extraordinarily poor internet etiquette to repost someone’s post at another site *without their permission*?
I didn’t repost his post. Others do that with my posts all the time, and I don’t mind, but I almost never repost someone else’s post, and if I do, I would never pass it off as my own work.
If it is simply “here is a rebuttal” type re-post then that is much better etiquette than what Roy has apparently endured directly from Postma.
barry, please stop trolling.
Greenhouse effect theory says Earth on average gets about 240 watts and If something gets 240 watts and emits 240 watt {per square meter] it should be -18 C [if in a vacuum of space].
And reasons that something must be warming earth by 33 K because Earth average temperature is 15 C.
And Joe is saying one can get 240 watt constantly if on Flat world if 2 AU from the Sun. And such world would have average temperature of about -18 C
I would say if flat world at 2 AU and temperature was -18 C, and ask question is there anything you change that would cause the -18 C flat world to have average temperature of 15 C?
Greenhouse effect theory says Earth would have average temperature of -18 C if not for greenhouse gases.
So greenhouse effect theory suggest that if greenhouse gases are added the flat world could increase in temperature from -18 C to 15 C.
Now on the flat world one is always getting 240 watts and always radiating 240 watt, so everywhere and at all time one is getting and emitting 240 watts per square meter.
With our spherical world, roughly speaking one is always emitting about 240 watts everywhere, but it’s closer if you say, on average everywhere emits 240 watts per square meter.
But on our world, one is not normally receiving 240 watt per square meter. Most of the time you getting near 0 watts. And larger 50% of the time one gets less than 240 watts or 0 watts. And for about 25% of the day one normally or broadly speaking one gets more than 600 watts per square meter.
There are serious differences between a world getting a constant of 240 watt per square meter of sunlight and our world.
For instance with the flat world getting 240 watts constantly, solar energy would be good way to get electrical energy.
Or big problem with earth is you only get solar energy for about 25% of the time- which makes solar energy pretty useless.
So if you are deluded fool who imagines Earth gets 240 watts per square meter constantly, you might imagine solar energy is good idea.
Now Earth is currently in an Ice Age. A Ice Age has a cold ocean and has polar ice caps. We have a cold ocean and ice caps.
But we in the interglacial period of our Ice Age, which roughly means we don’t have 1 mile ice covering New York. And also means sea levels are not 100 meters lower.
There quite dramatic difference between a glacial period and an interglacial period and they are not caused by different levels of greenhouse gases. The differences of greenhouse gases are the effects of these conditions rather than the cause of these different conditions.
The cause of these different conditions is broadly called “natural variations”, or specifically they thought to be related to:
“The glacials and interglacials also coincided with changes in Earth’s orbit called Milankovitch cycles.” – wiki
So, our average ocean temperature is about 3.5 C, and glacial period have ocean average temperatures of 1 to 3 C and interglacial period have average ocean temperature of 3 to 5 C.
Or the cold oceans of our Ice Age are about 1 to 5 C.
Now Earth is currently in an Ice Age.
No longer. We’re now in the anthropocene — 0.2 C/decade surface warming.
There is no long term trend of 0.2 C/decade, nor is there any reason to guess there will a long term trend of 0.2 C/decade.
There is nothing particularly notable about the last 100 years.
The trend of 0.2 C/dec has existed for several decades now.
And it’s fully expected to continue, unless you deny physics.
gbaikie
“There is nothing particularly notable about the last 100 years.”
One thing. An industrial civilization has pumped enough CO2 into the atmosphere to increase the CO2 concentration by 40% and the temperature by 1C in one century.
Don’t recall it happening at that rate before. There have been natural changes of that size, but they took tens of thousands of years.
I think it’s fair to say that we have more accurately measured the last 100 years as compared to any 100 year period in history.
Though most all agree there was more error in the measurement 100 years ago as compared to say, 50 years ago.
One can get sidetracked with a focus of possible error in temperature record of last 100 years but it seems rather foolish to assume temperature guesses centuries or millenniums ago are better than modern record with all the possible flaws related to it.
And using trees as thermometers obviously quite silly in this regards- even if one were to use the “tree thermometers” in correct way.
Or what proxies are good at is comparative changes in temperatures.
And this might seem “good” because that how measure current global temperatures.
I don’t think it is good, I think it might be practical.
I also don’t think air temperature is a good metric for indicating global temperature, but it has a practical aspect.
So I think average volume temperature is a good measure of global temperature- though I agree it’s practical or it’s hard to measure. Or it has not been measured at all, really.
So, I limited to saying the average ocean temperature is about 3.5 C.
And it’s also similar to being limited to saying the global average surface air temperature is about 15 C.
And 30 years ago, I was saying global average surface air temperature was about 15 C, and other people hundred year ago were saying it was about 15 C, and in next 30 years, I have no reason to have a expectation of not saying Earth’s average surface air temperature is about 15 C- And would not matter whether in next 30 years, we get warming or cooling.
Currently someone could claim global average surface temperature is closer to 14 C or could say it’s closer to 16 C. But a problem with saying either is that there could be expectation to provide evidence of either.
Now, I would say, that Berkeley Earth made some effort to determine that global land surface air temperature was presently about 10 C.
{And they provided evidence and I am unaware of any attempt to dispute it- probably mostly due to lack of much interest [or chance to make much money compared work required].}
It seems possible it’s wrong, but more interesting to me, is they failed to likewise arrive a guess of global ocean surface air temperature.
Though it’s common to say that global ocean surface temperature is about 17 C. As common as saying hundred years ago that global temperature is about 15 C.
Anyhow I think average temperature of the entire ocean is about 3.5. And this temperature could only change by about a hundredth of degree per century. And such changes of temperature are severely lacking in entertainment value.
And I have noticed that it is common for people to fail to understand why changes in this ocean temperature could any effect upon their lives or the world we live in.
What I don’t know, is what the average temperature of all surface water: entire Ocean and entire glacial masses on the earth surface.
You could make it easier, by limiting it to entire ocean [about 3.5 C] plus all ice on Antarctica and Greenland.
Or maybe easier if include just the ice not on the land- polar sea ice and including ice not on the shore [beach].
A more complicated though maybe practical would be present ice temperature, that could flow into ocean to make sea level rise by say 10 meters [from Antarctica and/or Greenland]
So if sea level rose 10 meter from Antarctica and/or Greenland you have to first decide from portion of 10 meter was from Antarctica
or Greenland. Though you could dismiss the temperature of ice and just count the latent heat required melt ice.
That last is pretty easy. 10 tons of ice per square meter.
How much 3.5 C water is required to melt it.
+cubic meter of ice weighing 1000 kg vs how many tons of water at 3.5 C.
334 KJ/Kg , 1000 kg: 334,000 KJ
4.2 KJ/kg, 1000 kg: 4200 KJ
So lower 3.5 water by 3 K
a ton of ice [per square meter] turns 26.5 cubic meters of 3.5 C water into .5 C water.
So get 265 meters of .5 C water over entire ocean surface.
It seems to me, that that much colder water would not change the average volume temperature of the ocean which about 3.5 C.
Or it’s a massive amount of cooling of entire ocean, but since we don’t know how the exact temperature of ocean, it remain around 3.5 C, but it changes by hundreds of degree.
Or 265 meter at .5 C mixed with 265 meter at 3.5 C gives 530 meter at 2 C.
And 530 meter at 2 C mixed 530 meter at 3.5 C gives 1060 meters of water at 2.75 C
1060 meter at 2.75 mixed 1060 meters at 3.5 C give 2120 meters of water at 3.125 C
Now, don’t have 4240 meter of ocean, but if adding 2120 meter of water at 3.5, you get 4240 meter water at 3.3 C.
So you could say it lower entire ocean by about .25 C, but due to lack measurement you couldn’t say the ocean was around 3.25 C.
Or as far as we know it’s possible the ocean is presently at 3.25 C or at 3.75 C.
Though if lower ocean temperature by .25 C, can say how much ocean has lower in sea level due the contraction due to lower temperature.
I would guess less than 6 inches and compared gain 10 meter, understandably one could say it’s lost in the noise.
Of course in real world one can’t mix the heat of entire ocean.
If ice fell in water and cooled ocean, it cool surface waters and with comparitively small region [close to polar region]. Or a reasonable assumption if that ice could rapidly [within a century] fall in the ocean, it should create a massive amount of polar sea ice. Or by itself would prevent that much ice falling into the water. And all that fresh water in winter would freeze quite fast.
@Entropic man:
“One thing. An industrial civilization has pumped enough CO2 into the atmosphere to increase the CO2 concentration by 40% and the temperature by 1C in one century.”
WRONG! .. According to the UN-IPCC, over the course of the past 100 years our atmospheric CO2 concentration has increased approximately 100ppm. Of that 100ppm increase, humans are responsible for approximately 3% – 4% of that rise. That is, the human contribution to the rise in CO2 concentration over the past 100 years has been approximately 3ppm – 4ppm.
NOT 40% !!!
Where did the rest of this CO2 come from then? .. I don’t know where you pulled your numbers, but they are not even remotely close to reality.
‘Of that 100ppm increase, humans are responsible for approximately 3% – 4% of that rise. ‘
Nope. IPCC never said any such thing.
Where do you get this faux fact?
Nate, please stop trolling.
The Radiative Greenhouse Effect Theory
1. The atmosphere warms the earth, to wit: 288 K 255 K = +33 C. (Wrong! It cools the earth) Just how does that work?
2. There are magical GHGs constituting a pitiful .04% of the atmosphere that trap/absorb/re-emit/back radiate some kind of 100% perpetual looping extra energy thereby warming the atmosphere, to wit: 333 W/m^2. (Wrong! Thermodynamic nonsense.) Well, just where does this magical extra energy come from?
3. The surface radiates LWIR as an ideal 1.0 emissivity BB, to wit: 289 K = 396 W/m^2 (Not Possible!)
1 + 2 + 3 are absolute horse manure.
0 RGHE = 0 CO2 warming = 0 man-caused climate change.
The evidence for the greenhouse effect is easy to see:
https://www.giss.nasa.gov/research/briefs/schmidt_05/curve_s.gif
It is “Easy to See” if you are brainwashed.
gallopingcamel
The correct wording of your post would it is “Not easy to see” if you are brainwashed.
Empirical evidence of many types demonstrates there is indeed a GHE (meaning the Earth’s surface, with solar input, is warmer because of DWIR from GHG than without the DWIR).
Below I used actual measured values of solar energy input at Los Vegas (the link has numerous other cities which would show the same thing). You can get the average amount of solar energy input to an area (that which does the potential heating) and see it is way too low to maintain the actual temperatures of that region.
One needs to totally ignore actual data to come to the conclusion that a GHE is not working to keep temperatures at higher levels.
I think you have a pressure theory to explain this. This theory a continuous input of energy from pressure. Pressure needs to change to generate new energy.
The classic real world evidence is that when you first compress air it will indeed heat up but then the compressed gas cools to room temperature and does not keep a higher temperature. Check a gas cylinder at 2000 psi. It is cool to the touch. I would hope, with your intellect, you would reject this false diversion and start to examine the actual evidence.
Norman,
The GHE is real but it is not 33 Kelvin as David Appell seems to think.
My analysis suggests between 77 and 89 Kelvin depending on whether you factor in the rate of rotation.
I have linked the three posts I did a few years ago on the subject several times on this blog. Here is the first of them:
https://tallbloke.wordpress.com/2014/04/18/a-new-lunar-thermal-model-based-on-finite-element-analysis-of-regolith-physical-properties/
that link is dead
By reflecting away 30% of the incoming solar energy the atmosphere/albedo make the earth cooler than it would be without the atmosphere much like that reflective panel behind a cars windshield.
https://www.linkedin.com/feed/update/urn:li:activity:6503085690262216704
Greenhouse theory has it wrong.
The non-radiative processes of a contiguous participating media, i.e. atmospheric molecules, render ideal black body LWIR from the surface impossible. The 396 W/m^2 upwelling from the surface is a what if theoretical calculation without physical reality. (And, no, it is not measured!) (TFK_bams09)
https://www.linkedin.com/feed/update/urn:li:activity:6507990128915464192
https://principia-scientific.org/debunking-the-greenhouse-gas-theory-with-a-boiling-water-pot/
Greenhouse theory has it wrong.
Without the 396 W/m^2 upwelling there is no 333 W/m^2 GHG energy up/down/back loop to warm the earth. (TFK_bams09)
https://www.linkedin.com/feed/update/urn:li:activity:6457980707988922368
Greenhouse theory has it wrong.
These three points are what matter, all the rest is irrelevant noise.
No greenhouse effect, no CO2 global warming and climate change neither caused nor cured by man.
No greenhouse effect
It’s trivial to see the overwhelming evidence for the greenhouse effect:
https://www.giss.nasa.gov/research/briefs/schmidt_05/curve_s.gif
Given 2 molecules (of any matter), what requirement is necessary for molecule A to further excite molecule B ???
This should be an extremely easy question to answer. If you are the intellectual you portend (pretend) to be, you should be able to answer this swiftly and succinctly. I suspect however, you will either not answer or your will obfuscate.
* The so-called RGHE is impossible within the known universe.
another dead link it seems
Your link is for ToA not the surface.
The atmosphere clearly cools the earth which contradicts GHE which says the atmosphere warms the earth. GHE loses.
One popular geoengineering strategy proposed for countering imaginary global warming/climate change is reducing solar heating by increasing the earth’s albedo.
This increase is accomplished by various physical methods, e.g. injecting reflective aerosols into the atmosphere, spraying water vapor into the air to enhance marine cloud brightening, spreading shiny glass spheres around the poles with the goal of more reflection thereby reducing the net amount of solar energy absorbed by the atmosphere and surface and cooling the earth.
More albedo and the earth cools.
Less albedo and the earth warms.
No atmosphere means no water vapor or clouds, ice, snow, vegetation, oceans and near zero albedo.
Zero albedo and much like the moon the earth bakes in that 394 K, 250 F solar wind.
These geoengineering plans rely on the atmosphere cooling the earth and expose the error and delusion of greenhouse theory which says the atmosphere warms the earth and with no atmosphere the earth becomes a -430 F frozen ball of ice.
A failure of greenhouse theory means no CO2 global warming and no man caused or cured climate change.
The obvious evidence for the GHE:
https://www.giss.nasa.gov/research/briefs/schmidt_05/
Not obvious to me, that is not an evidence.
Nick,
One of the many facts ignored in your posts is whether or not GHE theory has been tested or not.
It has.
In fact, it is tested every day via numerical weather models, which must incorporate the GHE to calculate and predict tommoorrow’s or the weekends’s weather.
These successful tests, every day, every week, improving all the time, show, demonstrate, that our standard understanding of the atmosphere is excellent.
Now, if you want to somehow remove the GHE from our understanding of the atmosphere, but keep all the rest, because weather models are working for you, well sorry, it doesnt work that way.
Can’t be done.
Nate,
The atmosphere cooling the earth trashes GHE.
Nick Schroeder
The atmosphere cooling the Earth does not trash GHE.
The atmosphere cools the Earth far less than the surface would if you suddenly removed the atmosphere.
The surface emits and average of 390 W/m^2. The atmosphere emits less than 240 W/m^2 (the rest comes from the IR window through the atmosphere from the surface).
That the surface has to heat up to the point it needs to emit 390 W/m^2 to get 240 W/m^2 out at the TOA is the GHE.
If you remove the atmosphere the surface only needs to emit 240 W/m^2 to get 240 W/m^2 out. Much cooler average surface temperature.
I think it is time for you to crack open a textbook on heat transfer and start to read. There are a few good free online textbooks available on the Internet.
So wrong, Norman.
Nothing new.
JDHuffman
Looks like you are trolling all my comments. Nothing new with you. Some people might think you are making a point. However, the truth is you just troll to troll. No higher purpose, no deeper motivation.
You still don’t know a thing about science or physics. I guess if you pretend long enough you fool a couple of gullible posters. The rest know you are an ignorant troll.
Norman, if you could face reality, you would realize it is you that is the troll.
But, we both know you abhor reality.
Nothing new.
Nick,
‘Nate,
The atmosphere cooling the earth trashes GHE.’
Assertion.
Experiments, done every day, trash assertion every time.
And a follow up to Nate’s point…it’s tested everyday by observational meteorology as well. The radiometers like the ABI instrument on the GOES-R satellites exploits the infrared spectrum behavior of both H2O and CO2 to produces their products. If the GHE wasn’t real then the GOES-R (and the like) satellites would be mostly useless.
“Your link is for ToA not the surface. The atmosphere clearly cools the earth which contradicts GHE which says the atmosphere warms the earth.”
The only location the Earth cools is ToA. If no radiant energy ever left the atmosphere the Earth would return to molten rock.
Nate, bdgwx, Craig T, please stop trolling.
Here is a link of actual measured average solar flux in certain cities of the world.
https://www.pveducation.org/pvcdrom/properties-of-sunlight/average-solar-radiation
This should end Postma’s poor reasoning. I know it won’t for the “true believers” but those who want actual facts may find it of value. Empirical evidence.
To convert the units given in the link to W/m^2 use this calculator:
https://barani.biz/apps/solar/
I did a calculation of Las Vegas yearly average measured solar flux and it came out at 236.8 W/m^2. Northern locations are much lower.
People want to believe Joe Postma for some reason. You can’t reason with this person. He twists and distorts clear points and plays dumb semantic games when wrong. I have strong doubts he never studied any heat transfer physics and when you point this out to his adoring sycophants he turns into a raging lunatic. He might sound calm and rational on his videos. That is not the case if you try and comment on his blog.
Here is one sample of the lunatic that has strong followers. A couple of them post on this thread.
https://climateofsophistry.com/2016/08/11/roy-spencer-ae-language-warning/
Here is a sample. Roy is not the only one he has attacked for questioning his terrible science.
“With your buddy Moncktons self-demolition on my blog yesterday, and this, you guys have totally exposed what you are.
You come on here accusing us of what you did. Well right back at you you lying, sophist, son of a bitch.
Go fuck yourself Roy, you god-damned sophist piece of shit. You peoples days of defending the fraud of climate alarm are soon coming to an end. You deserve no quarter and no respect in this matter any longer.
You fucking asshole.
Somehow he gets this strong following from a few people. Honestly I don’t know why. Robert Kernodle seems to think he is some type of genius.
Yes, Postma is a donkey.
No one should be spending the slightest amount of time on him. That includes Roy, who really should know better.
David Appell
I am sure Roy Spencer does not want to spend any time with the lunatic. Postma is unscientific, irrational and unstable. It is most impossible to have an intelligent discussion with him. The troll JDHuffman is annoying but not even close to this level.
Roy stated he keeps having people ask him about what Joe Postma claims so he thought he would post an article on it.
David Appell
I have seen where you attempt to convince the deluded followers of Postma that a GHE exists.
I can provide you with the empirical evidence of just such a reality. It won’t ever convince the irrational unscientific skeptics that haunt this blog nor the trolls who don’t care anyway.
Above I calculated the total measured yearly input to Las Vegas (it is measured with actual instruments) at 236.8 W/m^2 (others could get different results with different rounding).
The available energy reaching Las Vegas would get it steady state of 254 K. What the Joe Postma followers never understand is how cold the surface would get without a GHE during the night. Roy Spencer had a thread specifically dedicated to this point. Yes the Sun will warm is above the 254 K for a short time but the temperature will fall very fast at night so you will average 254.
Las Vegas actual air temperature average temperature is 69.3 F or 293.87 K.
https://www.usclimatedata.com/climate/las-vegas/nevada/united-states/usnv0049
In order to maintain an average temperature of 293.87 K it would have to have an input energy of 422.87 W/m^2. The Sun can only supply 236.8 W/m^2 where does the remaining energy come from?
People have to understand that these are real temperatures and real solar measured values.
The energy emitted by the atmosphere DWIR makes up the difference and is why Las Vegas is at 293.87 K instead of 254 K. The DWIR comes from GHG present in the atmosphere. The DWIR is actually more than the 186 W/m^2 difference since you have energy lost by convection as well as radiant energy and some evaporation that also have to be added to maintain the 293.87 K temperature.
It is very logical, scientific and empirical but the deniers will not accept the evidence.
You’re right, Norman.
Postma deserves to laughed back to his mommy’s house. It’s a complete waste of time to spend any time on him whatsoever.
Norman,
“In order to maintain an average temperature of 293.87 K it would have to have an input energy of 422.87 W/m^2. The Sun can only supply 236.8 W/m^2 where does the remaining energy come from?”
Using your Las Vegas numbers, assume a 20K difference in high/low temperature on an average day. My town is averaging about a 14K difference now. The hypothetical Las Vegas low temperature would be 283.87K and the high 303.87K. To go from the low to high would only require 115.25 W/m^2 of solar input. I don’t find your analysis very logical, scientific and empirical. I think the simplistic flat-earth static-sun energy balance diagrams lead to your misconceptions.
Meanwhile do you have any irrefutable evidence that increasing atmospheric CO2 will result in any concomitant increase in global temperature?
Norman demonstrates the problem with “average” flux.
He gets bogus results.
Nothing new.
Norman claims: “Above I calculated the total measured yearly input to Las Vegas at 236.8 W/m^2 . The available energy reaching Las Vegas would get it steady state of 254 K.”
Norman demonstrates the dangers of averaging flux. He uses the Las Vegas value to claim the temperature of 254 K means anything.
(I guess it means Norman doesn’t have a clue about the relevant physics.)
Chic Bowdrie.
I am not sure what a daily cycle would matter as to what I posted. You call it illogical but you are clueless about the nature of the post.
The 293.87 K is the yearly average temperature for Las Vegas. It includes the highs, lows, summer, winter temperatures all averaged over the year. I really do not know what you are trying to say. Sounds like a stupid point to make. Are you trying to be an ignorant troll like JDHuffman or will you bring up a good point.
I also do not know why you are asking me for evidence CO2 raises the Global temperature. That is not my point at this time. I am providing actual evidence of a GHE. You don’t have to accept it. I think logical people will be quite able to see my point.
‘. I dont find your analysis very logical, scientific and empirical.’
IMO, its clearly all of those things, Chic.
You can’t just throw random adjectives without specificity about what flaws you see.
‘the low to high would only require 115.25 W/m^2 of solar input.’
This doesnt make sense to me, you need heat capacities and times.
Norman,
I wish I had time for a more detailed explanation. The gist of it is that 300K temperatures don’t just appear after a single day of 459 W/m^2 energy input. The globe started out hot and cooled to roughly the temperatures they are today. On “average,” places only need 160 W/m^2 or so to maintain their average temperatures. You using SB equations to make your claims is not empirically scientific. How is it logical to claim otherwise?
Nate,
Same comment, except are you asking or stating that heat capacities and times are involved? I agree they definitely are, as I have written earlier.
‘The globe started out hot and cooled to roughly the temperatures they are today.’
Irrelevant to today.
Today all that matters is the balance in and out.
‘You using SB equations to make your claims is not empirically scientific.’
He is using it to calculate surface emission. Emissivity for dirt is ~1. So if for 293.87 get 423 W.
This is empirical.
You can complain about varying Temp over the day or year, but that will only change things a little from just using the average.
The solar insolation is 236.8 W/m^2, again empirical average over the year.
He is not considering energy lost due to convection. But only makes the deficit worse.
Nate,
Get your head out of the fictional flat-earth SB simplified diagram calculations and move into the realm of reality. SB calculations are not empirical (based on, concerned with, or verifiable by observation or experience rather than theory or pure logic).
If you radiate a surface with 500 W/m^2, what will its temperature be? You don’t know unless you measure it, after which you can use diffusion coefficients, heat capacities, emissivities, etc. with a suitable model to try to explain your result.
Norman is using black-body-in-a-vacuum numbers and wondering why, there is a 423 – 237 = 186 W/m^2 difference. It is not empirical science to conclude back radiation must be what makes up for it. It might sound logical, but I would call it a flawed hypothesis at best. Unless you can produce some actual data that shows an increase in CO2 will result in a detectable temperature increase, you are just running a con job.
Chic says: “Unless you can produce some actual data that shows an increase in CO2 will result in a detectable temperature increase, you are just running a con job.”
Chic, you’ve been studying….
The SB law is a law of physics. That means it has been tested thousands of times. That means it is empirically verified. Stop declaring things that are just plain False!
The SB law works for surfaces like dirt, etc whose emmisivity can be empirically determined. No need for vacuum.
The Flat Earth is a red herring here. This is one city with all parameters measured.
Why not make an effort to understand the argument before knee-jerk rejecting it?
Chic Bowdrie
YOU: “On “average,” places only need 160 W/m^2 or so to maintain their average temperatures. You using SB equations to make your claims is not empirically scientific. How is it logical to claim otherwise?”
The first part makes zero sense. I hope you do elaborate. What average temperature needs only 160 W/m^2 to maintain?
My claims are based upon empirical science. Empirical science is based upon things that can be measured and observed. You have countless actual temperature measurements that go into the Las Vegas average temperature. It is done by taking all your temperatures (highs, lows, winter, spring, fall, summer variations) adding them together and then dividing to come up with an average temperature for this location. The solar energy that is available is directly measured by instruments. The value I have is that averaged over the whole year. You can do the calculation yourself, I put the link in a post above.
The energy that a square meter of Las Vegas ground over the course of a year is not enough to maintain the average temperature.
I do have other empirical evidence I have shared with you on other threads. Actual measured values. I have given you a Downwelling IR spectrum which isolates the contribution of CO2. I have linked you to graphs with Downwelling IR.
Why do you waste my time and act like we have not already gone over all of this. You are starting to act the part of a troll. I would hate for this to be a reality. It seems to be when you act like we have not discussed all this in the past.
In case you forgot, here is empirical data (measured values). You can do what you want with them. Don’t say you did not see them.
https://www.esrl.noaa.gov/gmd/webdata/tmp/surfrad_5cf884227d576.png
Norman,
“What average temperature needs only 160 W/m^2 to maintain?”
The average temperature of a hypothetical point on the planet. And 160 is only the Trenberth average input. It varies depending on where the point is. Take your Desert Rock figure you linked to. I estimate the total day’s net W/m^2 to be about 2600 or 108 W/m^2 average for the day. You don’t provide the specific temperatures, but it is likely to be about the same from one morning to the next. 108 is a lot less than 160, but the point is a contribution from back radiation is not needed because Desert Rock was already at the temperature it was when the sun came up that day. It didn’t need 400+ W/m^2 to get the temperatures to what they were that day. There was already a baseline energy present at zero hour.
“My claims are based upon empirical science.”
No they aren’t. None of the figures you quote come from the Desert Rock data you linked to. You took the yearly average 293.87K for Los Vegas and played SB equation games with it. That is not empirical science. Using a scientifically determined equation to calculate numbers without measurements is not empirical. Now I can explain this over and over again, but why should I bother if you are not interested in understanding it?
You are not providing evidence of any GHE. All you are doing is regurgitating an hypothesis that has not been proven. Without evidence that increasing CO2 increases global temperatures, you have no point.
Chic Bowdrie
First I am discussing the GHE NOT AGW. Different topics with different points. It is not necessary to prove to you that Carbon Dioxide added to air is causing current warming trends to demonstrate the existence of GHE.
I am focusing only on the GHE here nothing more. There are lots of articles linking CO2 with warming. You can research the topic if you really are interested but that is not the POINT for this discussion. It is only the GHE.
First you can use SB equations on empirical data. The data used is empirically derived. It is based upon measured values.
I really do not think you are understanding the point at all. Not sure you ever will and I am not sure you want to.
You try to use some sideline points that really do not matter. Emissivity of materials, heat capacity, rate of conduction. None of those will change the outcome.
If you have a polished metal and asphalt, they will reach the same temperature at steady state when exposed to the same EMR source (provided no other energy exchange mechanisms are in place). The reason is that even though the two have very different emissivities. The polished metal will only absorb a small fraction of the available EMR. It will emit much less IR but it will be at the same temperature. Also heat capacities don’t matter. It just means one will take longer to reach a steady state.
Chic Bowdrie
Your explanation of my point: What average temperature needs only 160 W/m^2 to maintain?
In your explanation you have to use the GHE to get to where you are going. The reason the Desert Rock location only needs the 108 Watts/m^2 you estimated is because of the GHE. Roy Spencer spent an entire thread covering this issue. The nighttime temperature of Desert Rock or any location would fall drastically during the night. Look at the linked graph again. The DWIR continues during the night keeping the surface from getting super cold.
I am hoping we can try to work out the points. I remember you to be a thoughtful poster bringing up some interesting points. I am not sure you are doing this with your current points. I do not find them very thoughtful. It is not like you are spending any time with the ideas, more like you are just posting some reaction with little thought behind it.
Here is a link to the graph I made, make you own and use this tool to research your ideas. You can add temperature to the graphs.
https://www.esrl.noaa.gov/gmd/grad/surfrad/dataplot.html
Read Roy’s thread on the topic. He will explain it for you. No GHE, very cold night time temperatures.
http://www.drroyspencer.com/2015/04/why-summer-nighttime-temperatures-dont-fall-below-freezing/
‘Using a scientifically determined equation to calculate numbers without measurements is not empirical. Now I can explain this over and over again, but why should I bother if you are not interested in understanding it?’
Well, if that’s what you think, Chic, then science, and its empirically tested laws, don’t have much of a point-if they can’t be applied.
He can use well tested equations, ones that satellites use every day to determine sea surface temperature.
IR sensors use SB law every day to measure temperature. Not very different from using a mercury thermometer. Empirical.
As Norman discussed, there are lots of measurements at the site. Empirical.
Your objections are just plain ridiculous at this point.
Norman, Nate, please stop trolling.
Postma is kinda like Rush Limbaugh.
He says the most outlandish things in order to get attention.
But at least Limbaugh makes millions from his gullible, uneducated listeners. (And laughs all the way to the bank.)
Postma hasn’t been smart enough to earn any money from pretending to be stupid. But he keeps hoping!
David,
Think of it, there are no clients in the climate “business.” In theory, no one should make money out of the climate. Those who did, were just activists one way or another. They used politicians to make money.
Climate scientists don’t get paid any more than any other scientists. The only groups making money out of the climate sell products that increase CO2 levels. Even they are looking for new products that are carbon neutral.
David, Craig T, please stop trolling.
” When the debate is lost, slander becomes the tool of the loser ”
Socrates may or may not have said this, but if he did, he was spot on.
aido
Perhaps a lawyer will correct me, but IIRC it is not slander if it is true.
Entropic Man, please stop trolling.
This remembers me a discussion with a few of these people I call the Pseudoskeptics.
They pretend that the energy balance is wrong because the sunlit over the hemisphere (pi R^2) cannot be correctly compared with Earth’s IR radiation involving its entire surface ( 4 pi R^2).
They say the sunlit reaches one hemisphere at any time; thus it should be 2 pi R^2, and not pi R^2.
What they forget all the time is that sunlit is maximal at the Equator and zero at the Poles.
One must therefore operate with a latitude weighting based on the square of the cosine of the sunlit’s incidence angle.
Intergating cos^2(x) from 0 to pi/2 gives exactly 0.5.
And that is the reason why we work with pi R^2 in the energy balance equation, and not with 2 pi R^2…
Fundamental error 1:
Greenhouse theory says the atmosphere warms the earth. Since this is incorrect an entire field of bogus physics, i.e. up/down/back and BB LWIR, cold to hot, perpetual 100 % efficient, etc. climate science must be fabricated to explain how this non-existent process works.
Fundamental error 2:
Dividing ISR by 4 to average/spread it evenly over the entire spherical ToA 24/7 is really dumb.
As far as the sun cares ISR sees the earth as a flat disc where ASR = (1-α) * ISR (1,368 W/m^2)
To model the perpendicular energy distribution over the hemispherical lit side apply the following equation: ASR = ISR *(1-α) * Cos Latitude. Any engineer siting solar panels can explain why.
Reality:
The atmosphere cools the earth and Q = U A (Surface T ToA T) explains why the surface is warmer than ToA. No GHG hocus-pocus physics needed.
Nick gets it.
‘ Any engineer siting solar panels can explain why.’
Yes, and so can any meteorologist or climate scientist. And they indeed do that.
Nate, please stop trolling.
Any engineer siting solar panels knows that the most energy is gathered when the panel is facing the sun at noon. Half the time the panel will supply no energy because it is night. During daylight hours the percent of power depends on the angle to the sun. Batteries charged from the solar panel could supply over a 24 hour period at most 1/4 the watts that the panel collected at noon.
Craig T, please stop trolling.
Here is my video response:
https://www.youtube.com/watch?v=S-ezv5AckDk
I will do another follow-up today I think to address one of the central comments made in the OP over at WUWT, which will help clarify things.
BTW a true energy balance is conducted in kJ/h not W/m^2.
Nick, you have removed the per-area portion of the units. Surely you must realize that the rate of energy input “per area” is important. Using your suggestion, Jupiter receives much more kJ/hr of energy than does Earth, yet it’s atmosphere is far colder than Earth’s atmosphere. By ignoring area, you are ignoring that the solar constant at the distance of Jupiter is very weak, leading to low temperatures (yet the total energy input to the entire planet is large because Jupiter is so much larger than Earth).
Dr. Spencer, the point Nick is making is that “energy” and “flux” are not the same, which is the point you made.
So, a true “energy balance” must be in units of energy, not flux.
JDHuffman
Indeed. The difference between total energy in and total energy out is 3*10^22 Joules/year.
JDHuffman
IIRC Earth is gaining 3*10^22J/year.
E-man, are these “facts” similar to the “facts” in your unsolved murder?
JDHuffman
You don’t need to take my word for it. You can calculate it yourself four different ways, from the energy imbalance at TOA, from the change in bulk ocean temperature, from the change in sea level due to thermal expansion or the change in ocean heat content.
I’ve done all four and they agree around 3*10^22J/year.
Now E-man, if I know not to take your word for it, I would most assuredly know not to trust any nonsense you got from pseudoscience.
Got anything close to reality?
EM, I think you need to make that “most people could calculate it themselves four different ways.” JD has to argue without calculating anything.
Craig T, please stop trolling.
‘So, a true energy balance must be in units of energy, not flux.’
It is in units of energy. Energy per second per m^2, ie flux.
Per m^2 because nobody wants to work with 10^22 W.
I don’t think anyone expects you to understand, Nate.
Certainly not me….
True. BS is not my specialty.
Nate, please stop trolling.
I think the reason the power flux balance is selected is to toggle between power flux and temperature using the S-B equation.
Since hardly anybody understands how to apply emissivity properly the results are garbage.
Trenberth says the oceans have 0.97 emissivity. Because of the contiguous participating media at the interface, i.e. turbulent conduction, convection, advection and latent heat transfer processes, that is just flat wrong. Might be as little as 0.15. And IR temp/power flux measurements are incorrect because of emissivity comment above. Ask Kipp-Zonen, Apogee and Eppley. They won’t reply to me.
Converting 289 K surface average to 396 W/m^2 assuming 1.0 emissivity is also incorrect. Based on temperature surface emissivity is 0.16 = 63 actual/396 ideal. Based on surface distribution emissivity is 0.40 = 63 actual/160 balance.
kJ/h
Discular ISR = PI* 6371000^2 m^2 * 1,368 W/m^2 * 3.6 kJ/W = 6.280E17 kJ/h
Discular ASR = ISR * .7 = 4.40E17 kJ/h
Incoming energy is on the lit hemisphere only.
Outgoing leaves in all directions both lit and dark sides per 4.40E17 kJ/h = 1/R * A * (16 C (ToA – 40 C)).
No different from calculating the energy balance on an insulated house in winter.
Dont understand what Jupiter has to do with this. Comparing earth and moon is the most informative.
Without an atmosphere earth would be like the moon, blazing hot on lit side, bitter cold on dark, a barren rock receiving 25% to 45% more kJ/h and hotter than with an atmosphere. Nikolov and Kramm got lost in the weeds and wandered past this obvious comparison.
This lunar/earth model refutes, negates and nullifies the GHE.
“Because of the contiguous participating media at the interface, i.e. turbulent conduction, convection, advection and latent heat transfer processes, (ocean emissivity 0.97) is just flat wrong.”
Nick, measurements of wide swaths of ocean from airplanes, and observations of small areas of ocean near the shore from tall instrumented poles, in many published papers, all including natural turbulent conduction, convection, advection, latent heat transfer processes, wind, waves, tides, storms, and calm the ocean emissivity keeps being measured at ~0.97.
You can prove this to yourself by setting your inexpensive IR thermometer at emissivity 0.97 (or factory set at 0.95) pointing it at some ocean water with an imbedded thermometer. The IT gun will read the same as the thermometer. Heck, you don’t even need to travel to the ocean shore or rent an airplane, just draw a glass of tap water with an imbedded thermometer. Point your laser guided IR gun at the water and read out will be the same as the thermometer. Calibrate your IR gun using ice water & boiling water. IR gun will read 32F and 212F, they are actually very good these days.
“This lunar/earth model refutes, negates and nullifies the GHE.”
Don’t use a model, use measurements. The moon’s GHE is in the top few centimeters of the regolith and arguably comes out to about zero with global 255K steady state equilibrium based on Apollo and Diviner observations. Same as the earth observed from moon at 255K, they are both in same ~solar orbit. Given earth with atm. (instead of regolith) at global mean annualized steady state near surface T observed by thermometer at ~288K find Earth GHE confirmed by observation ~ 33K. Similarly find Venus GHE at ~500K, Mars GHE clear atm. ~5.1K.
Here again are those useless and misleading averages.
The moon is 390 K on the lit side, 95 K on the dark. UCLA Diviner data. So the average is lower. BFD!!!
That’s what the earth would be without an atmosphere.
Lit side: 270 K, dark side: 230 K, range: 40 C, average: 250 K. Habitable.
Lit side: 350 K, dark side 150 K, range: 200 C, average: 250 K.
Uninhabitable.
Identical averages, entirely different worlds.
Reply 2:
Emissivity & the Heat Balance
Emissivity is defined as the amount of radiative heat leaving a surface to the theoretical maximum or BB radiation at the surface temperature. The heat balance defines what enters and leaves a system, i.e.
Incoming = outgoing, W/m^2 = radiative + conductive + convective + latent
Emissivity = radiative / total W/m^2 = radiative / (radiative + conductive + convective + latent)
In a vacuum (conductive + convective + latent) = 0 and emissivity equals 1.0.
In open air full of molecules other transfer modes reduce radiations share and emissivity, e.g.:
conduction = 15%, convection =35%, latent = 30%, radiation & emissivity = 20%
The Instruments & Measurements
But wait, you say, upwelling LWIR power flux is actually measured.
Well, no its not.
IR instruments, e.g. pyrheliometers, radiometers, etc. dont directly measure power flux. They measure a relative temperature compared to heated/chilled/calibration/reference thermistors or thermopiles and INFER a power flux using that comparative temperature and ASSUMING an emissivity of 1.0. The Apogee instrument instruction book actually warns the owner/operator about this potential error noting that ground/surface emissivity can be less than 1.0.
That this warning went unheeded explains why SURFRAD upwelling LWIR with an assumed and uncorrected emissivity of 1.0 measures TWICE as much upwelling LWIR as incoming ISR, a rather egregious breach of energy conservation.
This also explains why USCRN data shows that the IR (SUR_TEMP) parallels the 1.5 m air temperature, (T_HR_AVG) and not the actual ground (SOIL_TEMP_5). The actual ground is warmer than the air temperature with few exceptions, contradicting the RGHE notion that the air warms the ground.
reply 3
Venus, we are told, has an atmosphere that is almost pure carbon dioxide and an extremely high surface temperature, 750 K, and this is allegedly due to the radiative greenhouse effect, RGHE. But the only apparent defense is, Well, WHAT else could it BE?!
Well, what follows is the else it could be. (Q = U * A * ΔT)
Venus is 70% of the Earths distance to the sun, its average solar constant/irradiance is about twice as intense as that of earth, 2,602 W/m^2 as opposed to 1,361 W/m^2.
But the albedo of Venus is 0.77 compared to 0.31 for the Earth – or – Venus 601.5 W/m^2 net ASR (absorbed solar radiation) compared to Earth 943.9 W/m^2 net ASR.
The Venusian atmosphere is 250 km thick as opposed to Earths at 100 km. Picture how hot you would get stacking 1.5 more blankets on your bed. RGHEs got jack to do with it, its all Q = U * A * ΔT.
The thermal conductivity of carbon dioxide is about half that of air, 0.0146 W/m-K as opposed to 0.0240 W/m-K so it takes twice the ΔT/m to move the same kJ from surface to ToA.
Put the higher irradiance & albedo (lower Q = lower ΔT), thickness (greater thickness increases ΔT) and conductivity (lower conductivity raises ΔT) all together: 601.5/943.9 * 250/100 * 0.0240/0.0146 = 2.61.
So, Q = U * A * ΔT suggests that the Venusian ΔT would be 2.61 times greater than that of Earth. If the surface of the Earth is 15C/288K and ToA is effectively 0K then Earth ΔT = 288K. Venus ΔT would be 2.61 * 288 K = 748.8 K surface temperature.
All explained, no need for any S-B BB RGHE hocus pocus.
Simplest explanation for the observation.
Nick, the no atm. moon rotation period is far different than Earth with atm., so hi lo readings are far different. However, global annualized mean T measurements (sparse thermometer, global brightness) result in the expected GHEs (expected from simple 1LOT calculations).
And, yes, the earth surface would be cooler, more like the moon surface steady state equilibrium, with no atm. and a transparent atm. at 1bar.
“and ASSUMING an emissivity of 1.0.”
No, their emissivity is closer to 0.95 and set by calibration to known BB radiation targets at known thermometer temperatures. Again, do the experiment with ice water and boiling water, your IR gun (brightness temperature) will read the thermometer temperature of the water. So the radiometers read the thermometer temperatures of the objects they are pointed at. This is how the UWIR and DWIR are known to within certain reasonable CIs at 95% significance levels, on a daily, 24/7 continuous basis at various earth locations.
“The Venusian atmosphere is 250 km thick as opposed to Earths at 100 km.”
No. Those numbers were chosen after the fact to make the math ratios approximate the already known, calculated and measured Venus mean global surface temperature. Cherry picking at its finest. A priori there is nothing fundamental to choose those numbers while the Venus temperature profile can be estimated from 1st principles and were reasonably predicted so the measurement instruments could be reasonably built. As I recall, the thermometer measurements nearly went off the instrumental scale high though surprising many.
Nick,
The average Temp of just the lit side of the Moon is well under 390 K, because in the polar regions the temp is very low.
But is going to satisfy 1360*cos(latitude)(1-alpha) = e sigma T^4.
Lets call this Tsb.
As you know if we do the same calc for the whole Earth we get Tsb = 255K, but the Earth average is in fact 288K.
The Earths surface radiates (T/Tsb)^4 = 1.62 x TOA, IOW 62% more than it should based on its Tsb
While the Moon radiates (T/Tsb)^4 ~ 1.
‘he thermal conductivity of carbon dioxide is about half that of air, 0.0146 W/m-K as opposed to 0.0240 W/m-K so it takes twice the ΔT/m to move the same kJ from surface to ToA.’
Thermal conductivity is irrelevant.
Convection, radiation overwhelm it.
Nick, this is gobbldegook.
The Moon is cooler than the Earth because it has no GHE.
My reply to you ended up in Ball4’s thread.
I don’t know what credentials Postma has, but credentials mean nothing if you get the basic physics wrong.
So he’s not much different from many here, such as Norman, Nate, DA, “Skeptic-Gone-WIld” (who keeps stealing my name), et al, who try to make up for their inability to learn by constant insults, false accusations, and misrepresentations.
Nothing new.
So says the troll that won’t do any experiments, won’t support any of his claims with valid textbook physics, makes up his own junk science and pretends to be an expert when he has never studied college physics. I have one more semester of physics than the troll.
He will troll this blog like a ghost haunts a house. Can’t get rid of this one.
People do not worry though. You will never convince him he is wrong and he WILL Troll you.
Norman, your insults, false accusations, and misrepresentations just make my case for me.
Thanks.
JDHuffman
But it will not stop your trolling or your pretending to be an expert at physics when you know you really don’t know any.
ibid.
A follow up vid:
https://www.youtube.com/watch?v=3Y9wV4cciXA
Joe Postma says:
June 5, 2019 at 8:45 AM
Here is my video response:
https://www.youtube.com/watch?v=S-ezv5AckDk
Yes. It is pseudo science.
I also prefer to call it cargo cult.
A cargo cult religion.
But it is based upon a model called a ideal thermal conductive
blackbody.
So a cargo cult is also based upon something which the believers
didn’t understand very well. Which was the US military landing airplane during WWII. And it was wonderful time and the native wanted more cargo planes landing on their island.
Now, Ideal thermally conductive blackbody is basically impossible, or I prefer to call it magical- or technology we don’t understand.
And one could characterize an Ideal thermally conductive blackbody
as magical passive refrigerator.
It take a surface which should be around 120 C and makes it 5 C and does this due to magical ability to conduct heat uniformly or a sphere. 5 C is a bit warm for a refrigerator, I believe you don’t want to have refrigerator exceed 4 C. But you also don’t want a refrigerator to go below 0 C. So it works as pretty good refrigerator because it doesn’t go below 0 C.
Anyhow the greenhouse effect theory starts with this model of a ideal thermally conductive blackbody and says it has uniform temperature of about 5.3 C.
And then like a bunch of monkeys they mess with this magical machine.
The machine was designed for vacuum, and how it would function under an atmosphere, is anyone’s guess.
Sorry Joe, but you are competing with a box of hammers.
Joe long ago used to comment regularly on this blog under screen name he used today. After taking much well deserved flak and having to deal with Dr. Spencer’s experimental results countering Joe’s comments, Joe stopped commenting regularly here. Joe then created the Climate Sophistry blog which is aptly named & totally devoid of experiments.
bobdroege, Ball4, please stop trolling.
Someone help me understand what Postma is saying here. I watched both of his videos, and is seems like his argument is like this:
Consider first the total radiance of the sun in W. (so, it’s time dependent and a huge number).
First, ask, “How much of that reaches earth?”
The obvious answer is: (pi*r^2) / (4/3*pi*R^3) where R = 93M miles, and r=4000 miles (approx radius of earth) times whatever the value of the total radiance is.
Now, for reasons which are beyond me, the interesting number is W/m^2, not the total W. So, we take the number we got above, and divide it by the same area we used above, and that gives 1370 W/m^2.
As far as I can tell, this is a number which everyone agrees on. Including Postma.
At this point there seems to be a divergence of thought processes.
So, help me understand this correctly:
Spencer, et al, say: We are interested in long term processes (which require more than a day to occur), so we actually want to use a number which is “per unit surface area spread around the entire globe” and therefore we will divide this by 4 to make up for the difference between the surface area of the earth, and the cross sectional area which the sun is radiating directly. And, thus is borne 342 W/m^2 (+/- rounding error). And, we will proceed from there because what we are interested in is long-term averages of “Radiation in” as opposed to “Radiation out” or “Radiation shared among other pieces, ie, from clouds to ground and so forth”.
As far as I can understand, Postma’s actual argument in his 2 videos goes like this:
Hey!! Maybe taking a time average like that isn’t actually valid. The differences from day to night and because of the earth shadowing itself, which affect how much sun-radiation falls per m^2 on a piece of the earth are so great that to average them in such a way is make mistakes. Perhaps the heat loss due to radiation from the ground and the ocean is dependent on cycles which are of smaller time than 1 day, and thus such averaging is incorrect. It is as if to say, “To smear the solar radiative incidence all over the earth is to say that there is No rotation, and no difference between pole and equator at all”
Again, as far as I can tell, that’s all he’s really saying:
That, for the sake of incoming solar radiance, the assumption of a valid averaging is equivalent to saying that the earth is flat and has twice the radius it actually has.
But, then, once you do that, you can’t treat the rest of the energy budget calculations as if they come from a globe, because you have already averaged the globe away.
For myself, with no degree but a very great interest in physics in general, I feel like that time scale of climate processes of long enough that a global average should be a fine approximation here. Without it, you are actually doing , not , and so you need to know instantaneously a whole bunch of things like cloud cover in detail all over the earth, and CO2 levels in each square km all over the earth, and etc, which seems quite difficult.
Again, am I missing something in his arguments here? They seem like picking at a fine point, which fine point, in the end, doesn’t really affect anything.
Sorry, I wasn’t sure which punctuation to use. In the 2nd last paragraph, I meant to write, “you are actually doing weather, not climate….
His whole argument is nonsense and a strawman.
No one is saying that the suns input is spread out instantaneously over the whole Earth. No one but Joe Postma.
No one thinks the Earth has uniform solar input.
Climate science averages total energy input from the sun over the daily cycle. So what?
They can do that if they want to study balance of energy in and out over the whole day, or longer.
They sometimes average over the seasons. Again they can do that if they want.
They can do these things because they are useful.
And no one but Joe Postma is putting in the wrong number for calculating the distance to the sun. That is totally on him.
Poor Nate gets tangled in his own web, again: “No one is saying that the suns input is spread out instantaneously over the whole Earth.”
http://www.climatechange2013.org/images/figures/WGI_AR5_Fig2-11.jpg
Nothing new.
All numbers shown are averages JD, not instaneous.
“Figure 2.11: Global mean energy budget under present-day climate conditions. Numbers state magnitudes of the individual energy fluxes in W m^2, adjusted within their uncertainty ranges to close the energy budgets. Numbers in parentheses attached to the energy fluxes cover the range of values in line with observational constraints.”
Nate and Craig get to argue.
May the most incompetent win.
JD is so dumb, he cant even tell that Craig and I agree!
Nate, please stop trolling.
–Rich IL says:
June 5, 2019 at 1:56 PM
Someone help me understand what Postma is saying here. I watched both of his videos, and is seems like his argument is like this:
Consider first the total radiance of the sun in W. (so, it’s time dependent and a huge number).–
At 1 AU distance [which is earth distance from the Sun. Or “exactly 149,597,870,700 metres, or about 150 million kilometres (93 million miles)” wiki, AU
There is about 1360 watts per square meter area of sunlight before the sunlight enters Earth atmosphere. But Postma and others are giving the number of 1370 watts per square meter. So say 1360 to 1370 watts per square meter.
Total radiance of sun depends on how close you are to sun and/or if talking striking the large area of the Earth and that’s pretty big number. It’s total disk area of earth in square meters times about 1360 watts or Earth’s radius in meters squared times pi times 1360 watts.
In terms of times it’s 1360 watts per square meter per second.
Wiki:
“Total energy from the Sun that strikes the face of the Earth each second: 1.7×10^17 joules” [or watts].
https://en.wikipedia.org/wiki/Orders_of_magnitude_(energy)
“First, ask, “How much of that reaches earth?”
The obvious answer is: (pi*r^2) / (4/3*pi*R^3) where R = 93M miles, and r=4000 miles (approx radius of earth) times whatever the value of the total radiance is.”
Yeah, above.
Of course you could asking how sunlight reaches the surface of Earth. The above is amount reaching the Top Of Atmosphere [TOA].
The amount of sunlight reaching the surface, depends where the sun is in the sky. When sun is directly overhead [zenith]:
“If the extraterrestrial solar radiation is 1367 watts per square meter (the value when the Earth–Sun distance is 1 astronomical unit), then the direct sunlight at Earth’s surface when the Sun is at the zenith is about 1050 W/m2, but the total amount (direct and indirect from the atmosphere) hitting the ground is around 1120 W/m2.”
https://en.wikipedia.org/wiki/Sunlight
“Now, for reasons which are beyond me, the interesting number is W/m^2, not the total W. So, we take the number we got above, and divide it by the same area we used above, and that gives 1370 W/m^2.
As far as I can tell, this is a number which everyone agrees on. Including Postma.
At this point there seems to be a divergence of thought processes.”
Yes agree it’s 1360, 1367, or 1370 watts per square meter- all are close enough.
“So, help me understand this correctly:
Spencer, et al, say: We are interested in long term processes (which require more than a day to occur), so we actually want to use a number which is “per unit surface area spread around the entire globe” and therefore we will divide this by 4 to make up for the difference between the surface area of the earth, and the cross sectional area which the sun is radiating directly. And, thus is borne 342 W/m^2 (+/- rounding error). And, we will proceed from there because what we are interested in is long-term averages of “Radiation in” as opposed to “Radiation out” or “Radiation shared among other pieces, ie, from clouds to ground and so forth”.
As far as I can understand, Postma’s actual argument in his 2 videos goes like this:
Hey!! Maybe taking a time average like that isn’t actually valid. The differences from day to night and because of the earth shadowing itself, which affect how much sun-radiation falls per m^2 on a piece of the earth are so great that to average them in such a way is make mistakes. …”
I don’t think Postma mentioned anything about time average.
But 342 W/m^2 in vacuum and if blackbody surface has temperature of -18 C.
And greenhouse effect theory says:
However, because Earth reflects about 30% of the incoming sunlight, this idealized planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) would be about −18 °C (0 °F).”
https://en.wikipedia.org/wiki/Greenhouse_effect
So I would say [and I would guess Postma agrees] that if planet only receives 342 W/m^2 of sunlight, the planet would be -18 C
but Earth receives more than this, and only by averaging it, does it seem that Earth only gets as much a constant source of 342 W/m^2 of sunlight.
Or constant source of 342 watts will never warm above -18 C but because Earth get higher amounts, it has higher average temperature than -18 C .
But anyhow, I roughly agree with:
“Hey!! Maybe taking a time average like that isn’t actually valid.”
“but Earth receives more than this, and only by averaging it, does it seem that Earth only gets as much a constant source of 342 W/m^2 of sunlight.”
Averaging can’t make the amount received by Earth less.
It does get 340W on average per m^2, no more.
“constant source of 342 watts will never warm above -18 C but because Earth get higher amounts, it has higher average temperature than -18 C .”
No, doesnt get more. So it needs a GHE to reach 288 K.
–Nate says:
June 6, 2019 at 3:19 PM
“but Earth receives more than this, and only by averaging it, does it seem that Earth only gets as much a constant source of 342 W/m^2 of sunlight.”
Averaging can’t make the amount received by Earth less.–
Well, as Joe Postma says a constant source of sunlight at 342 watt per square meter is available at about 2 AU distance.
And that constant source of sunlight not magnified will only cause a blackbody surface to reach about -18 C.***
[[You can make the sunlight more intense by magnifying the sunlight and it can quickly reach a much higher temperature.
With large enough magnification you could make sunlight melt a brick at 2 AU.
But per square meter of sunlight, the magnification, doesn’t add energy, it merely makes the sunlight more intense.
Or with magnification of sunlight you can boil water, without magnification the 1050 watts per square meter of sunlight can’t cause water to be 100 C [and boil water].]]
So we are not on flat earth 2 AU from the Sun, we at 1 AU and we have more intense sunlight at 1 AU as compared to 2 AU.
The 1 AU distance can warm a sidewalk to about 60 C. And only thing preventing sidewalk to warm up to around 80 C, is the convection heat loss to cooler air above it.
Or if prevent convection heat lost to the colder air, it will reach about 80 C.
And that occurs in parked car with window rolled up or insulated box or greenhouse. If air is about 50 C and sun is directly over head the surface can warm to about 70 C, but with 30 to 40 C air, it warms to about 60 C.
And if in car with window rolled up and at 2 AU, the temperature is limited to about -18 C.
*** Correction:
In terms of blackbody in vacuum
342 watts is about 5 C
242 watts is about 255 K [-18 C ]
So flat world at 2 AU would need to reflect as much sunlight a Earth does lower amount it absorbs to about 240 watts per square meter. And it of course it would also emit about 240 watts of IR.
Ref:
“The distance from the Sun to the spacecraft would be 2 AUs so… d = 2. If we plug that into the equation 1/d^2 = 1/2^2 = 1/4 = 25% The spacecraft is getting only one quarter of the amount of sunlight that would reach it if it were near Earth. ”
https://tinyurl.com/y4ae9tkj
So if around 2.4 AU then sunlight would about 240 watts and could warm to -18 C [255 K].
Vesta is around 2.4 AU.
4Vesta, wiki:
Aphelion 2.57138 AU
Perihelion 2.15221 AU
Semi-major axis 2.36179 AU
https://en.wikipedia.org/wiki/4_Vesta
And Dawn spacecraft went there and measured it’s surface temperature.
Here it is:
“Temperature: 85 to 255 K ”
https://www.space.com/12097-vesta-asteroid-facts-solar-system.html
So if Vesta were a flat planet and always faced the sun, it would have temperature of about 255 K [-18 C].
Next thing you said:
–“constant source of 342 watts will never warm above -18 C but because Earth get higher amounts, it has higher average temperature than -18 C .”
No, doesnt get more. So it needs a GHE to reach 288 K.–
Well I would say a flat world at 2 AU and had atmosphere which reflected the sunlight so that only 240 watts was absorbed and 240 watts emitted, it could have warmer temperature than -18 C.
So it might have warmer temperature than -18 C without any greenhouse gases, but since I am lukewarmer, if it had greenhouse gases it could be a bit warmer than compared not having greenhouse gases.
But I generally tend to think an atmosphere would cause spherical world with a day and night to have larger warming effect [as compared airless world].
Dr Spencer,
You claim it is legitimate to divide solar INPUT by 4, i.e. spread it over the whole earth. But the sun shines on 1/2 the Earth, while the OUTPUT is over the whole Earth. The output is 2x the input.
We can test out your theory by putting a 1 m^2 ice cube onto a 1 m^2 hibachi grill set to 300F. According to your “science”, the ice cube never melts. Why? 300F -> 1800 W/m^2, Divided by 6 sides -> 300 W/m^2 = <0C.
This is exactly what you're claiming, though you will deny it.
Now what do you have to say about this?
Peace. -Zoe
‘We can test out your theory by putting a 1 m^2 ice cube onto a 1 m^2 hibachi grill set to 300F. According to your science, the ice cube never melts. Why? 300F -> 1800 W/m^2, Divided by 6 sides -> 300 W/m^2 = <0C.'
No, thats just dumb. Nobody's 'science' is saying that.
Nobody in science is saying it. But everyone in pseudoscience is saying it.
“1360/4 = 340”
The 1800 W are all in excess of what the ice cube is normally receiving from a room temp environment on all sides.
It melts.
1360 W/m^2 is in excess of what Earth normally receives from space.
Does Earth melt?
(Since you won’t understand, Nate, the point is your “logic” is a big fail.)
JD as a pure troll, you dont need to follow the discussion or even make sense.
So go troll your mom.
Does the ice cube melt from the top or from the bottom?
Nate, bobdroege, please stop trolling.
Dr. Spencer does not claim that the output is 2x the input. He is very much in agreement that *on average* the incoming flux is 340 W/m^2 and the outgoing flux is 340 W/m^2 at TOA. Or in terms of the energy over one orbital cycle it is 340 W-years/m^2 incoming and 340 W-years/m^2 outgoing. The Earth is, for the most part, in energy balance.
Here is a list of Postma’s strawmen based on the bullets points in his second video.
1. The diagram in question forms the basis of climate physics. False. This diagram does not form the basis of climate physics. This is just a simple 3-layer box model used to facilitate discussion regarding the flux at each layer and Earth’s energy budget.
2. The Sun cannot create the climate. False. Climate scientists fully acknowledge the Sun’s role in modulating the climate.
3. The climate creates itself ex-nihilo. False. Climate scientists fully acknowledge that the climate is the product of many factors including but not limited to solar radiation, albedo, aerosol optical depths, infrared active gas species, shortwave active gas species, ocean currents, convection, advective processes, continental positioning, volcanic activity, and the list goes on and on.
4. Climate science is based on flat Earth physics. False. No one believes the Earth is flat. More complete models that explain and predict the climate are fully coupled, 3-dimensional, based on observational/experiment evidence and the laws of physics, and are represented by a round Earth.
5. Climate science is refuted by real greenhouses. False. The fundamental principals in play in a structure greenhouse are not the same as the greenhouse effect as it occurs in the atmosphere.
I’m not sure what “Fake version of greenhouse effect” or “output reversed for input” even means so I can’t comment on that.
And although they aren’t really strawmen these points are not correct either.
1. The Sun does NOT create the climate. The Sun plays a role in modulating the climate, but it is THE agent that creates it.
2. Climate is NOT the transfer of solar heat. Climate is the prevailing conditions or average values of various atmospheric properties over large spatial and temporal domains.
Postma, if I have in any way misrepresented your position then please correct any inaccuracies. If you feel that I have unjustly presented your bullet points as strawman then please provide justification as to why you think these are genuine positions held by the scientific consensus.
Typo…that should read…
1. The Sun does NOT create the climate. The Sun plays a role in modulating the climate, but it is *not* THE agent that creates it.
bdgwx, please stop trolling.
My comment disappeared?
bdgwx,
Can you read? The output area is 2x the input area.
The ice cube doesn’t melt, according to Dr. Spencer.
ZP,
The 340 W/m^2 incoming is the average over 510e12 m^2 and one orbital cycle. The 340 W/m^2 outgoing is the average over 510e12 m^2 and one orbit cycle. The incoming and outgoing areas are the same.
Exposing 1 m^3 cube of ice to > 0C temperatures will cause it to melt. Adding 1800 W/m^2 to the bottom face will cause it to melt faster. The fact that the 1800 W/m^2 to the bottom face averages out to 300 W/m^2 over the entire surface area does not change the fact that it will still melt. Nor does it change the fact that the mass is accumulating energy at a rate of 1800 joules per second.
bdgwx recognizes that you can’t divide 1800 W/m^2, but he still wants to divide incoming!
(I love it when they get caught in their own pseudoscience.)
JD objects to averages. He never took advanced Stats.
Statistics is for girls that can’t understand physics.
And he’s sexist too…
What a catch!
That’s what she said….
Dr. Spencer:
The model of the atmosphere that you’re showing in the picture is incorrect. There is no radiation between the surface and atmosphere because they are in intimate contact. Only conduction and convection heat transfer may exist between two bodies in intimate contact. This is what we learned at school, and they are basic heat transfer facts that cannot be compromised.
As for the solar constant, we see and measure 1370 W m-2 on a circle as viewed from the sun, and we go by what we see and measure. The circle has an area that is 1/4 of the surface of the sphere of the earth. These are facts that should be adhered to for good modelling of the atmosphere.
Nate,
The hibachi grill is in a vacuum chamber. Did you not understand the purpose of my example?
Oh, and vacuum matters why?
The ice melts. Go figure out why.
Nate, please stop trolling.
JDHuffman,
You got it. Their stench of hypocrisy is overbearing, right?
Right!
bdgwx,
“The incoming and outgoing areas are the same.”
Uhm, no.
Sorry but in real life the ice cube heats to a full 1800 W/m^2, and that’s what it will emit on all “sides” … as it melts to a pool of water.
The lesson: We don’t divide input area by output area. Not for Earth, not for the Cube. Welcome to Physics.
”Sorry but in real life the ice cube heats to a full 1800 W/m^2, and that’s what it will emit on all “sides” … as it melts to a pool of water.”
Best laugh I’ve had today. Even given some of the above comments. So, uhm, no.
“Sorry but in real life the ice cube heats to a full 1800 W/m^2, and that’s what it will emit on all “sides”…as it melts to a pool of water.”
Best second laugh I’ve had today. Even given some of the above comments. So, uhm, no. If you think not, run the test. Freeze a thermometer in your ice cube, let us know the results.
‘Sorry but in real life the ice cube heats to a full 1800 W/m^2, and thats what it will emit on all sides as it melts to a pool of water.’
Zoe, is that really what you think?
You’re fine with the cube receiving 1×1800 W, while emitting 6 x 1800 W?!
Ball4, Nate, please stop trolling.
One of the problems with this issue is that it involves “flux”. And a “flux” appears to confuse many.
A flux is not a normal scalar quantity. Typically it cannot be processed with simple arithmetic. A flux cannot be added, subtracted, divided, averaged, etc., except in special cases.
Flux is not conserved. Energy is conserved, but not flux. A flux obeys the inverse-square law, meaning that it changes with distance. For example, the solar flux at Sun’s surface is 64,000,000 Watts/m^2. But the flux is reduced to about 1365 W/m^2 when it reaches Earth. (Some silly pseudoscience clown might ask “Where did the missing 63998635 W/m^2 go?”)
So the concept of dividing the solar flux by 4 is wrong from the start. It quickly leads to pseudoscience, as seen with the GHE nonsense.
Ok, JD 6:43pm I’ll bite on flux. Laughing a little along the way, it’s been a humorous blog read today.
The total flux of a property is the product of 1) the number density of carriers of the property, 2) the speed of the carriers, and 3) the amount of the property each one carries. Many define Q/A as a heat flux but get stymied right away in 1) finding the carriers of such a flux, in this example: heat.
Q/A in a gas cannot be decomposed into the product of the number density of molecules, times their speed times the quantity of heat each carries.
Understanding starts to rise, the number of stymies reduces, if one instead will specify the flux of kinetic energy (KE) because each molecule DOES have a definite KE. You then properly can compute a kinetic energy flux and there is NO need to relabel it as a heat flux.
So wherever the term heat flux is used or meant (as you do simply by flux), it can be replaced by energy flux and as you may know, energy is conserved.
As I indicated, a “flux” confuses many.
JD, confused like many over flux, 6:43pm also opines: “So the concept of dividing the solar flux by 4 is wrong from the start.”
Given Re for the mean radius of the earth as a sphere, (So) for the solar irradiance at earth orbit (say 1369 W/m^2 annualized or the latest TSI as measured by SORCE), the amount of solar radiant energy intercepted by Earth is So*pi*Re^2.
Probably not going to do an experiment the size of the Earth to prove this to yourself so take a golf ball illuminated by a sun like source far enough away the rays intercepted are sufficiently collimated (you know, like the sun while standing on the ISS properly suited). Hold up a white sheet of paper right behind the golf ball radius Rgb. See the black shadow area? That missing energy is the amount intercepted by the golf ball and it is equal to So*pi*Rgb^2.
Spherical Earth total mean surface area is 4*pi*Re^2.
So the gross solar irradiance spread uniformly over the entire globe mean surface area is (So*pi*Re^2/4*pi*Re^2) = (1369/4) W/m^2. Fraction of that is reflected/scattered to space (albedo), as the oceans appear dark in satellite pictures, brightened here and there by clouds. This yields a net solar irradiance, with albedo=0.3, of S = So(1-albedo)/4 ~ 240 W/m^2.
It is humorous to me why this basic geometry/albedo/collimated illumination is controversial at all. Now you can argue the Earth is an oblate spheroid, so when the annual cycle in Earth’s declination angle and Earth-sun distance are accounted for, the well-known So/4 expression for the mean solar irradiance of a spherical Earth becomes So/4.003 for an oblate spheroid, where So is the TSI at earth annual orbit.
Ball4, please stop trolling.
Ball4 is very confused!
Its not in any physics book like that!
And he assumes if energy is conserved then so must be the flux, which is where he goes wrong!
Point proven!
You don’t have to divide the solar constant by 4 to get 340 W/m^2. This can be computed by integrating 1360 W/m^2 with respect to the sine of the incident angle of the radiation for the lit hemisphere and the repeating the same procedure with 0 W/m^2 for the unlit hemisphere and sum them together. You’ll get 340 W/m^2 either way. It’s just easier to use established geometrical principals. But, if that offends you then by all means do it the long way and compute the integrals.
bdgwx, you remain confused.
It is not your geometry, or math, that is the problem. it is your knowledge of the relevant physics.
“A flux is not a normal scalar quantity. Typically it cannot be processed with simple arithmetic. A flux cannot be added, subtracted, divided, averaged, etc., except in special cases.”
The total flux of a property is the product of 1) the number density of carriers of the property, 2) the speed of the carriers, and 3) the amount of the property each one carries.
If the speed of the property carriers goes down by a factor 4, divide the total flux by 4.
If the number density of carriers of the property is added to, just do the addition and recompute the property total flux. It’s not that hard JD, even you can do it, well, maybe, I’d need to see proof.
This is a general result for all kinds of fluxes of quantities to which physical reality can be assigned.
Sometimes I almost feel sorry for fluffball.
Almost….
‘Flux is not conserved. Energy is conserved, but not flux. A flux obeys the inverse-square law, meaning that it changes with distance. For example, the solar flux at Suns surface is 64,000,000 Watts/m^2. But the flux is reduced to about 1365 W/m^2 when it reaches Earth. ‘
Very good JD. The total energy of the sun, at the distance of the Earth has been divided by the total area of a sphere with the Earth-sun distance as its radius.
The energy gets divided by area that it is spread over to find average flux.
But you see you are doing EXACTLY what we are doing when we divide the total energy hitting the Earth by the surface area of the Earth, and getting 340 W/m^2 as average flux over a full day.
Wrong Nate. The inverse-square law is valid. Flux actually decreases with distance. Trying to average that flux over Earth’s surface is invalid.
It’s just one more instance of your not being able to accept reality.
Nothing new.
JD, if you want to be serious for a brief moment, what exactly in my post do you object to?
“We” are not doing what you are doing.
“We” are obeying the laws of physics.
You are not.
Nothing new.
IOW, nothing. You cannot point to any science that you object to.
Ur just trolling as usual.
I don’t object to science.
Your problem is you never recognize any….
W/m2 have nothing to do with temperatures.
Ice emitting 300 W/m2 will melt if exposed to a polished silver teapot containing water at 90 C, emitting less than 30 W/m2.
According to pseudoscientific climate cult types, the fact that the ice is emitting more than it receives means it will get colder, and the teapot receiving an excess of 270 W/m2 means the water will get hotter!
Oh well, all part of the rich tapestry of pseudoscientific stupidity and ignorance.
Misusing the same radiative equations, same inputs, the pseudoscientists calculate the Earth’s surface temperature as over 1300 K (when the surface was molten), 373 K (when the first liquid water appeared), 288 K (God alone knows why), or anything in between! A miracle of climatological proportions!
And still, nobody can actually describe the GHE in any useful way. No wonder.
Cheers.
Mike Flynn points out: “According to pseudoscientific climate cult types, the fact that the ice is emitting more than it receives means it will get colder, and the teapot receiving an excess of 270 W/m^2 means the water will get hotter!
Mike Flynn
Why do you want to demonstrate you lack of any understanding of heat transfer by posting your point?
If you knew any real physics and actually cared to make a valid point you would know that the situation you describe means your polished pot at 90 C radiating only 30 W/m^2 has an emissivity of around 0.03.
That means it will absorb less than 9 watts/me^2 from the ice.
I am not sure why you think the ice would melt if all there was was the pot and ice?
I really do not think you thought this through at all and it shows.
I also notice when you get approval from the ignorant blog troll you are way on the wrong track of a valid point. If that one supports you run the other way!! In a hurry!
N,
You wrote –
“I am not sure why you think the ice would melt if all there was was the pot and ice?”
Do you think the reason you are not sure, might be that you are stupid and ignorant?
Put some ice up against a teapot containing water at 90 C. Convince yourself that the ice isn’t melting, and that the water is getting hotter because it is absorbing energy from the ice!
Good luck if you can make a teapot full of water at 90 C even hotter by surrounding it with ice! You must be either incredibly stupid, or incredibly delusional if you believe this is possible.
You still cannot even describe this GHE that you worship, can you?
Ah, the ineffable joy of worshipping and invisible deity? Who can prove it doesn’t exist?
Carry on dreaming.
Cheers.
Mike Flynn
YOU: “Ice emitting 300 W/m2 will melt if exposed to a polished silver teapot containing water at 90 C, emitting less than 30 W/m2.”
Never in your initial post do you state that the ice will touch the post and you infer quite the opposite. You use the word exposed meaning it is in the view of the IR emitted by the pot and you lead a reader to believe this is what you meant as you intentionally put the radiant energy emitted by the pot. If you meant touching it say so.
Now you can do the test both myself and Ball4 ask you to do but you refuse. Why is this?
Add heat to the water until it reaches a steady state temperature with the environment and the heat source (temperature no longer changes). Now put ice on the pot and yes the temperature goes down even with the heat source. Now put dry ice against the pot (careful use tongs) and it cools more. Maintaining the same amount of heat touch ice to the pot again. The temperature goes up. Why? The ice is adding more energy to the pot than the dry ice could so with the heat source you get more internal energy. Do the test and quit the nonsense of your ignorant computer generated posts.
Norman, if you knew the relevant physics then you would understand changing from dry ice to water ice proves nothing about “cold” warming “hot”.
You just keep confirming you don’t understand thermodynamics, and can’t learn.
Nothing new.
JDHuffman
Since I have a far greater understanding of real thermodynamics (textbook physics kind not your made up fake version) that your point is invalid. Changing from dry ice to water ice proves everything. The energy of the colder ice transfers to the hotter heated fluid. You are just plain wrong about most everything. Too bad you will never understand how ignorant you actually are, it is easy to stay that way for you. Just avoid the textbooks or doing any experiments you can remain ignorant as long as you want.
Yes Norman. Keep imagining.
You understand the relevant physics, and there is a Santa.
Reality is just for adults, huh?
JDHuffman
No imagining necessary I do know so much more actual physics than you will ever be able to learn. You are good at trolling. That is all you know. You know zilch textbook physics. You never have and you are not able to learn now. Carry on with some more trolling. It is all you are able to accomplish.
Norman, if you really knew more physics than me, you should be able to demonstrate such.
Instead, your constant insults, misrepresentations, and false accusations just reveal your massive insecurity.
Nothing new.
JDHuffman
I believe the approximate count on demonstrating valid physics to you is well over 1000 demonstrations. I have linked you many times to valid physics textbooks. As to date you have not linked to even one valid textbook physics. I have linked you to measured empirical values demonstrating your fake physics is total junk. You, on the other hand, have demonstrated zero empirical evidence to support any of your wild conjectures that are not even logical.
Yes I am the one who demonstrates actual physics. You do nothing but troll. If I post a link to a valid physics textbook your only reply (and not just to me but others) is to state we do not understand the link. A fairly pointless opinion as you never even attempt to describe what is not understood.
You do pretend to be an expert. You look up a few physics words from time to time and post them like an expert. You don’t know what enthalpy or entropy are but you post them as if you do. You could not explain either concept if you had to.
I still want you to demonstrate how Poynting Vectors prove fluxes don’t add. I think I asked you to explain this several times and you avoided an answer each time.
You have not demonstrated even a basic level of understanding of even simple physics. But you will pretend you are an expert.
At this time I believe over 90% of the posters know you are the blog troll. Purpose only to bait people into senseless posts that never resolve anything (like the thousands on Moon rotation).
One thing has not changed about you. When you went as ger.an you were the blog troll and now you changed to JDHuffman you are the same blog troll. Nothing new and always expected.
Norman, if you really knew more physics than me, you should be able to demonstrate such.
(Long typing exercises mean nothing, but you can’t learn.)
Instead, your constant insults, misrepresentations, and false accusations just reveal your massive insecurity.
Nothing new.
JDHuffman
To make it easy for you to follow, I have given you countless links to real and valid physics. You have never supported your false physics with any valid physics. You offer your own made up cartoons.
YOU: “Norman, if you really knew more physics than me, you should be able to demonstrate such.”
So yes I have demonstrated I know far more physics than you. Also I have this in my favor. My physics is real and tested. Yours is made up and NEVER tested. I think that about settles this line of pointless debate. Yes I know much more physics than you.
I know a lot less than other posters on this blog but I do know quite a bit more than you.
Norman, if you really knew more physics than me, you should be able to demonstrate such.
Instead, your constant insults, misrepresentations, and false accusations just reveal your massive insecurity.
JDHuffman
Troll tactic “Rinse and repeat” when you can’t come up with a useful answer or point just repeat the same post over and over.
Let the rest of the posters see the troll that you are.
Troll on repeat some more, much easier than thinking isn’t it.
ibid.
Flynn,
Water appeared in liquid form before the earth cooled to 373 K because the atmospheric pressure was higher.
bobdroege, please stop trolling.
An ice cube has 6 sides. It gets heated from one. So according to Spencer’s religion it couldn’t possibly melt in my example.
Remember kids, we divide the input flux by ratio of output flux area over input flux area. 1800/6 = 300. Oh we don’t? Hypocrites!
“An ice cube has 6 sides. It gets heated from one.”
What is the illumination and conductive/convective situation on the other 5 sides of your ice cube? Do you not agree that info. is important?
Zoe Phin
I am thinking you are joking around trying to be funny with your ridiculous ice idea. Please say you are not trying to make an actual scientific or logical point with this!
Norman, Zoe is ridiculing the nonsense espoused by pseudoscience. You just don’t get it.
Nothing new.
Nate,
“The globe started out hot and cooled to roughly the temperatures they are today.
Irrelevant to today.
Today all that matters is the balance in and out.”
This right here^^^^ displays the fundamental flaw in the thinking of the GHE paradigm…
Nate,
i take to equal sized turkeys and place them on the counter under equal heat lamps..
I throw a terry cloth over one of them…
30 mins later i measure their skin temp…
The bare one is 29 C
Room temp is 18 C
The one under the cloth is 49 C
Explain the difference….
Phil J,
Your example seems to have no connection to my point that the what the Earth’s Temperature history was 4 Billion years ago is irrelevant to today.
And your turkey’s temperatures, like that of the the Earth, are determined by balancing input and output energy flux.
Nate,
“Your example seems to have no connection to my point that the what the Earths Temperature history was 4 Billion years ago is irrelevant to today.”
well lets see if i can make the connection for you…
the uncovered turkey was a raw room temp bird which had its surface temp raised by the heat lamp..
the other bird was removed from the oven with an internal temp around 200 C, placed on the counter and covered with the cloth… it cooled despite the input from the heat lamp…
The GHE hypothesis takes a cold black body earth (like the cold room temp turkey) and warms it with the solar input to 255K then tries to raise the temp by adding atmosphere and back radiation… This is of course backward thinking…
The real Earth is a hot ball of magma and gases that has cooled for 4 billion years despite the solar input (like the hot baked turkey)
Entropy dictates that the Earth will continue to cool until (like the Moon) it approaches thermodynamic equilibrium with its surroundings
to sum up, the reason the Earth surface is 33K warmer than its computed BB temp, is because it has not yet cooled to that BB temp. It may take billions more years to do so….
Far from being irrelevant, the Earths total internal energy and temp are crucial to understanding how it has cooled, and how it continues to cool, short term oscillations in energy input/output notwithstanding…
Phil, your turkey was removed from the oven days ago. The Earth cooled from its molten state 3.5 B y ago, when liquid water was present and life.
Today the internal geothermal heat flux reaching the surface is measured to be 90 mW/m^2 on average. That is neglible compared to the GHE or the solar input.
So No, you have not made the case that the Earth is 33K warmer because it ‘has not yet cooled to that BB temperature’
Nate, please stop trolling.
“What is the illumination and conductive/convective situation on the other 5 sides of your ice cube?
Do you not agree that info. is important?”
It’s so important that you don’t care to ask those questions about the Earth. You just divide by 4. Now when dealing with an ice cube, suddenly it becomes important … hypocrite!
“Please say you are not trying to make an actual scientific or logical point with this!”
I am enjoying the sophist and hypocrite comments I’ve received.
Zoe, the illumination and conductive/convective situations on all sides of the earth are well known & continuously monitored, not ignored. You need to do so for your ice cube also. The simple geometry results in a divisor of 4 for the rotating earth. Again, it is humorous to me why this basic geometry/albedo/collimated illumination is controversial at all.
Again, if you think not, run your test. Freeze a thermometer in your ice cube, let us know the results. Hint: an IR thermometer can be used to measure the ice cube emission on all 6 sides, then the puddle of water.
Ball4, please stop trolling.
It is obvious to anyone except a pseudoscientific GHE cultist that the Earth’s surface has cooled since its original molten state.
Neither four and a half billion years of sunshine, nor an atmosphere containing far greater amounts of CO2 than now, nor anything else of any sort has managed to stop the surface cooling.
No heat accumulation, multiplication, addition, or anything similar has stopped the surface from cooling. Open your eyes, and look down between your feet – if you don’t believe your lying eyes, that is not my problem.
Work away feverishly at your stupid calculations, and your endless reanalysis of history. Nature doesn’t care. The future remains as inscrutable as ever. Apart from the climate cultists, the only people liable to believe prophecies of CO2 induced doom are the usual crop of dimwits – mainly journalists and politicians!
Onwards and upwards, eh?
Cheers.
If I look at the sun, I can see half of it. Did you know that? There’s a square meter shining 1361 W/m^2 at me, and look, another one, and another one, and another one, … . Surely we can add all those square meters I see into a really big number of W/m^2.
ZP,
That is the precise method Al Gore used to get the “Extremely Hot, Several Million Degrees” temperature of the interior of the Earth.
A thousand here, a thousand there, add them all up, and pretty soon you’ve got several million! No doubt Al Gore is a GHE true believer.
Cheers.
Dont look at the sun too long, you may go blind.
And you could bake your brain, which Im afraid may have already happened.
Nate, please stop trolling.
The bottom line here is that Joe Postma is right and Dr. Spencer is wrong. After all, anyone using the figure 255 K (-18C) for surface Tav in absence of radiative atmosphere is wrong.
Joe’s argument “flat earth physics” may be poorly worded. He is not debating whether dividing solar flux by 4 gives an average value for solar illumination of a rotating sphere. It is unfortunate that his online exchanges degenerated into bad language, but that is likely the product of people not just misinterpreting, but willfully misinterpreting his argument.
Joe’s argument as I see it is that using average solar insolation in conjunction with instantaneous radiative balance calculations like the Steran-Boltzmann equation cannot possibly yield an accurate figure for “surface Tav without radiative atmosphere” for this ocean planet. In this, all the empirical evidence supports Joe’s position, not that of the believer + lukewarmer 255 K crowd.
Undeniable proof of this is given by comparing the S-B estimate for lunar average temperature to the empirical results returned by the DIVINER radiometer flown on the Lunar Reconnaissance Orbiter. Using an albedo of 0.15 and solar irradiance of 1370 w/m2, the S-B calculation (flat moon physics) returns 268 K for surface Tav. Compared to the empirical radiometer data, this is around 80 K too high.
If the S-B calculation won’t work for the simple surface of the Moon, there is zero chance it will work for the far more complex surface of the Earth. It is worth researching the subsequent papers on modelling Lunar temperatures that followed the DIVINER results. The Moon had to be treated as a rotating sphere, surface texture taken into account, actual radiative properties modelled and critically surface conduction. Modelling the Moon as a simple grey-body and ignoring the critical variable time simply didn’t work.
How far out is the 255 K figure for “Surface Tav in absence of radiative atmosphere” for our planet? It’s out by around 57 K. It’s 57 degrees too low. And that means our radiatively cooled atmosphere acts to cool the surface of our planet not warm it. And they wonder why the climate models that include CO2 as a positive forcing don’t work …
Adding 57 to 255 gives 312 K.
Applying S/B to the “disk” flux (480 Watts’m^2), and using 0.95 emissivity, gives 307 K.
Very close, even with no other adjustments.
Clearly the 255 K has serious problems.
“Using an albedo of 0.15…”
Thats a Lambertian albedo, Konrad, where the moon regolith surface is known to be non-Lambertian.
Once the sparse Apollo thermometer measurements are used to calibrate the Diviner brightness temperatures, the tested non-Lambertian regolith albedo used, and the tested amount of diffraction in the regolith accounted for in its emissivity, then the global multi-annualized moon steady state equilibrium surface thermometer Tav can be reasonably estimated at 255K, give or take.
That 255K estimate has wide CIs at 95% significance level, an estimate which cannot be checked against a moon global thermometer field until one is installed. There is no motivation for this currently so research funds in the field have dried up & researchers moved on to other current fields of interest.
The ~33K GHE for Earth is observed & measured to reasonably small CI at 95% significance level and computed from 1st principles along with many other solar system objects. There are plenty of research articles to catch up on all of this to form an opinion if one has the chops to go to the library (or online) & reasonably read enough of them.
Sorry “Ball4′, but you can never defeat me. Nor can Dr. Spenser.
To defeat me, you would need to provide empirical evidence in support of the crazed claim that the sun alone could only heat the complex materials of this planet to a frozen average of -18C (255 K).
You can never provide such evidence. All you can depend on is frantic censoring of those like myself who got the physics right.
But despite the concerted efforts of GooGooPlex, Spacechook and Twitface, the permanent Internet record remains.
“Konrad” said the net effect of the atmosphere was to cool the surface of the planet by 26 degrees, not warm it by 33 degrees.
Thrash, flex, squeal and moan! Whatev’s darl …
I got it right. I said the atmosphere acts to cool the surface of this planet. I said it first. And given this is the age of the Internet, that is forever.
To defeat me, you need to destroy the scientific method. What price your ego … ?
Providing empirical evidence in support that the sun alone could only heat the complex materials of objects at earth’s orbit to a reasonably estimated multi-annual mostly frozen near surface average of -18C (255 K) is fairly easy for an IR transparent atm. optical depth or no atm. at all regolith as it exists on-line and in the stacks at your local college library to find for free. You need the chops to find it and do the work to understand it.
In fact, similar observations exist for many objects in other sun centered orbits in the solar system. You do have to do the work though. Google maps can be used to find your local college library and even estimate its distance to reasonable CI at 95% confidence level. The librarians there will be more than pleased to exercise their craft.
Poor fluffball is lost without his pseudoscience. He has no knowledge of the relevant physics, and is unable to think for himself.
Nothing new.
313K is close.
I have solid empirical data for 335 K for 71% of the planet’s surface. I use 255 K for the remaining 29% even though I know it must be wrong.
What you need to have and use is solid empirical data for the moon’s surface. That moon regolith empirical data exists on line and in the stacks at your local college library.
I have the DIVINER data. Did you miss that I directy referred to it in my original comment?
The DIVINER data shows the S-B estimate for Lunar surface Tav is in error by around 80 degree.
S-B estimates for the Moon, Mars and now Pluto have all been shown to fai against empirical measurements. If it won’t work for those worlds, it won’t work for Earth.
“Did you miss that I directy (sic) referred to it in my original comment?”
No miss.
So, tell us, Konrad, what surface emissivity were the radiometers on DIVINER set to as sourced from the builders of the instrumentation (like your own radiometer IR thermometer setting) to convert to brightness temperature? Does their data take into account that ~25% give or take of the moon’s regolith particle diameter is on the order of the wavelength of the illumination being studied causing diffraction to be material & brightness temperatures measured lower than surface thermometer? How does the data correct for the known non-Lambertian albedo & how was all this calibrated to Apollo thermometer data?
One that relies on such data should readily know these answers.
Turns out that when these effects are accounted for the 33K Earth GHE is confirmed by DIVINER data within as I wrote: “That 255K estimate has wide CIs at 95% significance level, an estimate which cannot be checked against a moon global thermometer field until one is installed.”
“The DIVINER data shows the S-B estimate for Lunar surface Tav is in error by around 80 degree.”
No, the DIVINER data, when applied correctly, confirms the S-B work & GHE as measured for Earth at Earth’s orbit. To understand that, please see your local college librarian with a list of subjects to read/study papers for this purpose. Your librarian will be very excited to get a chance to practice library science.
Then you will understand how to make S-B estimates work for the Moon, Mars, Venus (they worked to build the first insitu instruments to a reasonable scale) and now Pluto and can be applied to UltimaThule, Planet 9, asteroids and exoplanets, exo-objects. Have fun!
“So, tell us, Konrad, what surface emissivity were the radiometers on DIVINER set to as sourced from the builders of the instrumentation”.
Ball4, I don’t miss a Trick.
You have just completely beclowned yourself. If you’d done the slightest study you would know that the DIVINER radiometer was calibrated in the lab on Earth by using lunar surface samples at controlled temperatures within a vacuum chamber with cryo cooled walls to simulate the 3 K background of space. Perhaps it is you who should learn some new Tricks, possibly “library science” as you call it.
The DIVINER data did not confirm the validity of the S-B estimate of 255 K for Earth’s surface without radiative atmosphere. Rather it proved conclusively that the S-B approach would fail for all intermittently illuminated planetary surfaces.
“If you’d done the slightest study you would know that the DIVINER radiometer was calibrated in the lab on Earth by using lunar surface samples at controlled temperatures within a vacuum chamber with cryo cooled walls to simulate the 3 K background of space.”
Lunar samples of what particle diameter? What was the emissivity setting in the radiometers as a result of these experiments? How was the non-Lambertian albedo corrected as result of these experiments? How was the Apollo data used for calibration? You don’t know, do you?
You have nothing Konrad, and get the wrong results because you didn’t bother to understand. Once you do dig into and understand the data you’ll find DIVINER results do not show Tav is in error by 80 degree, S-B does not fail as you incorrectly write & is useful for all the solar system objects. For the moon DIVINER results: “That 255K estimate has wide CIs at 95% significance level, an estimate which cannot be checked against a moon global thermometer field until one is installed.”
“How was the Apollo data used for calibration? You dont know, do you?”
Yet again you beclown yourself. The Apollo temperature data wasn’t used for calibrating the DIVINER instruments, the returned surface samples were used.
Come on fluffball, I’m just toying with you. It doesn’t matter how many times you change your screen name, you are too old to change your brain. You’ve lost every scientific debate you ever engaged me in from 2011 to today. A changed screen name will never hide you from me.
All you have achieved is a permanent Internet record of the tactics AGW propagandists would stoop to in an effort to defend a failing hoax. Look to how you have behaved. Would any true defender of the scientific method descend to Alinsky techniques or Gaslighting?
Your permanent Internet record shows you held some political cause higher than defence of the scientific method. That record is permanent.
“The Apollo temperature data wasn’t used for calibrating the DIVINER instruments, the returned surface samples were used.”
Of course, Apollo temperature data wasn’t used for calibrating the DIVINER instruments Konrad, the Apollo surface thermometer data is way too sparse. Their brightness temperature results have to be calibrated to known thermometer data locally, there is no global thermometer field for the moon. That’s the same process used for Earth observations.
Averaging for the moon extremes needs to be done in the radiation domain due the S-B nonlinearity then converted to temperature domain. Your internet record shows you have done NONE of this work, you don’t fully understand and don’t use the existing published papers in the field on the subject, and continually make unfounded pronouncements. I’m glad the internet record is there, that way anyone familiar with the field can plainly find you have not done your homework and will know not to rely on your comments.
Ball4, please stop trolling.
‘How far out is the 255 K figure for “Surface Tav in absence of radiative atmosphere” for our planet? Its out by around 57 K. ‘
How did you arrive at this number? When I do it by proper spatial average of SB Temp I get 251 K.
It is true that for the Moon, if you use average temperature in calculation with average flux, you get disagreement with SB law.
Thats because (T^4)ave not/= Tav^4 when T varies by as much as it does on the Moon.
For Earth, T variation is much less, due to atmosphere and fast rotation, so (T^4)av is much closer to agreeing with Tav^4, still not exactly.
Nate, I rely on empirical experiments.
The S-B calculation is an instantaneous radiative balance calculation. You can’t use it for calculatind solar thermal gain in the oceans.
‘Nate, I rely on empirical experiments’
What experients?
You said “How far out is the 255 K figure for ‘Surface Tav in absence of radiative atmosphere’ for our planet? It’s out by around 57 K. It’s 57 degrees too low. ”
you implied that you have a calculation for Earth that is 57 K higher than 255K.
How do you get that?
Konrad,
Roy Spencer’s next post shows what you actually get with a proper calculation for Earth with no radiative atmosphere.
It is 252 K. So 3 K lower, not 57 K higher.
Nate, please stop trolling.
DREMT, pls stop being a hypocrite.
Nate, stop being a Nate.
DREMT,
Stop being paranoid.
“Zoe, are you by any chance a JD sock puppet?”
Like JD:
Zoe says ‘Learn some physics’
Zoe is ignorant.
Zoe tosses lots of insults: ‘Are you retarded?”
Your point, if any?
Nate, stop being paranoid.
DREMT, stop being a 5th grader.
Nate, pls stop being a hypocrite.
Nate,
“Youre fine with the cube receiving 11800 W, while emitting 6 x 1800 W?!”
The cube receives 1800 W/m^2 and emits 1800 W/m^2, not 300 W/m^2. Learn physics.
If you heat a substance to 300F, then all of the substance will become 300F, and therefore emit 1800 W/m^2. That’s just what 300F objects do.
‘The cube receives 1800 W/m^2 and emits 1800 W/m^2, not 300 W/m^2. Learn physics.’
Zoe, you seem to believe that 6 sides with a total area of 6 m^2, all emitting 1800 W/m^2 will in total emit only 1800 W!
The total emitted is 6 m^2 x 1800 W/m^2 = 10,800 W!
Meanwhile, the incoming heat from the hibachi is 1800 W/m^2 x 1m^2 = 1800 W.
Incoming 1800 W. Outgoing 10,800 W? Ok with that, Zoe?
Pls tell me which part you disagree with, and why.
ZP,
Do you understand that a 1 m^3 cube has a surface area of 6 m^2?
If the cube is receiving a 1800 W/m^2 input on one face then it is receiving 1800 W/m^2 / 6 m^2 = 300 W averaged over all faces. The cube is still receiving 1800 j/s of energy.
A cube that is emitting 1800 W/m^2 is emitting 10800 j/s because 1 m^3 cube has a surface area of 6 m^2.
Remember, a cube where all sides are 1 m in length has a volume of 1 m^3 and an area of 6 m^2.
“If the cube is receiving a 1800 W/m^2 input on one face then it is receiving 1800 W/m^2 / 6 m^2 = 300 W averaged over all faces.”
No. If the cube is receiving an 1800 W/m^2 input on one face then it is receiving an 1800 W/m^2 input on one face. You don’t average the input over the whole cube. Assuming the cube heats through and then emits evenly from all six faces, then it emits 1800 W/m^2 / 6 m^2 = 300 W averaged over all faces. This is what the “output reversed for input” comment meant, that you were unable to understand, but were certain was wrong anyway. You people, for some reason, want to average input over all six faces when only one is receiving the flux! No, the output is what you average over all six sides, assuming it does leave from all six sides.
The final key to understanding Zoe’s point is that an ice cube is not going to “heat through”, it’s just going to melt.
Get it yet? For crying out loud.
We all agree its going to melt.
Zoe’s point is lost down a rabbit hole.
DREMT,
Yes. We already know that the cube melts. That’s what we’re trying to explain. What we are saying is that it is incorrect to say that 1800 W applied to one face is equivalent to 1800 W/m^2 because the surface area of the cube is actually 6 m^2.
And it’s doesn’t matter if it’s 300 W/m^2 over the whole thing or 1800 W/m^2 over one face. The melt rate is the same assuming the ice and ambient are at 0C to start with and the liquid is evacuated at the same rate regardless of whether it is one face exposed or all 6. Why…because the amount of solid turned into liquid is still 1800 W / 333 j/g = 5.4 grams/s in either scenario. Notice that I can use either 1800 W/m^2 * 1 m^2 = 1800 W or 300 W/m^2 * 6 m^2 = 1800 W to determine the input power.
And it also doesn’t matter if the heat source is cycling at 3600 W 50% of the time and the cube is rotating such that each face gets equal time toward the heat source. The cube is still receiving 1800 W with an average of 300 W/m^2. In other words, if you place a radiometer on each face they will all agree that the average flux is 300 W/m^2.
The point…ZP’s assertion that averaging the flux over spatial and temporal domains doesn’t change the fact that the cube still melts. The statement “So according to Spencer’s religion it couldn’t possibly melt in my example.” is simultaneously false and a strawman. It’s a strawman because Dr. Spencer makes no such claim and it’s false because…well…the cube still melts.
The original comment, in its entirety:
“Dr Spencer,
You claim it is legitimate to divide solar INPUT by 4, i.e. spread it over the whole earth. But the sun shines on 1/2 the Earth, while the OUTPUT is over the whole Earth. The output is 2x the input.
We can test out your theory by putting a 1 m^2 ice cube onto a 1 m^2 hibachi grill set to 300F. According to your “science”, the ice cube never melts. Why? 300F -> 1800 W/m^2, Divided by 6 sides -> 300 W/m^2 = <0C.
This is exactly what you're claiming, though you will deny it.
Now what do you have to say about this?
Peace. -Zoe”
How you can still miss the point, is beyond me. Maybe you are just beyond help.
“What we are saying is that it is incorrect to say that 1800 W applied to one face is equivalent to 1800 W/m^2 because the surface area of the cube is actually 6 m^2”
Are you retarded? The 1800 W is only applied to one face. So you don’t average it over the whole cube. You average it over the surface area that receives the flux. 1800 W over 1 m^2 (that one face only) = 1800 W/m^2.
The rest of your comment is missing the point.
You people are so unbelievably thick.
One other point…you do realize that the solar constant of 1360 W/m^2 is the input received perpendicular to the surface of Earth at TOA and is itself an *average* over one full orbital cycle. It’s just an average over a temporal domain instead of a spatial domain. So if you’re really that hung up on this averaging idea then by all means you can derive the total energy that Earth receives by integrating the actual flux received at every square meter and every second through one orbital cycle. Or…just throwing this out there…you can use already established canonical geometrical ideas and divide the average perpendicular flux by 4 to get the average spatial flux and then multiple that by that by the total area of Earth and the total number of seconds in a sidereal year. Both are yield the exact same result.
I’m not “hung up” on anything, I’m not making whatever point you seem to want me to be making, I am simply explaining to you Zoe’s point. It is very, very, simple and easy to understand.
‘You people are so unbelievably thick’
You guys are raising utterly stupid strawmen, then double down and triple down on stupid, until you start saying things like 1800 W is input and 10800 W is output.
Then have people like you jump on the pile to defend what was stupid from the start.
Honestly DREMT, never mind me. You really think Dr. Roy and all Earth scientists believe in physics that does not allow heated ice to melt, and gives the wrong answer for the Earth-sun distance?
Or instead, could it be that you are misunderstanding what he is saying?
‘It is very, very, simple and easy to understand.’
No, it is completely garbled and wrong.
“According to your ‘science’, the ice cube never melts. Why?
300F -> 1800 W/m^2, Divided by 6 sides -> 300 W/m^2 = <0C."
If 1800 W is input and 6 sides emitting equally, than that is 300 W/m^2 emitted. It cannot be anything else.
Because of simple math: 6 x 300 = 1800.
'300 W/m^2 = <0C.' FALSE
300 W/m^2 is NET flow of heat, not the SB temp. Once again we need the rad heat transfer equation!
“If 1800 W is input and 6 sides emitting equally, than that is 300 W/m^2 emitted. It cannot be anything else.”
My earlier, ignored comment again, to show you I’m not arguing otherwise:
“No. If the cube is receiving an 1800 W/m^2 input on one face then it is receiving an 1800 W/m^2 input on one face. You don’t average the input over the whole cube. Assuming the cube heats through and then emits evenly from all six faces, then it emits 1800 W/m^2 / 6 m^2 = 300 W averaged over all faces. This is what the “output reversed for input” comment meant, that you were unable to understand, but were certain was wrong anyway. You people, for some reason, want to average input over all six faces when only one is receiving the flux! No, the output is what you average over all six sides, assuming it does leave from all six sides.”
Followed by:
“The final key to understanding Zoe’s point is that an ice cube is not going to “heat through”, it’s just going to melt.”
‘Assuming the cube heats through and then emits evenly from all six faces, then it emits 1800 W/m^2 / 6 m^2 = 300 W’
Well, then why are you arguing with us?
Clearly you need to be arguing with Zoe, who said all sides are emitting 1800 W/m^2
Try reading the comment again. Maybe the point will sink in.
‘Try reading the comment again. Maybe the point will sink in.’
Try reading other people’s comments, then argue with the right people.
N “If 1800 W is input and 6 sides emitting equally, than that is 300 W/m^2 emitted. It cannot be anything else.
D: “My earlier, ignored comment again, to show you Im not arguing otherwise”
Z: “The cube receives 1800 W/m^2 and emits 1800 W/m^2, not 300 W/m^2. Learn physics.”
Bdg “And its doesnt matter if its 300 W/m^2 over the whole thing or 1800 W/m^2 over one face. The melt rate is the same..”
D: “The final key to understanding Zoes point is that an ice cube is not going to ‘heat through’, its just going to melt.”
Z: “If you heat a substance to 300F, then all of the substance will become 300F, and therefore emit 1800 W/m^2. Thats just what 300F objects do.”
Yes, Nate. Zoe went on to say some things I don’t agree with. Which I have been perfectly clear about, and I am only supporting her original point, that you and bdgwx seem programmed to miss at all costs.
DREMT,
‘her original point’ was that averaging, as Roy does, means ice won’t melt. That was FALSE, dumb, and a strawman.
You want to now say she meant something else?
Your point seems to be 1800 W/m^2 input is not the same thing as 300 W/m^2 input on all sides.
What we are saying is, yes, of course they are not the same, but they result in the SAME amount of heat transfer.
For a spinning Earth, they result in the same amount of heat transfer to Earth from the sun over a day.
For understanding climate change, this is a perfectly fine simple model that is useful to understand energy imbalance.
An analogy. I get paid twice a month. I pay my bills and do the budget accounting once a month.
You guys are saying, no that’s wrong, you have to do the accounting twice a month!
Puleez..
Nate, if I kept responding to you, would this discussion continue until this thread closes for comments?
Honestly just wondering. As for your last comment, there is nothing worth responding to. I have made my point, and if you don’t understand it, that really isn’t my problem. Go and troll somebody else.
‘As for your last comment, there is nothing worth responding to. I have made my point, and if you dont understand it, that really isnt my problem. ‘
And we all know that your points, now matter how useless, are the only ones that really count.
Nate, please stop trolling.
I have approached the problem with a different thermodynamic question. Rather than ask “what is the energy balance of electromagnetic radiation efflux and influx?”, I ask, “what is the amount of heat captured and transferred by a carbon dioxide molecule to the rest of the atmosphere and surface?” I have spent since august 2015 pondering this question. In finding my answer, I have exhausted all my other questions that are automatically generated, once a possible answer is generated to my original question. I have been chasing clouds and parabolas, trying to catch up on finding that answer. In my dreams, I have ridden on the resonant vibrations of a carbon dioxide molecule, as it absorbs an IR photon. I have imagined the perfectly elastic collisions that CO2 molecule, in a state of superposition, makes with lower energy molecules and higher energy molecules, and how this energy is transferred to or from other molecules in the atmosphere, or the surface. Any surface weather station is obtaining data on the amount of upwelling radiation that is captured by greenhouse gases and then transferred to other gases and the rocky/aquatic surface. I have spent a lot of time chasing the wrong parabolas, such as the ones reached when temperature is plotted as a function of latitude and time of day. It is kind of neat what I learned when traveling after that answer in my mind and armed with a spreadsheet, surface weather stations anywhere on the globe, a clock and a map. But when I found the temperature of a weather station recording nocturnal increases in temperature, I realized it is much more of a different problem. It is truly a problem of, what is the amount of temperature increase, from a blackbody temperature/background temperature for any weather station on the map? It is the greenhouse gases that trap this energy and spread it to the other molecules in the atmosphere and the rocky/aquatic surface of the planet. Moreover, water in the atmosphere and the rocky/aquatic surface can exist as a solid, liquid, or a gas. There is lots of latent heat energy exchanged when water travels through this system, powered by convection. It is a very difficult journey, chasing these parabolas. I just kept steady on my course, trusting in the laws of quantum mechanics and thermodynamics. The signals that the greenhouse gases send us are all recorded by surface weather stations around the globe, and there are tens of thousands of those. I think that the climate model I have put together will support the radiative forcing model of the climate. I have finally answered my ultimate question: What is the amount contribution to temperature rise by carbon dioxide, in terms of degrees per part per million? My calculator is shared on my website: http://www.chemsnippets.com The least fancy I can claim, is that I don’t have a PhD. But for what it is worth, I do have a Bachelor of Science in Biology and a Master of Arts in Secondary Education. Besides lots of zoology, botany, genetics and embryology courses, I have lots of chemistry, physics, earth science and ecology courses behind that bachelor’s degree. Another proud claim I can make is that of receiving a US Patent for a cleaning formula. I am a science teacher, credentialed in beginning to advanced courses in Biology, Chemistry, Physics, and introductory Earth Science. I teach Aerospace Engineering. I have taught Advanced Placement Physics and Advanced Placement Chemistry. I have taught “The Physics of Sports”, as a summer gig for Johns Hopkins CTY program. I would like to say I petted Schrodinger’s cat, but….. well there are a lot of legit buts in that claim. But before I am judged/misjudged by my lacking P.H.D., I will proudly sit in the hall with Michael Faraday and share what I have seen, through a different lens that never was used to defend a PhD study. But please, feel free to use my calculator on http://www.chemsnippets.com. It was a long journey in being able to find all the derivatives and integrals that pop up out of the graphs, that allowed me to make such a calculator. I think you will be amazed at the answer to how many degrees per part per million that CO2 contributes to obtaining higher than blackbody temperature. Or what is the blackbody temperature at any weather station location?
Rudolph, avoiding reality is nothing new.
But why do you avoid paragraphs?
Huffman, reality comes in many forms, and one form is in equations that are generated from graphs of carefully collected data. When that data comes from any of the tens of thousands of weather stations on the surface of the planet, I don’t feel I am avoiding reality. But that seems your M.O. on here. You get this one kind response from me. I have read many of your retorts on this blog. Now everyone gets to know you through your prose. Sorry I did not make paragraphs. If that is your second argument against my model, only second to the ill conceived notion that I ignore reality, then the credibility that you earn from people who read your responses is all earned and cannot be taken back.
Sorry Rudolph but reality does not come in many forms. There is only one form–REALITY.
People obsessed with themselves often prefer “alternative reality”, usually just to escape their failed lives. They exhibit such characteristics as the inability to accept constructive criticism.
Nothing new.
One form of reality huh? Really? Wow! I think the world found another science illiterate in the reality of this blog. Of course the reality of life behind the writer of the blog is different from what others perceive as an idiot. But prior to the idiot letting observers experience the science illiterate, behind the keyboard, convincing others that nothing is new, is very intriguing.
Very good, Rudolph.
Rambling incoherently is a valuable trait, in pseudoscience.
–JDHuffman says:
June 6, 2019 at 11:13 AM
Sorry Rudolph but reality does not come in many forms. There is only one form–REALITY.–
True, but possible future reality comes in many forms.
The D-Day 75th Anniversary:
https://twitter.com/hashtag/DDay75thAnniversary?src=hash
Illustrates that.
Whether CO2 levels could cause a significant increase in global temperature is possible.
But it’s obvious that most climate projections over estimated increases in global temperature.
–Christy and McNider estimate that when atmospheric carbon dioxide concentrations double, global warming will reach 1.1°C—a quantity called “transient climate response.” Christy comments:
This is not a very alarming number. If we perform the same calculation on the climate models, you get a figure of 2.31°C, which is significantly different. The models’ response to carbon dioxide is twice what we see in the real world. So the evidence indicates the consensus range for climate sensitivity is incorrect.–
https://cei.org/blog/bjorn-lomborg-and-john-christy-shred-climate-alarmism
I would say that if we double CO2 levels [560 ppm] that possible
such levels of CO2 might indeed add about .5 C to global temperature. And it’s possible it doesn’t.
And even if we get the full amount of 1.1 C of warming by increasing from 280 to 560 ppm, other factors might nullify such warming or might add to it.
But not sure we will get to 560 ppm of CO2.
And I am certain that we could cool Earth by more than 1 C, if we wanted to do this.
And should be able to do this for less than 1 trillion dollars, but I don’t think we would want to spend 1 trillion dollars or even less money, lowering global temperature by 1 C.
And We have already spent more than 1 trillion dollars doing foolish and non workable solutions to prevent increasing CO2 levels.
It seems we can rule out possibility of 100 ppm increase within 20 years. And 20 years in the future it seems unlikely we add 100 ppm in next 20 years.
In US we have long past peak CO2 emissions and only way regain it, would having to use a lot coal. US has huge amounts of coal to use but natural gas is simply a better fuel to use.
And what would even better to reduce CO2 emission would using more nuclear reactors.
So if political forces stop fracking and/or stop the use of nuclear energy, we could reverse the downward trend of CO2 emissions.
If political forces allow and technology advances we could have significant increase of nuclear energy use, and sharply decrease US CO2 emissions further.
Another possibility in terms of global CO2 emission could be dramatic increase in natural gas use which fracking could enable.
But we could also get to the point of cheaply mining methane hydrates and getting a vast amount natural gas from the ocean.
The mad and foaming alarmists “should be” pushing the use of ocean methane hydrate- should be if they were not crazy, and irrational.
Instead the mad and foaming alarmists favor the dumb things like wind mills and solar energy, which to date have done nothing to reduce CO2 emissions. Instead these things have increase the price of electricity prices, caused global pollution, and litter natural landscapes with these stupid things- and wasted time, and trillions of dollars. And have only benefited large global corporations- or they are pawns of international corporate brainwashing.
There some signs that we moving in direction of using more nuclear power to reduce CO2 emissions. One can see it US political landscape, and obviously well under way in China and India.
But there doesn’t seem like much hope in Germany.
Really sorry that the reason that the sphere/disk area ratio of 4 enters the energy balance calculation is so hard to understand. Makes me sad for the rest.
Everybody understands why it’s done. The question is, should it be? Does the math correctly represent the actual physics involved?
‘Everybody understands why its done.’
Clearly not. Read the comments.
‘Does the math correctly represent the actual physics involved?’
If you are asking that, then you don’t understand the why.
It’s sad that an astrophysicist cannot understand a basic 3-layer box model. But to then arrogantly employs his misunderstanding to claim that he’s figured something out that has somehow baffled the world’s leading experts is beyond belief. It’s not common to see Dunning-Kruger like behavior from an educated scientists.
“It’s sad that an astrophysicist cannot understand a basic 3-layer box model”
It’s also untrue to say that he doesn’t understand it. As I said, everybody understands the why. The question is, should it be done? Does the math correctly represent the actual physics involved?
The actual physics involved is that in each instantaneous moment, on a second by second basis, the sun shines on half the Earth, while the output leaves over the whole Earth.
‘The actual physics involved is that in each instantaneous moment, on a second by second basis, the sun shines on half the Earth, while the output leaves over the whole Earth.’
If you read the discussion, you will see that no one is claiming otherwise. That’s why JP’s argument is a strawman.
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356192
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356206
I’ve read the discussion, thanks. And thank you for your recent determination that no matter what comment I write, on whatever thread, and whoever I’m talking to, you must respond. It is always nice to know that you have a fan.
That’s right, nobody is claiming otherwise. Of course they wouldn’t. But still, the math does not reflect that physical reality.
You may notice that I responded to MANY people here, DREMT.
It wasn’t my point to suggest otherwise. Let’s see if you are capable of correctly interpreting one single, simple thing:
I have written some comments on various threads. Not just under this article, and to various people. In almost every case, you have chosen to respond to me, out of the blue. I am not saying you don’t respond to other people. I am saying, at the moment, wherever and to whomever I choose to respond, you pop up.
Well, of all the ingredients in posts that invite a response from me, I would say hypocrisy usually works best.
And you exhibit that trait more often then most. Let’s see,
‘Please stop trolling’ followed by days of baiting and trolling, then again ‘pls stop trolling’
Bringing up the same two tired issues for months on end, that have been discussed ad-nauseum.
Then saying: “As if these discussions havent all been had several times before. I’m just laughing…”
Continually whining that people have ‘missed the point’ while thoroughly missing points by others, that rebut your claims.
Then claiming a faux ‘debunking’ having forgotten all rebuttals, or misrepresenting previous discussions.
Nate, “baiting and trolling” would suggest that I want a response, or a discussion. Whereas the only response I want is for you people to finally admit you are wrong.
I do not bring up the moon issue. Others always broach that subject first.
Noting that all these discussions have been had before, is kind of the point. They have. You people lost. You never admit that. So they come up again. There is no “hypocrisy” there, in my statement, because I only bring things up that you people just won’t admit you were wrong about.
Your final two points are you doing your usual thing of falsely accusing me of doing what you do yourself.
Probably the reason you act so upset by your perceived hypocrisy in others is that somewhere deep down you are aware of what a colossal hypocrite you are yourself.
But anyway, back to my point…
…go back through our discussions and count the number of times I engage you in a discussion compared to the number of times you engage me. Overwhelmingly, it is you jumping into a discussion I am having with somebody else.
And why on Earth would I choose to engage with you in a discussion, when I know that all you will do is falsely accuse, misrepresent, and insult (as JD puts it)? And he’s right. That’s all you do.
‘Whereas the only response I want is for you people to finally admit you are wrong.’
Well, DREMT, you and I have had lengthy discussions on these topics. I have tried to explain to you the basic physics of these issues several times, but in the end, you always ‘miss the point’.
How do I know for certain that you are wrong about these issues, and that you have ‘missed the point’?
Because, as it should be obvious to you by now, I am very well educated in physics, and have years of expertise in it.
This is why I keep saying that I did the GPE as a homework problem years ago. Because I did!
You can continue to believe what you want.
But if you expect someone like me (or Tim, or Eli, or Swanson) to ‘admit we’re wrong’ about basic physics and agree with your fake physics, which is obviously not based on any formal physics education, sorry that’s never gonna happen.
But for you to hope or expect that to happen is a deep misunderstanding of both physics and humans.
You are an anonymous internet commenter, who appeals to his own authority.
“OK, Nate”.
‘anonymous internet commenter’
Yes, I am.
But Eli, Tim, David, and Swanson are not anonymous. You can look up their Bios.
They agree with me, and often make the same points.
Yet you don’t believe them, either.
If you yourself understood physics, you would be quite certain that I do as well.
“OK, Nate”.
‘Ok, Nate’
And here is how you always respond when you (often intentionally) miss the point.
#2
“OK, Nate”.
DREMT said…”It’s also untrue to say that he doesn’t understand it.”
Well, his implication is that he believes the model says that 340 W/m^2 is the solar constant. He even drives the point home by calculating (incorrectly I might add) the orbital distance of the Earth using that figure. It would be nice to get Postma to respond so that he can explain what went wrong.
You people are physically incapable of correctly representing someone’s argument.
No, the flat earth refers to thinking of the model as equivalent of a flat sheet (of size surface(earth) if you want) orbiting the sun at twice the distance, having a normal pointed towards the suns’ center.
Now it is obviously silly to recalculate the distance to the sun of such a thing, because it is 2 times the distance by design.
Whether or not someone literally believes the eath to be flat has strictly nothing to do with that someones willingness to employ the model.
“No, the flat earth refers to thinking of the model as equivalent of a flat sheet (of size surface(earth) if you want) orbiting the sun at twice the distance, having a normal pointed towards the suns center.”
The “flat sheet” is the same distance from the sun as the Earth. The virtual circle has the diameter of the Earth and all the sunlight striking the Earth goes through it.
The solar constant is the power per square meter carried by sunlight when it hits the Earth. It is roughly 1,370 W/m^2. The power carried by all of the sunlight striking the Earth is the solar constant times the area of the virtual circle. If r is the Earth’s radius, the area of that circle is πr^2.
The area of the entire Earth is 4πr^2. The average power reaching one square meter of the Earth is the entire power reaching the Earth divided by the area of the Earth:
Solar constant X πr^2 / 4πr^2
or
Solar constant / 4
Craig T, please stop trolling.
Natw said:
“Incoming 1800 W. Outgoing 10,800 W? Ok with that, Zoe?”
Tell us the temperature the ice cube reaches!
Don’t stare at the sun, but do have a quick glance. Do you see 1 sq meter sending you 1361 W/m^2 to the 1 sq meter you’re standing on?
The sun’s closest 1 sq meter has 8 adjascent square meters that are second closest, the angle from direct normal is negligible. In your religion of adding fluxes, you woule have to concede that sun sends you AT LEAST 9*1361 = 12,249 W/m^2.
‘Incoming 1800 W. Outgoing 10,800 W? Ok with that?’
Zoe, these are numbers you gave me. 1800 W/m^2, from 6 m^2, gives 10,800 W, output.
1800 W/m2 x 1 m^2 = 1800 W input.
I am just doing simple arithmetic with these numbers that you gave us.
Your numbers dont work, regardless of what the temperature is!
BTW,
”
If you heat a substance to 300F, then all of the”
As you stated the problem, the Hibachi was set to 300 F. Ever cook a steak?
That doesnt mean the ice cube reaches 300 F.
ZP said..”The suns closest 1 sq meter has 8 adjascent square meters that are second closest, the angle from direct normal is negligible. In your religion of adding fluxes, you woule have to concede that sun sends you AT LEAST 9*1361 = 12,249 W/m^2.”
What are you talking about here? No one thinks the solar constant is 12249 W/m^2. No one thinks you need to multiple 1360 by 9. I don’t even know what that would represent.
What we think is that the solar constant is 1360 W/m^2. This means the average flux received perpendicular to the surface at TOA over one orbital cycle works out to 1360 W/m^2.
The most important part of the solar constant that people seem to gloss over is the fact that it represents the flux *perpendicular* to the surface. But because Earth is a sphere solar radiation spreads out over a larger and larger area as the surface bends away from the zenith.
bdgwx, the most important part of the “solar constant”/4 (340 W/m^2) that people seem to gloss over is the fact that it does NOT represent the flux *perpendicular* to the surface. By dividing by 4, it is no longer “flux”.
In fact, after adjustments, your bogus “energy balance” ends up with only 161 W/m^2! 161 W/m^2 has zero ability to warm a surface at 288 K.
Pure pseudoscience.
‘In fact, after adjustments, your bogus energy balance ends up with only 161 W/m^2! 161 W/m^2 has zero ability to warm a surface at 288 K.’
Very good JD. Then please do tell us how the Earth is able to warm to 288 K.
It’s the Sun, stupid.
Leave out ‘the sun’ and it makes more sense.
You must ignore reality for your false religion to make sense to you.
Nothing new.
You probably mean the flux through 1m^2 having its normal pointed towards the heart of the sun.
Everyone, try focusing this issue on the relevant point. All these arguments are about CO2’s impact on climate change. The calculation isn’t really relevant. No matter what the calculation is, the earth is a certain temperature. The physics of the CO2 molecule are constant. CO2 is transparent to incoming visible radiation. By allowing Joe Postma to get everyone focused on the minutia, we aren’t asking the relevant questions like HOW CO2 causes warming. Why Antarctica isn’t warming with an increase in CO2. How does CO2’s logarithmic decay result in linear warming? Joe is using an old legal trick of getting people to take their eyes off the prize.
In the bigger picture, it may be that most skeptics are independent thinkers. That means, unfortunately, that they do not always agree.
There are many ways to defeat the CO2 hoax. Physics is just one. Dr. Spencer has chosen to use his knowledge of weather and climate. Some prefer to use basic common sense, pointing out the billions and billions of dollars wasted on fraudulent “scientists”, as revealed by Climategate.
It would be better if all skeptics could organize, but that does not appear to be in our DNA.
“It would be better if all skeptics could organize, but that does not appear to be in our DNA.”
Sorry, I certainly didn’t intend any offense, and I greatly respect and appreciate Dr. Spencer’s work. I was just trying to point out that incoming radiation isn’t the issue, and if it is, CO2 certainly isn’t the cause of the warming.
If this is the issue: “Next in that article, Joes (mistaken) value for the solar constant is then used to compute the resulting Earth-Sun distance implied by us silly climate scientists who believe the solar constant is 342.5 W/m2 (rather than the true value of 1,370 W/m2).”
My point is simply that regardless if the constant is 342 or 1,370, the earth is a certain temperature. The actual temperature of the earth and outgoing LWIR is what the significant issue is. I was simply looking at the result of the incoming radiation, not the calculation.
Once again, sorry if I offended anyone, I was just making an observation.
No offense taken here.
I was just lamenting this rift between Spencer and Postma. I don’t know who started it, but at this point, both are wrong and continuing the dispute just makes them look worse.
Nate seems to believe that 6 times the substance can’t hold 6 times the heat. And therefore can afford to cool 6 times as much.
He never learned that q = m * Cp * T in his middle school chemistry class.
That’s why he believes more ice can only cool as much as less ice.
“That doesnt mean the ice cube reaches 300 F.”
What temperature does the ice cube/water reach?
“Incoming 1800 W. Outgoing 10,800 W? Ok with that?
Zoe, these are numbers you gave me. 1800 W/m^2, from 6 m^2, gives 10,800 W, output.
1800 W/m2 x 1 m^2 = 1800 W input.
I am just doing simple arithmetic with these numbers that you gave us.
Your numbers dont work, regardless of what the temperature is!”
Nate seems to believe that 6 times the substance cant hold 6 times the heat. And therefore can afford to cool 6 times as much.
He never learned that q = m * Cp * T in his middle school chemistry class.
Thats why he believes more ice can only cool as much as less ice.
Sorry dummy, but more ice can emit more power. 6 times ice surface = 6 times power. Why are you so shocked about this?
‘Sorry dummy, but more ice can emit more power. 6 times ice surface = 6 times power. Why are you so shocked about this?’
More ice can emit more, yes, if the energy is provided to it.
The Hibachi is only providing 1800 W to the ice.
That’s not enough to account for 10,800 W you want the ice to emit, is it?
There’s a law called the 1st Law of Thermodynamics. Look it up. you break that law you get in serious trouble.
This is not hard, Zoe.
Do you have an income, bank account, pay bills? Same thing.
Can’t pay $10,800 every month in bills for fancy cars and crack, if I’m only earning $1800/mo.
Well you can, briefly, with savings (somewhat like your heat capacity). But once the savings runs out, thats it.
Pretty sure you dont have good credit!
‘Nate seems to believe that 6 times the substance cant hold 6 times the heat. And therefore can afford to cool 6 times as much.’
We are not trying to cool the ice, we’re heating it, remember?
Its can’t emit 1800 W/m^2, by cooling, while simultaneously heating!
Zoe, are you by any chance a JD sock puppet?
The best theory I have seen on global warming
Glowal warming is due to the following
First as the earth is flat, and actually there is more people that are sinners, this people when they die go to Hell . so hell is overcrowded, and burning all this souls , as Hell is under the Earth , a pan effect
Heating all the planet
I love this theory
Although the energy supplied per second is important whatever value you disagree on, surely you also need to look at the frequency of the radiation that is supplying it? Different frequencies will interact with different parts of the atmosphere in different ways. The output of the sun in terms of diffferent frequencies will change over time.
Just looking at the joules per second to me appears to be an oversimplification of the models which makes their results meaninglesss..
Nate,
Why don’t you tell us the temperature of the ice cube @ equilibrium since you’re so “smart”.
I love how you made a $ analogy showing your belief that radiation and temperature are extensive properties. LOL
Didnt say temperature = $, zoe.
Energy = $ makes more sense.
T of the ice is the temp at which 1800 W is net output. Thus 300 W from each face.
Sigma*(T^4-293^4) = 300 W/m^2. Solve for T. Assuring surroundings @ 20C.
Nate, please stop trolling.
This post would be for Joe Postma.
I will attempt rational discussion with you. I failed on your blog a while back.
https://www.esrl.noaa.gov/gmd/grad/surfrad/dataplot.html
This is a link to actual measured values of EMR reaching various sensors at some specific locations.
https://www.esrl.noaa.gov/gmd/webdata/tmp/surfrad_5cf9cb7dd78e7.png
This is a graph made from Desert Rock, Nevada sensors.
There are two lines on the graph. One is total net solar radiation (the amount of radiation a sensor receives from the Sun directly minus that which is reflected upward and not absorbed by the surface). The other line is the IR emitted by the surface (hotter than the air temperature).
You can see from the graph the solar input could not come close to adding enough energy to the surface to allow it sustain the rate it is emitting at. The solar input only exceeds the emission rate of the surface for a few short hours. If you integrate under the curves you can get the total amount of energy received from the Sun on one meter square of surface area at Desert Rock location and also the amount of energy lost by continuous emission. The amount of total energy in a 24 hour cycle emitted greatly exceeds the amount of energy received from the Sun in a 24 hour cycle.
You do not accept the GHE for reasons you have concluded are rational. I think not at all. This empirical information shows you are very wrong and Dr. Roy is quite correct. You have the blind and science illiterate following you. It works the same for all talking heads that manipulate people. To gain actual knowledge of heat transfer requires reading textbooks on the subject and working through problems to gain knowledge of how it works (effort and hard work). You make your lazy minded followers think they are genius if they just listen to your ideas. They don’t have to study, all the scientists are wrong only you can provide them with the truth so you feed them your nonsense and they buy it. It is the lazy minded that follow you. Those who falsely think they are so much smarter than everyone else. I see the comments made by your followers.
Norman, please stop trolling.
bdgwx,
“What are you talking about here? No one thinks the solar constant is 12249 W/m^2. No one thinks you need to multiple 1360 by 9. I dont even know what that would represent.
What we think is that the solar constant is 1360 W/m^2. This means the average flux received perpendicular to the surface at TOA over one orbital cycle works out to 1360 W/m^2.”
Are you saying you can’t see at least 9 sq meters of the sun hitting your eye?
The answer you gave is not sufficient.
The cross sectional area of the Sun is 1.5e18 m^2. That is what I see. What does that have to do with anything?
Joe Postma
From the graph:
https://www.esrl.noaa.gov/gmd/webdata/tmp/surfrad_5cf9cb7dd78e7.png
If you take the area under the solar input (using the area of a parabola 2/3(Height)(width). 750 peak and 12 hours width times 2/3 gives a value of 6000. To convert this to joules take this amount and multiply by 3600 to get 21,600,000 joules (that is it, that is all you will have available with solar input). If you look at the amount of energy emitted it is over 500 W/m^2 for the whole 24 hours. Harder to get as close with this line as the parabola but it would be even greater than if you use just 500 for your value.
500 watts/m^2 in 24 hours will give you an energy value in joules of
500 joules/second-m^2 times 24 hours times 3600 seconds/hour with a total of 43,200,000.
Using this actual values it is obvious that the surface is emitting far more energy away than it can possibly receive from the Sun.
If you add the GHE it makes rational sense. Your view makes no sense at all.
https://www.esrl.noaa.gov/gmd/webdata/tmp/surfrad_5cf9d2ea7e5f3.png
With the GHE in place you get a NET IR of around -200 watts/m^2 for the 24 hour cycle. The surface is gaining 21,600,000 watts in 24 hours from the Sun and losing by radiant energy around 17,000,000 joules. The remaining loss of energy from the surface are from convection and evaporative cooling of the surface. Now things balance the energy in will equal the energy out in the 24 hour cycle.
Here is the measured values that determined the NET IR.
https://www.esrl.noaa.gov/gmd/webdata/tmp/surfrad_5cf9d4baca0e1.png
You have your measured Upwelling IR (the IR sensor is pointed down to the surface) and you have the measured Downwelling IR (the sensor is pointed upward). The NET is the UP minus the Down.
The GHE is a real empirically provable reality. You and your followers can deny it. That is why phony skeptics like you get the label of science deniers. The evidence is real, logical, empirical and valid yet you are not able to accept it. Too bad for you.
I really don’t care about your lazy followers, I am sad a person who actually studied real physics gets such a distorted mind.
Anyway Joe Postma, you are plainly wrong and it is easy to demonstrate you don’t know what you are talking about at all. Go make some more phony videos so you lazy followers can each the garbage you feed them.
Norman found some more links he can’t understand.
Nothing new.
N,
You wrote –
“Using this actual values it is obvious that the surface is emitting far more energy away than it can possibly receive from the Sun.”
Really? Like at night? That is why the surface cools, and the temperature drops.
It also shows why averages in this case are the refuge of the pseudoscientific incompetent. A pack of bumbling fools, who can’t even usefully describe this supposed GHE!
At night, the surface loses all the heat it received during the day, plus a wee bit of the Earth’s internal energy. Baron Fourier (Fourier series and so on, agrees). I choose to agree with Fourier, rather than you. Who would you choose?
CO2 heats nothing. No GHE. So sad, too bad.
Cheers.
Mike Flynn
The bot can’t even respond with a logical thought. Just some programmed response that actually is quite stupid.
Example: The Mike Flynn bot: “Really? Like at night? That is why the surface cools, and the temperature drops.”
At night, without GHE keeping the surface from rapid cooling, the surface would be emitting far more energy than the Sun added during the day.
Learn some math, learn some physics. Your programmer needs to take you off-line and do some more work. You are very lacking in many areas.
Norman, please stop trolling.
Joe Postma
Oh yes before I am done I just wanted you to know that if you took all the available solar input and spread it out over 24 hours for this location you would end up with a value of 250 w/m^2 continuous input really similar to what the Climate Scientists (people who actually do the hard work to solve the problems, unlike yourself that make stupid videos for lazy brain dead follower that believe they are the smartest people on the planet but have no ability to read an actual science textbook). You reject the calculation of dividing the solar input by 4 and spreading the energy to each square meter uniformly to get some sense of comparison between two states (a GHE and no GHE) but the actual measured value would be ball park to what they are telling you.
I wonder how long you will continue to reject real science and keep your cult going?
Norman, the links you can’t understand prove you wrong.
That’s why no one takes you seriously. And that’s probably why you are so frustrated.
Nothing new.
JDHuffman
I kind of assumed you would troll my comments with your basic nonsense that means absolutely nothing. It is easy to write those few words for you. They mean nothing.
Prove that I do not understand the links. What don’t I understand about them. You use this same troll tactic on many people who post links. You fail to ever state what is not understood.
In troll world just saying a mindless opinion without any detail is enough for troll followers.
Too bad you aren’t smart enough to remotely grasp the concept I developed.
Norman attempts a question, but omits the “question mark”: “What don’t I understand about them.”
Norman, if you behave I will try to help. But if you resort to obfuscation, insults, false accusations, or misrepresentations, then you get to stay in your pseudoscience swamp.
Take one of your first links:
https://www.esrl.noaa.gov/gmd/webdata/tmp/surfrad_5cf9cb7dd78e7.png
If you really understood that link, then you should be able to answer a couple of simple questions:
1) What does “Solar Net” refer to?
2) Where do they show the spectra for the two lines on the graph?
Please keep your answers as short and direct as possible. Again, act responsibly and I will help. Behave like an impudent child and you lose.
JDHuffman
Since you use the endless repeat mode of the troll you are.
What do the links show you? I know you don’t have the knowledge of physics to even remotely understand any of the graphs and I also know you are too much a troll to be interested in what thy do show.
For those that are not blind cult followers of Joe Postma (nothing will change their thoughts) and those that are not trolls like the JDHuffman blog troll. The graphs show very clearly that the energy of the Sun alone is not even close to enough energy to maintain the emission rate of the surface. The GHE is required to make this work. The links show this, the trolls and blind followers will not be able to see this.
Whoops almost forgot about the Crackpot Gordon Robertson. He might show up and make up a pseudoscience explanation for things he never learned and can’t learn today. He pretends to have studied college physics. The evidence in his posts suggest he only has read blog physics from crackpot web sites and now he knows more than the entire world of physics even though he will not do even one experiment he just knows he is right.
Norman, you commented here while I was commenting above.
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356530
Deal with the 356530 comment.
JDHuffman
YOU: “1) What does “Solar Net” refer to?
2) Where do they show the spectra for the two lines on the graph?
Please keep your answers as short and direct as possible. Again, act responsibly and I will help. Behave like an impudent child and you lose.”
1) It is the amount of energy from the Sun hitting a sensor pointing up minus the energy (solar not the surface IR so mostly visible range) hitting a sensor pointing down. It is the solar energy that the surface is then able to absorb (it would probably be sand in this case).
2) They do not need to show a spectra for the two lines on the graphs. They are the total amount of energy. The emissivity of sand being about 0.9 for IR so it will absorb 90% of the energy of DWIR regardless of the spectra.
That should suffice for now.
1) The reason I asked was to make you realize that the visible was being reflected. That is, the photons were NOT absorbed. All photons are NOT always absorbed. You probably heard that before….
2) No, you DO need to see the spectra. That way you won’t fool yourself, as you have done so many times before. Find the relevant spectra, or your nonsense just fails, as usual.
JDHuffman
1) Yes most posters know that not all photons are absorbed. Depends upon the surface.
2) I have linked many times to spectra of DWIR and UPIR. The spectra does not seem to matter much. It will definitely NOT make any difference at all. The desert sand will reflect so much solar energy which you can see if you make other graphs and it is a measurable amount. The sand will absorb 90% of the IR spectrum. It will emit a spectrum 90% of a blackbody at the same temperature.
The points you are bringing up in no way explain anything about what is being pointed out. The surface is emitting far more energy than it is receiving from the Sun. This is a measurable amount. Your demand for a spectrum is non-issue only designed to make people look at something insignificant instead of the “elephant in the room!”. There is not enough solar input to maintain the emission rate. The energy must come from the DWIR or the surface will cool until it equalizes with the solar input, that would be about 250 Watt/m^2 in the summer at Desert Rock on a clear day. Nothing more complex than this. Quit with the Red Herring arguments and stick with the measured values.
Yes Norman, I keep forgetting the relevant physics is just a red herring to you.
You can’t understand the physics. And you can’t think for yourself.
Even without an understanding of the physics, you could plot the temperatures with the fluxes and see that your conclusions are wrong.
But reality just frustrates you, and then you start lashing out like a toddler having a temper tantrum.
Nothing new.
JDHuffman
First you could not maintain a scientific discussion. You had to revert to being a troll.
YOU: “You can’t understand the physics. And you can’t think for yourself.”
You offer your unwanted and unsupported opinion with no explanation. You do not explain what I don’t understand at all nor do you provide the slightest evidence about my thinking ability (which is far above your meager thought process).
Again another stupid troll opinion. No thought, no explanation just stupid trolling endlessly.
YOU: “Even without an understanding of the physics, you could plot the temperatures with the fluxes and see that your conclusions are wrong.”
No it would not prove me wrong at all. You don’t know enough physics to rationally discuss the subject with. If I wanted to learn the latest troll tactics you would be the one to consult. But physics no way. You are far to ignorant in physics to hope to get a valid answer. Maybe you should type Poynting Vector when all else fails. As always you don’t know a thing about any physics and you can’t even fake your way out of the obvious. But you will troll comments continuously. It is what you do, it is all you are able to do. You don’t know enough physics is offer a scientific debate nor will you ever be able to.
Nothing new with the blog troll.
Can I predict Norman, or what?
“But reality just frustrates you, and then you start lashing out like a toddler having a temper tantrum.”
‘All photons are NOT always absorbed.’ somehow morphs into NO photons are absorbed.
Lets see, if the emitting object is 1 degree cooler than the absorbing object, JD seems to believe that its BB spectrum will have shifted SO much that NO photons are absorbed.
JD NEVER has explained this bit of illogic.
More false accusations and misrepresentations from Nate. That’s all he has.
Nothing new.
Nothing false or misrepresented, JD.
You claimed: The BLUE reflects ALL 200 W/m^2 from the Green, even though green and blue are nearly the same temperature.
Your justification “All photons are NOT always absorbed.”
You fail at logic. Nothing new.
Nate, please stop trolling.
“””Hi all, Norman or anyone.
You all are taking the average flux received, you say its 340 W/m^2, below it says its 1366 W m−2. So you are dividing this my 4. WHY ARE YOU DIVIDING THIS BY FOUR ??? Does not the Sun give out energy/heat for the full 24 hours that we are counting in the debat ??? So why divide by 4 ??? I don’t understand that ??? Also, when the Suns not shining on a certain area, there will still be heat in that area, from the land, seas, convection and other, this heat was from the Sun, so we have to count this in, thus you must add in this heat. I need this more in layman’s terms please. ARE SOME OF YOU THINKING, that when I stand outside on a hot day and the Sun is hitting me at 30c, that, thats not the Suns heat hitting me, but the GHG’s ??? As some claim it would only be -18 without the GHG’s, “IF” that was the case, then why on a summers day in England would it be 25c, but in Dubai 40c ??? If the Sun only gave us -18, then the GHG’s would heat the Earth up constantly in all Countries but my England Dubia proves that “WRONG” “””
The resulting flux density, 1380 W m−2, is an estimate of the extraterrestrial ‘solar constant’, which is very close to the currently accepted value of 1366 W m−2, also known as total solar irradiance (TSI; see below).
Wayne
“WHY ARE YOU DIVIDING THIS BY FOUR ???”
Basically the Earth thermometer field is spread out around the whole globe, sparse in some places, denser in others. So the global thermometer near surface mean air temperature is known within reasonable significance levels.
To calculate that global mean temperature from 1st principles, need to find the global mean total solar illumination of the whole spheroid which when done properly employing simple geometry basics and Planck radiation principles, the factor of 4 arises on a rotating object. When that calculation is properly performed as shown in textbooks, the results compare reasonably well with the observed global mean temperature.
Calculated global mean T ~ 288K and observed global mean T ~ 288K. Once that calculation was proven out from measured inputs, other solar system objects and exoplanets global mean near surface temperatures could be estimated from measured inputs which is of interest in many fields.
Ball4, please stop trolling.
bdgwx,
“The cross sectional area of the Sun is 1.5e18 m^2. That is what I see. What does that have to do with anything?”
Not a sufficient answer.
“The sun’s closest 1 sq meter has 8 adjascent square meters that are second closest, the angle from direct normal is negligible. In your religion of adding fluxes, you woule have to concede that sun sends you AT LEAST 9*1361 = 12,249 W/m^2.”
All the sunlight hitting the Earth travels in parallel rays. There are no overlapping beams hitting the Earth at different angles traveling from different parts of the Sun like a bank of lights.
That is why all the light striking the Earth passes through the equivalent space of a circle with the Earth’s radius. And at any given time every square meter of that virtual circle has 1,370 watts passing through it. Over a 24 hour period all sides of the Earth receives light passing through that virtual circle. When the tilt of the Earth’s axis keeps part of a polar area in the dark, the equivalent area on the other pole gets 24 hours of light.
CT,
You wrote –
“The equivalent area on the other pole gets 24 hours of light.”
Maybe you mean 6 months of sunlight? It makes no difference. The flux does not accumulate or multiply. After 6 months of continuous sunlight, the Polar regions remain very, very, cold.
No GHE. No CO2 warming. All nonsense.
Cheers.
“Maybe you mean 6 months of sunlight?”
Not during 24 hours.
Maybe you can clarify your question then? I know for certain that my eye is not receiving 1360 W/m^2 * 1.5e18 m^2 = 2e21 W. Imagine what 2 zettawatts would do to something as small as an eyeball.
Nate,
“Energy = $ makes more sense.”
You’re using watts, not energy.
“Sigma*(T^4-293^4) = 300 W/m^2. Solve for T. Assuring surroundings @ 20C.”
I didn’t tell you the temperature of surroundings. The surrounding temperature is below zero degrees C.
“T of the ice is the temp at which 1800 W is net output. Thus 300 W from each face.”
OK, here you admitted that the hibachi grill can’t melt ice.
BTW, how about you trim the edges of the cube an turn it into a dodecahedron. Now the cube magically gets even colder.
Zoe,
“OK, here you admitted that the hibachi grill cant melt ice.”
You are simply are very very clueless about heat transfer.
Heat flows from hot to cold (you know why right?).
If your cube is emitting 300 W/m^2 of heat, then clearly it is WARMER than its surroundings!
“I didnt tell you the temperature of surroundings. The surrounding temperature is below zero degrees C.”
1. Put ice on a Hibachi. You say it won’t melt!
2. Oh, BTW, my Hibachi is in vacuum.
3. Oh, BTW, my Hibachi is in a very large freezer.
4. Oh, BTW, I just graduated from 6th grade!
Use whatever you want for Ts, Zoe.
Sigma*(T^4-Ts^4) = 300 W/m^2. Solve for T.
Nate, please stop trolling.
Craig T,
“All the sunlight hitting the Earth travels in parallel rays. There are no overlapping beams hitting the Earth at different angles traveling from different parts of the Sun like a bank of lights.”
LMAO!!!
I can clearly see at least 9 sq. meters of sun rays hitting my eye. But according to you, they all travel parallel, so the sun should appear as 1 sq meter (pretend you could see that), and if I move over a meter, I would see a different sq. meter of the sun. LOL
No, the answer to my puzzle is much simpler than that.
So does Zoe need to go back on the meds, or is Zoe JUST a troll?
Whats your vote?
“I can clearly see at least 9 sq. meters of sun rays hitting my eye.”
The Sun and Moon both appear to be about the same thickness as your thumb held out at arm’s length. As bdgwx pointed out you see the entire Sun when you look at it.
The light from the east side of the Sun reaches my eye at an angle of 1/2 degree to the light from the west side. I’ll let you do the math on how the angle changed when you step over a meter.
The solar constant is in reference to the average flux hitting the square meter at TOA that is most perpendicular to the Sun. It has nothing to do with the size of the Sun or the apparent size of the Sun as observed from Earth.
bdgwx says:
June 4, 2019 at 6:00 PM
We are interested in the average for a few reasons. First, the actual solar flux changes throughout the year. 1360 W/m^2 is itself an average computed over one full orbital cycle. The actual flux changes because Earth’s orbit is elliptical. Second, this allows us to mentally compute the total energy received by Earth over a period of time. Third, most of the other energy budget constituents are themselves yearly averages.
Knowing that 340 W/m^2 is the average flux received at TOA we can immediately conclude the energy received per square meter over one year is…wait for it…340 W-years/m^2! Over a 10 year period it is 3400 W-years/m^2. And, of course, you can multiple these number by 500e12 m^2 to get the total over the entire Earth.
“””Hi all, bdgwx or anyone.
You all are taking the average flux received, above you say its 340 W/m^2, below it says its 1366 W m−2. So you are dividing this my 4. WHY ARE YOU DIVIDING THIS BY FOUR ??? Does not the Sun give out energy/heat for the full 24 hours that we are counting in the debat ??? So why divide by 4 ??? I don’t understand that ??? Also, when the Suns not shining on a certain area, there will still be heat in that area, from the land, seas, convection and other, thus you must add in this heat. I need this more in layman’s terms please “””
The resulting flux density, 1380 W m−2, is an estimate of the extraterrestrial ‘solar constant’, which is very close to the currently accepted value of 1366 W m−2, also known as total solar irradiance (TSI; see below).
Wayne
The solar constant is the average **perpendicular** flux at TOA. It is NOT the amount received everywhere on Earth. We can exploit canonical geometrical principals and divide by 4 to normalize this to the average spatial flux over a sphere. Or you can do it the hard way and integrate the actual flux at every square meter over the course of one year and take the average. Either way gets your 340 W/m^2.
Hi there bdgwx, or anyone,
Sorry, but I still don’t get why you divided be 4 ??? Let me see if I have this right. 340 W/m^2 is the average flux received at TOA. Does this mean the averge flux hitting that area of the Earth, say one quarter, that at that moment for one second, we get 340 W/m^2. Or is the averge flux hitting the area of the Earth, say one quarter, that at that moment for 6 hours, quarter of a day ???
Wayne
The amount of radiation received is related to Earth’s cross sectional area; not Earth’s surface area. One of the easiest ways of understanding this is to visualize the shadow cast by the Earth. That shadow is represented by Earth’s “disk” area. This is the area etched out if you intersect a 2D plane through the center of a 3D sphere. This forms a 2D circle with area pi*r^2. Note that the area of a sphere is 4x the area of a circle with the same radius.
Another way to visualize what is going on is to do a fun experiment in your home. Shine a flashlight onto your floor while holding it perpendicular to the floor. Make a note of the area that is being lit. This works better if done on a tile floor so that’s easy to work out the area. Now tilt the flashlight at an angle and make note of the area that is being lit this time. Notice that the same amount of light is now being spread out over a larger area. Like the flashlight demonstration the spherical shape of the Earth causes the surface to tilt away from the Sun so that the radiation is hitting at a shallower angle.
Yes. 340 W/m^2 is the average flux received TOA.
No. 340 W/m^2 is not being received everywhere. Some areas will receive 1360 W/m^2 while others receive 0 W/m^2. It depends on how far away the location is from the zenith. The relationship is 1360 * cos(theta) where theta is the angle separating the location and the zenith. When you integrate this cosine relationship on the lit hemisphere you get an average of 680 W/m^2 while the unlit hemisphere receives 0 W/m^2. And (680 + 0) / 2 = 340.
The zenith only occurs between 23.5N and 23.5S latitude with the exact latitude changing throughout the year. In other words the 1360 W/m^2 figure is only valid for locations in which the Sun is directly overhead.
Hi bdgwx,
Thank you for all that.
HOWEVER is 340 W/m^2 is the average flux received by TOA, for a second, or 6 hours quarter of a day, or for how long in time is this average worked out please ???
Wayne
It is averaged over one orbital period (365.24 days). The reason for this is that the zenith flux of 1360 W/m^2 varies by 3.5% above and below that figure because Earth’s orbit is elliptical. The peak zenith flux occurs in January.
Hi there,
Ok get all that.
HOWEVER why do you divide by four ??? The Suns heat constantly hits the Earth, so why divide ??? You should actually be adding more heat to this, as of the heat that’s left in the seas and earth when the sun moves around the Earth.
Wayne
“The Suns heat constantly hits the Earth, so why divide ???”
Only on one hemisphere and there are two hemispheres reporting thermometer results over multi-annual periods so to compare measured thermometer mean temperature results to analytical mean temperature results the divide by 4 is found necessary for a rotating object as explained by Dr. Spencer and in text books.
If one doesn’t divide by 4, the analytical global mean temperature results do not match observations
Ball4, please stop trolling.
“So why divide by 4 ??? I don’t understand that ???”
The solar constant is the average rate energy reaches the Earth from the sun.
r = Earth’s radius in meters
S = solar constant in watts per square meter
πr^2 = area of virtual circle all sunlight passes through to reach Earth
Sπr^2 =total watts reaching Earth at any given time
4πr^2 = Earth’s surface area in square meters
S/4 = average watts per square meter reaching the Earth
That number is only useful when looking at the Earth as a whole. For any location you need to look at the solar energy reaching the ground during a 24 hour period. Here is a measurement of the sunlight reaching Desert Rock Nevada on August 18, 2018.
https://www.esrl.noaa.gov/gmd/webdata/tmp/surfrad_5cf9cb7dd78e7.png
“Also, when the Suns not shining on a certain area, there will still be heat in that area, from the land, seas, convection and other, thus you must add in this heat.”
The solar energy reaching the Earth is still on the planet in some form until it radiates back into space. An energy budget tracks the average energy reaching and leaving the Earth.
https://en.wikipedia.org/wiki/Earth%27s_energy_budget#/media/File:The-NASA-Earth's-Energy-Budget-Poster-Radiant-Energy-System-satellite-infrared-radiation-fluxes.jpg
Again it is much more complicated when looking at a single location. For example the gulf current carries energy that reached the Earth in the tropics northward to England and Europe. Cloud cover lets less energy radiate away from the Earth than clear dry air.
Hi there Craig T, or anyone,
Sorry, but I still dont get why you divided be 4 ??? Let me see if I have this right. 340 W/m^2 is the average flux received at TOA. Does this mean the averge flux hitting that area of the Earth, say one quarter, that at that moment for one second, we get 340 W/m^2. Or is the averge flux hitting the area of the Earth, say one quarter, that at that moment for 6 hours, quarter of a day ???
Wayne
wayne, it is all part of the hoax. Eventually they get it down to 161 W/m^2. That way they can claim the Sun can’t heat the planet, so it must be CO2.
Some people fall for it.
Yes it does seem like. 340 W/m^2 is the average flux received at TOA. as this average is hitting the Earth all the time as its spins in 24 hours. So you “CAN’t divide by 4 ??? Why would you ???
“HERE”, LETS USE A LAYMANS TERMS ON THIS.
I hit the Earth with a average force of 340. I hit the Earth every second, and put a small dent in it. in 24 hours I have made a small dent all around the Earth. THUS, you CANNOT divide my hitting force by 4 ???
Wayne
Wayne, see David Archer divide by four:
https://tinyurl.com/y3pnvmmb
Yes it does seem like. 340 W/m^2 is the average flux received at TOA. as this average is hitting the Earth all the time as its spins in 24 hours. So you “CAN’t divide by 4 ??? Why would you ???
“HERE”, LETS USE A LAYMANS TERMS ON THIS.
I hit the Earth with a average force of 340. I hit the Earth every second, and put a small dent in it. in 24 hours I have made a small dent all around the Earth. THUS, you CANNOT divide my hitting force by 4 ???
Wayne
A watt is a measure of energy per second. The solar constant is the power density of sunlight, a measure of the watts per square meter reaching the Earth.
If your punch is the solar constant your average punch is 1360 not 340. There are 2 problems with comparing sunlight to you hitting the Earth.
After 24 hours the dent would be around the Earth but it wouldn’t be as deep as if you hit the same spot for 24 hours punching once per second. The damage would be distributed. It would be a ring of dents where each location was hit once during 24 hours.
The bigger problem is hitting the Earth north or south of the equator. Would those punches be more glancing blows that do less damage? Those locations never receive the full solar constant because the light rays striking them are not perpendicular to the ground.
Craig T, please stop trolling.
Wayne, I don’t know if this will help but the link is to an image showing how sunlight is distributed over the Earth. At noon (during equinox) the equator receives the full solar constant or 1360 watts per square meter. The farther north, south, east or west you are from that spot the more indirectly sunlight strikes the Earth. The average flux striking the Earth is 1/4 the solar constant.
https://i.imgur.com/5oHAZRp.jpg
Craig, were you able to understand this:
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356255
No JD, I don’t understand why you are comparing the change in flux over distance to the distribution of light over a sphere.
The reason the edge of a sphere receives less light than the center has more to do with the angle of the beams to the surface than the edge being farther away from the light source. The edge of the Earth where the sun rises or sets is only 0.46% farther from the sun than the closest part of the Earth.
This doesn’t even have anything to the greenhouse effect. It does seem to be another area where people like yourself made mistakes in the calculation and aren’t wise enough to listen to the people that study these things as a profession.
I wasn’t “comparing the change in flux over distance to the distribution of light over a sphere”. The reason you can’t understand reality is that you’re too busy misrepresenting others.
A flux cannot be treated as a simple scaler, especially a power flux.
Where did I make “mistakes in the calculation”?
That’s just a desperate false accusation.
“scalar”
Stupid auto-correct!
Is watts per square meter a scalar or vector?
https://physics.stackexchange.com/questions/360987/why-is-power-considered-to-be-a-scalar-quantity
(Don’t let poor Norman see that link. He hates the Poynting vector, as he hates all physics.)
Now, again, Where did I make “mistakes in the calculation”?
Where did you make any calculations? How about you use the inverse square law to show the right way to distribute energy around the Earth?
Craig, did you get caught making a false accusation?
“It does seem to be another area where people like yourself made mistakes in the calculation…”
AGAIN, where did I make “mistakes in the calculation”?
You need to take responsibility for your comments.
It was wrong of me to say you made a mistake in calculation. You made a mistake in the formula to use but never made the calculation. Use inverse square to figure out how much energy reaches the Earth and then we can talk.
Craig, admitting you falsely accused me of “mistakes in the calculation” started you up out of your hole, but then you dug yourself back down with “You made a mistake in the formula to use…”
Now you have to show me where I made “a mistake in the formula to use”.
Already done.
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356673
Use the formula then we’ll have more to talk about.
Craig, once again you seem to be avoiding your own words:
“You made a mistake in the formula to use but never made the calculation.”
Show me the “mistake in the formula”, or admit you have falsely accused me.
There are cranks on this forum that believe my ice cube example will lead to a thermal gradient from 1800 W/m^2 (hibachi side) to 300 W/m^2 (out-sides). They refuse to believe in the phenomena of resonant heating, yet there’s dozens of youtube videos debunking them. Here’s one:
https://youtu.be/AiZWFrb76pM
We see a constant heat source coming in on the top right. The input surface area is <1% of the output surface area.
According to their junk science, we should see a thermal gradient from RED to room temperature, and yet … the whole radiator becomes RED.
All the molecules are resonanting with the input (equilibrium), and hence all ice cube sides will emit 1800 W/m^2 as the radiator example, and long established science proves.
Watch these twerps deny reality again as they insult their betters.
Zoe, reality comes from running the experiment and using an IR thermometer on the 5 sides of your ice cube not illuminated by the Hibachi to determine the W/m^2 they really do emit.
Get the experiment done & report results if you really want to know reality. If you do not care about reality like JD, DREMT, Joe P., then avoid properly running the experiment at all costs.
For example, Dr. Spencer cares about reality and thus runs proper experiments & reports results as do some other commenters on this blog who really do care about reality.
“If you do not care about reality like JD, DREMT, Joe P”
Once again I’m dragged into something…
fluffball is just spinning fluff again.
Nothing new.
Nothing new?
Well, there is the new screen name Ball4. But it is still the same Alinsky tactics and Tricks from 2011.
However in some ways I do find the comments from fluffball entertaining. The tactics used in lieu of valid scientific debate telegraph loud and clear that someone knows they are on the losing side of the debate.
He has been caught several times with entirely false statements. He has no regard for truth.
“Alinsky tactics”, indeed.
JD, you are right, fluffball has no interest in the truth. Every comment is an attempt to kick up dust, deflect and delay the inevitable end of the hoax.
Fluffball has had several screen names since first trying to attack me in 2011. Back then it was hard. The original screen name and IP address was being used to post comments from 3 separate alarmists. Thankfully stylometrics software helped me identify the separate individuals involved. When you discover 3 persons doing that, you have discovered 3 persons who provably know they are doing the wrong thing. No one who had science on their side would ever stoop to such behaviour.
8 years later, fluffball is on his own. Still trying Alinsky techniques, still trying to “gaslight”, just using new screen names. I don’t even need to use stylometrics software to identify any more, the tactics and language are like a fingerprint.
I’m happy for fluffball to continue to post comments. This is the age of the internet, the age of permanent record. Fluffball is leaving a permanent record of how low the AGW propagandists would go to delay the collapse of their hoax.
‘According to their junk science, we should see a thermal gradient from RED to room temperature, and yet the whole radiator becomes RED.’
Ok, Zoe, keep telling us things that our science is NOT saying.
But, meanwhile, you need to fix your own violations of thermal physics laws:
1LOT: 1800 W input cannot produce 10,800 W output. And BTW ice will never reach 300 F!
2LOT: Ice emits 300 W/m^2 but is colder than its surroundings?
Radiative heat transfer law: Ice emits 300 W/m^2 but is < 0 C ???
Nate, please stop trolling.
Nate,
“Use whatever you want for Ts, Zoe.
Sigma*(T^4-Ts^4) = 300 W/m^2. Solve for T.”
OK, Ts = 0, T < 0C, Ice Cube doesn't melt. Thanks for playing.
Oh Ts = 0 Kelvin now???
Oh that’s a really good freezer, and that will really keep the ice from melting!
And that’s a really BIG movement of the goal posts!
If you meant Ts = 0 C, then you need to review your 6th grade math that you just learned, Zoe.
That gives T = 323 K = 50 C. The ice melts.
Nate, please stop trolling.
Here’s another video:
https://youtu.be/JknpBPCAiiU
At 2:00, we see output over the entire radiator matches input over a tiny region.
We don’t see a gradient from hot to ambient.
Case closed.
Craig T,
“The light from the east side of the Sun reaches my eye at an angle of 1/2 degree to the light from the west side. Ill let you do the math on how the angle changed when you step over a meter.”
Sure. 1361 + Slightly Less than 1361 > 1361.
Why still 1361?
We don’t have to go to the far east side.
The angle between the closest 1 sq. meter of the sun and the second closest 8 sq meters is so small as to be NEGLIGIBLE. Why not 1361*9?
Come on now, the answer is not hard.
Hi Roy and other readers,
The following long comment will maybe be read by a few because there have already been 417 comments.
In the 16th Century there was a Swiss naturalist who saw evidence that glass windows allowed (caused) the temperature of whatever interior space to warm above the temperature (ambient temperature) of the exterior environment. So he concluded that glass traps the energy of solar radiation. So based on this idea he began to construct a device to see to what extreme temperature that glass could trap the energy of solar radiation.
He insulated a box, as best as he could, and initially glazed its window with 5 spaced glass panes. After observing the maximum temperature that this box reached shortly after midday when the atmosphere appeared cloudless, he began reducing this glazing one pane of glass and observing what the new maximum temperature might be. It seemed he tested this modification and until the double glazing resulted in maximum temperature less than that he had observed with the previous triple glazing. And the maximum temperature he observed, by pointing this devise toward the sun, was 230F. Which he obviously did not know is the maximum surface temperature that has been observed for the moon which has no atmosphere.
We should not ignore that Horace obviously had reasoned, or observed, what the 5 spaced glass panes would accomplish. I can imagine the Horace, a ‘real’ experimentalist might have started by placing one pane of glass upon another, without spacing, using the logic that if one pane glass traps the solar radiation, 2 panes might trap it better. And, if he did, we cannot know what he might have observed. For, unless we actually do this; we cannot know what the result might have been. Regardless, of the preliminary constructions of his ‘hot box’, Horace reported what design produced the maximum temperature and what this maximum temperature was.
Did his observation prove that glass traps the energy of solar radiation?
I ask: When do you imagine that Horace stopped monitoring the temperature of his ‘hot box’? My answer: When he observed that the temperature was decrease beyond any ‘shadow of doubt’. Pun intended. Because I have observed, because I finally constructed a ‘hot box’ triply glazed with 3 panes of glass, what happens when a cloud casts its shadow on the hot box. The temperature being observed begins to decrease. And because better insulating matter, and experimentation which caused me to modify the dimensions of my hot box, which I prefer to call a very simple radiometer, my radiometer achieved a temperature of 230F, without pointing it at the sun, as the Styrofoam insulation began to melt.
So, to use my radiometer to measure the solar radiation during a variety of atmospheric conditions, I began to cover the glass with a sheet of Mylar when the temperature increase to 212F, and observed that the temperature decreased from 212F to about 140F in 10min. From which I conclude that glass does not trap the energy of solar radiation, it only slows the rate of the longwave IR radiations transmission through the glass. And for comparison purposes, I have constructed another radiometer with the same dimensions which is glazed with three films of polyethylene (food wrap). It requires a significantly greater time to acquire a temperature of 212F, because the polyethylene, which does not strongly absorb longwave IR radiation, does hindered this radiation’s transmission (loss) through the window. But when the temperature does reach 212F and the polyethylene film is cover with a sheet of Mylar, it cools at the same rate as the radiometer with a glass window. Because I have a little trouble ‘explaining’ this observed fact, I have repeated this cooling experiment several time and it is reproducible. So whether I can explain it or not, is not important. What is important in science is that which is observed.
Have a good day, Jerry
“In the 16th Century there was a Swiss naturalist…”
200 years later, in the 18th Century, there was another “Horace” that did the same thing.
Amazing coincidence, huh?
I believe you may be referring to Horace Bndict de Saussure’s solar flask experiment from the 18th century.
What is interesting about this experiment is that Joseph Fourier makes reference to it in his articles. Joseph Fourier was the first to make the flawed calculation that the surface temperature of the earth was higher than it should be for an object this distance from the sun. Fourier had made the same mistake that others later made revisiting his calculation after the time of Stefan and Boltzmann. He failed to account for the true surface properties of the planet.
Unaware of the flaw in his calculation, Fourier speculated that cosmic rays or some sort of physical greenhouse effect was responsible for the higher than calculated surface temperatures. Fourier considered de Saussure’s experiment, but because the flask had glass barriers to convection and the real atmosphere did not, he rightly dismissed it as a model of an atmospheric greenhouse.
Sadly Fourier was a better mathematician than he was a physicist. If he had thought more laterally, he might have realised that de Saussure’s solar flask experiment, while inapplicable to the atmosphere, had great relevance to how the sun heats the oceans. Just like the glass panes in the solar flask, water is transparent to solar radiation, opaque to LWIR and has a slow speed of conduction. Further the viscosity of water limits the speed of convection.
If Fourier had recognised that the effects of solar heating in de Saussure’s flask would be partially replicated in the oceans, he might have realised why his original surface temperature calculations for this ocean planet were in grave error.
In studying solar thermal gain in the oceans, I conducted experiments on similar lines to de Saussure. Initially I used clear acrylic blocks so as to replicate a material transparent to SW and opaque to LWIR, while eliminating the complication of convecting fluids. With clear blocks painted black on their lower surface and insulated on their underside, I found that solar exposure could drive the lower surface to 120C, 10C higher than de Saussure achived.
Yes, so I was quite prescient:
“Watch these twerps deny reality again as they insult their betters.”
Idiots,
“1800 W input cannot produce 10,800 W output. And BTW ice will never reach 300 F!”
Really? The observed radiator example demolishes your stupid religion.
We see a constant heat source (~60C) coming in on the top right. The input surface area is <1% of the output surface area. And then we see the whole thing emit ~60C.
I'll stick with observable science, and you can have your stupid religion.
"Oh Ts = 0 Kelvin now???"
"And thats a really BIG movement of the goal posts!"
"Use whatever you want for Ts, Zoe."
LMAO. Looks like I like I hit the edge of your goal post. No shift, moron.
"Sigma*(T^4-Ts^4) = 300 W/m^2. Solve for T."
This is the heat flow equation, not energy conservation. Why are you conserving heat flow? As a hot object heats a cold object, the heat flow is gradually reduced to ZERO.
In reality the hibachi heats the ice cube, then the air. The very close surrounding air will go to 300F, and the heat flow directly from ice to air will be 0.
You can touch the radiator that got heated from a small tube, and it will be ~60C all around.
‘LMAO. Looks like I like I hit the edge of your goal post. No shift, moron.’
Oh, really? You stated that the Hibachi was surrounded by 0 K (or C??) in the original problem?? No you didnt.
Thats called moving the goal posts, moron.
‘This is the heat flow equation, not energy conservation. Why are you conserving heat flow? As a hot object heats a cold object, the heat flow is gradually reduced to ZERO.’
If you don’t understand why I use this equation, (conserving heat flow, Huh??) well you are quite confused, Zoe.
‘In reality the hibachi heats the ice cube, then the air. ‘
No, you said it was in vacuum, dimwit!
Nate, please stop trolling.
‘We see a constant heat source (~60C) coming in on the top right. The input surface area is <1% of the output surface area. And then we see the whole thing emit ~60C.'
This is hot water flowing into a radiator. And….. it gets hot!
What do you think this is showing us, that is in any way relevant to our previous discussion?
“This is the heat flow equation, not energy conservation. Why are you conserving heat flow? As a hot object heats a cold object, the heat flow is gradually reduced to ZERO.”
It’s dishonest, but it convinces those that don’t know any better. That’s the sort of people we’re dealing with.
‘Its dishonest’ Oh? How so?
Is it the wrong equation, DREMT? You seem to imply you know better.
“OK, Nate”.
IOW, you have no idea, do you?
And yet, somehow, you know I’m both wrong AND dishonest.
#2
“OK, Nate”.
OK, Nate, means DREMT is missing a point, again.
#3
“OK, Nate”.
340 W/m^2 is the average flux received at TOA. as this average is hitting the Earth all the time as its spins in 24 hours. SO YOU “CAN’t divide by 4 ??? WHY WOULD YOU ??? NOW ONE IS EXPLAINING WHY THEY THINK THEY NEED TO DIVIDE BY 4 ???
“HERE”, LETS USE A LAYMANS TERMS ON THIS.
I hit the Earth with a average force of 340. I hit the Earth every second, and put a small dent in it. in 24 hours I have made a small dent all around the Earth. THUS, you CANNOT divide my hitting force by 4 ???
Wayne
Just to be clear…no one is suggesting that you divide 340 by 4. It’s 1360 that you divide by 4. Remember, 1360 is the average zenith flux. 340 is average spatial flux over the entire Earth.
Right get you. THUS, you should “NOT” divide the 1360, the average zenith flux then ??? “why” do you think you should divide this number ??? Also at what time is this 1360 is the average zenith flux added up against ??? 1 second, 1 hour ???
WHY WOULD YOU ??? NO ONE IS EXPLAINING WHY THEY THINK THEY NEED TO DIVIDE BY 4 ??? The Sun does not stop producing this average zenith flux of 1360, so how and why do you think you need to divide it ???
“HERE”, LETS USE A LAYMANS TERMS ON THIS.
I hit the Earth with a average force of 1360. I hit the Earth every second, and put a small dent in it. in 24 hours I have made a small dent all around the Earth. THUS, you CANNOT divide my hitting force by 4 ???
Wayne
wayne said…”why do you think you should divide this number ???”
Because the solar constant (1360 W/m^2) itself doesn’t tell us a whole lot. It does not represent the effective flux hitting the Earth. It cannot be used to compute the energy received over a period of time.
wayne said…”Also at what time is this 1360 is the average zenith flux added up against ??? 1 second, 1 hour ???”
1 sidereal year (365.24 days).
wayne said…”The Sun does not stop producing this average zenith flux of 1360, so how and why do you think you need to divide it ???”
Because the zenith flux is only observed at exactly one point on Earth. It is not observed everywhere. For example, where I live here in the middle of the United States I will never observe the solar constant…EVER.
wayne said…”I hit the Earth with a average force of 1360. I hit the Earth every second, and put a small dent in it. in 24 hours I have made a small dent all around the Earth. THUS, you CANNOT divide my hitting force by 4 ???”
If your hit were to occur on June 21st then it will make the deepest dent along 23.5N latitude. For each degree of latitude away from the Tropic of Cancer the dent your hit is making gets shallower becomes there is less punch. Once you get up to the Arctic cycle your hit packs no more punch and the depth of the dent is now 0. The deepest part of the dent is 1360 units, but the average depth is 340 units.
bdgwx, dividing solar flux leads to pseudoscience.
You say the averge is 1360 W/m^2. So lets say I dropped a bomb on the Earth where the Sun was shining for that day, I would get a hit or dent in the ground of 1360 W/m^2 in that area, So how and why could you divide by 4 ??? It does not make any sense ??? Or make it simple, you drop a 1 pound brick down and calculate the force hitting the ground, you don’t divide by 4 ???
Yes I know its not observed everywhere, but where its observed, its putting out a power of 1360 W/m^2, so you cant divide this ???
I have a car traveling quarter a way around the Earth, my averge speed is 80MPH, you would not divide this by 4 ??? Or would you ??? It seems you would, but does that make sense to you this car speed being devude by 4 ???
Wayne
Hi there,
The more I think of this the worst it gets. The 1360 W/m^2 is the averge energy/heat the Sun hits the whole the Earth in one year. So why are you dividing it ??? That’s like saying I wrapped the whole Earth up to sell it, but your saying I only wrapped up one quarter ??? Its like saying the Earth went around the Sun and orbited, ne full orbert, and heated the Earth to 15c, but your saying the Sun only heated it to 3.75 ??? Or I gave you an averge of 4 apples, but your saying I only gave you one ??? Dividing does not make any secoe whatever ???
If the Sun only hit a quarter of the Earth in 24 hours, and gave 1360 W/m^2, then, and only then could you say the averge temp. of Earth was a quarter of 1360 W/m^2. Then that would be wrong.
Wayne
wayne said…”You say the averge is 1360 W/m^2. So lets say I dropped a bomb on the Earth where the Sun was shining for that day, I would get a hit or dent in the ground of 1360 W/m^2 in that area,”
NO. That’s the thing. The dent won’t be 1360 units…everywhere. That’s what we’re trying to say.
wayne said…”Yes I know its not observed everywhere, but where its observed, its putting out a power of 1360 W/m^2, so you cant divide this ???”
Dividing by 4 is just a shortcut using canonical geometrical principals. The long way is to integrate the function 1360 * cosine(theta) where theta is the zenith angle.
wayne said…”I have a car traveling quarter a way around the Earth, my averge speed is 80MPH, you would not divide this by 4 ??? Or would you ??? It seems you would, but does that make sense to you this car speed being devude by 4 ???”
No you wouldn’t. The reason is because there is no relationship between your average speed and the ratio of the cross sectional and total areas of Earth. Speed does not have units of m^2 in it, but flux does.
wayne said…”The more I think of this the worst it gets. The 1360 W/m^2 is the averge energy/heat the Sun hits the whole the Earth in one year.”
No. That is not correct. 1360 W/m^2 is the zenith flux average over 1 sidereal year. Nothing more. It is most definitely NOT “the averge energy/heat the Sun hits the whole the Earth in one year.”
waynes,
Let’s do an exercise…using the 1360 W/m^2 zenith flux average compute the total amount of energy Earth receives in one year. I want to see how you are doing it.
bdgwx. dividing solar flux by 4 puts you on the road to pseudoscience.
Is that where you want to go?
JD, same question for you…how much solar energy does Earth receive in one sidereal year? Show your work.
Answering a question with a question is bad form, bdgwx. Such distractions are often used in pseudoscience.
“dividing solar flux by 4 puts you on the road to pseudoscience.”
Then please give us the scientific answer using the inverse square rule or whatever you see as correct.
Craig and bdgwx, start here first:
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356255
If you still can’t understand, a simple example might help:
An electric motor is rated at P Watts. P is the exact power needed to lift a fully loaded elevator one floor of a building, in time T. The energy required is E Joules.
For the elevator to go to 2nd floor:
Power = P Watts, Energy = E Joules
For elevator to go to 3rd floor:
power = P Watts, Energy = 2E Joules
For elevator to go to 4th floor:
power = P Watts, Energy = 3E Joules
For elevator to go 5th floor:
power = P Watts, Energy = 4E Joules
The mistake you are making, by dividing solar flux by 4, is analogous to saying: “P Watts will take the elevator up 4 floors, so P/4 will take the elevator up 1 floor”.
See why dividing solar by 4 is pseudoscience?
If it takes 1 sec to go one floor, it takes 4 sec.s to go to 5th floor. Do you understand why your bogus analogy is a straw man that you then stab?
JD,
For your analogy to be apt, you need to bring in the time factor, and something analogous to area factors.
Until you do, it just another way to obfuscate.
For example, I can go up one floor by applying P for time T,or by applying P/2 for time 2 T.
This is analogous to the sun shining T = 12h on each side with input power P, then for 12h with input power 0.
Or shining for 2T = 24 h, with average power P/2.
Same result.
Nate, I’d say better to give JD a chance to think on his analogy than just do the thinking for JD. Though I do get JD will never change; JD’s humor flows continuously with JD avoiding the scientific method at all costs – only JD & the cartoon method are observed hereabouts.
Now give me a little more time to warm up the BBQ for Zoe’s ice cube scientific method later. The poor ice cube is forming in the ‘fridge unaware of its fate.
Nate and fluffball rush in to pervert reality.
They want to ignore the info from the simple example: “P is the exact power needed to lift a fully loaded elevator one floor of a building, in time T.”
Nothing new.
“A flux obeys the inverse-square law, meaning that it changes with distance. For example, the solar flux at Suns surface is 64,000,000 Watts/m^2. But the flux is reduced to about 1365 W/m^2 when it reaches Earth.”
So let’s employ the inverse square rule. The part of the Earth closest to the sun receives 1365 W/m^2 but the power would drop to 1364.8 W/m^2 upon reaching the point on Earth farthest from the sun, if that location wasn’t in the shadow of the Earth.
JD, how do you propose using that information to determine the average watts per square foot striking the Earth?
Very good JD, you are making progress. Yes time T is 1sec., power is full “on” for each floor transit in the 1 sec. Do you understand NOW why your bogus analogy is a straw man that you then stab?
Hint: a correct analogy would not have the power on for the full 1sec. Solar power only illuminates 1 hemisphere at a time on a rotating planet or rotating ~spherical object, now let’s see if you can run with that.
It’s interesting to watch how they work.
Nate and fluffball are into complete perversion and corruption.
Craig T is avoiding his own words:
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356856
And bdgwx has slipped out the backdoor.
A group of geese is a “gaggle”.
What’s the term for a group of clowns?
bdgwx = “BackDoor Guy Worshipping ‘X‘”
Yes, it’s fascinating. As we know, “Natan: Master of Lies” and “Testicle4” are entirely bereft of any human decency, honesty, or integrity, and will simply say anything to deceive others. Lacking empathy helps these “people” in their daily grind of perverting and disrupting the arguments of their intellectual and moral superiors.
But equally intriguing, and revolting, are the new avatars that the GHEDT (Green House Effect Defense Team) dreamed up. They decided on adding a coupla good ol’ boys that would hold back on the usual ad hom onslaught and try to stick to just discussing (read: distorting) “the science”. They both started off in this manner, however it didn’t take long for Kreg T to devolve into a typical alarmist “Snark Thug”, whilst carrying on with the usual endless misrepresentations and false accusations.
bdgwx continues to amuse and disgust with his “I’m just here to educate these lesser morons” attitude whilst he relentlessly misrepresents every word you say. Always playing “the teacher”, he even sets his victims little tests as he wildly distorts their arguments and obfuscates at every opportunity. Professional deceivers, sent to hold back the progress of mankind. Oh well, never mind.
It is never my intent to be condescending. If anyone has gotten that impression then I apologize. I learn more from posting here then anyone could possibly learn from me. And I’ll be the first to admit that mistakes are part of learning. So if I make a mistake please bring it to my attention. But remember, I’m convinced more by math with work shown and by the abundance of evidence than by nuh-uh statements.
bdgwx, to your credit, you have not insulted, misrepresented, or falsely accused me.
But, to your discredit, you have rambled incessantly in support of pseudoscience. Then, when shown the physics that squashes your pseudoscience, you rapidly exit by the backdoor, only to pop up later as if you were never exposed to the truth.
You behave as if you are some wimp that is afraid to face reality.
Am I not understanding you correctly?
Dr Roys Emergency Moderation Team says:
Like Dr. Roy Spencer, Ph.D.
The top post is in defense of the Green House Effect.
Don’t worry too much, bdgwx, it’s all said with tongue firmly placed in cheek. You are definitely one of the best-behaved of the Decepticles. They created an interesting false persona this time, “the concerned teacher, politely correcting straw men arguments”. Probably a good idea to generate a new avatar that focuses on “the science”, so I can see why you were generated.
‘are entirely bereft of any human decency, honesty, or integrity, and will simply say anything to deceive others. ‘
DREMT:
Good argument for anononymity
The problem with this paranoid scenario is that you never seem to be able to point out exactly what is or isn’t true that I say, and why, given your ignorance of the subject matter.
Such as in this example. What have I said that is lie intended to deceive people?
The solution to this problem of yours is NOT to get more and more paranoid, but instead, to go crack open some textbooks or take a course and LEARN the subject matter.
Only then will you be able to tell what is truth or not, and argue based on the facts, rather than paranoia.
‘bdgwx, to your credit, you have not insulted, misrepresented, or falsely accused me.’
While being on the receiving end of very much of that from you!
Yes, he is remarkably tolerant.
Meanwhile we have JD, who simply slings snide insults, “Nate and fluffball are into complete perversion and corruption.”, while never pointing out what is ‘corrupt’ or ‘perverse’ in the ordinary physics we show.
We, know, of course, thet he is simply a troll, who doesnt need to make sense.
Yes, I’m thinking I got the reference wrong…it should have been, “Natan: Father of Lies”.
That’s better.
As far as I can tell, DREMT thinks I’m dishonest because I have the audacity to disgree with his opinions, and I explain why I disagree using facts and science that he cannot dispute.
It just galls DREMT, that he is not able to understand these facts and science.
But still, he DESPERATELY wants people who DO understand these things, to inexplicably, admit they’re wrong and HE is right.
And it just infuriates DREMT that people who DO understand science have the audacity to insist that no, he is still not correct, and that they will NEVER accept fake science over real science.
See?
Nate claims: “…while never pointing out what is ‘corrupt’ or ‘perverse’ in the ordinary physics we show.”
Here’s an example from this very thread:
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356863
P/2 would not raise the elevator, as required. Nate tries to pervert and corrupt my simple example.
Nothing new.
P/2 for 2T, of course it can, JD. Why not?
‘P is the exact power needed to lift a fully loaded elevator one floor of a building, IN TIME T.’
There’s your own words explaining it.
Suppose the elevator weighs 2000 lbs. T = 10 s. And 1 floor = 10 feet. How much power is needed?
Then same, but with T = 20 s, how much power is needed?
P/2 would not raise the elevator, as required. Nate tries to pervert and corrupt my simple example.
Nothing new.
It can’t be that he’s that dense, so it’s got to be dishonesty.
When I use ordinary basic physics, you guys think I’m making it up!
Hilarious! Proof that you guys don’t speak the language.
Obviously JD couldnt answer my question, no surprise there.
‘Suppose the elevator weighs 2000 lbs. T = 10 s. And 1 floor = 10 feet. How much power is needed?’
P = 2000lbs x 10ft /10s = 2000 ft-lbs/s old english unit!
Then same, but with T = 20 s, how much power is needed?
P =2000 lbs x 10 ft/ 20 s = 1000 ft-lbs/s.
Half the power, but twice the time.
DREMT,
‘It cant be that hes that dense, so its got to be dishonesty.’
Perfectly illustrating my earlier point:
You need to “go crack open some textbooks or take a course and LEARN the subject matter.
Only then will you be able to tell what is truth or not, and argue based on the facts, rather than paranoia.”
Maybe he actually is that dense. At least, that’s the way he’s going this time.
‘Maybe he actually is dense..”
There is no ‘maybe’ with you guys when it comes to this subject matter.
If DREMT is still unsure if I am correct or dishonest about science this BASIC, ‘maybe’ he should reconsider his man-splaining to those who CAN understand it, that they have it all wrong.
Are you assuming I am disputing your math!?
Can you dispute my math or physics, DREMT? Pls do show us.
I’ll take that as a “yes”.
Since DREMT never misses an opportunity to tell someone, not in his squad, theyre wrong, clearly in this case he can’t.
Then JD must be wrong. Ok.
Not wise to be misled on science by JD.
JD already told you exactly how you were corrupting his example.
Instead of listening, you pretended we didn’t understand your math or physics.
“Nothing new?”
‘JD already told you exactly how you were corrupting his example.’
He said ‘P/2 would not raise the elevator’.
I showed him this is wrong, and gave a clear example of how it could.
Is that what you call ‘corrupting’?
‘Instead of listening, you pretended we didnt understand your math or physics.’
Heres the problem DREMT. I listened, and heard no evidence that you DO understand it.
If you did understood it, and could tell me why I’m wrong, and how I’m ‘corrupting’ things, you most certainly would have.
You obviously have no idea, so you obfuscate, distract.
And if JD could easily show me why Im wrong, he would most certainly would not hesitate to do so.
Instead he slinked away with no response. That should be a clue.
No Nate, as always you miss something out.
JD said, “P/2 would not raise the elevator, as required”.
What is required by his example, in order to make his point? That the elevator lifts one floor in time T. By corrupting the example so that at half power the elevator lifts one floor in time 2T, you’ve already missed the point, and are trying to change things to make your own point.
‘What is required by his example, in order to make his point? That the elevator lifts one floor in time T. By corrupting the example so that at half power the elevator lifts one floor in time 2T, youve already missed the point, and are trying to change things to make your own point.’
OMG, how dumb you guys are. Any analogy if poorly framed, as this one was, is simply misleading people.
What you call ‘corrupting’, I call making it relevant to the problem at hand.
Relevance:
“For your analogy to be apt, you need to bring in the time factor, and something analogous to area factors.
For example, I can go up one floor by applying P for time T,or by applying P/2 for time 2 T.
This is analogous to the sun shining T = 12h on each side with input power P, then for 12h with input power 0.
Or shining for 2T = 24 h, with average power P/2.
Same result.”
Bad analogy made relevant. Both of you missed the point. No surprise.
BTW, JD: ‘P/2 would not raise the elevator, as required’
Is pretty clear that JD, as per his usual, cannot understand power and energy.
BTW, grammar is clearly not your forte either.
“P/2 would not raise the elevator, as required”
The comma in there can only mean one thing:
P/2 would not raise the elevator–Pause–as required.
IOW P/2 would not raise the elevator. Raising the elevator is what is required.
That the elevator needs to move goes without saying. So it wouldn’t make sense for the “as required” to refer to that.
“P/2 would not raise the elevator, as required”.
I think it more likely that the comma is superfluous, than the words “as required”. Unless JD wishes to correct me, I will go with my reading.
As has already been seen, you jump to whatever conclusions you need to, to put down the people you are talking to. Whether this is a deliberate tactic, or just a part of your personality, who can say? But your other comment is yet another example of this. And it also seems, once again, that you are guilty of something you falsely accused me of, earlier:
“And we all know that your points, now matter how useless, are the only ones that really count.”
History repeating. I find myself defending a straightforward example, that makes a clear and simple point, from somebody whose sole purpose seems to be to bury that point. That’s how it comes across to me. Maybe that’s “jumping to conclusions”, maybe not. There certainly are a lot of examples of it happening, so…
‘As has already been seen, you jump to whatever conclusions you need to, to put down the people you are talking to’
‘Put down’?
So lets see, JD posts an elevator analogy.
I respond to it. And suggest a change to make it more relevant to the discussion.
Nate and Ball4: “are into complete perversion and corruption.”
“Yes, its fascinating. As we know, Natan: Master of Lies and ‘Testicle4’ are entirely bereft of any human decency, honesty, or integrity, and will simply say anything to deceive others.”
Apparently, you guys are so fragile and insecure that you think the normal back and forth of debate is putting “people down”, and all the rest..
BTW, this was JDs conclusion from his original elevator post:
“The mistake you are making, by dividing solar flux by 4, is analogous to saying: P Watts will take the elevator up 4 floors, so P/4 will take the elevator up 1 floor.
See why dividing solar by 4 is pseudoscience?”
Whats wrong with his analogy is that he is not considering TIME, and he NEEDS to, because E = P x Time.
“P Watts will take the elevator up 4 floors, P/4 will take the elevator up 1 floor”
sounds very dumb, as JD wants it to.
But if include TIME: you could say it this way
P watts applied for 4T gives 4E, enough to take the elevator up 4 floors.
P/4 watts applied for 4T gives 1E, enough to take the elevator up 1 floor.
Not dumb after all!
That’s what I mean by
‘Any analogy if poorly framed, as this one was, is simply misleading people.’
“Apparently, you guys are so fragile and insecure that you think the normal back and forth of debate is putting “people down”, and all the rest…”
So you take one phrase and run to a whole marathon of false conclusions with it. None of what you’ve said at 6:31am is in any way addressing what I was trying to get at. Nate, nobody cares about any of the insults that fly either way in these discussions. All I’m trying to say is that your false accusations that “we don’t understand this, or that”, seem deliberate, and the intent is to try to undermine credibility of your opponent. I call you dishonest because that’s what I really think of you. If you want to pretend that you really think JD is unable to understand your point about half power and double the time then go ahead, but you’re not convincing me, and I doubt you’ll be convincing anybody else.
JD is</b already incorporating time into the analogy. You are again corrupting his analogy, missing the point he made, in order to make your own point. You are also just repeating yourself, over and over again.
It’s also abundantly clear, as always, that there is no Universe in which you will not respond, so I will await the next comment chock full of misrepresentations and false accusations, and if I keep responding to you, we will be here until the discussion closes for comments.
DREMT,
“So you take one phrase and run to a whole marathon of false conclusions with it.”
Nope, many phrases by you in this thread accusing me of all sorts of ridiculous things.
“None of what youve said at 6:31am is in any way addressing what I was trying to get at. Nate, nobody cares about any of the insults that fly either way in these discussions. All Im trying to say is that your false accusations that we dont understand this, or that, seem deliberate, and the intent is to try to undermine credibility of your opponent.”
I see LOT^S of evidence that both you and JD do not have a very good understanding of the science.
I tell you I have a background in Physics. You are doubtful.
If you understood basic physics, you should be able to tell immediately from my posts that I’m not BS-ing you.
I see confusions from you both about concepts like 1LOT, 2LOT, entropy, heat flow, power vs energy, angular momentum, etc etc.
Your arguments are mostly about your intuition, rather than equations and laws. You must admit that!
IMO one should be aware of ones limitations. You could learn a lot from this blog if you wanted to.
But instead you insist on man-splaining to a bunch of obvious science nerds, that they are all wrong about SCIENCE!
It makes about as much sense as me going to a blog full of historians and telling them they’ve got the Renaissance all wrong!
Once again, Nate: who are you trying to convince? Me, others, or yourself?
You’re not fooling me. So why do you bother?
‘Youre not fooling me. So why do you bother?’
Im not fooling you about what?
Dispute anything I’ve said.
As if I haven’t disputed anything you’ve said already!
“OK, Nate”.
Good example, DREMT.
JD thoroughly confused about 2LOT, and has no answers.
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-357484
Feel free to help him out!
Typical Nate. Every single thing I say seems to irk him, yet he’s always desperate to bait me into a discussion (or should I say, more accurately, into repeating a discussion that’s already taken place). If I refuse to participate, he will just increase the baiting.
‘Irk him’??
No. These are comments you broght up DREMT.
My response was, you guys need to read them to see how your ‘debunking’ was debunked.
Case in point, JD falsely implicating 2LOT, and quite confused about it.
So, keep up the man-splaining!
See what I mean?
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356661
You guys don’t seem to be talking physics ere, your just running numbers out.
Lets do some thought experiments, one I did today.
1,
Temp. today was 16c, we had lots of moving clouds. So your telling me that when the clouds moved away, and in the full Sun it was so much warmer, I had a thermometer and it started to go up as soon as the clouds went and full Sun hit us.
Your saying that’s not the Sun’s heat, as your saying he Sun can only warm us up to -18. So what was this so much warmer effect that when the clouds moved out of the was and the Suns full heat hit as ??? Its not the Sun, but what ???
2,
Lets do my other thought experiment. Your saying the Sun can only warm us to -18c, and its the GHG’s that heat us up to 16cish. So why in Wales UK today inn the full Sunlight was it 16c, but over in Dubai it was 40c ???
3,
Why do you think you have to divide by 4 ??? Please explain, as you can not divide by 4 ???
You say the averge is Sunlight reaching the Earth for one year = 1360 W/m^2. So lets say I dropped a bomb on the Earth where the Sun was shining for that day, I would get a hit or dent in the ground of 1360 W/m^2 in that area, then I dropped bombs 3 more on the 3 other quarters of the Earth. So how and why could you divide by 4 ??? It does not make any sense ??? Or make it simple, you drop a 1 pound brick down and calculate the force hitting the ground, you dont divide by 4 ???
Yes I know its not observed everywhere, but where its observed, its putting out a power of 1360 W/m^2, so you cant divide this ???
I have a car traveling quarter a way around the Earth, my averge speed is 80MPH, you would not divide this by 4 ??? Or would you ??? It seems you would, but does that make sense to you this car speed being divided by 4 ??? if the car traveled quarter a way or all the way around the Earth, and its averge speed was 80MPH, then it will always be 80MPH.
Wayne
1) Think thru the change in instantaneous albedo. Then think thru global mean albedo.
2) Think about:
Wales 52.1307degrees N
Dubai 25.2048degrees N
3) The observed mean global near surface T is ~288K.
Using geometry’s well known divide by 4 the calculated mean global T ~288K.
If you don’t correctly divide by 4 the answers do not compare.
“I have a car traveling quarter a way around the Earth, my average speed is 80MPH, you would not divide this by 4 ???”
You had to stop for gas at 0 mph, somewhere you had to divide by a factor other than 4 to get your avg. speed.
I stated Wales and Dubai, as if you think the Sun can only heat up the Earth to -18c, then there would be not heat from the Sun like we feel it both these places ???
What does 1360 W/m^2 actually mean ??? Is it the Suns heat hitting one square meat or ??? What does the mean global near surface T is ~288K mean ???
You have not explain why you divide by 4 ???
Wayne
You say; “The observed mean global near surface T is ~288K.
Using geometrys well known divide by 4 the calculated mean global T ~288K.
If you dont correctly divide by 4 the answers do not compare.”
However your not saying why you divided by 4 ??? Why 4 ??? There are not 4 Earths, I just don’t understand the 4 ???
Wayne
Re=radius earth
So=annualized total solar irradiance (TSI) input as continuously measured by SORCE satellite in the time period ~1369 W/m^2
The total annual measured amount of solar radiant energy intercepted by Earth is So*pi*Re^2 watts.
~Spherical Earth total mean surface area is 4*pi*Re^2 m^2.
So the gross solar irradiance spread uniformly over earth entire global mean surface area used to compare to the global thermometer field mean temperature is (So*pi*Re^2/4*pi*Re^2) = (1369/4) W/m^2.
Fraction of that is reflected/scattered to space (albedo), as the oceans appear dark in satellite pictures, brightened here and there by clouds.
This yields a net solar irradiance, with albedo=0.3, of S = So(1-albedo)/4 ~ 240 W/m^2 which is used as input to the first law energy balance to compute the mean temperature of earth (~288K) to compare to the measured thermometer field mean (~288K).
Got it? Get it.
Ball4, please stop trolling.
“So why in Wales UK today in the full Sunlight was it 16c, but over in Dubai it was 40c ???”
(For everything below the calculations are as if it was the Equinox where the sun is directly above the equator.)
Dubai’s latitude is 25 degrees North and Wales is 52 degrees North. The sunlight in Wales is less intense than in Dubai because of the curvature of the Earth.
The ground on the equator is perpendicular to the rays of sunlight at noon. One square meter of sunlight will hit the Earth square on so the land receives the full solar constant of 1370 watts per square meter.
Dubai is 25° North of the equator. At that latitude the Earth is only slightly slanted toward the North compared to the equator. One square meter of sunlight will be spread over 1.04 square meters of land. Not quite the solar constant but close.
For Wales the curvature of the Earth puts it at more of an angle to the sun. One square meter of sunlight is spread over 1.23 square meters. That means a square meter of land only receives 80% of the solar constant at noon.
Norilsk, Siberia has a latitude of 69° North. A square meter of land in Norilsk only receives 63 percent of the solar constant at noon during the equinox.
There are a lot of other factors that makes the temperature of Dubai different from Wales besides the amount of solar power it receives but you can see why the farther from the equator the less intense the sunlight.
I stated Wales and Dubai, as if you think the Sun can only heat up the Earth to -18c, then there would be not heat from the Sun like we feel it both these places ???
What does 1360 W/m^2 actually mean ??? Is it the Suns heat hitting one square meat or ??? What does the mean global near surface T is ~288K mean ???
You have not explain why you divide by 4 ???
Wayne
Assuming 12 hours day and 12 hours night the equator averages 436 watts per square meter for 24 hours while Wales averages 349 watts per square meter over the same period.
However the Earth is not static, its rotating. Your saying above you divide the heat coming from the Sun by 4 ??? If the equator averages 436 watts per square meter for 24 hours while Wales averages 349 watts per square meter over the same period. That’s fine.
Lets just say the whole Earth has 100 watts over the whole Earth in 24 hours. What you seem to be doing is dividing this by 4 ??? This is what I cant understand you are doing ???
Wayne
“Lets just say the whole Earth has 100 watts over the whole Earth in 24 hours.”
Re=radius earth
So=your TSI W/m^2
Your total 24hr. imagined amount of solar radiant energy intercepted by Earth is So*pi*Re^2 = 100 watts.
~Spherical Earth total mean surface area is 4*pi*Re^2 m^2.
So your imagined gross irradiance spread uniformly over the entire globe mean surface area used to compare to the global thermometer field mean temperature when steady state equilibrium is achieved in 24hours is (So*pi*Re^2/4*pi*Re^2) = (1oo/(4*pi*Re^2)) W/m^2.
Fraction of that is reflected/scattered to space (albedo), but your imagined case albedo is unknown (way higher than 0.3) as net illuminated only by a 100W light bulb Earth is very cold frozen solid even at the equator.
Ball4, please stop trolling.
“Your saying that’s not the Sun’s heat, as your saying he Sun can only warm us up to -18. So what was this so much warmer effect that when the clouds moved out of the was and the Suns full heat hit as ??? Its not the Sun, but what ???”
Imagine an actual greenhouse. The greenhouse is warmed by the Sun but it is warmer inside the greenhouse. Something is keeping more heat in the greenhouse than outside. Actual greenhouses don’t work the way “greenhouse” gasses work but it’s a good place to start.
It’s always the Sun’s heat. Greenhouse gasses let less of that heat leave the Earth. That’s what Dr. Spencer, Ball4, Nate, bdgwx and I are saying.
Greenhouse gasses let less heat in, thus cool the Earth ??? As in the Moon day time temp.
But lets not go there yet, I need to understand why your dividing by 4 ???
Wayne
wayne,
What we’re saying is that although GHGs are not a source of heat they do impede the transfer of it. GHGs do not strictly warm the Earth, but they slow the rate of heat loss thus allowing the Sun to warm the Earth to a higher temperature than it would be otherwise.
The best analogy would be the behavior of a furnace in your home. It is capable of warming your home. But, your home can be warmed to a higher temperature with insulation as opposed to without it. Like GHGs the insulation in your home does not provide any heat on its own, but it does slow the rate at which your home is losing the heat thus allowing the furnace to warm it more effectively and thus bring it to a higher temperature.
bdgwx, if repeating that same nonsense, over and over, is not working, you might want to face reality.
I made another image that might help.
https://i.imgur.com/FgVwIYh.jpg
The Earth’s shadow shows how much sunlight reaches the Earth. Watts are a measurement of energy over time, of joules per second. The power of sunlight reaching the Earth is 1380 Watts per square meter. The total power of light reaching the Earth is the area of the Earth’s shadow times 1380 Watts per square meter.
The area of the Earth’s shadow is the Earth’s radius squared times PI so the total power reaching the Earth is 1380 watts times the Earth’s radius squared times PI.
Half of the Earth is exposed to sunlight and half isn’t at any time. But because of the curve of the Earth not every square meter in the Sun receives the same amount of power. The Earth directly under the Sun gets the full 1380 watts per square meter but the farther on the globe from there in any direction the less powerful the sunlight.
The surface area of the Earth is found with the equation 4 times the Earth’s radius squared times PI. Some of the Earth gets all 1380 Watts per square meter, other gets less and half of the Earth is dark. The entire power – the energy per second – reaching the Earth is 138 watts times the Earth’s radius squared times PI.
The average energy per second a square meter of the Earth receives is 1380 watts times (the Earth’s radius squared times PI) divided by 4 times (the Earth’s radius squared times PI) or 1380/4 watts per square meter.
So on average one square meter of Earth receives 345 watts or 345 joules per second. That won’t tell you the temperature in any one location on Earth but if you’re looking at the big picture of the power reaching the Earth and the power leaving the Earth you use the number 345. The grid in the picture may help you see how the sunlight striking the Earth is spread over the surface unevenly.
Craig T, please stop trolling.
Wayne,
Do you know anybody with solar panels? You can ask them about it.
They may have a panel that can generate 1000 W = 1 kW in full sun at noon in the summer.
They will tell you that they don’t certainly don’t get that at night, they get 0.
They also may tell you that they don’t get that much at 5 pm or 8 am, they may only get 200 W at those times, even on a clear sunny day.
Thus during an average sunny day they may get only the equivalent of 6 hours of full sun on their panel, or 1 kW x 6 hours of energy, IOW 6 kWh.
That is referred to as 6 kwh of solar insolation.
As you can see that is about 1/4 of the energy you would get with 24 h of full sun.
Thus the factor of 4.
Clouds, of course reduce this even more. So UK solar panels may only get 4 kW/day on average.
You said, “That is referred to as 6 kwh of solar insolation.
As you can see that is about 1/4 of the energy you would get with 24 h of full sun.
Thus the factor of 4.”
RIGHT I GET YOU NOW. “HOWEVER”, here is where you are going wrong. “WHY” are you just counting the 6 hours of heat from the Sun that comes in in 6 hours, on say one quarter of the Earth ???
There are 4 quarters of the Earth and 2 hours, thus you “HAVE” to count them all up, RIGHT ???
Wayne
The 6kWh is the average energy obtained from the sun on a square meter on Earth in a day. The average insolation
Solar panels show that this is a REAL effect, that is measured by how much energy they collect in a day.
If we take all locations in the world, and find the daily insolation (for clear sky), the average will be 6 kWh, or 1/4 of what full sun could provide in 24 h.
Of course cities closer to the poles get less and those near the equator get more, but the average is 6 kWh, which is 1/4 of the full sun for 24 h.
Nate, please stop trolling.
wayne, what is the total amount of solar energy received by Earth in one sidereal year (365.24 days)?
Not sure, you tell me ???
Wayne
The answer is 173e15 W-years. This works out to…340 W-years/m^2.
The total amount of energy received per second at the top of Earth’s atmosphere per second. Because the surface area of a sphere is four times the cross-sectional surface area of a sphere, the average TOA flux is one quarter of the solar constant and so is approximately 340 W/m.
WHY WOULD YOU DIVIDE BY 4 ??? Let me see if I have this right ??? You have the energy received for one second for one quarter of the Earth, thus why divide ??? as there are 4 quarters, and the Sun will heat all four quarters in a 24 hour orbit, you must x by 4, as the Suns energy does not go on and off, it stays on giving out energy for 24 hours. Why divide, “why” ???
So, if you had a square meter surface in space directly facing the sun, it would be receiving the 1340W/m2 of energy from the sun (strength of sunlight, 93.5 million miles from the sun).
You divide the figure by 4, because a sphere has 4 times the surface area of one side of a disc.
Wayne
wayne rowley says:
“You divide the figure by 4, because a sphere has 4 times the surface area of one side of a disc.”
Eureka moment, but subtract the albedo first (30%).
340 W/m^2 is the average flux at TOA.
240 W/m^2 is the average flux at the surface from the 30% albedo reduction.
The other 30% gets reflected back into space and never reaches the Earth.
Svante, bdgwx, Craig T, please stop trolling.
Nate,
“This is hot water flowing into a radiator. And.. it gets hot!”
Water falls down. This gets heated top down. Probably gas.
Here’s an electric one:
https://youtu.be/IlHLKixwqDQ
“What do you think this is showing us, that is in any way relevant to our previous discussion?”
Deny deny deny. Only pathetic people wish to live in unreality.
‘Deny deny deny’ what?
That hot water transfers heat? Nope, I agree it does.
Heat transfer happens in various ways? Yes.
What am I denying?
And, I repeat the simple question,
‘What do you think this is showing us, that is in any way relevant to our previous discussion?’
Can you help us understand what point is being made by this video? How does this relate back to the ice cube thought experiment?
“No, you said it was in vacuum, dimwit!”
Doesn’t matter. There is no heat flow to space, because space is not matter. The right hand side can’t be 300, it’s 0.
The proper equation to use is the SB equation.
‘The right hand side cant be 300, its 0.’
Huh?
Before you said it was 1800 W/m^2 x 6 m^2 = 10,800 W flowing out!
Now it is 0 ??!
Zoe, you really need to get your story straight!
Nate, please stop trolling.
The idiots are now playing dumb.
Dr Spencer
I think your proposing a simple model to exaggerate a point is entirely disingenuous. I don’t think that anyone believes that the atmosphere does not have an effect on the temperature of the Earth.
I believe that the entire point of the JP posts concerning flat-earth science and physics, is to point out that the entire Kiehl-Trenberth energy budget is conceptually and factually false.
I challenge any atmospheric scientist to actually prove their work through physical experimentation and physical observation. Measurement of LWR with current pyrometers, for example, does NOT qualify as vigorous scientific observation. The results are derived and not actually measured, due to the misapplication of the SB constant.
The false and misleading application of the Stefan-Boltzmann constant is at the root of virtually all nonsensical climate science. You are as guilty as any, in perpetuating these false paradigms.
‘ The results are derived and not actually measured, due to the misapplication of the SB constant.’
What is the misapplication of the SB constant that you are talking about, Dan?
‘Measurement of LWR with current pyrometers, for example, does NOT qualify as vigorous scientific observation.’
Sure it does. No different from using a mercury thermometer or a light meter, a pressure gauge, anemometer, humidity gauge.
All rely on deriving results from a calculation that takes what the level on the meter is and turns it into temperature, pressure, humidity etc.
And of course testing them many times.
PATENT:
Apparatus for calibrating and testing infrared detection devices is provided. The apparatus comprises a substrate which supports a target pattern of dielectric material which is at least partially absorbing to infrared radiation. A heater is used to supply heat to the substrate. Since the substrate and dielectric material have different emissivities, an apparent temperature difference is perceived by an IR detection device. As a consequence, temperature differences as low as about 0.02 C. and below can be generated for calibrating and testing IR imaging devices.
https://patentimages.storage.googleapis.com/9b/1d/72/23afdbd1a57677/US4387301.pdf
Can you provide any example of pyrometers giving inaccurate measurements of infrared radiation?
Nate, Craig T, please stop trolling.
Nate,
The RHS is 0 in your heat equation. Equilbrium is reached.
Why are you so retarded?
https://cdn.britannica.com/66/149866-004-95ACE9E9.jpg
The red cube is the hibachi
The blue cube is the ice cube
Only one side has thermal contact.
According to you, there will be a gradient from 1800 W/m^2 to 300 W/m^2, but according to physics there will be equilibrium.
In this example the red cube is not a constant heat source, so it goes to purple. But if it was, they would both end up red cubes.
Why don’t you join reality?
‘Why are you so retarded?’
Zoe, why do you keep changing the problem and your answers, until they no longer even relate to your original point?
There is nothing wrong with your blocks in contact reaching equilibrium.
But it involves no heat INPUT or heat OUTPUT.
That has little to do with your original point about flux, area, INPUT, OUTPUT.
‘You claim it is legitimate to divide solar INPUT by 4, i.e. spread it over the whole earth. But the sun shines on 1/2 the Earth, while the OUTPUT is over the whole Earth. The output is 2x the input.’
Nate, please stop trolling.
Nate,
“Zoe, why do you keep changing the problem and your answers, until they no longer even relate to your original point?”
As has been explained several times: because the ambient temperature has nothing to do with our discussion. We are discussing how hot object effects cold object. You are claiming surface area matters, I’m claiming it doesn’t.
“There is nothing wrong with your blocks in contact reaching equilibrium. But it involves no heat INPUT or heat OUTPUT.”
Heat input and output makes equilibrium impossible?
Remember, dumbass, the red cube only touches ONE face of blue cube, and according to your religion, the RED should be divided by 6, and so the end result should be a double cube that’s much closer to blue. And yet the result is purple, a perfect halfway point between red and blue.
Did you miss that, dumbass?
Here’s another example:
https://i.gifer.com/9vmM.gif
The input area is 5, the output area is 46.
According to your religion, the result should be a slightly purplish blue.
In reality it’s all red, and by analogy there’s no gradient from an analogous 1800 to 300 W/m^2, it’s all 1800 W/m^2.
Why don’t you face up to reality?
‘according to your religion, the RED should be divided by 6, and so the end result should be a double cube thats much closer to blue.’
Nope, not at all. Another weird strawman.
Its pretty clear that you are either very very ignorant and confused, or some kind of troll, or both, Zoe.
Trolls don’t need to make much sense, and you certainly don’t.
Trolls throw lots of ad-hom grenades, as you do.
If you are not a troll, you still qualify as one.
Go troll your mom, she may put up with it.
Nate, please stop trolling.
Nate,
You need more proof?
https://teachuphysics.files.wordpress.com/2015/01/heat-passing-through-rod.gif?w=780
Why dont you face up to reality?
“Nope, not at all. Another weird strawman.”
Filthy liar.
Filthy language.
Svante, your judgements are rather selective.
I wonder why that is….
Nate,
“Today the internal geothermal heat flux reaching the surface is measured to be 90 mW/m^2 on average. That is neglible compared to the GHE or the solar input.”
The geothermal heat flux that actually heats the surface is insignificant but the ‘back radiation’ that doesnt is?
Ludicrous…
Apparently so. See Roy’s next post to see the evidence that without a GHE, the Earth is much cooler..
‘The geothermal heat flux that actually heats the surface is insignificant’
But at least you seem to agree that the geothermal part is too small to produce the 33K warming?
Nate, once again you avoid reality.
The 33K nonsense arrises from comparison to an imaginary object.
Imaginary = not real = avoiding reality.
More nonsense, please.
JD’s cartoon arises from comparisons to an imaginary object JD calls a “blackbody”. So JD admits JD’s cartoons: Imaginary = not real = avoiding reality.
Good job JD. You always live up to all my expectations.
http://www.drroyspencer.com/2019/03/uah-global-temperature-update-for-february-2019-0-36-deg-c/#comment-345722
“JD’s separated version would have the middle (blue) plate increasing in temperature with no change in incoming or outgoing energy, which IS a clear violation of 1LOT law of physics, typical for JD as an entertainment specialist:
244K…290…244K”
Yes, good reminder clip DREMT, thank you for your support, showing JD’s imaginary 244K…290…244K solution which IS a clear violation of 1LOT law of physics.
Imaginary = not real = avoiding reality. More humor JD, please.
Now all you need to do is be honest enough to admit that:
Plates together: 244 K…244 K…244 K
Plates slightly separated: 244 K…290 K…244 K
is not JD’s position, but is in fact the one defended by Swanson, Norman, Svante, Nate, Tim Folkerts, bobdroege and the rest of the “Team”.
JD wrote:
“3 plates together, at equilibrium:
244K…244K…244K
3 plates slightly separated, at equilibrium:
244K…290K…244K
Exactly same energy in/out.”
So, no DREMT it is you that has it wrong, JD was perfectly clear about his position supports 244K…290K…244K, repeatedly. You will not find 244K…290K…244K written defended by Swanson, Norman, Svante, Nate, Tim Folkerts, bobdroege and the rest of your imaginary “Team”, only JD writes it out in JD’s capacity as an entertainment specialist.
Thank you for proving your dishonesty to anybody that has followed that particular debate. Another point proved.
All you have to do is find 244K…290K…244K written, defended by Swanson, Norman, Svante, Nate, Tim Folkerts, bobdroege or the rest of your imaginary “Team”. In fact it is only JD supports and writes it out in JD’s capacity as an entertainment specialist.
Please, stop trolling here DREMT. I’m sure the crowd over at Climate Sophistry will welcome you and JD with open arms.
fluffball has no idea what fluffball is talking about.
Nothing new.
Thanks for all the entertainment JD.
All JD has to do to know what I’m talking about is find 244K…290K…244K written, defended by Swanson, Norman, Svante, Nate, Tim Folkerts, bobdroege.
fluffball, the incompetence of you clowns has no limit.
Here’s Swanson claiming the green plates would be even cooler than 244 K!
“Both Green plates will be cooler than 244 K…
http://www.drroyspencer.com/2019/05/uah-global-temperature-update-for-april-2019-0-44-deg-c/#comment-353119
Not even close JD, E. Swanson does NOT write 244K…290K…244K in your link like you do as DREMT points out. Thanks for the entertainment though, do learn to read and do learn some physics.
fluffball, you only have 3 choices:
1) Accept the incorrect solution, offered by pseudoscience:
244 K…290 K…244 K
2) Accept the correct solution:
244 K…244 K…244 K
3) Or waffle aimlessly and incoherently, as you usually do.
I bet I can predict your choice….
I accept the solution that 1LOT determines. Try figuring it out for an excercise JD showing your work, it will help you learn some physics. With your self awarded expertise, you ought to be able to do it.
Well, I was right again, fluffball.
What flavor syrup do you want with your waffle, “aimless” or “incoherent”?
I knew you weren’t capable figure it out according 1LOT JD, I was only joking. Thanks for the entertainment.
Yes, it’s well known what a joke you are, fluff.
Nothing new.
Nate defending the 244 K…290 K…244 K:
http://www.drroyspencer.com/2019/03/uah-global-temperature-update-for-february-2019-0-36-deg-c/#comment-345647
Tim defending it:
http://www.drroyspencer.com/2019/03/uah-global-temperature-update-for-february-2019-0-36-deg-c/#comment-345140
bobdroege defending it:
http://www.drroyspencer.com/2019/03/uah-global-temperature-update-for-february-2019-0-36-deg-c/#comment-344836
Now Ball4’s dishonesty is proven even to people who hadn’t followed that particular discussion.
Good work DREMT, you actually found that Tim DID write “244 K 290 K 244 K” but none of your other links did so. Though as I wrote Tim did not actually defend solving the diffy q: “It would be an interesting differential equation to solve” so Tim was just guessing. Funny that you didn’t find anyone actually solving the problem including you.
Why don’t you and JD collaborate to solve your own problem statement possibly with Tim’s interesting “differential equation” using 1LOT consistent with 2LOT and show us your work?
Demonstrate you understand whether Tim made a good guess at the solution. My bet: not going to happen, you won’t be able to show you can properly solve your own problem statement. Better yet, back your solution up with a proper experiment!
Ball4, if somebody quotes the “244 K…290 K…244 K”, or even just mentions the blue plate rising in temperature to 290 K, and goes on to write words in defence of that proposition, then they are defending the 244 K…290 K…244 K. In all three links, the people involved are defending it, not just Tim.
That is NOT what you wrote DREMT: “Plates slightly separated: 244 K…290 K…244 K is not JD’s position, but is in fact the one defended by Swanson, Norman, Svante, Nate, Tim Folkerts, bobdroege and the rest of the “Team”.”
Only TIM “244K-290K-244K” & JD wrote “244K…290 K…244 K” in your clip and links but the others did not as you claimed. I take your comment to prove I will win my bet; you cannot solve your own problem statement & experimentally back up your solution to demonstrate you understand whether or not Tim made a good guess.
Ball4 is doing a great “dishonesty display” for his captive audience. The audience roars with laughter as they can see from the links provided that Ball4 was present during the discussion and so Ball4 is fully aware of what position the people I have referred to are defending, Ball4 continues to entertain as Ball4 is an entertainment specialist.
bobdroege, again:
http://www.drroyspencer.com/2019/03/uah-global-temperature-update-for-february-2019-0-36-deg-c/#comment-346009
“I want to know why you think it’s an obvious error to have the 244 290 244 solution.”
Another example for Nate:
http://www.drroyspencer.com/2019/03/uah-global-temperature-update-for-february-2019-0-36-deg-c/#comment-346185
In response to me saying: “Anybody prepared to acknowledge that the 244K…290K…244K is wrong, like Ball4 has?”
Nate responds with: “Nah”.
More please, Ball4, thank you for your support, you always live up to all my expectations. The audience roars with approval.
JD/DREMT might want to reread carefully all of those posts to be reminded of how their ‘debunking’ was, well, debunked.
The lack of substantive answers to these rebuttals, other than generic insults, and JD crying ‘2LOT violation’ but never able to explain why, are good indicators.
Nate falsely accuses: “JD crying ‘2LOT violation’ but never able to explain why…”
Nate, if I provide examples of where we have explained the violations, will you agree to not comment here for one year per each example?
Good point Nate. No substantive science, just words from DREMT.
So, thanks for the added entertainment DREMT, all you have to do to know what I’m talking about is find 244K…290K…244K written, defended by Swanson, Norman, Svante, Nate, Tim Folkerts, bobdroege.
DREMT has failed to do so despite great effort which is humorously obvious to the crowd. DREMT can’t even solve his own problem statement, show Tim’s guess is good or not, or back any DREMT solution with experiment.
fluffball, can you make a decision, or just eat waffles?
1) Accept the incorrect solution, offered by pseudoscience:
244 K…290 K…244 K
2) Accept the correct solution:
244 K…244 K…244 K
3) Or waffle aimlessly and incoherently, as you usually do.
Ball4 continues his “dishonesty display”, to the delight of the crowds.
Actually the substantive 1LOT solution 4) have DREMT/JD show their work for the correct solution to DREMT’s problem statement, back it with experiment, determine if Tim’s guess is good or not.
I’m still waiting along with the crowd laughing at the lack of substantive science from DREMT/JD. Nothing new.
The crowds chant Ball4’s name as he is raised on the shoulders of his “Team” mates, as Ball4 receives “Liar of the match” for Ball4’s relentless attempt to rewrite the history of a discussion that anybody can read for themselves. Ball4 now wants us to “do the math” for the “244 K…290 K…244 K” solution, which is the one Ball4’s other “Team” mates defend, but which Ball4 disputes, as do we, yet Ball4 is trying to pretend that this is the solution we think is right! Ball4’s mission is to generate as much confusion as possible, to cover up for Ball4’s blunder where Ball4 inadvertently agreed with us that the “244 K…290 K…244 K” violates laws of physics, and knowing that I will continually bring this up every time Ball4 attacks JD on the “plates” issue, Ball4 must do his best to muddy the waters, to the widespread approval of his “Team” mates and the cheers of the crowd! “Ball4! Ball4! Ball4! Ball4!”
“Nate falsely accuses: ‘JD crying ‘2LOT violation’ but never able to explain why”
Not False at all.
Here’s a thread where you claim 2LOT violations. Despite being asked to explain several times where it is, you can’t, and don’t.
http://www.drroyspencer.com/2019/03/uah-global-temperature-update-for-february-2019-0-36-deg-c/#comment-344836
“If clowns cant see the blatant violation of 2LoT, then they are clearly admitting they are clueless about the relevant physics.”
3 plates together, at equilibrium:
244K244K244K
3 plates slightly separated, at equilibrium:
244K290K244K
Exactly same energy in/out.”
Nothing to do with 2LOT at all!
Hilarious!
On the contrary, your ‘244K244K244K’ with lots of heat flow between the plates is an obvious violation of the 2LOT!
Still waiting along with crowd for your proper solution and experimental backing.
“Ball4! Ball4! Ball4! Ball4! No de-cen-CY! No de-cen-CY! No de-cen-CY! No de-cen-CY! No hon-es-TY! No hon-es-TY! No hon-es-TY! No hon-es-TY! No integ-ri-TY! No integ-ri-TY! No integ-ri-TY! No integ-ri-TY! No em-pa-THY! No em-pa-THY! No em-pa-THY! No em-pa-THY! Just like DREMT SAID! Just like DREMT SAID! Just like DREMT SAID! Just like DREMT SAID! Ball4! Ball4! Ball4! Ball4!…”
The crowds are really getting into the chants now, as Ball4 proudly continues his “dishonesty display” for his legions of devoted fans…
I see there will be no substantial defense from DREMT/JD as expected so I’ll get back to them when they present a proper solution workout backed by experiment. The crowd applauds.
As if I’m the one with something to defend!
The crowd cheers on Ball4’s refusal to admit to his deliberate deception even though they can all read it and see for themselves…
DREMT trolling and trolling and just for good measure, trolling some more!
Nate, please stop trolling.
hypocrisy
/həˈpkrəsē/
Learn to pronounce
noun
the practice of claiming to have moral standards or beliefs to which one’s own behavior does not conform; pretense.
“his target was the hypocrisy of suburban life”
synonyms: sanctimoniousness, sanctimony, pietism, piousness, affected piety, affected superiority, false virtue, cant, humbug, pretense, posturing, speciousness, empty talk; More
Nate, you can’t seriously be defending Ball4’s obvious dishonesty!
Your only purpose here is humiliation, nothing else.
Ball4 wouldn’t need to humiliate himself if he would just acknowledge the reality that he is at odds with the rest of you. I mean…how difficult is it to admit that!?
Ok, DREMT.
OK, Nate.
Svante says:
June 9, 2019 at 3:12 PM
wayne rowley says:
You divide the figure by 4, because a sphere has 4 times the surface area of one side of a disc.
Eureka moment, but subtract the albedo first (30%).)))))))))
Svante, why do you thing you need to divide by 4 ???
You seem to be doing it here as well, I mean dividing by 4 ??? As the cross sectional area of a sphere is 1/4 of its total area, this is the area you are taking for the solar constant. Then as the Sun is heating each one of these cross sectional areas, you must x by 4, not divided by 4 ???
Let’s go for one day 48 hour. The Suns heat goes all around the Earth. So let’s just say in one second, one quarter is heated up 1, in 6 hours this Sun heats this quarter up 21,600, this Sun goes onto the next quarter, and heats it up 21,600, and for the next 2 quarters, totaling 86,400. The next time around there is going to be heat left in the seas and earth, thus the total will be more.
“HOWEVER” do you get that you need to multiply not divide ??? Your way of thinking would be a static Earth that does not rotate, and just one quarter of the Earth is heated up, thus to get the averge temp. on Earth, we have to divide, by 4, but with a rotating Earth, we have to multiply by 4.
Wayne
“So lets just say in one second, one quarter is heated up 1, in 6 hours this Sun heats this quarter up 21,600, this Sun goes onto the next quarter, and heats it up 21,600, and for the next 2 quarters, totaling 86,400.”
A watt is the measure of energy per second so we don’t multiply by the time the Earth is exposed. We average the watts received over the exposure time.
Let me get something out of the first. 340 W/m^2 is the average flux at TOA. 240 W/m^2 is the average flux at the surface from the 30% albedo reduction. We are focused on what happens at TOA right now. We can talk about the average surface flux if you want, but that needs to wait until TOA is fully understood.
1360 W/m^2 is what is hitting the cross sectional area. This is an abstract concept because the Earth isn’t actually flat. If you want to know the REAL flux you have to project this VIRTUAL cross sectional flux onto a sphere. You can do this two ways. You can either integrate the solar constant with respect the cosine of the zenith angle along the surface of the lit hemisphere and then repeat the procedure for the unlit side except instead of using the solar constant use zero…because ya know…the unlit side is dark. Or you can use already established canonical geometrical principals and just divide the VIRTUAL cross sectional flux by 4 to get the REAL flux. Note that this geometrical principal is itself derived using calculus and integrals.
Also, you’ve been talking about a quarter of the Earth being heated by the Sun in several posts. This is not correct. The Sun is continuously warming 1/2 of the Earth. The big epiphany here should be that the 1/2 that is lit is not lit evenly. The zenith receives the maximum flux while the dawn and dusk edges receive 0 W/m^2.
backdoor guy claims: “Or you can use already established canonical geometrical principals and just divide the VIRTUAL cross sectional flux by 4 to get the REAL flux.”
FALSE, bdgwx. As explained several times, you get the math right, but your physics is wrong. You cannot divide a power flux.
I think the backdoor is awaiting you….
I think here is where you are going wrong. But first, yesSun is continuously warming 1/2 of the Earth. I was talking about something else, no need to get into that.
You stated; “repeat the procedure for the unlit side except instead of using the solar constant use zero…because ya know…the unlit side is dark.”
This is where you are going wrong. Your thinking that the Earth is not rotating, and there are not 24 hours in a day. You are taking the average for the lite side, and the unlite side ??? Your “NOT” working on a rotating 24 hours Earth are you ??? Yes or no, well you cant be.
Here is what you should do. Integrate the solar constant with respect the cosine of the zenith angle along the surface of the lit hemisphere, then work out how heat is on the unlite side, so far we have worked out 12 hours. Now do the exact same thing, as we will say the Sun has lite the other side, and the now unlite side is cooler, but still warm as the Sun has been shining there. What you have done is to work out the solar constant for a Earth that does not rotate, and only has 12 hours. Then thou you maybe still working out the unlite side to a temp. that has no heat on it. As if the Sun never shines on one side of the Earth it would be very cold, but it des shine on that side and warmers it up quite a lot.
In all of our chats, you “still” have “not” explained why you divided by 4, do you know why you divide by 4 ??? Virtual cross sectional flux by 4. That’s why I stated this on I thought, you thought the Sun was just heating up quarter of the Earth, that’s why I thought you divided by 4, but now I don’t understand why ??? BUT, I think this is why you do this ??? The total amount of energy received per second at the top of Earth’s atmosphere (TOA) is measured in watts and is given by the solar constant times the cross-sectional area of the Earth. Because the surface area of a sphere is four times the cross-sectional surface area of a sphere (i.e. the area of a circle), the average TOA flux is one quarter of the solar constant and so is approximately 340 W/m².
As the solar constant = 340 W/m² for one quarter, “WHY” do you divide by 4 ??? Are you working on say 1 hour and a no rotating Earth ??? Dose not the 340 W/m² go from one quarter to the next until it heats up and then the unlite quarter cools down very slightly, then the 340 W/m² moves to the next area and so on until in a full orbit on 24 hours the 340 W/m² has gone on to each and every quarter ??? If the Earth was static, and the Sun shines just on one quarter and it was 340 W/m², yes you would divide by 4, but you have to x 340 W/m² by 4, not divide ??? Do you get what I am saying, and do you agree you are working things out for a static noone rotating Earth ???
Wayne
wayne said…”This is where you are going wrong. Your thinking that the Earth is not rotating, and there are not 24 hours in a day. You are taking the average for the lite side, and the unlite side ??? Your “NOT” working on a rotating 24 hours Earth are you ??? Yes or no, well you cant be.”
The math doesn’t care if the Earth is rotating or not. The solar constant is 1360 W/m^2 regardless. Remember, the solar constant is function of only the average distance between the Earth and Sun.
wayne said…”Here is what you should do. Integrate the solar constant with respect the cosine of the zenith angle along the surface of the lit hemisphere, then work out how heat is on the unlite side”
There is no solar flux on the unlit side. None whatsoever. It’s value is therefore 0 W/m^2 everywhere along the unlit said. It’s only the lit side that is receiving a solar flux. The fact that the Earth rotates such that points cycle between lit and unlit in way no changes the fact that exactly 1/2 is lit and 1/2 unlit…ALWAYS.
wayne said…”so far we have worked out 12 hours. Now do the exact same thing, as we will say the Sun has lite the other side, and the now unlite side is cooler, but still warm as the Sun has been shining there. What you have done is to work out the solar constant for a Earth that does not rotate, and only has 12 hours. Then thou you maybe still working out the unlite side to a temp. that has no heat on it. As if the Sun never shines on one side of the Earth it would be very cold, but it des shine on that side and warmers it up quite a lot.”
Remember, the solar constant has nothing to do with rotation. It’s going to be 1360 W/m^2 regardless of the spin rate. The fact that the Earth rotates such that different faces move into and out of the field of solar flux is completely irrelevant to what the zenith flux is and what the average real flux is over Earth’s entire surface area. In other words, the real flux is going to be 340 W/m^2 no matter what.
wayne said…”In all of our chats, you “still” have “not” explained why you divided by 4, do you know why you divide by 4 ??? Virtual cross sectional flux by 4. That’s why I stated this on I thought, you thought the Sun was just heating up quarter of the Earth, that’s why I thought you divided by 4, but now I don’t understand why ???”
You divide by 4 because the Earth is NOT flat. You divide by 4 because the zenith flux is NOT received everywhere. You divide by 4 because that’s easier than integrating the real flux.
I do NOT think the Sun is heating 1/4 of the Earth. It is heating 1/2 at any given time. But, flux is NOT distributed evenly within the 1/2 that is lit.
wayne said…”As the solar constant = 340 W/m² for one quarter, “WHY” do you divide by 4 ???”
340 W/m^2 is NOT what is hitting 1/4 of the Earth. It is the average hitting the entire Earth. However, this average flux value is not homogeneous over every square meter. Equatorial regions get more than 340 while polar regions get less. But, it averages out to be 340 W/m^2.
wayne said…”Dose not the 340 W/m² go from one quarter to the next until it heats up and then the unlite quarter cools down very slightly, then the 340 W/m² moves to the next area and so on until in a full orbit on 24 hours the 340 W/m² has gone on to each and every quarter ???”
Actually the average flux hitting only the lit said is 680 W/m^2 while simultaneously the unlit side is getting 0 W/m^2. The combined average flux over the lit side + unlit side is thus (680 + 0) / 2 = 340 W/m^2. It’s the 680 W/m^2 that is “moving” to different faces of the Earth. Or more precisely the Earth is rotates faces into the 680 W/m^2 field of view.
wayne said…”If the Earth was static, and the Sun shines just on one quarter and it was 340 W/m², yes you would divide by 4, but you have to x 340 W/m² by 4, not divide ???”
That’s a difficult question to answer because I need to know exactly which 1/4 of Earth you are thinking of. However, either way at no time do you ever divide 340 W/m^2 by 4. Nor would you ever multiple by 340 W/m^2 by 4 unless the intent was to convert the average real flux into the average zenith flux.
wayne said…”Do you get what I am saying, and do you agree you are working things out for a static noone rotating Earth ???”
The reasoning behind dividing the average zenith flux of 1360 W/m^2 by 4 to get the average real flux of 340 W/m^2 is in no way dependent upon the rotation of Earth.
backdoor guy says “In other words, the real flux is going to be 340 W/m^2 no matter what.”
big, you keep trying the same thing over and over. 340 W/m^2 is NOT a “real” flux. It is a calculated value by people that do not understand physics.
(Exit via the backdoor, please.)
Are saying that you’ve figured out something out that has somehow eluded thousands of experts for decades?
If so this is your opportunity to tell the world.
How much energy does Earth receive in one orbital cycle?
Show your work.
What I’m saying is that you are WRONG about the 340 W/m^2 being a “real” flux.
You are incorrect, and you won’t stop spouting pseudoscience.
What should be done is focus the 340 W/m^2 on JD’s forehead which ought to quickly teach JD some reality.
If I’m wrong then you should be able to compute the total amount of energy Earth receives in one sidereal year starting from the solar constant. Give it a shot. Show your work.
Ball4, bdgwx, please stop trolling.
bdg, your arithmetic is correct, but your physics is incorrect.
Nothing new.
Wayne wrote; The irradiance of the sun on the outer atmosphere when the sun and earth are spaced at 1 AU – the mean earth/sun distance of 149,597,890 km – is called the solar constant. Currently accepted values are about 1360 W m-2.
Big wrote; The math doesnt care if the Earth is rotating or not. The solar constant is 1360 W/m^2 regardless. Remember, the solar constant is function of only the average distance between the Earth and Sun.
Wayne wrote; The distance in our debate does not matter. The total amount of energy received per second at the top of Earth’s atmosphere (TOA) is measured in watts and is given by the solar constant times the cross-sectional area of the Earth. Because the surface area of a sphere is four times the cross-sectional surface area of a sphere (i.e. the area of a circle), the average TOA flux is one quarter of the solar constant and so is approximately 340 W/m
OF COURSE it matters if the Earth is rotating or not. LETS GET THIS RIGHT BEFORE WE MOVE ON. For one second, the energy received from the Sun on one quarter of the Earth is 340 W/m for the whole Earth for 1 second its 1360 W/m^2, do you agree with this ??? If so, this is what you have done, and wrong in my opinion.
You have taken the energy from the Sun falling on one quarter of the Earth to be 340 W/m, HOWEVER thats for one second on one quarter of the Earth. NOW, as I said the Earth is NOT static, its rotating, so you can NOT divide by 4.
LET ME EXPLAIN IN LAYMAN’S TERMS.
1,
None rotating unreal senario,
We have a chicken on a spit that’s not rotating, the heat/energy hitting one quarter of this chicken for 4 second is 1360 W/m^2, the quarter gets very cooked, the other three quarters do not get any heat from the cooker thus not cooked at all, thus the averge heat/energy on all the chicken = 340 W/m. And lets just say the total heat on the 4 side of this chicken = 1360. Average heat of this chicken = 340.
2,
Rotating real scenario,
We have a chicken on a spit that’s rotating, the heat/energy hitting one quarter of this chicken for one second is 340 W/m^2, it rotates once every 4 seconds, all the quarter get hit with 340 W/m^2 per second, and all 4 quarters get cooked. Thus the average heat/energy on all the chicken = 340 W/m {BUT THAT’S THE WRONG AVERAGE} And lets just say the total heat on the 4 side of this chicken = 1360c. Average heat of this chicken = 340c.
NOTE,
The huge difference in the rotating chicken, all for sides get cooked. Note how in this case averge means “nothing”, and should not even be considered, or should it be. On your Earth as you say it does “not” matter if its rotating or not, one quarter would be 1360c, and the other three quarters 0c. on my Earth where its rotating, each quarter would be 340c. {Or actually 50% hotter see below.
NOTE,
On the rotating chicken, each 4 quarters that were heated, that moved away from the heat of the cooker, as it was rotating, will stay quite hot, how hot ??? {big your should be able to work this out} lets say half as hot. So we have the total heat of the rotating chicken at 1360c, add half of this as the retaining heat as of rotation = 680c + 1360c = 2040c, Average heat of this chicken = 510c.
None rotating Earth = 340c averge heat all over.
Rotating Earth = 510c average heat all over. 50% hotter than the none rotating Earth. And the more the Earth rotates, the more the seas and earth will heat up after the Sun moves to the next area of the Earth. This is why a Earth without an atmosphere, would NOT be -18c, it would be SO much hotter than it is today, the GHG’s shade us from the full force of the energy/heat of the Sun.
A LOT hotter than what you calculate a flat none rotating Earth, as the rotating Earth will “always” got far hotter than the none rotating. That’s why I said you can not divide by 4, it does not add up right in this real World issue.
Wayne
As someone said; use the divide by four calculation who are effectively moving Earth from its current orbit to an orbit twice the distance from the Sun.
You either have to accept that solar flux exists at one fourth power on the dark side of Earth where zero solar power actually exists, or you have to accept that the intercept disc for the half sphere remains flat AND its power divided by four, which places it twice the distance from the sun. There is no other rational way to interpret this combination of numbers and pictures Roy or the others provide.
In my reality, solar power falls on a hemisphere, in real time, at the given power for the hemisphere. Solar power does NOT fall on the entire sphere, all the time, at one fourth power.
So as showed, if the Sun heated half of the Earth up to an average of 20c, when the Sun moved to the other half, this first half would cool, but not cool to zero, say to 10c, then slowly as the days goes by, the to areas that were not in sunlight, would cool less, and they then would stay at an averge temp. of whatever. You cant say one half have 20c and the other half 0c and call the averge temp. 10c ??? I mean that’s like saying I have £1,000,000, and the other guy has £0, but our averge = £5000,000 Thats not how real World works.
Wayne
I mean thats like saying I have 1,000,000, and the other guy has 0, but our average = 500,000 That’s not how real World works.
Wayne
“use the divide by four calculation who are effectively moving Earth from its current orbit to an orbit twice the distance from the Sun.”
No, the divide by 4 calculation is from simple geometry and is used to compute the correct measured global mean surface temperature. An orbit twice the distance would result in a much cooler mean surface temperature calculation than has been measured in the correct current orbit.
fluffball believes: “the divide by 4 calculation is from simple geometry and is used to compute the correct measured global mean surface temperature.”
WRONG fluffball.
You can divide by 4 and get the arithmetic right. But, in this case, your physics is wrong. You can not divide solar flux. And the resulting computation is NOT the “correct measured global mean surface temperature”. The resulting computation is for an imaginary black body.
You don’t even understand your own pseudoscience.
“I mean thats like saying I have 1,000,000, and the other guy has 0, but our average = 500,000 Thats not how real World works.”
It is if you want to know the amount of money the average person has. That’s how they find the mean income for a country.
The only thing you learn when you take the total power striking the Earth and divide by the surface area is the average watts per square meter the Earth receives. You don’t do that to find out the average watts per square meter a country gets.
“The resulting computation is for an imaginary black body.”
That’s wrong JD, the resulting 1LOT computation from measured data of the global mean temperature is for a real body called Earth with a real L&O surface and a real atm. That computation when done with measured input results in the same global mean temperature as measured (~288K). Do learn some science, look it up! You could learn how to do a simple analogue correctly instead of your bogus elevator analogue.
wayne said…”As someone said; use the divide by four calculation who are effectively moving Earth from its current orbit to an orbit twice the distance from the Sun.”
NO. Actually the exact opposite occurs. If you don’t divide the solar constant by 4 you are effectively treating the Earth as if it were 2x closer than it really is. How Postma botches this simple geometric fact is beyond me.
wayne said…”In my reality, solar power falls on a hemisphere, in real time, at the given power for the hemisphere. Solar power does NOT fall on the entire sphere, all the time, at one fourth power.”
Correct. And the average flux at which it falls on the lit side, and only the lit side, is 680 W/m^2. And, of course, the average flux on the unlit side 0 W/m^2.
wayne said…”You cant say one half have 20c and the other half 0c and call the averge temp. 10c ???”
Well, yes, actually you can. And it’s a valid thing to do in this case because each half of the Earth has the same surface area (mostly anyway). This is essentially what we call the global mean temperature. In modern reanalysis systems the Earth is actually divided up into a million or more equally sized cells each with their own cellular mean temperature that are then combined to form a global mean temperature at a specific moment in time. And once you have that completed you can keep doing this day in and day out and then combine that to form global mean temperatures over a specific period time like a month or a year.
Craig T, Ball4, bdgwx, please stop trolling.
wayne said…”OF COURSE it matters if the Earth is rotating or not. LETS GET THIS RIGHT BEFORE WE MOVE ON. For one second, the energy received from the Sun on one quarter of the Earth is 340 W/m for the whole Earth for 1 second its 1360 W/m^2, do you agree with this ??? If so, this is what you have done, and wrong in my opinion.”
bdgwx says…NO! I absolutely do not agree with this. 1360 W/m^2 is what is received on the cross sectional area. 340 W/m^2 is what is received on the whole Earth. Note that the cross sectional area is 127,500,000 km^2 and the whole Earth is 510,000,000 km^2. The cross sectional area is an abstract concept because the Earth is not flat.
wayne said…”You have taken the energy from the Sun falling on one quarter of the Earth to be 340 W/m, HOWEVER thats for one second on one quarter of the Earth. NOW, as I said the Earth is NOT static, its rotating, so you can NOT divide by 4.”
bdgwx says…NO! If by “quarter of the Earth” you mean the cross sectional area then this figure is 1360 W/m^2. This is the zenith flux. This has absolutely nothing to do with the rotation of Earth. Likewise, diving the zenith flux by 4 to get the effective flux over the entire sphere is compulsory regardless of whether the Earth is rotating or not.
wayne said…”1,
None rotating unreal senario,
We have a chicken on a spit thats not rotating, the heat/energy hitting one quarter of this chicken for 4 second is 1360 W/m^2, the quarter gets very cooked, the other three quarters do not get any heat from the cooker thus not cooked at all, thus the averge heat/energy on all the chicken = 340 W/m. And lets just say the total heat on the 4 side of this chicken = 1360. Average heat of this chicken = 340.”
bdgwx says…Yep. Agreed.
wayne said…”2,
Rotating real scenario,
We have a chicken on a spit thats rotating, the heat/energy hitting one quarter of this chicken for one second is 340 W/m^2, it rotates once every 4 seconds, all the quarter get hit with 340 W/m^2 per second, and all 4 quarters get cooked. Thus the average heat/energy on all the chicken = 340 W/m {BUT THATS THE WRONG AVERAGE} And lets just say the total heat on the 4 side of this chicken = 1360c. Average heat of this chicken = 340c.”
bdgwx says…Nope.
Assume the chicken is 1 m^2. Each side is 0.25 m^2.
340 W applied to one quarter (0.25 m^2) is 1360 W/m^2.
The chicken is always receiving 340 W.
The total amount of energy received in one revolution is (340+340+340+340) = 1360 joules in 4s.
The average flux over the entire chicken is 340 W / 1 m^2 = 340 W/m^2.
The average flux over the “lit” area is 340 W / 0.25 m^2 = 1360 W/m^2. We can call this the “heat constant” and is analogous to the solar constant for Earth.
Because the average flux over the entire chicken is 340 W/m^2 and because the chicken is 1 m^2 and because this occurred in 4s the total amount of energy is 340 * 1 * 4 = 1360 joules.
If you want to start with our “heat constant” of 1360 W/m^2 you need to first understand that this in reference to 0.25 m^2. So the energy received in 1s on a single side is 1360 W/m^2 * 0.25 m^2 = 340 W. And since this happened for 4s the total energy is 340 W * 4s = 1360 joules.
And since we know the total energy of 1360 joules occurred over 4s and 1 m^2 the effective flux is 1360 / 4 / 1 = 340 W/m^2.
bdg…”For one second, the energy received from the Sun on one quarter of the Earth is 340 W/m for the whole Earth for 1 second its 1360 W/m^2, do you agree with this ???”
I don’t.
You cannot take an intensity on the solar sphere at the radius of the Earth and represent it as a straight-line vector like 340 W/m^2. It is not possible to calculate that value given the Earth’s dynamics.
Just to clarify…that is not my quote.
“You cannot take an intensity on the solar sphere at the radius of the Earth and represent it as a straight-line vector like 340 W/m^2.”
Of course not, because 340 W/m^2 is not intensity. Gordon simply bungles one of the seven SI base units.
Ball4, please stop trolling.
Hi there,
I thought you may say something like the below.
The non rotating chicken will receive the same overall heat as the rotating chicken, BUT, the whole of the rotating chicken will be receiving the heat, also the rotating chickens area that’s not receiving heat as the rotation has moved passed the heat source, as like on Earth this area will stay warmer, and slowly warm up agreed ??? The chicken that’s not rotating, might get more heat in the one area, but as the chicken {Earth} is so thick, and heat source not strong enough, it never can penetrate the more than say 1/1000ths.
In the case of the rotating chicken, the heat left on the chicken when the heat moves around as rotation, will build up more, thus the overall temp. after the same amount of time will be higher, yes ??? As the heat will be retained more, and slowly built up.
I know this to be true, see my Web-Site, as when I warm steel up, I dont just hold the heat to one area, I slowly spread the heat all over the whole steell. As this way the whole steel heats up faster and the whole glows red, the otherways and it takes over twice as long. There are other examples of using say the same force, that you can get things done faster, same as the heating things I just talked about. Say you have to cut a piece of steel 100mm x 10mm, if you put the 10mm up, facing the saw, it will cut the steel far faster than laying it flat and the 100mm facing the saw. same force used, same aea of steel, but one was cuts the steel so much faster. same as the rotating Earth, it “WILL” heat up so much more, and faster agreed ??? If not pop into my workshop and I will prove it to you.
Wayne
More later.
Wayne
wayne said…”1,
None rotating unreal senario,
We have a chicken on a spit thats not rotating, the heat/energy hitting one quarter of this chicken for 4 second is 1360 W/m^2, the quarter gets very cooked, the other three quarters do not get any heat from the cooker thus not cooked at all, thus the averge heat/energy on all the chicken = 340 W/m. And lets just say the total heat on the 4 side of this chicken = 1360. Average heat of this chicken = 340.”
bdgwx says…Yep. Agreed.
wayne said…”2,
Rotating real scenario,
We have a chicken on a spit thats rotating, the heat/energy hitting one quarter of this chicken for one second is 340 W/m^2, it rotates once every 4 seconds, all the quarter get hit with 340 W/m^2 per second, and all 4 quarters get cooked. Thus the average heat/energy on all the chicken = 340 W/m {BUT THATS THE WRONG AVERAGE} And lets just say the total heat on the 4 side of this chicken = 1360c. Average heat of this chicken = 340c.”
bdgwx says…Nope.
Assume the chicken is 1 m^2. Each side is 0.25 m^2.
340 W applied to one quarter (0.25 m^2) is 1360 W/m^2.
The chicken is always receiving 340 W.
The total amount of energy received in one revolution is (340+340+340+340) = 1360 joules in 4s.
The average flux over the entire chicken is 340 W / 1 m^2 = 340 W/m^2.
The average flux over the “lit” area is 340 W / 0.25 m^2 = 1360 W/m^2. We can call this the “heat constant” and is analogous to the solar constant for Earth.
Because the average flux over the entire chicken is 340 W/m^2 and because the chicken is 1 m^2 and because this occurred in 4s the total amount of energy is 340 * 1 * 4 = 1360 joules.
If you want to start with our “heat constant” of 1360 W/m^2 you need to first understand that this in reference to 0.25 m^2. So the energy received in 1s on a single side is 1360 W/m^2 * 0.25 m^2 = 340 W. And since this happened for 4s the total energy is 340 W * 4s = 1360 joules.
And since we know the total energy of 1360 joules occurred over 4s and 1 m^2 the effective flux is 1360 / 4 / 1 = 340 W/m^2.
Wayne
Wayne wrote; For one second, the energy received from the Sun on one quarter of the Earth is 340 W/m for the whole Earth for 1 second its 1360 W/m^2, do you agree with this ??? If so, this is what you have done, and wrong in my opinion.”
bdgwx says…NO! I absolutely do not agree with this. 1360 W/m^2 is what is received on the cross sectional area. 340 W/m^2 is what is received on the whole Earth. Note that the cross sectional area is 127,500,000 km^2 and the whole Earth is 510,000,000 km^2. The cross sectional area is an abstract concept because the Earth is not flat.
Wayne wrote; The energy received from the Sun on one quarter = 1360 W/m^2, averge energy received by the whole earth with 4 quarters is 340 W/m^2. Your saying the whole Earth gets 340 W/m, then you contradict yourself, and say one part of the Earth, call this one quarter, get more energy on one quarter of the Earth, than the whole earth ???
THIS is odd, lets see if we agree here.
1360 W/m^2 is what is received on the cross sectional area. Average this out on the whole Earth, and say the cross sectional area is one quarter. Thus each quarter = 340 W/m^2. how can one area of the Earth have 1360 W/m^2 energy, but the whole same Earth have 340 W/m^2 energy ???
Wayne
wayne said…”BUT, the whole of the rotating chicken will be receiving the heat, also the rotating chickens area thats not receiving heat as the rotation has moved passed the heat source, as like on Earth this area will stay warmer, and slowly warm up agreed ???”
bdgwx says…I absolutely agree with this. And it’s a very important topic worthy of a discussion. The implications of this impact how the heat is distributed through the climate system. But it has zero impact on how much energy Earth actually receives.
wayne said…”In the case of the rotating chicken, the heat left on the chicken when the heat moves around as rotation, will build up more, thus the overall temp. after the same amount of time will be higher, yes ??? As the heat will be retained more, and slowly built up.”
bdgwx says…Again, I’m okay with that. In fact, that sounds like a very realistic hypothesis. But, once again, that has nothing to do with the fact that the “heat constant” in this thought experiment is 1360 W/m^2 while the effective flux over the entire chicken is 340 W/m^2.
wayne said…”The energy received from the Sun on one quarter = 1360 W/m^2, averge energy received by the whole earth with 4 quarters is 340 W/m^2.”
bdgwx says…Let’s clear up some terminology here, because I think it is causing confusion. The 1360 W/m^2 is in reference to the more abstract idea of cross sectional area. Saying “on one quarter” is ambiguous because I have no idea if you’re talking about the more abstract cross sectional area idea or if you’re talking about a special subarea on the surface that happens to 1/4 the total. From context clues I eventually figured out that you’re talking about the more abstract cross sectional area. But, just know that there aren’t 4 cross sectional areas so saying things like “the whole earth with 4 quarters” is a bit weird.
wayne said…”Your saying the whole Earth gets 340 W/m, then you contradict yourself, and say one part of the Earth, call this one quarter, get more energy on one quarter of the Earth, than the whole earth ???”
bdgwx says…And I think this is where the breakdown occurs. Yes, on average the Earth receives 340 W/m^2 relative to the full area of 510e12 m. I’m not claiming that one part of Earth gets more energy than the whole thing. I’m saying that the solar constant of 1360 W/m^2 doesn’t mean what you think it does. It is NOT the flux that is occurring on the surface of the sphere. Therefore in this context it is a non-physical or abstract quantity that is relative only to the more abstract concept of the cross sectional area of Earth which is 127e12 m. Note that even though 127e12 is mathematically 1/4 of the real value of 510e12 it has no real meaning in the context of the surface area of Earth. This what I mean when use words like virtual and abstract to describe this quantity.
wayne said…”1360 W/m^2 is what is received on the cross sectional area. Average this out on the whole Earth, and say the cross sectional area is one quarter.”
bdgwx says…I think so yes. But, let’s be careful with terminology. 1360 W/m^2 isn’t what is received *ON* the cross sectional area. It is what is received in *REFERENCE* to the cross sectional area. When you say “on” there is an implicit assumption that it is real thing that is happening. It is not. Remember, the cross sectional area of Earth is abstract in the sense that it is not accessible because a curved hemispherical dome of mass is what the actual flux is landing on. The flux doesn’t actually land *ON* the cross sectional area.
wayne said…”Thus each quarter = 340 W/m^2.”
bdgwx says…Again, that’s be really careful here. If by “one quarter” you really mean the more abstract cross sectional area then understand that there isn’t 4 cross sectional areas so it doesn’t even make sense to say “each quarter”. But, 340 W/m^2 is the average flux over the entire Earth. This is a “real” value the sense that the surface of Earth is real and that is what the actual flux is really landing on.
wayne said…”how can one area of the Earth have 1360 W/m^2 energy, but the whole same Earth have 340 W/m^2 energy ???”
bdgwx says…It can’t. These are fundamentally different metrics. It’s like comparing apples and oranges. They mean different things. The 1360 W/m^2 figure is more abstract in this sense that it doesn’t really happen (except in a special case that I can talk about later). The 340 W/m^2 is real in the sense that it represents the actual flux that really occurs averaged over the entire surface area through the course of one sidereal year.
Hi there big, and thank you for taking the time to explain. We did agree on a few things :–)))
Here is where I was going wrong, I thought the 1360 W/m^2 was the heat/energy coming from the Sun in one second, for a cross sectional area of the Earth, and the cross sectional area was a quarter of the Earth’s surface. As Roy was dividing by four. As the surface of a sphere has an area four times as great as the area of a disk of the same radius. So the 1,368 W/m2 is reduced to an average of 342 W/m2 over the entire surface of our spherical planet. I thought he was dividing the 1360 W/m^2 by 4 as of 4 quarters of a static Earth, and he was not taking into account, the Earth rotates, and as we agreed, the Earth will stay warm long after the Sun moves around the Earth, and warm up more each time.
Ok, we on the same page now ???
Ok, if you got time, could you explain why Roy was devideding by 4 ???
I don’t go for the Earth being -18c cos of GHG’s, I say these so called GHG’s actually shade us, and col us slightly, it makes so much more sense. As your saying if your one of the -18c guys, that what I said before, that I am not actually feeling 20c in thUK, or 40c from Dubai, that this heat I feel is not from the Sun, but from the GHG’s ??? If your right about the -18c and GHG’s, then the curvature of the Earth, and the point in the Dubai receives more direct sunlight will not hold, as your saying the Sun only gives -18, so the area on Earth where you feel the heat more, should not work in your -18c theory, but it does work, and ho so well ???
The moon, which plainly does not have a greenhouse effect {according to your logic}, you say yes the lite side gets very close to 123c, but the dark side is so cold the average is -18C.
The moon has a very low surface heat capacity, compared to earth. So, it heats up quicker, and higher than the earth, it also cools quicker and lower than the earth. The obvious explanation for this is the differences in the length of day, AND surface heat capacity between the earth and the moon. No GHG’s effect required, no P / 4 required.
Wayne
wayne said…”Ok, we on the same page now ???”
YES! And you’re absolutely correct that the rotation of the Earth effects the distribution of heat. This has huge implications on the minimum and maximum temperatures, weather patterns, and climate in general. It matters a lot for a lot of other reasons.
I’ll try to comment on the other stuff later.
wayne…”Here is where I was going wrong, I thought the 1360 W/m^2 was the heat/energy coming from the Sun in one second…”
I worked this value out in a previous post several months ago. The 1360 W/m^2 is the solar intensity on a sphere around the Sun with a radius at the distance of the the Sun to the Earth (1 AU or about 93 million miles).
If you imagine that sphere around the Sun with radius 1 AU, the 1360 W represents the intensity per square metre on that sphere. It has nothing to do with the Earth’s shape, it is strictly a property of solar radiation.
The argument in this article is whether the value can be spread over a sphere with a radius from Earth to the top of Earth’s atmosphere by simply dividing by 4. Roy claims it can ‘over time’, but therein lies the problem. Joe claims you cannot spread it over an entire sphere.
I think both arguments are missing the point that this problem is very complex. For one, there is not just one square metre of the solar intensity radiating to Earth at the 1 AU radius from the Sun. the 1360 W/m^2 represents only one square metre on that sphere.
I am willing to bet that Roy’s model showing cooler temperatures on the Earth due to solar radiation alone is due to that assumption.
By the way…try to work backwards. Think in terms of why the total amount of energy received in one year is 173e15 joules. Note that 173e15 joules over 510,000,000 sq km is…340 W-years/m^2!
Hi there,
Why don’t we concentrate on my last answer to you first please ??? Bring the odd bit back and state what you think and why you are doing things.
Wayne
bdg…”Think in terms of why the total amount of energy received in one year is 173e15 joules. Note that 173e15 joules over 510,000,000 sq km is340 W-years/m^2!”
The problem is infinitely more complex than your awareness is allowing. The Solar flux field on a sphere at 93 million miles in 1360 W/m^2. That flux is being intercepted by the Earth’s sphere as it rotates at 1000 mph at the equator.
The Earth is also tilted and receives solar energy unequally and it’s orbit changes the energy it receives at each location on Earth.
Don’t forget cloud cover.
This problem cannot be solved simply, if at all, and dividing 1360 W /m^2 by 4 is even more simplistic than the GHE itself.
Nah, it’s not that hard. The solar constant or zenith flux or perpendicular flux or whatever you want to call it at TOA is completely independent of the rotation rate of Earth, the size of Earth, the tilt of Earth, clouds, or even whether Earth has an atmosphere at all. As long as the average orbital distance is 150 million km, solar luminosity remains constant, and the orbiting body is a sphere the solar constant will be 1360 W/m^2.
bdgwx, please stop trolling.
Hi Big,
You said when you have time you would comment of this ???
Are you saying without GHG’s that the Earth would be 18c ??? Lets see now, your saying the Sun does not heat us at all then basically. So why on a summer day is it 20c in the UK and 40c in Dubai, if as you claim it’s only the GHG’S warming us, then most people would say Dubia is so much warmer cos of its position on Earth, but if you are right, then the Sun would not heat different Countries like it does, it would basically be the same temp. all over the Earth !!!
Ok, if you got time, could you explain why Roy was dividing by 4 ???
I don’t go for the Earth being -18c cos of GHG’s, I say these so called GHG’s actually shade us, and col us slightly, it makes so much more sense. As your saying if your one of the -18c guys, that what I said before, that I am not actually feeling 20c in thUK, or 40c from Dubai, that this heat I feel is not from the Sun, but from the GHG’s ??? If your right about the -18c and GHG’s, then the curvature of the Earth, and the point in the Dubai receives more direct sunlight will not hold, as your saying the Sun only gives -18, so the area on Earth where you feel the heat more, should not work in your -18c theory, but it does work, and ho so well ???
The moon, which plainly does not have a greenhouse effect {according to your logic}, you say yes the lite side gets very close to 123c, but the dark side is so cold the average is -18C.
The moon has a very low surface heat capacity, compared to earth. So, it heats up quicker, and higher than the earth, it also cools quicker and lower than the earth. The obvious explanation for this is the differences in the length of day, AND surface heat capacity between the earth and the moon. No GHG’s effect required, no P / 4 required.
Wayne
You’d better tell Dr. Roy Spencer he got it all wrong at the top of this page, how embarrassing for one of you.
Svante, please stop trolling.
Roy…”Joe refuses to accept that the S=1,370 W/m2 solar constant energy that is intercepted by the cross-sectional area of the Earth must then get spread out, over time, over the whole (top-of-atmosphere) surface area of Earth”.
The confusion here comes from taking the solar constant and dividing it by 4 to get a surface radiance. Simply put, you cannot take solar intensity and treat it as a single line vector with a magnitude.
I don’t think the solar constant is clearly understood. It has nothing to do with the Earth or the shape of the Earth, it is the radiation measured from the Sun at a sphere with a diameter of nearly 93 million miles. Each square metre on that sphere has 1360 watts of solar power.
This problem is far more complex than either of you are conceding. In their paper debunking the GHE, Gerlich and Tscheuschner cover the topic well. Gerlich specialized in mathematics related to thermodynamics and in their paper they explain the problem using complex math.
It should be noted that Gerlich, an expert in thermodynamics and mathematics claims the problem of distributing the solar energy over the surface is impossible.
In one of the diagrams in your article, it still infers a heat transfer from an atmosphere with a temperature equal to or less than the surface. No heat transfer is possible under such conditions no matter how much physics is abstracted to infer it.
“It should be noted that Gerlich, an expert in thermodynamics and mathematics claims the problem of distributing the solar energy over the surface is impossible.”
Yet using measured input data the 1LOT surface energy balance result is the same mean global temperature as is measured by Earth’s installed thermometer field. Gordon is simply wrong.
“In one of the diagrams in your article, it still infers a heat transfer from an atmosphere with a temperature equal to or less than the surface. No heat transfer is possible under such conditions no matter how much physics is abstracted to infer it.”
Gordon invokes his basic misunderstanding of Clausius’ heat term. Actually:
The diagram shows an energy transfer from an atmosphere with a temperature equal to or less than the surface. Energy transfer is possible & measured under such conditions no matter how much physics Gordon abstracts to counter it.
ball3…”Yet using measured input data the 1LOT surface energy balance result is the same mean global temperature as is measured by Earths installed thermometer field”.
The 1st law has nothing to do with surface energy balance, it is a relationship between work, heat and internal energy.
“The diagram shows an energy transfer from an atmosphere with a temperature equal to or less than the surface. Energy transfer is possible & measured under such conditions no matter how much physics Gordon abstracts to counter it”.
Your delusions are good, I’ll give you that. Show me an example of heat being transferred from a colder body to a warmer body without an extravagant system driven by external power.
“Show me an example of heat being transferred from a colder body to a warmer body without an extravagant system driven by external power.”
Per Clausius heat is a measure of the KE of the body’s constituent molecules.
Per the Maxwell-Boltzmann (M-B) distribution of particle velocities in KE of the constituent molecules, they do not all have the same KE, the molecule’s KEs are spread randomly around an avg. which they computed.
Two different containers of gas at different temperatures are placed in contact. A faster molecule than avg. in the colder container bangs against a slower than avg. molecule in the warmer container increasing universe entropy.
Boom! That’s an example of a measure of the KE of the object’s constituent molecules (heat) being transferred from a colder body to a warmer body without an extravagant system driven by external power in accordance with 2LOT. And this process happens all the time, continuously.
Gordon wouldn’t know about this because Gordon has never demonstrated learning about the work of M-B which followed the work of Clausius and improved upon that work.
Ball4, please stop trolling.
Regardless of which divisor you may choose, the simple fact is that analysis of empirical data shows that temperature is independent of CO2 concentration and the rate of release of CO2 into the atmosphere is determined by the climate. This is a mathematical synthesis of what has actually happened over the past few decades, regardless of any hypotheses. See :
https://www.climateauditor.com
Bevan…”the rate of release of CO2 into the atmosphere is determined by the climate”.
Not only that, surface temperature is determined by the interaction of solar energy with LOCAL masses. Certainly, solar energy is the driver, but the IR given off is solely dependent on the local mass receiving that energy.
There is no proof of a generalized, one-to-one solar input to surface IR output since certain masses store solar energy better than others. Furthermore, the heat created by solar energy is dissipated by conduction and convection. At terrestrial temperatures, radiation is a minor player.
I see no one solved my sun puzzle. Let’s try again:
The closest 1 m^2 of the sun sends 1361 W/m^2 to my eye. The 2nd closest 8 m^2 of the sun also sends us ALMOST 1361 W/m^2 to my eye. Why is not the solar constant AT LEAST 1361*9 W/m^2? -Zoe
The cross sectional area of the Sun is 1.5e18 m^2. That means on average each square meter of the Sun is only sending 1361 / 1.5e18 = 8.9e-16 W/m^2.
WRONG!
Way Wrong.
Seriously wrong.
(But your pseudoscience is always good for a laugh, backdoor guy.)
bdgwx is some sort of a moron.
“The cross sectional area of the Sun is 1.5e18 m^2. That means on average each square meter of the Sun is only sending 1361 / 1.5e18 = 8.9e-16 W/m^2.”
He took W/m^2 and divided it by m^2 to get W/m^4.
To be fair I should have been more careful in the wording. I should have said…On average each square meter of the Sun contributes 8.9e-16 W/m^2 to the solar constant of 1361 W/m^2. Though I don’t think “sending” was entirely inappropriate since the directionality is from the Sun to a point on a sphere of radius 1 AU. So this works out as 1 m^2 * 8.9e-16 W/m^4 = 8.9e-16 W/m^2 or about 600 sextillionth of the solar constant.
Yep. And the interpretation here is that the contribution to the solar constant of 1360 W/m^2 from each square meter of the Sun is 8.9e-16 W/m^4.
What this means is that we can now do the following.
8.9e-16 W/m^4 * 1.5e18 m^2 * 127.5e12 m^2 = 173e15 W
Note that 1.5e18 m^2 and 127.5e12 m^2 are the cross sectional areas for the Sun and Earth respectively.
Let’s play around with the 173e15 W figure.
173e15 W / 510e12 m^2 = 340 W/m^2
or if want to know how much energy this represents in one sidereal year…
173e15 W * 1 y = 173e15 W-year which is equivalent to 340 W-year/m^2!
bdgwx is some sort of moron.
“And the interpretation here is that the contribution to the solar constant of 1360 W/m^2 from each square meter of the Sun is 8.9e-16 W/m^4.”
Google for “W/m^4” = ZERO results. You’ve went outside the bounds of science.
“8.9e-16 W/m^4 * 1.5e18 m^2 * 127.5e12 m^2 = 173e15 W
Note that 1.5e18 m^2 and 127.5e12 m^2 are the cross sectional areas for the Sun and Earth respectively.”
LMAO.
8.9e-16 W/m^4 * 1.5e18 m^2 = 1361 W/m^2
Then you multiplied this by the cross sectional area of Earth? LOL
What if instead of planets, we had concentric spheres? Your stupid math would mean the further we go out from the sun, the more energy! LOL
Sorry, dummy, but the sun emits 63 MEGAWatts/m^2, and it gets reduced to 1361 W/m^2 via inverse square distance law.
1361 W/m^2 is direct normal irradiance. My sun puzzle is simple: Why don’t we add NON-direct normal radiation? Or more cluefully, why do we ignore the addition?
Google for “NON-direct normal radiation” = ZERO results. You’ve /went gone outside the bounds of science.
“Google for NON-direct normal radiation = ZERO results. Youve /went gone outside the bounds of science.”
You can find “DNI Direct Normal Irradiance”. If DNI was the only type of radiation, “direct normal” would be redundant. What do you call radiation that is not direct normal?
You not only escaped google, you escaped all 3 physical dimensions with your retarded “m^4”.
I didn’t ask you to calculate how much sun shine hits the earth, I asked you how much sun shine hits my eye. Read again.
Oh that’s nothing. m^4 isn’t even remotely close to being the weirdest units encountered in physics problems. Case in point…the Boltzmann constant…take a stab at interpreting W/m^2/K^4. Seriously…what does the 4th root of temperature even mean anyway?
One of my favorite strange units is m^2/s^2. This is encountered in atmospheric sciences in relation to a quantity called helicity. Give this one a stab as well. How would you interpret this? Hint…think outside the box here. What is this equivalent to?
Typo…4th root should have read 4th power
“What do you call radiation that is not direct normal?”
In atm. science, radiation from the zenith is the term for what you are terming direct normal. In westerns, they sometimes use somewhat looser term “high noon”. The non-zenith rest of the incident irradiation standing on Earth surface comes from a hemisphere of directions.
Didn’t you want an answer to how much each square meter of the Sun contributes to the solar constant?
Also, my math doesn’t in any way imply that the further out the Earth the more energy it receives. In fact, my math implies the exact opposite. The Sun and Earth cross sectional areas are constant, but because of the inverse square law the solar flux decreases.
By the way, I’m not in any way implying that knowing the contribution of the solar flux at TOA from each square meter of the cross section of Sun is useful. It’s actually not AFAIK. That’s doesn’t mean there isn’t an answer to that specific question though. If that wasn’t the question you were asking and if it isn’t a bother would you mind clarifying what the question is?
ZP said…”I didnt ask you to calculate how much sun shine hits the earth, I asked you how much sun shine hits my eye. Read again.”
If your eye is pointed directly at the Sun and the solar flux at the point of your eye is 1360 W/m^2 and the radius of your iris is 5mm then 1360 W/m^2 * 3.14159 * (0.005 m)^2 = 100 mW is what is entering your eye. Do you agree?
Ball4, bdgwx, please stop trolling.
“W/m^2/K^4”
“m^2/s^2”
None of this violates Physics – the study of motion in our 3 physical dimensions.
“Seriouslywhat does the 4th root of temperature even mean anyway?”
A much smaller temperature. Duh!
“Didnt you want an answer to how much each square meter of the Sun contributes to the solar constant?”
I didn’t want yoy to derive a meaningless number with physics-less units.
“Also, my math doesnt in any way imply that the further out the Earth the more energy it receives. In fact, my math implies the exact opposite. The Sun and Earth cross sectional areas are constant, but because of the inverse square law the solar flux decreases.”
Yes, that’s exactly what your math implies.
You didn’t use inverse square law: You multiplied 3 numbers.
X * As * Ap
X = 8.9e-16 W/m^4
As = Sun cross area
Ap = Planet cross area
“That means on average each square meter of the Sun is only sending 1361 / 1.5e18 = 8.9e-16 W/m^2”
You claimed that’s what the sun sends, not what the Earth receives.
Given X*As*Ap, and given that Ap obviously grows as distance from sun increases, you are claiming more energy for outer planets.
ZP said…”You claimed thats what the sun sends, not what the Earth receives.”
Ok, sure. Busted. You definitely got me there. I should have said the contribution that the Earth receives from each square meter of the Sun is 8.9e-16 W/m^2…my bad.
As and Ap are both constants. The Earth does not magically get bigger as you move it further away from the Sun; at least I was assuming we were both in agreement that the Earth would stay the same size. I suppose if we are magically moving it further away would could magically increase it’s size too; I just didn’t think we were considering that.
Now, if you want to talk about a bigger planetary body…say Jupiter…then yes, Jupiter does, in fact, receive more energy from the Sun because it is bigger. It has a larger cross sectional area so it intercepts more of the sunlight. Therefore, it receives more total energy.
Here’s the cool thing. Flux is measured in W per square meter. So two different sized bodies can receive the same flux, but receive different amounts of energy. The best known example of this…the Earth and the Moon of course.
I think you’re incredulous about m^4 because you cannot find meaning in it. I’m demonstrating that finding meaning in strange units is often difficult by giving you other examples.
What is the meaning or interpretation underlying K^4?
What is the meaning or interpretation underlying m^2/s^2?
And FWIW m^4 does have a specific meaning or underlying interpretation. In fact, it has multiple interpretations depending on the context. One is area moment of inertia. Another is second moment of area.
https://en.wikipedia.org/wiki/Second_moment_of_area
bdgwx is some sort of moron.
Math is nice, but it’s not physics. The factor is 1/m^4, not m^4. What does it mean to pack a Watt into four dimensional space?
Let’s go back to my original point:
“The closest 1 m^2 of the sun sends 1361 W/m^2 to my eye. The 2nd closest 8 m^2 of the sun also sends us ALMOST 1361 W/m^2 to my eye. Why is not the solar constant AT LEAST 1361*9 W/m^2? -Zoe”
So far all you have managed is a regression into:
The closest 1 m^2 of the sun sends 1361 W/m^2 to my eye. The 2nd closest 8 m^2 of the sun also sends us ALMOST 1361 W/m^2 to my eye. Why is not the solar constant AT LEAST (8.9e-16 W/m^4 * 1.5e18 m^2 = 1361 W/m^2)*9?
Brilliant. Would you like to regress some more and divide by As and then multiply by As so we can discuss the meaning of W/m^6?
What a low grade imbecile.
“If your eye is pointed directly at the Sun and the solar flux at the point of your eye is 1360 W/m^2 and the radius of your iris is 5mm then 1360 W/m^2 * 3.14159 * (0.005 m)^2 = 100 mW is what is entering your eye. Do you agree?”
My iris has lots of rods and cones, and each one is receiving radiation from more than 1 m^2 of the sun, although I can only process one “pixel” at that resolution.
The question: is why you can’t you add non-zenith to zenith radiation?
Remember, dummy, I already know the answer. I’m just waiting for you to realize the crackpot nature of your religion.
ZP said…”What does it mean to pack a Watt into four dimensional space?”
I don’t know. Good question. Similarly what does it mean to pack a meter into two dimensional time? Or, more bizaarely, what does it mean to pack a square meter into two dimension time? The joys of interpreting units…
ZP said…”The closest 1 m^2 of the sun sends 1361 W/m^2 to my eye. The 2nd closest 8 m^2 of the sun also sends us ALMOST 1361 W/m^2 to my eye. Why is not the solar constant AT LEAST 1361*9 W/m^2? -Zoe”
The closest 1 m^2 of the Sun is not sending 1361 W/m^2 to your eye. Didn’t I just show mathematically that this is not true?
ZP said…”Remember, dummy, I already know the answer. Im just waiting for you to realize the crackpot nature of your religion.”
I’m all ears. If you have an insight here that I haven’t thought off then I definitely want to know what it is. This is your opportunity to teach me something that I don’t know. Ya know…I’m actually a pretty receptive person. If you’re getting a different vibe then I truly apologize.
“I dont know. Good question. Similarly what does it mean to pack a meter into two dimensional time? Or, more bizaarely, what does it mean to pack a square meter into two dimension time? The joys of interpreting units”
Time is not a dimension. These units are discussed in physics, yours are not. Why don’t you see your junk math when DNI is discussed?
“The closest 1 m^2 of the Sun is not sending 1361 W/m^2 to your eye. Didnt I just show mathematically that this is not true?”
No. You divided flux by an area, then multiplied it again by that area.
“Im all ears. If you have an insight here that I havent thought off then I definitely want to know what it is.”
Place two ice cubes @ 0C, next to each other. Each ice cube sends and receives 315 W/m^2. According to your religion, you can add them, but according to reality, you can’t
Fluxes can’t be ADD()ed, They can only be OR()ed. Duh!
Place a molecule thick sheet of paper between the ice cubes. Does the paper get to 630 W/m^2.
Take 4 ice cubes. Does the center corner get to 1260 W/m^2
A sphere of ice is like N ice spikes lumped together. You think you can get to 315*N W/m^2
Fluxes are not additive, they are maximal.
We ignore those 8 sq meter of the sun that are 2nd closest to the 1 sq meter closest to us. Duh!
ZP said…”We ignore those 8 sq meter of the sun that are 2nd closest to the 1 sq meter closest to us. Duh!”
If you’re willing to ignore 8 sq meters of the Sun then what is stopping you from ignoring the other 3e12 m^2 that Earth faces? Do you really think a single sq meter on the Sun is sending Earth 173e15 W to Earth?
ZP said…”Time is not a dimension. These units are discussed in physics, yours are not. Why dont you see your junk math when DNI is discussed?”
Einstein would cringe at this statement. Anyway, my unit examples are none other than m/s^2 which is more commonly interpreted as acceleration and m^2/s^2 which is helicity. Helicity is a fun one to interpret because there are actually different ways to make the interpretation.
ZP said…”Place a molecule thick sheet of paper between the ice cubes. Does the paper get to 630 W/m^2.”
What is the surface area of this sheet of paper? Is the intent to assume that the sheet is sufficiently thin that we do not consider it to have two sides? If we are considering that it has 2 sides then the answer is 315 W/m^2, but if it only has 1 side then the answer is 630 W/m^2.
Here is how this works out. Assume the sheet is 1×1 m. If it is one side then the total surface area is 1 m^2. But, if it is two sided then the total surface area is 2 m^2. One side case…the power received from the left is 315 W and from the right is 315 W. This means the sheet is receiving 630 W and since the area is 1 m^2 the average flux is thus 630 W / 1 m^2 = 630 W/m^2. Two side case…the left and right and power is the same. But, the sheet is now 2 m^2 so this works out to 630 W / 2 m^2 = 315 W/m^2.
ZP said…”Take 4 ice cubes. Does the center corner get to 1260 W/m^2″
How are the ice cubes arranged? What is the center corner? Either way, I’m not envisioning any arrangement where the entire surface area would get 1260 W/m^2 nor am I envisioning any arrangement where even a single face could get 1260 W/m^2.
ZP said…”A sphere of ice is like N ice spikes lumped together. You think you can get to 315*N W/m^2″
Let’s work it out together. We have a sphere between two sources where the perpendicular flux is 315 W/m^2 on each face of the sphere. The effective flux over the entire sphere is thus (315 / 2) + (315 / 2) = 315 W/m^2. What I did here is was exploit a canonical geometrical principal to work this out. I could have done it the hard way and integrated the perpendicular flux down the curvature of each hemisphere.
If instead of a sphere we assume the shape is a cube then the average flux works out to (315 + 315 + 0 + 0 + 0 + 0) / 6 = 117 W/m^2.
The point…the shape of the object matters. The sphere ends up having a higher flux because it has a smaller surface area. A sphere is the perfect shape that minimizes the surface area to volume ratio. The concept in play here is called the square-cube law.
ZP said…”Fluxes are not additive, they are maximal.”
It depends on how you do it. For example, with the cube above I can’t just add the fluxes on each face as-is. But with a sphere that is receiving 100 W/m^2 from the outside and 100 W/m^2 from the inside then I can just add these fluxes to get the total flux of 200 W/m^2. The difference between these examples is that former is using fluxes from different sub-areas while the later is using fluxes from the exact same area.
“with a sphere that is receiving 100 W/m^2 from the outside and 100 W/m^2 from the inside then I can just add these fluxes to get the total flux of 200 W/m^2”
No, you can’t. Only in climate “science”. Only something greater than 100W/m^2 can raise 100 W/m^2.
A 200 W/m^2 object has lower and higher frequency photons (aside from higher intensity at all frequencies for such an object) that a 100W/m^2 objects lacks, or 2 100W/m^2 objects lack. Where did they come from?
If you don’t add them then you’d be breaking the 1LOT. Here is how this works. Assume you have a sphere with a surface area of 1 m^2. There is an energy source inside the sphere that is generating 100 W. There is an energy source outside the sphere that is sending 100 W to it. The sphere is receiving 200 W. And since the surface area is 1 m^2 that means the effective flux averaged over the surface is 200 W / 1 m^2 = 200 W/m^2. Notice that this is the exact same result you get if you just added the two 100 W/m^2 fluxes from the start. If you disagree then tell me what you think the average flux is on the sphere’s surface. Show your work and do not violate the 1LOT.
backdoor guy, your set up is not detailed enough. How is the energy being transferred?
You don’t even know how to pose an example that makes sense.
I’d ridicule you, but the challenge is gone….
“Einstein would cringe at this statement.”
If Einstein thought that ~9 billion vibrations of Cesium-133 occuring in 3D represents another dimension, then he’s a moron.
Ya know…4 spacetime dimensions…3 spatial and 1 temporal. No?
bdg…”Ya know4 spacetime dimensions3 spatial and 1 temporal. No?”
Since when is a temporal dimension real? Even the 3 spatial dimensions are not real, other than to the deluded human mind.
There is certainly no physical dimension representing the past, or the future, which is actually the past projected forward by the human mind. We are in the same physical space we have always been in on the Earth. All that has changed is physical properties, like mountains eroding, human structures rising and falling, etc.
If you imposed a Cartesian or Polar coordinate system on the Earth, centred at the Earth’s centre, which way would you point the +ve y-axis? Those of us in the Northern Hemisphere would think it better to point it towards the North Pole but folks in Australia might prefer it pointing toward the South Pole.
Have you ever looked at a star chart? East and west is reversed to those expecting to have the north celestial pole pointing up the way from the Northern Hemisphere. Some rocket scientist decided the directions should be aligned so that someone looking down on the Earth would see East-west in that manner.
The human mind is so deluded, and so screwed, it amazes me we have managed to come so far in science.
JDHuffman,
You must be amused by the amount of sheer stupidity some people engage in just to uphold something stupid.
You guys would make a cute couple.
Just like Bonnie and Clyde, Sodom and Gomorrah.
Just more adolescent nonsense from Nate.
Nothing new.
nate…”You guys would make a cute couple.
Just like Bonnie and Clyde, Sodom and Gomorrah.”
Or Tesla and Newton.
Tesla just called Einstein a moron, so..
Nate, please stop trolling.
“Ya know4 spacetime dimensions3 spatial and 1 temporal. No?”
No. Time doesn’t exist. There’s only motion in 3D, such as vibrations of Cesium-133.
Dimensions must be orthogonal – moving in one doesn’t effect the other.
If you traveled back in time to 1955, all the matter would have to rearrange itself just for you like it was in 1955.
Where would this information be stored?
Maybe god keeps an atomic snapshot of every planck moment of all the matter in the universe, but now we’ve left observable science.
Interesting. So is it safe to say you are a contrarian of modern spacetime theory?
By the way, if you reject that idea that space and time are connected then how do you explain the conservation of the spacetime four-vector such that faster movements through the spatial dimensions necessarily decreases the movement through the time dimension? This is an example of movement in one dimension affecting the movement through another. And it’s tested everyday by billions of people.
I’m just curious what your thoughts are here even though they aren’t terribly relevant to this blog post. At least it’s still a sciency topic though.
dbg…”No. Time doesnt exist. Theres only motion in 3D, such as vibrations of Cesium-133″.
Zoe said, “No. Time doesnt exist. Theres only motion in 3D, such as vibrations of Cesium-133”.
Zoe is talking fact, bdg, it seems you are the contrarian if you believe that pseudo-science of space-time.
zoe…”If you traveled back in time to 1955, all the matter would have to rearrange itself just for you like it was in 1955.
Where would this information be stored?
Maybe god keeps an atomic snapshot of every planck moment of all the matter in the universe, but now weve left observable science”.
*******
I have been trying to convey that obvious conundrum to space-time believers here but it’s tough to dislodge an ingrained belief about time.
Time is defined based on the constant angular velocity of the Earth, therefore time has to be a constant, albeit an imaginary constant. Seems Einstein was wrong and Newton was right. Time is absolute, albeit an imaginary absolute.
“If you dont add them then youd be breaking the 1LOT. Show your work and do not violate the 1LOT.”
1LoT concerns Joules, not Watts or Watts/m^2.
“And since the surface area is 1 m^2 that means the effective flux averaged over the surface is 200 W / 1 m^2 = 200 W/m^2. Notice that this is the exact same result you get if you just added the two 100 W/m^2 fluxes from the start. ”
Sphere has an inner m^2 and an outer m^2. It’s receiving 200 W over 2 m^2 = 100 W/m^2
If the sphere is one molecule thick, then you get photon interference. Unlike 1LoT, photons are not conserved. Either they are able to raise electron energy level and DIE, or they are unable to raise electron level (because frequency and intensity are the same) and DIE.
ZP said…”1LoT concerns Joules, not Watts or Watts/m^2.”
Well, a watt is a joule per second so…
ZP said…”Sphere has an inner m^2 and an outer m^2. Its receiving 200 W over 2 m^2 = 100 W/m^2″
Sure, we could certainly model that arrangement. And I agree. In that case the average flux would remain 100 W/m^2. That’s an example where fluxes cannot be added as-is. I’m totally fine with that. And it is a completely legit and valid way of thinking about it. The important thing is that you have defined the area and we both agree that the two fluxes are not in reference to the same area so they cannot be added as would be similar to my cube example above. I happen to be focused on the 1 m^2 interface area of the sphere. I figured that if you put a 100 W energy source inside the sphere an equilibrium would be achieved such that, eventually, 100 W/m^2 must interact at the 1 m^2 interface area. Otherwise, the energy would accumulate ad-infinitum.
bdg…”ZP said…”1LoT concerns Joules, not Watts or Watts/m^2.”
Well, a watt is a joule per second so…”
Just replying out of curiosity. It strikes me as being odd that electromagnetic energy would be expressed in joules since no work is being done. There is not even heat being transferred at that point. There is a potential for heat transfer if a mass is encountered but as it stands, solar energy at TOA is neither heat nor work.
A joule is a unit of work, it is a measure of the work done by a Newton of force over one metre. A watt, as in power, has a time element, therefore a watt is expressed in joules per second. You could express it in foot-pounds/second.
In the 1st law, the internal energy U becomes a sum of the external thermal energy, heat, and the external mechanical energy, work. Neither heat nor work have units in common therefore one of them has to be stated in equivalent units of the other.
If you see the 1st law stated in joules, it means the heat is being expressed in its mechanical equivalent. The natural unit of heat is the calorie. You could convert work to calories and state the 1st law in calories.
This is a repost of the derivation of the 1360 W/m^2 solar constant…
Presume distance to Sun is 150 x 10^6 km = 150 x 10^9 metres.
The area of a sphere is A = 4.pi.r^2
Note…this presumes a circular orbit for the Earth, as if it’s orbit follows a circular sphere at radius = 1 AU.
Work out r^2 = (150 x 10^9m)^2 = 2.25 x 10^22 m2
4.pi = 4 x 3.14 = 12.56
Area of sphere = 12.56(2.25 x 10^22)m^2 = 2.826 x 10^23 m^2
Radiation intensity at the Suns surface = 3.846 x 10^26 watts. We need to know how much that power will spread out onto the sphere we just calculated so we divide that power by the area of the sphere:
3,846 x 10^26 W/(2.826 x 10^23 m^2) = 1.3609 x 10^3 W/m^2
= 1361 W/m^2.
Anywhere on that sphere with radius 150 x 10^6 km, the power will measure 1361 W/m^2.
What does the 1360 W/m^2 mean physically. It means that on one square metre of a sphere around the Sun, the solar intensity will be 1360 W/m^2.
If you sum the power of all those square meters over the solar sphere at 1AU, you get the total power. So why are we taking one square meter of that total power and dividing it by 4?
How does that 1360 W/m^2 get to the surface? Through the atmosphere, which acts as a filter for different wavelengths and intensity (clouds). By the time the solar energy filter down to the surface it is much less than 1360 W/m^2. So, why are we using 1360 W/m^2 and dividing by 4?
As I said, this is a far more complex question than what is being considered. It is so complex, that two experts in thermodynamics (one also an expert in thermodynamics math) claimed it cannot be done.
The question being answered is what is the average effective flux over the entire surface area of Earth. Since 1360 W/m^2 is the zenith flux only it does not represent the flux received at TOA everywhere. To get the effective flux you need to integrate the solar constant with respect to the cosine of the zenith angle down the lit hemisphere and then repeat the procedure with the unlit hemisphere except using 0 W/m^2 because ya know its dark. Or you can exploit a well known geometrical principal and just divide the solar constant by 4 to get the average effective flux at TOA. Both give you the same result.
By the way…excellent derivation of the solar constant.
backdoor guy, dividing the solar constant by 4 does NOT provide “the average effective flux over the entire surface area of Earth”.
You got caught confusing arithmetic with physics, again. 1370 W/m^2 can produce temperatures close to 400 K. That’s way above the temperature of boiling water. 1370/4 could barely melt ice.
You have been drenched in pseudoscience, and swallowed every drop.
Time to head out the backdoor, huh?
Well, as the good Dr. showed in the next post:
“Here’s how the incident solar flux changes with time-of-day and latitude. This should not be controversial, since it is just based upon geometry. Even though I only do model calculations at latitudes of 5, 15, 25, 35, 45, 55, 65, 75, and 85 deg. (north and south), the global, 24-hr average incident solar flux is very close to simply 1,370 divided by 4, which is the ratio of the surface areas of a circle and a sphere having the same radius:”
So JD is, well, wrong again.
He needs to argue out with the spreadsheet.
Nate just doesn’t get it.
Solar 1370 W/m^2 –> 400 K –> Boils water
“Spreadsheet” 1370/4 W/m^2 –> 252.7 K –> Freezes water
So Nate is, well, wrong again.
Nothing new.
‘Spreadsheet 1370/4 W/m^2 > 252.7 K > Freezes water’
Exactly JD, no GHG, no liquid water.
So you’re proven clueless twice in one post!
Solar 1370 W/m^2 –> 400 K –> Boils water
No GHC needed.
Nate’s ignorance is compounded by his cherry-picking.
Nothing new.
No JD, as usual, you are all hot air. Just more declarations and zilch to support them.
Nate, as usual, you demonstrate your inability to face reality. The Moon, with the same solar constant as Earth, reaches temperatures close to 400 K.
Nothing new.
And what is the mean temperature of the Moon?
backdoor guy, DA likes to use the one-line distracting question, when his pseudoscience fails him. Is that where you learned the deceptive technique?
‘Nate, as usual, you demonstrate your inability to face reality. The Moon, with the same solar constant as Earth, reaches temperatures close to 400 K.’
So what? The Earth is rotating much faster. As Roy showed, but you ignore, the Earth does not reach 400 K!
So GHE needed after all, dorkus malorkus.
Nate, please stop trolling.
GR, I want you stumble upon the answer of the division by 4 for yourself. I think this will happen if you first calculate how much energy is received by Earth in one second.
Keep demonstrating your ignorance of physics, backdoor guy.
It’s great for laughs.
bdg…”GR, I want you stumble upon the answer of the division by 4 for yourself”.
It doesn’t interest me, the value is far too simplistic, like the GHE itself.
The 1360 W/m^2 represents only one square meter of the bulk solar intensity on a sphere around the Sun with with radius = 1 AU. Extrapolating the solar intensity of that square metre over the entire hemisphere of a planet rotating at 1000 mph at the Equator is not a trivial problem, not can it be justified. Especially, when the planet has an axial tilt of 22 degrees which affects local temperatures throughout the planet’s orbit.
BTW…you have to be careful when dealing with flux fields. The word flux comes from Newton’s fluxion, which he intended as equivalent to our current derivative. Therefore, the instantaneous change of solar intensity is a line of flux that needs to be integrated over a square metre using a double integral.
If you add fluxes, what is it you are adding? If you double the density of the flux field in one square meter then obviously the W/m^2 would increase, but how do you double the density? The current flux intensity at TOA is a result of energy at the solar surface being diluted over a larger area at 1 AU.
The current idea in this blog, is that 1 square metre of that intensity at 1 AU can be divided by 4. I don’t see the relationship between 1 m^2 on the solar sphere and the 4 factor derived from the Earth’s sphere.
Look at the derivation of the 4 factor. Interesting stuff. Not talking about the actual derivation but an aside.
If you take a sphere of radius r, it forms a circle at its widest point of pi.r^2. Consider a hemisphere with base = pi.r^2 with r = 5. The area of the unit-less circle is pi.(5)^2 = 78.4.
If you squash that hemisphere down into a circle, its radius is (root 2).r = 1.414r. Taking the area of that circle, we have pi.(1.414)r^2. Since r was 5, that’s pi x (1.414 x 5)^2 = 157.
Note…I used the entire value on the calculator for root 2, not just 1.414.
If you divide 157 by 78.4 it comes to just over 2. That is, the surface area of a sphere squashed down into a circle is twice the area of a circle at the sphere’s widest point.
If you now double that value to get a full sphere the entire sphere squashed down into a circle has a surface area 4 times the largest circle on the sphere.
There’s your 4. What can that possibly have to do with the square metre on the solar sphere representing 1360 W? There’s no connection.
If you want to get a connection, you extend the 1360W/m^2 to the surface and measure it there. Of course, the beam will encounter all forms of interference like scattering and diffusion to dissipation due to clouds and such.
I don’t think the 360 W/m^2 has any meaning. It is used only to justify the greenhouse effect which is non-existing IMHO.
I would venture that on a seriously hot day with little or no interference, the surface value would be much higher than 360 W/m^2.
Furthermore. Joe is arguing that 4 should not be used and I have demonstrated that above. If anything, 2 should be used as in a hemisphere.
GR said…”It doesn’t interest me,”
Know the amount of solar energy the Earth receives doesn’t interest you?
GR said…”Extrapolating the solar intensity of that square metre over the entire hemisphere of a planet rotating at 1000 mph at the Equator is not a trivial problem, not can it be justified. Especially, when the planet has an axial tilt of 22 degrees which affects local temperatures throughout the planet’s orbit.”
The amount of solar energy the Earth receives has absolutely nothing to do with the rotation or tilt of Earth…literally…nothing. Those things effect the distribution of the energy the energy throughout the climate system, but they in no way have any relevance to the amount Earth receives.
GR said…”There’s your 4. What can that possibly have to do with the square metre on the solar sphere representing 1360 W? There’s no connection.”
You’re almost there. The big epiphany should be that the TOA of Earth does NOT receive 1360 W/m^2 as it is integrated over the entire surface area. The answer is that it actually receives 340 W/m^2. And the amount of energy Earth receives in one sidereal year is…5.4e24 joules which works out to 340 W-years/m^2. That’s worth repeating…the Earth does NOT receive 1360 W-years/m^2 of energy, but 340 W-years/m^2!
GR said…”Furthermore. Joe is arguing that 4 should not be used and I have demonstrated that above. If anything, 2 should be used as in a hemisphere.”
And Postma is wrong on several points. This is why Postma gets the wrong answer for the orbital distance of Earth. And the reason he gets it wrong is because he conflates the zenith flux of 1360 W/m^2 with the effective flux of 340 W/m^2. Note that if Postma had attempted to compute the total amount of energy Earth receives using his incorrect understanding of these concepts he would also get the wrong amount in this regard too.
backdoor guy continues to believe arithmetic is physics.
“The big epiphany should be that the TOA of Earth does NOT receive 1360 W/m^2 as it is integrated over the entire surface area.”
Wrong bdg. Earth at TOA receives that amount, on average. That’s why it is called the “solar constant”. You’ve got so much to learn, and you show no progress.
“And the reason he gets it wrong is because he conflates the zenith flux of 1360 W/m^2 with the effective flux of 340 W/m^2.”
Wrong bdg. 340 W/m^2 would be the INEFFECTIVE flux. 1360 W/m^2 is the solar flux at TOA. Even NASA knows that.
Quit trying to make stuff up, and learn some physics.
bdg…”Know the amount of solar energy the Earth receives doesn’t interest you?”
Did not say that. I said the factor of 4 does not interest me because it’s a non-sequitar factor. The 4 applies to a complete sphere and at beast, we are dealing with a hemisphere. If you are going to divide by a factor, divide by 2 for the hemisphere.
Think about it this way. If you have the Earth’s hemisphere squashed down into a flat circle to intercept solar radiation, you’d multiply the Earth’s radius by root 2 = 1.1414 and use that radius to calculate the surface area of the circle.
We know that area is twice the area of the circle formed at the equator with radius 6371 kilometres. That area is pi.r^2 = pi.(6371 km)^2 = 127,451,473 km^2. Double that to get the area of a squashed down hemisphere = 254,902,946 km^2.
We want to know the total watts delivered by the Sun over that area so we convert the value to m^2 = 254,902,946 x 10^6 m^2.
We know there are 1360 w/m^2 over that area with a total power of 254,902,946 x 10^6 m^2 x 1360 W/m^2 =
sorry…hit the enter button by mistake.
….to be continued.
continued…
We know there are 1360 w/m^2 over that area with a total power of 254,902,946 x 10^6 m^2 x 1360 W/m^2 =
346,668,005,853 x 10^6 watts of power over that area.
The Earth, on a 22 degree tilt, is rotating in that total solar power. Furthermore, the radiation and the amount of it reaching certain locations on the Earth varies with the orbit as well as effect on solar radiation in the atmosphere.
We have not covered the atmosphere. How do we know how much of that power is absorbed by the atmosphere, heating it? The power is not at a single frequency, it is spread over a wide range of frequencies. We know N2 and O2 absorb solar energy somewhere in that range therefore they must be warmed by solar energy.
We don’t need an uber-simplified cartoon of the atmosphere vis-a-vis the GHE to explain the warming and heat retention of the atmosphere…not to mention the oceans.
If you want to apply the 4-factor, or the 2-factor for a hemisphere, apply it to that total power, not to a square meter of the power at TOA.
GR said…”We know there are 1360 w/m^2 over that area with a total power of 254,902,946 x 10^6 m^2 x 1360 W/m^2 = 346,668,005,853 x 10^6 watts of power over that area.”
No that’s not correct. 1360 W/m^2 is not received over the hemispherical area. You can only say that it is received over the cross sectional area which is 127e12 m^2.
An alternate way of dealing with this is to integrate the 1360 W/m^2 with respect to the cosine of the zenith angle. You can actually refer to that paper by Gerlich and Tscheuschner you posted today for details on how this is done.
There is a pretty simple demonstration of this effect. Hold a flashlight perpendicular to the floor and record the surface area of the light. Then tilt the flashlight at an angle and watch how the same amount of light now spreads out over a larger area. Earth’s surface bending away from the Sun has the same effect.
bdgwx, please stop trolling.
‘By the way, if you reject that idea that space and time are connected then how do you explain the conservation of the spacetime four-vector such that faster movements through the spatial dimensions necessarily decreases the movement through the time dimension? This is an example of movement in one dimension affecting the movement through another. And its tested everyday by billions of people.”
Cesium-133 vibrates at different rates at different speeds.Very little difference at low speeds, huge difference at high speeds. At speed of light it doesn’t vibrate at all, at least in a measurable way. The vibrations can’t reach an external measuring device, because all matter is traveling at the same speed.
Einstein’s additional mistake was to believe in a universal clock that is not effected by anything. All clocks are mechanical (at the atomic/sub-atomic level) and as such, are subject to external “stresses”.
Like I said, there is no time. There is only motion in space, such as such atomic vibrations occuring in 3D. How can you combine 3D with 3D? You’re folding space by space. It’s meaningless nonsense.
‘Its meaningless nonsense.’ Yep.
nate…”Its meaningless nonsense. Yep.”
Zoe is right. We are imposing an imaginary 4th dimension of time on and imaginary 3-D universe and calling it spacetime.
There is little doubt that cesium atoms, as in an atomic clock, vibrate in various directions. However, those directions have no meaning till we humans impose a coordinate system.
I am not implying the universe is imaginary, I am claiming the coordinate system we have imposed on it is imaginary. The 4 dimensions of spacetime have no existence other than in the minds of humans.
“The 4 dimensions of spacetime have no existence other than in the minds of humans.”
Perhaps. And the minds of humans are amazing. To imagine such things as Black Holes, then finding them. Then, to imagine them colliding and merging, and then observing just such an event with Einstein’s predicted gravity waves. Detecting these waves, in itself, was an incredible feat of imagination.
So what humans imagine, sometimes turns real, and amazingly can be observed.
‘However, those directions have no meaning till we humans impose a coordinate system.’
I wish you would apply this reasoning to the Moon’s motion and rotation, Gordon.
nate…”To imagine such things as Black Holes, then finding them”.
No one has located a black hole. They have only located areas where light seems to be absent.
The explanation for a black hole is still steeped in pseudo science. A few decades ago, black holes were thought to be the result of the dying stages of a star. If the star did not explode as a supernova it was thought to collapse into a neutron star. In some cases, those densely packed neutrons were thought to collapse further into a black hole.
No one has ever explained how. In the Big Bang theory, the entire mass of our universe is theorized to have appeared out of nothing. No one has ever explained that one wither. There is no science to back the theory, none whatsoever. Same with AGW.
The modern definition is even more steeped in pseudo-science. Black holes are now regarded as space-time anomalies. Pretty neat trick since time has absolutely no existence.
It’s the same with exo-planets. No one has ever seen one, all they have done is detect irregularities in the gas spectrum of stars and leapt to major conclusions.
“No one has located a black hole.”
Sure they have, and they took a picture of it!
“Pretty neat trick since time has absolutely no existence.”
You are wrong about that Gordon as we are interested in processes and ALL real processes go forward in time. Everything takes time.
In thermodynamics (DYNAMICS!) no practitioner pretends time is absent from considerations, even you write time derivatives of thermodynamic variables without fear nor apology. When a closed system interacts with its surroundings, its thermodynamic internal energy U changes with time dU/dt = Q+W where Q is also a rate as well as W. A rate absolutely involves the existence of time.
zoe…”Einsteins additional mistake was to believe in a universal clock that is not effected by anything. All clocks are mechanical (at the atomic/sub-atomic level) and as such, are subject to external stresses”.
In his GRT paper Einstein referred to time as the hands on a clock. Had he taken the time to check out its universal basis, as in the angular velocity of the Earth, I think he’d have seen his error.
It’s not clear to me whether time had been based on the Earth’s angular velocity in a formal sense in his day. It’s possible that Einstein was unaware of that and Louis Essen, the inventor of the atomic clock, has suggested Einstein lacked such skills in measurement.
Clocks are machines and as such they are prone to error. If a clock speeds up or slows down, it is not time changing, it is the machine. That’s why we synchronize all clocks to a clock at Greenwich, England.
Atomic clocks, like any mass-based clock are prone to error as well. If you heated the cesium you mention with a torch, I am sure the atoms would vibrate madly. I presume atomic clocks are maintained in environments free of such temperature variations.
Dr. Spencer,
you really need to step back to basic physics.
your response to Joe is:
“Joe refuses to accept that the S=1,370 W/m2 solar constant energy that is intercepted by the cross-sectional area of the Earth must then get spread out, over time, over the whole (top-of-atmosphere) surface area of Earth.”
The solar constant already spreads out the Wattage over time. It is already given in Watt which is a Joule per Second.
Now why would you spread out the power over a surface that does not get solar impact at all? The sun does only light up one side of Earth at any step in time. At any small step in time, which you could call a Second.
If you want to account for rotation you need to find another way of presenting the energy budget, but you should not use Power as a unit!
Sorry but the approach to spread power over a surface that never all at once (in one second) receives solar radiation is wrong.
Joe is correct!
Another Joe
You can easily do your own testing to see that Roy Spencer is correct and Joe Postma is a clueless cult leader of a bunch of people that have very little knowledge of heat transfer physics so he can easily manipulate their limited understanding and seem like a genius to them. He is just wrong.
Take a ball mounted so that a motor can spin it at variable speeds.
Take a heat lamp and place it close to the ball.
Turn the lamp on with no rotation and measure how hot the one exposed surface gets and measure the other side that receives no light. Now have it rotate at different speeds and see what happens.
The result of rotation will be that the surface never gets as hot as when it is not rotating. The energy of the lamp is the same in both cases but gets spread out over a larger area so never achieves the highest possible temperature.
You can also do this with meat on a rotating spit. If you do not rotate the meat over a fire, one side will burn and the opposite side won’t cook. If you rotate it you can evenly cook all the meat and avoid burning any.
The energy IS spread out over time because the rotation of the Earth. No one meter will ever get as hot as it would if the Earth did not rotate. I watch Joe Postma’s videos and they are lacking in any rational thought and his examples do not at all reflect what Roy is saying. Only the ignorant will think Joe Postma is clever and smarter than all the scientific community. If you don’t know this, that is how cults work, they make the followers think they are special and above every one else.
Anyway, do a simple experiment and you will find Joseph Postma is very lacking in understanding anything.
“Joe Postma is a clueless cult leader of a bunch of people that have very little knowledge of heat transfer physics…”
Whereas people like Norman think the heat transfer equation, where heat flow goes to zero at equilibrium, should be used as if there were an amount of heat still flowing at equilibrium! As Zoe correctly pointed out earlier, you don’t conserve heat. This leads them to all sorts of erroneous conclusions, and the misuse seems designed to mislead others. As they should know, but seemingly pretend not to, energy can be flowing through a system at equilibrium, but not heat.
Dr Roys Emergency Moderation Team
Please pay attention to the content of the post instead of bringing in your side points.
It is quite clear Joseph Postma is totally wrong about his inability to understand that the amount of energy received by the Earth’s surface WILL be spread out over a larger area, IN TIME, than it would be if the Earth were a flat disk in space that did not rotate.
I have stated so many times that I guess I will have to do so again. HEAT will not flow when two objects reach the same temperature. Energy will still flow back and forth between the two. EMR does not stop emitting from a surface just because the surfaces are at the same temperature. It is really clear and simple physics.
So if you have nothing of value to offer, take you own advise and Pleas Quit Trolling. It is of zero value. If you have something of value to say, say it. But if all you wish to do is troll, look elsewhere. We already have JDHuffman. One ignorant troll is enough for a blog.
“HEAT will not flow when two objects reach the same temperature”
There’s more to it than that, but glad you generally agree.
Why do you have to resort to insults, false accusations, and misrepresentations, every time someone corrects your pseudoscience?
Isn’t that a signal that you don’t have a clue about the relevant science?
Norman,
you fail to understand that W/m2 is IN TIME.
It is exactly per each second that you count the Joules.
If Each second the Earth is lit up by the sun on every square meter of its surface you are right.
If there is something like day and night you are wrong.
Wow, DREMT goes all in!
‘As they should know, but seemingly pretend not to, energy can be flowing through a system at equilibrium, but not heat.’
“This leads them to all sorts of erroneous conclusions”
You’re talking about Dr. Roy, all meteorologists, and Earth scientists here drawing these conclusions!
You, JD, and Zoe are brilliant blog posters who have noticed errors that all of these professional scientists have somehow missed!
Righhht!
Or, could it be instead that you, again, misunderstand basic science concepts, like equilibrium, heat, energy and flux.
Equilibrium: no energy or heat flow.
Sorry DREMT, quite OBVIOUSLY there is energy AND heat flow into and out of the Earth’s atmosphere. Earth is NOT in equilibrium!
Heat is simply energy flow driven by temperature differences.
The Sun, the Earth, its atmosphere, and space all are at different temperatures, hence energy and heat are flowing between them!
But please continue to man-splain to us how this is all wrong!
“Sorry DREMT, quite OBVIOUSLY there is energy AND heat flow into and out of the Earth’s atmosphere. Earth is NOT in equilibrium!l
Yes, agreed.
No heat flow, heat is only a measure of kinetic energy. Which is why they use terms energy budget & energy balance; not erroneously use heat budget or heat balance.
ball3…soon to be ball2…”No heat flow, heat is only a measure of kinetic energy”.
Temperature is a measure of kinetic energy, not heat. Heat is the kinetic energy measured by temperature.
Temperature of an object is a measure of LOCAL avg. kinetic energy (where the thermometer bulb is placed); Clausius’ sensible heat is a measure of TOTAL kinetic energy in the object. Very different concepts in thermodynamics, Gordon, which you ceaselessly confuse.
OK, Ball4.
“You can also do this with meat on a rotating spit. If you do not rotate the meat over a fire, one side will burn and the opposite side won’t cook. If you rotate it you can evenly cook all the meat and avoid burning any.”
Perfect real-world demo of whats going on, Norman.
Will the TEAM pay attention and get the point of it? Let’s see…
What TEAM?
You, jd, gordon, now zoe.
The only “TEAM” I see in operation in the comments under this article would be (in the main): Testicle4, Natan: Father of Lies, Kreg T, and Back Door Guy Worshipping “X”. Intentionally or not, you guys display that team spirit of working together to silence any opposition. Whereas the people you list are just a bunch of free thinkers who agree on some things, disagree on others. You lot are the ones that represent “one united voice”. Even if one of you contradicts the others on an unrelated subject, another one jumps in to defend and distract, never acknowledging there was any dissent from the group.
The thing my team has in common is that we like and respect science, and often use it in our work.
The thing that unites your TEAM, crew, it seems, is that you all lack respect for science and scientists.
You think the professional scientists have it all wrong, and the free-thinking amateurs have it figured out.
Free-thinking must be combined with real empirical knowledge. Then you get Einstein, Tesla, the Wright brothers.
Free-thinking combined with ignorance -you get Anti-Vaxxers, Flat-Earthers, Charles Manson, conspiracy-theorists of all kinds.
What TEAM?
All arguments with DREMT end the same.
He can’t win on the facts, he returns to 6th grade silliness.
What argument?
Another Joe,
One important thing to understand is that the Earth does not always receive a zenith flux of 1370 W/m^2 (or 1360 or whatever value you want to use for the solar constant). In fact, most of the time it doesn’t. The solar constant is just the average zenith flux over one orbital period. It is itself an average that includes enough time for the Earth to revolve 365 times (relative to the Sun). What this means is that the solar constant is a quantity that can be used to represent the solar impact over the entire TOA of Earth. You just have to know how to use it.
Postma is wrong on many points and presents many strawman arguments. The one most relevant to this blog post is his conflation of the zenith flux with the effective flux which is the reason why he does not get the right answer when he computes the orbital distance of Earth. Remember, that is entirely his doing. No reputable climate scientists would make such a trivial mistake including Dr. Spencer. Spencer and the rest of the scientific community is right and Postma is wrong.
bdgwx is wrong on many points and presents many strawman arguments.
I made a list (probably not complete) of the strawmen and inaccuracies in Postma’s argument.
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-356206
If you feel that I have presented a strawman or made a mistake then please bring them to everyone’s attention so that they can be reviewed.
If I see you correctly represent Postma’s argument, I’ll let you know.
Postma has not corrected bdwgx arguments here, so they still stand; let JP speak for himself. No need for DREMT to let blog readers know.
I love that that’s how you think it works.
bdg…”The solar constant is just the average zenith flux over one orbital period”.
I just laid out the derivation of the solar constant mathematically and you agreed. It has nothing to do with the Earth’s rotation or orbit. The solar constant is the solar intensity on the surface of an imaginary sphere at 1 AU radius from the Sun.
As EM spreads out from the solar surface, it is subject to the inverse square law. As the energy front spreads out and reaches the distance of the Earth, an imaginary sphere can be formed with radius equal to Sun-Earth distance (1 AU). By calculating the surface area of that sphere and dividing it into the solar intensity at the solar surface, it can be calculated how much the energy has dissipated.
The level the solar energy has dissipated due to spreading out on a sphere at 1 AU is the solar constant, 1360W/m^2.
I do agree with your derivation. It was excellent. And I’m glad you agree that it has nothing to do with Earth’s rotation. However, the solar constant has everything to do with Earth’s orbit. That’s because the solar constant is the average zenith flux at TOA of Earth specifically; not Mercury, not Mars, or any other planetary body. It is Earth’s solar constant. Mercury, Mars, etc. will have their own. The reason this is specific to Earth is because Earth intersects the solar radiation sphere at different distances. This is because Earth’s orbit is elliptical. This is why the solar constant is an average over one orbital period.
So what,
nobody disputes this.
The question is if the sun heats up the whole surface of Earth in each second or not.
And the simple answer is NO.
It is only ever each second a hemisphere that receives sunlight.
That’s right, and it comes around quite quickly so you get a good average temperature approximation if you divide the intercepted solar power by four.
Svante says:
“so you get a good average temperature approximation if you divide the intercepted solar power by four.”
He forgets that the average temperature is another crutch and mainly arrived at by using the power of the sun divided by 4 over the whole surface and voila: here we go again.
This is what is being disputed. Nothing new on the claim, whats missing is the correct explanation behind it!
As long as we cannot agree that the sun only lights up the Earth on a hemisphere the discussion is moot.
The rotation does not change the solar flux that hits the Earth surface. How is this not getting through, I have no idea.
Svante, before you average over the whole surface, what temperature has the sun lit side of Earth?
bdgwx, please stop trolling.
bdgwx,
Physically the solar constant is the same as the spectral radiance coming from the sun and when measured and compared to the Planck spectra it will have somewhat 1360 W/m2.
Look up and tell there is are any measured clear sky spectra that suggest that anywhere on this world the incoming radiation has not a similar value.
You can check polar spectra as well or some from any Latitude on Earth.
If you find any that is clear sky and around 350 W/m2 let everyone know or accept that the Solar power what it is: your averaged 1360 W/m2.
IF you want to express the amount of energy that is spread over Earth you better do not use Power. Because you will be wrong!
“If you find any that is clear sky and around 350 W/m2 let everyone know..”
June 13, about 5:00pm at Desert Rock, Nev. clear sky downwelling solar was measured ~350W/m^2. And also at about 6:45am. There, I let everyone know.
Link?
How measured?
Which adjustments have been done to account for the Latitude and receiving area?
Find the ESRL Surfrad data for that day, at that site. Measured by calibrated radiometers. If you know what you are talking about, this should only take a few minutes.
Ball4
so you throw a do it yourself in.
Thanks for a nice discussion.
Mind you I only said that the spectrum is the same.
The downwelling solar is already estimated for the area it will hit.
Here is the basic idea behind Postma’s argument. In the context of the 1st diagram Fs(1-A) is a flat surface concept while Fs(1-A)/4 is a spherical surface concept. And in the context of the 2nd diagram S is a flat surface concept while S/4 is a spherical surface concept. The diagrams are clearly labeled with Fs(1-A)/4 and S/4. That means they are, in fact, spherical Earth models. But, Postma’s tells his audience the exact opposite. I’m puzzled as to how or why an educated scientist (an astrophysicist nonetheless) would do this? Honest mistake? Intentional con? Something else?
backdoor guy, you keep going in the same circle. Your arithmetic is correct, but that is not the issue. Your physics is wrong. You can NOT divide a power flux, without changing the corresponding temperatures.
The backdoor is open and you are cleared for departure….
Just to be clear…I’m not claiming that 1360 W/m^2 over 510e12 m would result in the same temperature as 340 W/m^2 over 510e12 m. I’m also not claiming that 1360 W/m^2 over 127e12 m would result in the same temperature as 340 W/m^2 over 510e12 m. In either case the mean T would be different. And the instantaneous T at specific locations and times would obviously depend on a variety of factors as well. Everybody agrees…I hope. I have made no such claim otherwise. In fact, until now I haven’t even mentioned T at all. What I’m saying is that the Earth actually receives 340 W/m^2 and for it to be in equilibrium it would have to emit 340 W/m^2…on average over the entire TOA boundary through one sidereal year.
Wrong backdoor guy. You have claimed the 340 W/m^2 is the “effective flux”. You just got caught with a truckload of pseudoscience, and now you are heading for the backdoor.
So what do YOU think is the average or mean flux received over the entire TOA boundary of Earth over the course of one sidereal year? If not 340 W/m^2 then what is it? Show your work.
Nice try, bdg.
You can NOT average solar flux. It has no physical significance. You’re stuck with the solar constant. You’ll just have to live with it.
Learn some physics, and quit trying to twist reality.
“You can’t average the flux. It doesn’t reflect the physical reality. Here’s why.”
“OK. So here’s the math, the mistake you’re making is…”
“The math is fine. The physics isn’t.”
“OK. So how would you calculate the average?”
“You can’t.”
“Right. But what you’re missing is, you need to do this, and this, and this, mathematically”.
“The math is fine. The physics isn’t.”
“OK. But I think the big epiphany will come when you do this, and this, and this, with the math”.
“You’re not listening to me.”
“Sure. But you still just aren’t getting that if you do this and this and this with the math…”
“THE MATH IS FINE. THE PHYSICS ISN’T”.
“Got you. So you’re saying the math is wrong. So how would you calculate the average? Show your work, or I’m right.”
“No. You’re still not listening!”
“I hear you. So why is the physics wrong”
“[re-explains]”
“Right! I see where you’re going wrong now. You need to do this with the math…
“[puts gun to own head and pulls trigger]”
No, your physics is wrong too.
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-357742
Why have you linked to a comment where you are arguing against things I wasn’t actually saying, as was confirmed by my immediate response?
Puleez. Of course you were saying those things, and youve said them before, and Zoe said them.
You are very confused about equilibrium, and heat flow, and energy, and your comment demonstrates that.
Nate tells me what I’m saying. That’s new.
This is always how it goes with you!
D:”Whereas people like Norman think the heat transfer equation, where heat flow goes to zero at equilibrium, should be used as if there were an amount of heat still flowing at equilibrium!”
Norman was talking about the main issue in this whole post, the Earth and energy flow into and out of it.
You brought up “equIlibrium” as if it was relevant. But of course it is not, as I pointed out to you.
“As Zoe correctly pointed out earlier, you dont conserve heat. This leads them to all sorts of erroneous conclusions, and the misuse seems designed to mislead others. ”
Zoe is saying gobbledegook here. And was ALSO saying a system was in equilibrium when it clearly had heat flow in and out. Dumb, wrong.
What erroneous conclusions do Roy Spencer and all others come to?
“As they should know, but seemingly pretend not to, energy can be flowing through a system at equilibrium, but not heat.”
FALSE. Energy does not flow through a system at equilibrium. Why is this relevant?
Same thing you said about GPE, which, like the Earth, is NOT IN EQUILIBRIUM because it has ENERGY (heat) flowing through it.
I don’t know what your problem is. It was a bit of a distraction from the main point of this article, but I said what I said. Not whatever it is that you want to rewrite history to make it. It’s up there. So stop making stuff up.
You brought ‘equilibrium’ just cuz.
Because you knew it had no relevance to the discussion…
OK, DREMT.
Keep on man-splaining! Its going so well.
Nate, energy can flow through a system that is in equilibrium. The energy in must equal the energy out, however.
You have little understanding of thermodynamics, like Swanson, fluffball, Norman, bdg, and several others.
Nothing new.
Norman said:
“Joe Postma is a clueless cult leader of a bunch of people that have very little knowledge of heat transfer physics…”
…and then I wrote my comment, responding to that, specifically. As can be seen by scrolling upwards and reading. And I stand by what I said.
‘Nate, energy can flow through a system that is in equilibrium. The energy in must equal the energy out, however.
You have little understanding of thermodynamics, like Swanson, fluffball, Norman, bdg, and several others’
Both of you have stated the GPE is in ‘equilibrium’, ignoring all the energy (heat) flow between the plates, the 2LOT, and from the plates to space!
Another example of you guys being thoroughly confused about physics and just declaring things with nothing to back it up.
‘There are known unknowns. That is to say, there are things that we now know we don’t know. But there are also unknown unknowns. There are things we do not know we don’t know. D. Rumsfeld
The last sentence is talking about you guys!
Yes Nate, the 3-plates diagram is a good example of a system that is in equilibrium, AND has energy flow through it. Notice energy in equals energy out.
https://postimg.cc/KKx5hx4H
In the incorrect solution, there is no net energy increase to the system, yet enthalpy increases and entropy decreases.
That’s why the incorrect solution is solidly debunked.
JD said…”You can NOT average solar flux. It has no physical significance.”
Of course you can average a flux. I just did it. And it does have a physical significance as long as the interpretation is applied appropriately. For example, because we know the zenith is 1360 W/m^2 (which by the way is itself an average) we can use a geometric principal to quickly deduce the mean flux over the entire Earth of 340 W/m^2. Now I can use this value to conclude that Earth receives 340 W-years/m^2 of energy in one year which multiples out to 340 W-years/m^2 * 510e12 m^2 = 173e15 W-years or 5.4e24 joules. At no time have I broken any laws of physics or misused the mean flux of 340 W/m^2.
Postma, on the other hand, did misuse the 340 W/m^2 figure. But, that’s on him; not me, not Nate, not Ball4, not Norman, etc. So if anyone needs to be indicted on twisting reality then Postma should be at the top of the list.
“Nate, energy can flow through a system that is in equilibrium…the 3-plates diagram is a good example of a system that is in equilibrium”
No, JD gets all that wrong with a bogus x4H cartoon which is not unusual. JD’s statement is wrong in the case of strict thermodynamic equilibrium which is thermostatics where there are no changes in thermodynamic variables. Thus there can be no gradients in thermostatics, so no energy flow, the GPE does not qualify.
Clearly the earth LT is not thermostatic:
http://www.drroyspencer.com/2019/06/uah-global-temperature-update-for-may-2019-0-32-deg-c/
However, the black line has such a small anomaly over such a long time that the black temperature mean line (~288K give or take) can be usefully discussed as the measured & computed global steady-state equilibrium thermometer or brightness near surface air temperature.
A mean (or better yet, effective) temperature that does have an unambiguous physical meaning is the effective radiative equilibrium temperature Teff=255K of Earth system defined as the brightness temperature of blackbody radiation with a total emission equal to the measured net annualized solar radiation received by Earth system averaged over its entire ~spherical surface (240in=240out W/m^2).
Despite what JD writes. JD just needs to learn some physics. JD won’t ever do so hence JD’s one usefulness in entertainment value will be undiminished. Please, more cartoons JD. Knee slapping wrong but funny.
bdg, I think we’ve been here before. You understand the arithmetic, but your physics is flawed. Your made-up phrase, “effective flux”, is just one example of twisting reality.
And, speaking of “twisting reality”, here comes fluffball!
“Yes Nate, the 3-plates diagram is a good example of a system that is in equilibrium”
Heat into it, heat to space, and thus heat between the plates.
Not equilibrium. Goggle it. You guys are thoroughly confused.
Is the Earth in equilibrium? It has energy flowing thru it too?
Heat doesn’t “flow” to space, Nate. There is no matter there to “heat”. Energy flows from the green plates outwards into space, based solely on their temperature.
Heat only flows from the central plate to the neighboring green plates if the green plates are introduced at a lower temperature than the blue plate. Once they all are at the same temperature as the blue plate, since they are identical objects and we’re assuming the view factors between them are close to unity, heat is no longer flowing. Energy flows from the blue plate, through the green plates, and out to space. Equilibrium.
Dr Roys Emergency Moderation Team
YOU: “Heat only flows from the central plate to the neighboring green plates if the green plates are introduced at a lower temperature than the blue plate. Once they all are at the same temperature as the blue plate, since they are identical objects and were assuming the view factors between them are close to unity, heat is no longer flowing.”
The funny thing is that if you actually would do an experiment with three simpler plates. Heating the middle one and bringing the other close to, but not touching, you would find the outer plates do not get to the temperature of the middle plate. The middle plate will warm up as you move the green plates closer. If you did an actual test this is what you would find. You will argue endlessly about your false opinions and your total incorrect view of heat transfer and the 2nd Law of Thermodynamics. Nothing will change this.
Before E. Swanson did his vacuum test with two plates, the debate was Eli Rabbet was wrong with his understanding of physics. The debate went on for hundreds of comments. Back and Forth. So E. Swanson took it upon himself to end the foolish debate with actual science. His results verified Eli Rabbet and made fools of JDHuffman, you, Mike Flynn, Gordon Robertson. You anti-science skeptics could not accept what he has demonstrated and you still don’t. You can do the test yourself but you refuse to do so. You will endlessly believe you are right (you are not). Nothing will convince you. Textbooks won’t, experiments won’t.
Check again to refresh your memory. E. Swanson moves an unheated green plate close to the heated blue plate. The temperature of both go up. You are just wrong! That is a fact. Proven by real world evidence.
https://app.box.com/s/5wxidf87li5bo588q2xhcfxhtfy52oba
“Heat only flows from the central plate..”
Sorry, DREMT, all the experiments to find heat contained “in the central plate” will fail as experiments on anything else have failed. Try again to get the physics right.
‘Heat doesnt ‘flow’ to space, Nate. There is no matter there to ‘heat’. Energy flows from the green plates outwards into space, based solely on their temperature.’
Perfect example of you not understanding basic physics, making up your own, then man-splaining it.
Energy doesnt just ‘flow’ out of an object for no reason. If it could, there would be no point to 2LOT.
It flows to space because it is COLDER. It is 3K. Just as does from the Sun or Earth to space.
And that flow is heat. What else could it be?
“The funny thing is that if you actually would do an experiment with three simpler plates…”
The funny thing is when Norman starts talking about experiments that he hasn’t done, and that nobody has done (a “3-plate” experiment), as if he already knows the outcome. I don’t doubt that in real life the two green plates would drop in temperature when you separate them from the blue plate, because energy would be lost past the edges of the plates. However, in the thought experiment, you rule that out, by assuming very large, or even infinite-size plates, so that edge effects aren’t a problem.
“The middle plate will warm up as you move the green plates closer”
It’s very subtle how they manipulate things. Norman, your argument has to support the idea that the blue plate increases in temperature when the green plates are separated from the blue.
BTW, never heard of dark matter?
Its in space.
“Energy doesnt just ‘flow’ out of an object for no reason. If it could, there would be no point to 2LOT. It flows to space because it is COLDER. It is 3K. Just as does from the Sun or Earth to space.”
An object radiates based on its temperature and emissivity. So yes, there is a reason. No, heat isn’t flowing to space, because there is no matter in space to heat.
Not that matters..ha!
But seriously folks, space has T =3K. It has BB radiation in it. That came from……yep, matter.
“And that flow is heat. What else could it be?”
Energy. Radiation. Photons. Electromagnetic waves.
Nate, the earth system has a total thermodynamic internal energy U which reduces as it radiates to, increases as it absorbs from, space.
Earth system dU/dt=Q+W where W=0 as the atm. does no net work like on the gas giants. There is no such “thing” as heat ever found in the earth system, only energy in one form or another. Heat is only a measure ever since Clausius and contemporaneous experiments ruled heat out of existence in any object.
“But seriously folks, space has T =3K.”
But seriously folks, space has brightness T~3K. That is from the CMB which is as close to thermodynamic equilibrium as you can easily find.
Dark matter? You mean the matter that’s “invisible” to the entire electromagnetic spectrum? OK, Nate. That’s me done.
I started to collect a bunch of the ridiculous comments from Norman, fluffball, Nate, and the guy that went out the backdoor. (I especially enjoyed their confusion over equilbrium.)
But then I realized, there will always be more to come!
Nothing new.
‘ No, heat isn’t flowing to space, because there is no matter in space to heat.”
OMG.
How do we know heat is flowing to space? If a bucket of water is in space, its going to radiate and cool.
Things only cool by losing heat! By definition.
Eventually that radiation will hit something, a planet, gas or dust, and be absorbed.
Sorry folks, you don’t get to make up special rules like “No heat flow to space allowed!”
Feel free to find it and show me.
And your favorite law, 2LOT requires it.
So no, the plates are NOT in equilibrium.
‘Earth system dU/dt=Q+W’
Yes Ball4, and Q is heat flow to space!
Things only cool by net loss of thermodynamic internal energy!
Q is the rate of change of thermodynamic internal energy of system of interest as a consequence of a temperature difference between it and its surroundings.
Heat does not exist in an object therefore cannot be lost from that object so calling Q heat just means you (et. al.) aren’t reasoning well enough.
Nate, you act like you think that there needs to be heat flowing through a system at equilibrium! So you take into account the temperature difference between the green plates and space, and assume you also need a temperature difference between the blue plate and the green plates, at equilibrium, so you have heat flowing! You are conserving heat, not energy.
Your error is in thinking you need to apply the heat flow equation between the green plates and space, but that is absurd, because the green plates cannot heat space. All you need to consider is that the green plates will emit based on their temperature and emissivity, so for their outward energy flow to space you apply the SB Law, “one T” equation only. So the temperature difference between the green plates and space doesn’t enter into it.
When the green plates are at the same temperature as the blue, heat flow between them has gone to zero, energy is flowing from the middle plate with the electrical power supply, through the green plates, to space, and that is equilibrium.
‘You are conserving heat, not energy.’
This makes no sense to me. JP said this, Zoe said it, now you are saying it.
It is gobbledegook.
“temperature and emissivity, so for their outward energy flow to space you apply the SB Law, one T equation only. So the temperature difference between the green plates and space doesnt enter into it.”
False, made-up physics.
The temperature of the environment matters. This is a central principle of HEAT TRANSFER.
“Net Radiation Loss Rate
If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as
q = ε σ (Th4 – Tc4) Ah (3)
where
Th = hot body absolute temperature (K)
Tc = cold surroundings absolute temperature (K)
Ah = area of the hot object (m2)”
https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html
If the temperature of space was same as the plate, heat flow would then be 0.
If the temperature of space or whatever is surrounding the plates is higher than heat will flow TO THE PLATES.
This satisfies the requirements of the 2LOT that heat only flows on its own from hot to cold.
While your made-up-physics does not.
“When the green plates are at the same temperature as the blue, heat flow between them has gone to zero, energy is flowing from the middle plate with the electrical power supply, through the green plates, to space, and that is equilibrium.”
Again, completely made-up HEAT TRANSFER physics!
Please show me where this idea comes from. Show me an example worked out this way from anywhere.
Whether you call the energy HEAT or not, if it is ” flowing from the middle plate with the electrical power supply, through the green plates, to space”, it will NOT FLOW without a TEMPERATURE DIFFERENCE to drive it from hot to cold!
How the hell does the energy know where to flow to? That is the whole point of temperature.
Again, if it can just flow without a temperature difference, the second-law would be constantly violated.
Yes, Nate. That would be used where there are “surroundings”, i.e matter. Like gases, like “air”. You know, stuff that can be heated, to any extent, by the object.
Space is not “surroundings”, is it!? It’s the absence of “surroundings”. It’s “space”.
You seem to think you should solve the heat transfer equation between the plates as though there is still heat flowing at equilibrium. You think the result should be a value other than zero! That’s what we are getting at by, “you are conserving heat”. Zoe already explained that to you.
“it will NOT FLOW without a TEMPERATURE DIFFERENCE to drive it from hot to cold!”
Heat will not flow without a temperature difference to drive it from hot to cold. Energy does not need a temperature difference to flow. The green plates just radiate to space based on their temperature and emissivity. Energy is flowing out from the plates regardless of temperature difference.
I’m not going to waste too much more of my time explaining the obvious to you. No point.
“Energy does not need a temperature difference to flow. The green plates just radiate to space based on their temperature and emissivity. Energy is flowing out from the plates regardless of temperature difference.”
Why do you insist on making up your own facts? Show me where this is from, other than your own imagination.
L As I said you cannot avoid a 2LOT problem.
‘You seem to think you should solve the heat transfer equation between the plates as though there is still heat flowing at equilibrium. You think the result should be a value other than zero! That’s what we are getting at by, “you are conserving heat”. Zoe already explained that to you.’
I don’t think. I know. It is simply a law of physics. A real one that I showed you.
If energy is flowing from the blue plate to the green plate, then the Blue plate MUST BE HOTTER. Other wise there is no reason for it to choose to flow from BLUE to GREEN (2LOT!).
Just because you don’t feel like calling that energy flow HEAT (it is) does not make any thermodynamic difference, it is still NOT allowed to flow, in equilibrium, by the second law.
Energy can do work. If you can get work without heat flow from hot to cold, that is a perpetual motion machine.
Nobody on my side has suggested that we can ‘conserve heat’. That is a strawman and just plain weird!
‘Space is not ‘surroundings’, is it!? Its the absence of ‘surroundings’. Its ‘space’.
Again this is your feeling, not a fact. Show me a fact from somewhere.
A star radiates its energy into space. It doesnt matter if it hits anything or not.
If it hits a far away planet fine. The only difference is the planet will send back a bit of radiation to the sun according to its temperature. And you will call the NET energy transfer heat.
Space has 3 Kelvin BB radiation in it. It was emitted long ago by hot gases. Matter.
This BB radiation is what is returned to the plates which are also emitting radiation (SB). The difference, NET energy emission is heat flow.
It is no different than if the plates were emitting their heat to an asteroid @ 3 Kelvin.
Sure, Nate. There is a law of physics that says that there must be heat flow at equilibrium, and space isn’t “space”. You win. Well done, great debate as always. Worthwhile.
‘There is a law of physics that says that there must be heat flow at equilibrium, and space isnt ‘space'”
BS.
I never claimed plates are in equilibrium. You did. Based on what? Your feeling?
As I mentioned several times, but you IGNORED, space has matter in it and that matter has radiated.
Your feelings about ‘space’, equilibrium and heat are not facts, not science.
When you make claims, you need to show where they comes from.
If you can’t do that, why should anyone believe it?
Yes Nate. You don’t think the plates are at equilibrium at 244 K…244 K…244 K. You think they are at equilibrium at 244 K…290 K…244 K. Because you think there is a law of physics that says there must be heat flow at equilibrium, and space isn’t “space”. Plus, if I don’t address every single dumb thing you bring up, and connect all the dots for you, and do all your thinking for you, that means those points are “unchallenged” and therefore, you’re correct.
Now you can say something beginning with, “well, DREMT…” and write an insulting false summary of the discussion where you declare yourself the victor, claim you’re an expert at heat transfer physics and I’m “mansplaining” this and that to you, and that “once again, DREMT…” does this and that and the other, and blah blah blah. Off you go.
‘write an insulting false summary of the discussion’
No need, you took care of that, asshole.
‘You think they are at equilibrium at 244 K…290 K…244 K.”
BS, I never said they are at equilibrium!
“Because you think there is a law of physics that says there must be heat flow at equilibrium”
BS, never said that, liar.
” that means those points are ‘unchallenged'”
Unsupported feelings and straight up lies are no challenge to facts and physics.
Again, show us where any of these ideas come from.
Can’t do that? Then they are just your made-up BS.
‘Now you can say something yada yada”
Already been said. Come back when you when you’ve got actual facts.
Oh sorry, “steady state”, not “equilibrium”.
☺️
Bye Nate. Always a pleasure.
Ball4,
‘Things only cool by net loss of thermodynamic internal energy!’
Heat Loss is commonly used in engineering to refer to an object cooling-transferring heat, and yes losing internal energy.
https://www.engineeringtoolbox.com/heat-loss-pipes-tanks-t_15.html
Q is the rate of change of thermodynamic internal energy of system of interest as a consequence of a temperature difference between it and its surroundings.’
Yes and Q is also rate of Heat Transfer. Which is also the name of many courses and textbooks.
http://web.mit.edu/lienhard/www/ahtt.html
bdg…”Of course you can average a flux.”
Makes no sense. Newton derived the word flux as a fluxion, an instantaneous rate of change of a field. You can integrate that derivative to find a sum over an area, but I have no idea how you’d average instantaneous values in a field.
In the electrical field, when a conductor turns in a magnetic field, as a generator, the instantaneous voltages generated form a sine wave. You can average those values over half a cycle so the average voltage is 0.636 Vpeak. Or calculate the root means square average as 0.707 Vpeak.
It’s not the same with the solar flux at TOA. It’s already a constant, not an average. The constant is derived by dividing the surface area of a sphere with radius = 1 AU into the total power generated by the Sun at its surface.
That is a constant, it does not vary nor can it be averaged.
If you want to average the effect of that TOA power on the surface I’m afraid you’re in for a seriously complex calculation. Dividing 1360 W/m^2 by 4 does not do it.
If you want to see basically the complexity of this issue, read this explanation by Gerlich and Tscheuschner.
Start on page 58 under
“3.7 The assumption of radiative balance” and read from there.
The actual explanation of the average begins on page 62 under: “3.7.4 The average temperature of a radiation-exposed globe”
https://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf
Page 62 is “For a radiation exposed static globe” Gordon. Nonrotating. Maybe you can find G&T on a rotating globe. Have at it.
Page 58 “Climatologic radiation balance diagrams are nonsense, since they cannot represent radiation intensities, the most natural interpretation of the arrows depicted in Figure 23”
Here G&T expose their misunderstanding of “Climatologic radiation balance diagrams” by writing the SI base unit “intensity” incorrectly as you sometimes do. No such diagram I’ve seen makes their mistake, but I’m not claiming to have seen them all.
GR said…”It’s not the same with the solar flux at TOA. It’s already a constant, not an average. The constant is derived by dividing the surface area of a sphere with radius = 1 AU into the total power generated by the Sun at its surface.”
The solar constant is the average zenith flux at TOA of Earth specifically; not Mercury, not Mars, or any other planetary body. It is Earth’s solar constant. Mercury, Mars, etc. will have their own. The reason this is specific to Earth is because Earth intersects the solar radiation sphere at different distances. This is because Earth’s orbit is elliptical. This is why the solar constant is an average over one orbital period.
GR said…”That is a constant, it does not vary nor can it be averaged.”
Well, actually, it’s not even constant which makes “solar constant” a misnomer in the strictest sense. The biggest reason is because solar output itself is not constant.
GR said…”The actual explanation of the average begins on page 62 under: “3.7.4 The average temperature of a radiation-exposed globe””
Well, at least he gets the divide by 4 right. But what is up with the shenanigans going on in the derivation of eq. 82?
“But what is up with the shenanigans going on in the derivation of eq. 82?”
In eqn. 82, they are showing their result for their “static” i.e. nonrotating globe. Commonly for a non-rotating globe people just divide by two (instead of 4) averaging the 2 hemisphere T averages while forgetting that radiation should be averaged then converting that avg. to temperature. As Dr. Spencer points out, this is a big deal for the moon T extremes but is negligible for Earth GHCN. You had better check with a real planetary thermodynamics ref. though about their 2/5 factor instead of 2.5/5, these guys may not be correct.
Ball4, bdgwx, please stop trolling.
bdgwx
Just to be clearIm not claiming that 1360 W/m^2 over 510e12 m would result in the same temperature as 340 W/m^2 over 510e12 m.
No this quiet obvious. But this is the crux.
By claiming that the power that hits Earth is less you require some magical energy source to get the temperatures right.
What do you get when you calculate temperature with the actual power hitting Earth?
You might actually want to consider the curvature to get another average.
You might want to throw in that this is only valid for the day side on Earth.
And then you might want to find out what needs changing in the arguments and descriptions of the energy balance of Earth to understand temperatures a bit better.
Using power as energy is not wrong if you have the impact area correct.
norman…”Another Joe
You can easily do your own testing to see that Roy Spencer is correct and Joe Postma is a clueless cult leader of a bunch of people that have very little knowledge of heat transfer physics so he can easily manipulate their limited understanding and seem like a genius to them. He is just wrong”.
*****
A clueless cult leader with a degree in astrophysics. Who was the other cult leader you panned? Ah, yes, Claes Johnson, with a degree in mathematics.
You seem to have a serious hatred for people with degrees who think the 2nd law should be applied as stated by Clausius.
If you disagree with Postma or Johnson, could you try laying out scientific arguments along with your hysterical ad homs?
Gordon Robertson
I have done what you have requested many times. I don’t know how many times I have linked you to valid science articles. You know what you say when I do this? I do. You say the textbooks are wrong because they were when you went to school. I really doubt you have any college physics background. You make up so much bad stuff that is totally unsupported by anything.
Joseph Postma is an astrophysics person but he knows nothing about heat transfer and he will not debate any person with knowledge on the subject.
Claes Johnson knows less about physics than Joseph Postma. He just makes stuff up and works out some phony math to support his terrible ideas. He has zero credibility with physics. I can make up stuff like you do but then I would have zero credibility like you have. I prefer to support all my points with valid physics. One day you may do the same but it is most unlikely. You enjoy making up junk physics to much to actually learn the real material. You still think Mid-IR is generated by electron transitions. No one can tell you different. Not all the Chemists in the world that analyze compounds using molecular vibrations (not electron transitions) to identify unknown material. You are so far gone in pseudoscience you can’t recognize how far off both Joseph Postma and Claes Johnson are. Keep believing you understand physics. I already know it is a waste of time trying to correct your vast false knowledge. You will just continue to make up whatever you want and will refuse to accept even experimental evidence demonstrating you are wrong.
Norman, you are so uneducated you probably do not realize the trouble you can get into. You are insulting professionals, falsely. Maybe you believe this is fun, but someone may take serious offense. You should look up the word “defamation”, and the legal consequences thereof.
Of course you do not have enough funds to interest a suing attorney, but you also need to look up the phrase “deep pockets”. By using MidAmerica Energy time and resources, an attorney would claim that MidAmerica should have done something to stop you. They would likely settle out-of-court, to avoid the publicity. I mean, how terrible would it look if some employee of an energy company used company time and resources to constantly and repeatedly attack individuals.
Just because you have a dead-end job there in the Omaha/Council Bluffs area does not mean you get to waste so much time. Likely there are MidAmerica customers that would wonder how many other employees are wasting time and money like you.
https://www.midamericanenergy.com
Even though you are immature, you should try to think about the consequences. You’re like a 4-year-old, playing with matches.
Hope that helps.
JDHuffman
So states another ignorant poster. You are the blog troll and will be one forever it seems.
Do you have anything of value to contribute or you just trolling?
You are a primary reason people would not reveal who they are on the Internet. When you have no science, no logic and only endless stupid trolling you attempt doxxing because you have nothing else left.
You are truly a negative human. Too bad you troll this blog.
Maybe if you spend so much time researching me, you could possibly research some physics textbooks and have some valuable contribution on this blog instead of your junk physics.
Here maybe this will help.
https://drive.google.com/file/d/1Vl02ky5h40xDygiCIvDHFrj5c9hYvcNN/view
Here is the definition of defamation.
https://www.nolo.com/legal-encyclopedia/defamation-law-made-simple-29718.html
I think textbooks would clear me of your accusations.
I have debated with Joseph Postma on his own blog. I did offer him textbook data. He incorrectly tried to explain the heat transfer equation, I corrected him. I think he banned me shortly after that.
JDHuffman
Here is the interaction I had with Joseph Postma on his blog:
“Norman says:
2019/02/02 at 9:23 PM
Joseph Postma
I know why you disabled the comments on your videos.
You make a totally incorrect point that is false. You say the only way to increase the temperature of an object (from your bad physics video, the long one) is to add more heat or do work.
Wrong!
[JP: Thats the first law of thermodynamics, all things being equal. RIGHT!!]
If you add the same energy flow to the object but reduce the outward loss the temperature of the object increases.
[JP: Greenhouse gases do not reduce the emissivity of the emitting surface! Greenhouse gases do not change the chemical and electronic structure of the ground or water surface such that said surfaces reduce their emissivity.]
You seem to lack understanding of heat transfer when you have a continuous energy input.
[JP: Greenhouse gases are supposed to have higher emissivity than the other gases, and thus, with continuous energy input, a system with a chemical makeup which increases in emissivity will become cooler! Soz!]
It also seems you will not be able to ever correct your own internal false misunderstandings.
[JP: Cute. Cute little goblin troll. Still looking for a soul? Sorry, but youre not getting ours any longer, freak.]
You wont listen to me or any other poster who points out you many flaws.
[JP: Thats because youre an idiot; a soulless, mindless, programmed, machine idiot, whose every sentence is rebutted and whose experimental results debunks itselfbut being a machine, cant understand any of that anyway.]
You have you small group of devoted disciples to honor you.
[JP: Theyre not my disciples and honoring me has nothing to do with it. Ah, but I seethose are the terms a satanic little goblin would think inarent they. LOL Stupid goblin. God hates you.]
None of them know any physics and are easy to fool.
[JP: They have been free to weigh the debate and the evidence. The difference between you and them, is that they have souls, and you are a machine, a goblin, that God hates.]
I have not seen any of your long term posters show any signs that they can rationally understand dynamic heat flow.
[JP: We have not seen you be able to scientifically set up an experiment, or conclude one. Or refer to the laws of thermodynamics. Or to demonstrate any equation from any textbooks you might reference for purpose. Or understand what the debate is about. Or understand the difference between flux and Watts. You are quite a stupid little goblin troll, a machine that God hates. Man that must suck being hated by God.]
Dude, the mechanism of how the cooler atmosphere can cause the Earths surface to reach a higher steady state temperature is by changing the heat flow of the surface.
[JP: Heat is not a conserved quantity. Gases do not stop the surface from emitting from itself. LOST!]
Heat is defined as the NET energy flow between objects.
[JP: HENCE is not a conserved quantity.]
If you have a hot object and bring a less hot but warm object near the hot one, it will lose less HEAT.
[JP: You just stated that heat isnt a conserved quantity. Or if you dont understand that, then note that it was repeated again. Heat is not something even really lost. Thermal energy is lost. That energy can act as heat for something else if the conditions at the surface of the something else allow for it.]
It will radiate the same but it will not gain energy from the other object. This means it loses less heat than before. If the hot object has some continuous heat supply to it, it will get warmer with the other object present than if not there.
[JP: No, it will just warm the other object. Heat flow goes to ZERO, spontaneously, as energy frequency state information is shared between ojects until they equilibrate. Its automatic. Heat is not the conserved quantity, and the equations of heat flow can only be solved if heat is not conserved, and ends at zero. The entire definition (and empirical manifest existence) of thermal equilibrium would be impossible as we know it if heat was a conserved quantity.]
You really do not know anything about heat transfer. I suggest you read some actual textbooks on it or stay in your own field of expertise.
[JP: Yes, I reference, what was it, 5 or 6 textbooks on radiant heat flow in my book and in my papers and demonstrate their usage, etc.
I suggest you allow your battery to run down to zero. God will still hate you though, even once youre drained.]”
I don’t think I have ever insulted people this intently. I call you what you are, a troll. That is NOT an insult it is a factual statement. I have even pointed out your many troll tactics you use. When you can’t counter a point you copy and paste a previous post. When someone posts a link contradicting your point, you claim they do not understand the link (but you never attempt to explain what they do not understand).
You are a exposed angry troll and are going to levels even strange for the typical troll. Learn real physics and stop trolling.
Norman…”Heat is defined as the NET energy flow between objects”.
Now JP knows what we know, that you are a raving looney. Heat has never been defined as a net energy flow. Clausius defined it as the average kinetic energy of atoms. At no time did Clausius ever refer to heat transfer as a net.
“Clausius defined (heat) as the average kinetic energy of atoms.”
No Gordon, that’s Clausius’ def. of temperature, Clausius def. of heat is the TOTAL kinetic energy of the atoms/molecules in an object.
ball3…”thats Clausius def. of temperature, Clausius def. of heat is the TOTAL kinetic energy of the atoms/molecules in an object”.
You’re confused between reality and fantasy. Heat is a real phenomenon, an energy. Temperatures is a human invention designed to measure relative levels of heat. Since heat is the kinetic energy of atoms, and temperatures measures the average KE of atoms, temperature measures heat.
In order to develop the concept of temperature, humans needed set points. They found them in the freezing point and boiling point of water. In the Celsius scale, the FP was set at 0C and the BP at 100C. Later, the Celsius scale was extrapolated to 0C by Kelvin, to get the Kelvin scale.
In order to raise the temperature from 0C toward 100C, heat as energy must be added.
Actually is all began in the 18th century when Count Rumford noted that turning a dull drill bit in water heated the water. The large drill bit, used for drilling cannons, was turned by horses. Later, Watt developed the notion of horsepower as 1 HP being 33,000 foot-pounds force/minute of work done by a horse.
Later still, the scientist Joule noted that a little wheel turned in water caused the water to warm. He calculated from that the relationship between work and heat.
“Heat is a real phenomenon”
No, never been found to exist in reality. Temperature measures avg. KE.
Energy must be added to increase water avg. KE (temperature) & total KE (U), heat is only a measure of the total KE added, temperature is a measure of the avg. KE increase.
Gordon Robertson
Here:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node137.html
Here:
http://physics.bu.edu/~duffy/py105/Heattransfer.html
Both of these discuss NET transfer of energy.
http://www.mie.uth.gr/labs/ltte/grk/pubs/ahtt.pdf
Read page 495 of this textbook on heat transfer.
There are more that say this very same thing.
Gordon Robertson
Here is another:
https://www.sciencedirect.com/topics/engineering/radiation-heat-transfer
“The net rate heat transfer by thermal radiation is then given by:
(3.2)q=σA(T14−T24)”
And still one more:
https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html
Norman, you found some more links you can’t understand.
Nothing new.
Norman Grinvalds, you reveal yourself. There’s nothing to you. You just insult, falsely accuse, and misrepresent.
BTW, you still haven’t identified even one time my physics has been wrong. You just make stuff up.
Nothing new.
JDHuffman
I actually have identified your physics wrong. The cartoon of heat transfer you post is wrong. Experimental evidence from E. Swanson verifies it is incorrect. Textbook physics verifies it is incorrect.
Norman, you have IMAGINED my physics is wrong. But your imagination is not reality. You can’t scientifically prove anything I’ve said is incorrect. Everything I’ve said is from established physics. You don’t have any technical background so you must rely on internet links you can’t understand.
You keep clinging to the bogus experiment by Swanson, which even he no longer calls an “experiment”. And, which neither of you can state what you believe it verifies, let alone supply the relevant data. Yet you dismiss the simple correct solution, which clearly indicates all temperatures and fluxes.
You have no regard for truth and honesty. You continually misrepresent and falsely accuse.
Nothing new.
norman…”You say the textbooks are wrong because they were when you went to school”.
I have never inferred that all textbooks in thermodynamics are all wrong. I have referred only to their coverage of radiative transfer where they take liberties with the Stefan-Boltzmann equation, inferring that EM can be transferred both ways between bodies of different temperatures.
Actually, the texts are mysterious about it all. They seldom give temperatures and show only arrow indicating EM flows between bodies. I have yet to see such an example with units and a solution. The only place you ever see radiative transfer included in a problem in a thermodynamics textbook is in relation to heat conduction and/or convection from a hotter body to a cooler body.
I have never seen an example of two way radiative EM transfer, involving temperatures, with units and temperatures. I have NEVER seen an example in any thermodynamics textbooks of a net heat transfer between bodies of different temperatures.
Never? Just behind in your studies: Planck 1912 p9: A body A at 100degrees C emits toward a body B at 0degrees C exactly the same amount of radiation as toward an equally large and similarly situated body B’ at 1000degrees C. The fact that the body A is cooled by B and heated by B’ is due entirely to the fact that B is a weaker, B’ a stronger emitter than A.
bally…”Planck 1912 p9: A body A at 100degrees C emits toward a body B at 0degrees C exactly the same amount of radiation as toward an equally large and similarly situated body B at 1000degrees C. The fact that the body A is cooled by B and heated by B is due entirely to the fact that B is a weaker, B a stronger emitter than A”.
Nothing there with which I disagree. Planck is obviously talking about radiation TOWARD bodies, he is not inferring that radiation from a cooler body can warm a warmer body.
The only thing I find vague in his analysis is that body B ‘cools’ body A. He does not specify enough information to make that claim. Furthermore, he made the statement in 1912 just before Bohr produced his theory relating EM to heat via electron action.
If body A at 100C is in a room at 20C, the air in the room will aid the cooling. If the body B is a significant distance from A I cannot see how it can produce cooling in A. It would have to be able to cool the air, which would cool A.
“he is not inferring that radiation from a cooler body can warm a warmer body.”
Planck is doing so Gordon. Planck is inferring that radiation from a cooler body can warm a warmer body when that cooler body replaces an even cooler body. Dr. Spencer has run that test a couple of times, in the lab and outside on the real atm., to demonstrate that is what Planck inferred.
Yes fluffball, you can raise the system temperature by bringing in a warmer body.
You accidentally got something right!
That’s something new.
Sorry, JD I replaced a much colder body with a colder body to warm a warmer body, all consistent with 2LOT since dQ/T is positive so entropy increases just like the experiments Dr. Spencer put together. You should try experimenting, it’s way more fun than drawing bogus cartoons that more learned commenters laugh at but that mo.o. is just great entertainment, more please.
Sorry fluffball, you replaced a colder cold-body with a warmer cold-body. That’s called “adding energy to the system”.
If you’re still confused, draw yourself a simple diagram….
Ok, then JD now you get how the GHE works, how the GPE experiment works et. al. since as you write I replaced a colder cold-body (space) with a warmer cold-body (the atm., green plate) in view of the warmer warm surface body (L&O, blue plate) increasing blue plate T. That’s called “adding energy to the system” warming the planet surface above 255K to 288K for the global mean planet near surface T.
You will want to correct your bogus cartoons now that you understand.
Now space is a “body”!
“You couldn’t make it up”…
“Now space is a “body”!”
Yeah, a pretty big one. Outer space (great term) is an object too actually. Contains a lot of stuff that absorbs and emits, reflects/scatters, so they say.
“Now space is a “body”!”
Yeah, a fairly big one. Outer space (great term) is an object too actually. Contains a lot of stuff that abs_orbs and emits, reflects/scatters, so they say.
OK, Ball4.
Sorry fluffball, but energy already in the system does not count as new energy.
That’s why you need a simple diagram, so you can see how silly your ideas are.
Nothing new.
JD tries to change his story: “That’s called “adding energy to the system”.” Won’t work but very funny. More humor please.
“Contains a lot of stuff…”
Yes, “stuff” can be found in space. The funny thing is that the “stuff” isn’t “space”. Space is what is “inbetween” the “stuff”. That’s why it’s “space”.
See why you need a simple diagram, fluffball? You can’t figure it out otherwise.
Look how easy-to-understand this one is–all energy is accounted for, all fluxes and temperatures are clearly indicated.
https://postimg.cc/KKx5hx4H
ball3…”Planck is inferring that radiation from a cooler body can warm a warmer body when that cooler body replaces an even cooler body”.
All he said was:
Ta = 100C > Tb = 0C….B cools A
Tb’ = 1000C > Ta = 100C….B’ heat A
B emits same amount of radiation to both B and B’.
Absolutely nothing about B warming A or A warming B’.
sorry…type…
B emits same amount of radiation to both B and B.
should read:
A emits same amount of radiation to both B and B.
Gordon, read the entire post and the next one, and many comments to see why JPs science is wrong.
nate…”Gordon, read the entire post and the next one, and many comments to see why JPs science is wrong”.
I visited JP’s site for the first time to see what he is about. He admits in his ‘About’ section that he used to believe in the anthropogenic theory. After reading on it extensively, he changed his mind.
I never did believe in AGW, I knew nothing about it till I read the iconic statement from the IPCC that it is 90% likely humans have caused global warming.
I was bothered by them applying a confidence level to an opinion and began researching that. One of my first hits on Google was by Lindzen who claimed the majority of reviewers had not felt that way. They thought the research was incomplete and that more time was required before jumping to conclusions.
Lindzen, who has been associated with the IPCC proceedings went on to explain that the 90% statement came from 50 politically-appointed lead authors. He questioned the IPCC practice of issuing the Summary for Policymakers, written by the 50, before the main report by the 2500 reviewers who wanted to wait and see. The main report was then amended to reflect the views in the Summary.
That got me curious and I began to research further. The more I read, the more I encountered consensus and plain, bad physics. I have yet to find a valid explanation for AGW or the GHE. All explanations I have seen to date depend on presumptions that are clearly incorrect.
What really annoyed was the penchant of alarmists to discredit bona fide skeptical scientists just because they disagreed. The more I read the skeptics, the more I saw compelling arguments based on physics.
Dr. Lindzen agrees that carbon dioxide is a greenhouse gas, calling people who dispute that point “nutty.”
“Anyone who challenges the paradigms is immediately dismissed as a nutjob.”
norman…”Claes Johnson knows less about physics than Joseph Postma. He just makes stuff up and works out some phony math to support his terrible ideas”.
I just wish I understood math an nth as well as Claes Johnson. I find his physics to be pretty sound as well.
Have not had the opportunity to view anything from Joe. Roy’s link to him was the first I’d seen. I expected a raving lunatic but I thought he presented his case well with clamness.
I have heard he tends to cuss but I have worked in environments where every second word is a cuss word. Joe admits to having a farm background and he is from Alberta, where fools are not highly regarded.
I am more interested in what the guy says about science. I have a thick skin and don’t take cuss words seriously.
I did like his comment about greenhouses, that we build them to do what they atmosphere cannot do.
norman…” You will argue endlessly about your false opinions and your total incorrect view of heat transfer and the 2nd Law of Thermodynamics. Nothing will change this”.
No false opinion Norman, the 2nd law states:
Heat can NEVER (repeat NEVER) be transferred by its own means from a colder body to a warmer body.
Similar statements might be:
-water can NEVER by it’s own means flow uphill.
-a mass can NEVER by its own means raise itself onto a cliff.
-electrical current can NEVER by its own means flow from a lower potential to a higher potential.
In fact, no energy, by it’s own means, can be transferred from a lower potential to a higher potential.
Never means never, Norman, unless you are a desperate alarmist with the environmental religion disease.
There is nothing in the 2nd law or in it’s mathematical equivalent, the entropy equation, to indicate that heat can be transferred as a net energy between bodies of different temperatures. There is nothing to create a net when heat ALWAYS transfers, by it’s own means, from cold to hot.
No Gordon, heat being a measure of the kinetic energy of atoms/molecules means it CAN (repeat CAN) be transferred by its own means from a colder body to a warmer body. Just read Maxwell-Boltzmann distribution of particle velocites to find out why.
ball3…”heat being a measure of the kinetic energy of atoms/molecules means it CAN (repeat CAN) be transferred by its own means from a colder body to a warmer body”.
That’s ironic, Clausius stated clearly that heat can NEVER be transferred by its own means from cold to hot.
Maxwell’s and Boltzmann’s later work on distribution of particle velocities improved on Clausius earlier work. It happens. Science progresses.
ball3…”Maxwells and Boltzmanns later work on distribution of particle velocities improved on Clausius earlier work”.
Boltzmann tried to prove the 2nd law statistically and failed. He became so distraught over the failure of his work in that area that he took his own life.
Maxwell was involved with Boltzmann only in the average velocity of atoms in a gas.
To this date, anyone with a smattering of common sense knows intuitively that heat can never be transferred by its own means from cold to hot.
That’s right, heat can’t, energy can.
Fine, so long as you’re aware it won’t necessarily be absorbed.
I am, and that factor is stable with temperature until it melts.
OK, Svante.
Gordon Robertson
YOU: “Heat can NEVER (repeat NEVER) be transferred by its own means from a colder body to a warmer body.”
This is correct but it is not what I am stating.
Again from Clausius himself (he is NOT talking about refrigeration here!!): “The principle assumed by the author as the ground of
the second main principle, viz. that heat cannot of itself, or
without compensation, pass from a colder to a hotter body,
corresponds to everyday experience in certain very simple
cases of the exchange of heat. To this class belongs the
conduction of heat, which always takes place in such a way
that heat passes from hotter bodies or parts of bodies to
colder bodies or parts of bodies. Again as regards the ordi-
nary radiation of heat, it is of course well known that not only
do hot bodies radiate to cold, but also cold bodies conversely
to hot ; nevertheless the general result of this simultaneous
double exchange of heat always consists, as is established by
experience, in an increase of the heat in the colder body
at the expense of the hotter. ”
That is Clausius. The hot body radiates to the cold and the cold to the hot. There is a simultaneous double exhange!! His words.
Heat transfers only from HOT to COLD. Energy transfers both ways. A hot object emits more energy to the cold than the cold to the hot. The net (heat flow) is from hot to cold. The warmer the cold the more energy it sends to the hot. The less heat flows from the hot to the cold.
So now you have to argue against Clausius and textbook physics and the E. Swanson experiment.
Norman, Clausius and Gordon get it right, but you STILL don’t get it.
Just because a cold body radiates to a hot body does not mean that the hot body will absorb everything. All photons are not always absorbed. For absorp.tion to occur, the wavelengths must match. Photons with wavelengths that do not match will be reflected.
norman…”That is Clausius. The hot body radiates to the cold and the cold to the hot. There is a simultaneous double exhange!! His words”.
In your quote from Clausius, he is explaining what he meant by ‘by its own means’. He goes on to explain the compensation that is required to allow heat to be transferred from a colder space to a hotter space, as in a refrigerator.
With compensation, heat can flow both ways because external power is used to drive a compressor that changes the state of a low pressure gas to a high pressure liquid. As heat is withdrawn from a colder space it is compressed to a high pressure liquid from which that heat can be vented to the atmosphere. The HP liquid is then converted back to a low pressure gas that can absorb more heat from the cold space.
This is a description of a manipulation of heat through the change of state of a gas using the Ideal Gas Law. It does NOT describe a net transfer of heat under natural means.
Clausius described that process in terms of a heat engine. To understand what he meant you’d have to read elsewhere where he describes the relationship between pressure, temperature, and volume. You just happened to cherry pick him talking about the process in a heat engine and what would be required to reverse it.
When Clausius wrote that comment he surely presumed the reader had read his in-depth description of the relationship between P, V, and T in a heat engine.
Gordon Robertson
No he is not doing what you attribute to him at all.
I will give you the link to his own works. He goes off to talk about concentration of light and blackbodies. You are just making up stuff not said by the man at all.
Chapter 12 of the link. Read through and post where he makes the claim you do.
https://archive.org/details/mechanicaltheor03claugoog/page/n317
Norman, that’s not the way to do it.
Clearly identify the exact passage you are concerned about, and maybe we can help you.
Tasking Gordon to find his own exact wording in a link you can’t understand just indicates your extreme desperation.
The link you found is a translation of Clausius’ work. If you believe you have found something that supports your pseudoscience, then you’re just wrong again.
Gordon Robertson
I will try again with you. Not expecting much success.
This quote comes from a College Textbook on heat transfer.
Here is a link to the textbook:
https://drive.google.com/file/d/1Vl02ky5h40xDygiCIvDHFrj5c9hYvcNN/view
“In heat transfer by radiation, energy is not only transported from hot to cold
bodies; the colder body also emits radiation that strikes the warmer body and
can be absorbed there. An exchange of energy takes place, in contrast to the
transfer that occurs in heat conduction and convection. This radiative exchange
depends on the mutual position and orientation of the radiating surfaces, their
temperatures and there radiative properties.”
And then this: “It is important to note that when it is stated that energy will not spontaneously flow from a cold object to a hot object, that statement is referring to net transfer of energy. Energy can transfer from the cold object to the hot object either by transfer of energetic particles or electromagnetic radiation, but the net transfer will be from the hot object to the cold object in any spontaneous process. Work is required to transfer net energy to the hot object.”
They are not my ideas or concepts. They are established science used in everyday heat transfer work.
Your points about other systems do not compare to EMR. Your analogies do not work.
With EMR there is NOT a potential difference between two plates at different temperatures that generates EMR. The rate of EMR emission is based upon the temperature of each surface. The hot surface emits more EMR than a colder one so the net energy flow is from hot to cold. The energy flows in both directions. More flows from hot to cold so you have a heat transfer from hot to cold.
All that is correct, Norman. Just remember to always leave out your pseudoscience that photons from a colder object will always be absorbed by a hotter object.
norman…”And then this: It is important to note that when it is stated that energy will not spontaneously flow from a cold object to a hot object, that statement is referring to net transfer of energy”.
This is an example where your textbook is egregiously wrong.
It is never stated that ‘ENERGY’ will not spontaneously flow from a cold object to a hot object, it is only stated that ‘HEAT’ will not do that. That’s the 2nd law.
Your author obviously does not understand what Clausius wrote about heat and the 2nd law. He at no time referred to a net transfer of heat, he only claimed that heat can ‘NEVER’ be transferred by its own means from a cold object to a hot object.
Furthermore, Clausius stated it mathematically as entropy and he defined entropy so that heat can never flow by its own means from cold to hot.
Your author is just plain wrong and you are so naive you are willing to believe some hack who has never taken the time to find out for himself what Clausius actually said. The author obviously swallowed what he was told by someone else who did not take the time to understand the 2nd law.
And I’ll bet something else. I’ll bet you cannot find one problem or example in the book, with units, where heat can be transferred as a net transfer.
It cannot happen. Quantum physics has the explanation as to why the electrons that transfer heat through solids and liquids and which convert heat to EM cannot spontaneously allow heat to transfer in both directions at the same time.
G’Your author is just plain wrong and you are so naive you are willing to believe some hack ‘
So we should not be so naive to believe a textbook, but we should believe some guy saying crazy stuff on a blog!
Hmmmm, Im going to ponder that..
nate…”So we should not be so naive to believe a textbook, but we should believe some guy saying crazy stuff on a blog!”
Your naivete comes in your belief that everything written in a textbook is true. In electrical engineering textbooks throughout the world it is blatantly claimed that electrical current flows from positive to negative. They justify that as a convention on one hand then claim is doesn’t matter which way current flows as long as you are consistent with the way you sum voltage drops.
It’s a blatant lie no matter what fancy theory is offered in its defense.
It matters a heck of a lot which way current flows in real life. Otherwise you’d have lightning leaving the positively charged Earth and traveling upward into the negatively charged clouds. Rather than an electron beam being generated in the neck of an older TV picture tube and guided to the screen where it is focused as a dot of light on phosphors, you’d have positive charges flowing from the screen back to ‘who knows where’.
It’s a lie, a blatant lie, and it is in every electrical engineering text I have encountered. Paradigms are a curse on science, and the AGW and GHE paradigms are preventing good science being done in climate science. Anyone who challenges the paradigms is immediately dismissed as a nutjob.
It has been known since at least 1925 that electrical current is made up of electrons and that electrons MUST FLOW negative to positive. However, some rocket scientist once inferred the ‘positive test charge’ which he claimed had mass and moved positive to negative.
Later, Shockley developed hole theory for semiconductors. As a model, and only a model, he visualized electrons leaving their position in a valence band and leaving a hole behind. Naturally another electron filled that hole, and the hole moved in the opposite direction. Shockley inferred a current of holes, or a hole flow, but he made it abundantly clear that he was referring to a model and not to actuality.
Skockley intended the model only for semiconductors doped with P-type silicon, where holes outnumbered the electrons. In other words, atoms are doped so they have an excess of spaces in their valence bands compared to electrons. Electrons are said to be the minority carriers even though they are they ONLY CARRIERS.
Holes do not carry current. If I dig a hole in the ground then another one in a line, and I fill in each hole in order with the soil from the previous hole, that first hole seems to have relocated. Is the hole doing anything…no it’s not. It’s the same in semiconductors.
That positive to negative theory is absurd but it has stuck as a paradigm for more than a century. Emgineering faculties throughout the world still cling to that age-old, incorrect paradigm.
I have questioned an author in a textbook claiming heat can be transferred both ways between bodies of different temperature. That is a blatant lie as well and the 2nd law makes it clear it is a lie. Quantum theory makes it clear that it’s a lie.
Heat can only be transferred from a higher energy state to a lower energy state. It is not possible for heat to be transferred BY ITS OWN MEANS from a lower energy state to a higher energy state.
You should know that.
Authors are not experts in everything. If they have to add a section in a book that is not their specialty, they will consult, or look it up. Obviously the author in Norman’s book got the wrong information.
Nate
I agree with your comment. You have a pretend expert going by JDHuffman (he can never prove his expertise but we are supposed to accept it on faith) and then you have Gordon Robertson. Knows more than all scientist combined. Only his understanding of physics is correct, everyone else is wrong. He even makes up what Clausius says about transfer from cold to hot. Clausius clearly states there is a double exchange of (he calls heat). I copy the translated quote, send the link to the translated book and tell him the Chapter it is in. He claims Clausius but rejects what Clausius says. He makes up unsupported opinions and he thinks all the major science theories are wrong (although he has no explanation for all the things these theories explain).
When he can’t figure out things, like time dilation (even though it has been experimentally verified many times by many people) he goes into abstract philosophy about time being an illusion created by man.
‘Your naivete comes in your belief that everything written in a textbook is true.’
C’mon Gordon. What is written in the textbook also agrees with other textbooks, with what I have learned in courses, and used for decades in my work.
On the other hand, if I were to believe a guy on a blog telling me to ignore textbooks, that would be naive and quite silly.
“Energy can transfer from the cold object to the hot object either by transfer of energetic particles or electromagnetic radiation, but the net transfer will be from the hot object to the cold object in any spontaneous process.”
JD “All that is correct, Norman.”
Wow, JD agrees with textbook physics! First time evah!
“Just remember to always leave out your pseudoscience that photons from a colder object will always be absorbed by a hotter object.”
Then, sadly, he reverts back to his twisted illogic.
“Not always” magically becomes “NONE” with his perfectly reflecting black body plates!
Nate, please stop trolling.
ball3…”Sorry, JD I replaced a much colder body with a colder body to warm a warmer body, all consistent with 2LOT since dQ/T is positive so entropy increases just like the experiments”
Here’s your pertinent entropy equation…
delta S = Q(1/T2 – 1/T1)
T1 represents the hotter body and T2 the colder body. In order for S to be positive, T1 > T2. If T1 = T2, entropy is zero.
BTW…dQ/T can never be positive. It is an infinitesimal change of heat at temperature T as defined by Clausius. The integral of all those changes must be positive.
nate…you suggested I read all the comments. While attempting to do so I encountered this one from Rosco:
“Using the values from NASAs Planetary Fact Sheets you can easily calculate the solar constant of 1361 W/m2 as listed on their site using the average orbit and the inverse square law.
To arrive at 340 W/m2 requires an orbit of twice the distance and Postma has repeatedly stated this there is no error on his part treating insolation as 340 W/m2 is simply incorrect in my humble opinion”.
Rosco is correct. The 1360 W/m^2 at TOA is dependent on the inverse square law. To arrive at a value of 340 W/m^2 OVER THAT SAME SQUARE METRE the Earth would have to be twice as far away in its orbit.
nate…you replied to Rosco…”The factor of 4 is for the average flux hitting the average square meter of the Earth.
If you want to calculate the suns distance, thats a different calculation altogether, that doesnt require the solar constant to be divided by 4″.
If you have a beam of solar energy that is 1 metre square at TOA why should it suddenly be divided by 4 to get the equivalent square metre on the surface? That beam should extend straight to the surface after going through the atmosphere.
The Sun’s distance is critical to the solar constant. The latter represents the attenuation of solar surface intensity after it has traveled 1 AU. Why should that intensity suddenly drop by a factor of 4 by the time it goes through the atmosphere?
On the other hand, if you want an intensity of 340 W/m^2, you’d have to move the Earth much further back.
The 4 factor makes no sense.
from Robert Kernodle…
“Even if we agree that output equals input, the input does not come in at the average, nor equally all over the sphere instantaneously. Thats the mistake in this division-by-four move.
The input comes in on an area that is 2pi(r)^2 and goes out on an area of 4pi(r)^2. The areas over which these two figures occur are different.
Even more, there is never a moment in time when all areas of the Earth are receiving an average flux to drive the many processes that determine temperature.
Temperature is not just a function of surface area and flux”.
It’s a serious mistake to presume the Earth’s radiative temperature is related to Stefan-Boltzmann. Local temperatures are determined by local phenomena, how they absorb and radiate energy and other factors such as weather and climate that affect the means of absorp-tion/emission.
Robert is correct, the divide by 4 method is seriously is error.
Kristian said…
“Its an average energy/heat budget, Robert. The Earth is a rotating sphere in space with the solar flux coming in from one side only. And so, to get the average input from the Sun (over the full globe, and over the full day/year), you have to divide the solar constant by four. Why is this so hard to grasp?”
********
Kristian…as I have said before, I support your views in general. In fact, I have referred people to your site.
The reason the 4 factor is so hard to grasp is that no one can explain it. I gather that it is related to the mathematics of a sphere.
Surface area of a circle is pi.r^2
Surface area of a hemisphere is 2pi.r^2
Surface area of a sphere is 4pi.r^2
In other words, if you take a hemisphere and squash it into a circle, it’s area would be twice the area of the shadow circle cast by the hemisphere.
What the heck does any of that have to do with the average surface temperature created by solar energy at 1360 W/m^2?
The solar constant is called a constant because it is the solar intensity found at a sphere with radius 1 AU. Everywhere on that sphere the intensity is presumed constant. If you divide the surface area of the sphere into square metres, the area as an intensity is 1360 W/m^2.
The question is, how are we going to apply that 1360 W/m^2? If we sum all the m^2 over a circle cast as a shadow of the Earth’s sphere we have a very large amount of watts of energy. However, if we select only one metre, we must project that square metre onto the surface, not divide it by 4.
So take that square metre as the surface of a cone originating at the solar surface at TOA. Why would you not extend that surface to the Earth’s surface, allowing for dissipation in the Earth’s atmosphere? Sure, the square meter would be slightly larger by the time it reached the surface but what the heck?
Why the heck would you divide by 4? The 4 represents the surface area of the entire sphere of the surface, half of which is not affected by solar radiation at any one time. That’s the point I think is being made by Joe Postma.
His other point is equally valid. If you have 1360 W/m^2 at TOA and 340 W/m^2 at the surface, you have effectively moved the Earth to at least twice the distance from the Sun it is at present.
I get your point about the rotating sphere and averages but that problem is far more complex than dividing by 4. I just think the 4 should work both ways and it does not. Losing 1/4 of the solar energy from TOA does not explain dissipation due to the atmosphere or angles of incidence.
The 4 seems nothing more than a number pulled from a hat.
Gordon: Here is an alternative explanation for the factor of 4. The surface of the Earth facing the sun is a hemisphere. The energy transferred to a surface by radiation depends on the cosine of the zenith angle. This is Lambert’s cosine law. When the radiation arrives perpendicular to the surface and the sun is straight overhead, the zenith angle is 0 deg. Only a small fraction of the area of the hemisphere has the sun straight overhead and that only happens at noon. Everywhere else, the cosine of the zenith angle is less than 1. Everywhere around the circle dividing darkness from light, the zenith angle is 90 degC and no energy is absorbed. When you integrate the cosine of the zenith angle over the surface of a hemisphere, you get a factor of 0.5. In other words, a hemisphere with surface area 2*Pi*r^2 absorbs the same amount of energy as a disk of Pi*r^2. The dark hemisphere of the planet has area of 2*Pi*r^2 and also absorbs nothing. So the AVERAGE over the whole 4*Pi*r^2 surface is 1/4 the flux intercepted by a disk of Pi*r^2.
Gordon,
All I am saying is that in science there are different formulas used for different problems.
One problem is to find out what distance the Earth orbits and to do that you use the measured solar constant of 1360 W/m^2 atr Earths orbit, and use the inverse square law.
No factor of 4 needed! If JP puts in a factor of 4 for some strange reason, that is just stupid.
The second problem is: with a spinning, spherical Earth, to find the AVERAGE energy hitting a square meter on its surface, each second, DOES require a factor of 4 (Some square meters getting 1360, some < 1360, some 0).
DIFFERENT problem, different answer.
As Roy showed in the next post, when you numerically calculate the flux hitting Earth, its AVERAGE IS 1360/4 W/m^2.
GR said…”Robert is correct, the divide by 4 method is seriously is error.”
Would you mind justify the statement above with the following link you posted two days ago?
https://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf
See if you can understand this, before you hit the backdoor.
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-358018
svante…in response to a comment from Robert Kernodle you said,
“Sorry, get a better explanation from a professor of physics:
https://tinyurl.com/y3pnvmmb”
Why do you have to bs the credentials of the author at the link. He is a paleobiologist, hardly a physicist.
Furthermore, he is trying to model the Sun/Earth interaction and getting it wrong.
That’s all you alarmists have, bs science and scientists.
Nate,
“Things only cool by losing heat! By definition.”
No, things only cool by heating something else. Cooling is heating.
Thot goes down, Tcold goes up, and Q goes to ZERO.
“Eventually that radiation will hit something, a planet, gas or dust, and be absorbed.”
LOL
“So all of the matter in the universe would fit into about 1 billion cubic light years, or a cube that’s approximately 1,000 light years on each side. That means that only about 0.0000000000000000000042 percent of the universe contains any matter. The universe is a pretty empty place!” – HowStuffWorks
I’ll translate that for you: ZERO chance
zoe…”That means that only about 0.0000000000000000000042 percent of the universe contains any matter. The universe is a pretty empty place!”
Through the brain damage in my mind I recall a few things from a course in astrophysics decades ago. There is a lot of interstellar hydrogen and dust out there although I don’t dispute your numbers.
However, the temperature of that interstellar matter is claimed to have a temperature of +4K, or 4C above absolute zero. Therefore, it seems not much is being heated by all the EM radiated in space.
That same +4K is claimed as partial proof of the Big Bang theory. Some diehard theoretical physicists claim it is residual heat left over from the alleged BB.
Funny how heat persists for billions of years in an essential vacuum, where the collision between atoms is essentially unlikely.
‘0.0000000000000000000042 percent’
And thus no heat is allowed to flow into space?!
Riiiighht.
BTW, zoe, when i look up at space, i see an awful lot my emitted radiation can hit.
https://www.jetsetworldtravel.com/app/uploads/2017/09/Atacama-Admiring_the_Galaxy-2.jpg
Your calculation is way off.
Even looking at a dark spot, lots of stuff!
https://youtu.be/V-3ZzXCnyTE
Nate, please stop trolling.
Ball4,
“internal energy U changes with time dU/dt = Q+W where Q is also a rate as well as W. A rate absolutely involves the existence of time.”
Time IS a rate, or ratio. We define 1 second as ~9 billion vibrations of Cesium-133 at certain conditions.
When we say 1 Watt of power, we are saying 1 Joule flows WHILE Cesium-133 vibrated ~9 billion times.
The time it takes to do something is a ratio of Cesium-133 actions. By definition. We can define time by any other thing that has predictable motion in 3D space.
zoe…”The time it takes to do something is a ratio of Cesium-133 actions. By definition. We can define time by any other thing that has predictable motion in 3D space”
The Cesium-based atomic clock is irrelevant, it’s the second upon which we rely. The atomic clock is simply a better, more stable source of oscillations that keep the second more accurate.
The oscillations in the atomic clock are counted till you get the length of that 1 second interval.
The second is defined based on the rotation of the Earth. We defined one rotations wrt the Sun as a ‘day’. Of course, we also defined the hours, minute and second with 1 day = 24 hours, 60 minutes = 1 hour, and 60 second = 1 minute.
You have 24 hr/day x 60 min/hr x 60 sec/min = 36,400 seconds/day. Since the day = 1 rotation wrt the Sun, you divide that period by 36,400 to get the interval of 1 second.
It’s similar in an electronic wrist watch. The oscillator in a watch is a quartz crystal which is forced to vibrate at it’s natural frequency by applying a regular voltage to the crystal. That frequency is applied to electronic counters till it is broken down to the length of our pre-defined interval for the second. Then it’s a matter of counting second till you get minutes, hours and days.
I have actually built circuits to do exactly that in electronics classes.
Therefore time is based on the angular velocity of the Earth, which is essentially a constant, therefore time must be a constant that cannot dilate.
In fact, time doesn’t even exist. It a game humans play in their minds.
The angular velocity of Earth is not constant.
Here are some more facts for you, backdoor guy:
* CO2 is NOT a heat source.
* A racehorse does not rotate on its axis. Just as the Moon does NOT rotate on its axis.
* “Cold” can not raise the temperature of “hot”.
* You do not get to violate the laws of physics to fit your false religion.
* Reality is a freight train coming at you at the speed of light.
Hope that helps.
Gordon,
“However, the temperature of that interstellar matter is claimed to have a temperature of +4K, or 4C above absolute zero.”
It’s actually 2.725K, but that’s OK.
The idiots can claim an object in space emits to 2.725K, but it also receives 2.725K from space, and therefore heat flux is zero.
“That same +4K is claimed as partial proof of the Big Bang theory. Some diehard theoretical physicists claim it is residual heat left over from the alleged BB.”
LOL. That’s hilarious.
Eddington (1926) measured “temperature of space” using starlight and got the same temperature of ~3K. His 3K did not have same spectral distribution, but …
The Earth receives non-Planck Blackbody radiation from the sun, then it is converted on the way out to a real planck Blackbody (minus absorbing gases, and albedo).
Energy In = Energy Out
The instruments that measure Cosmic Background Radiation have an internal heater. This heater is auto-adjusted to a local temperature that creates
energy in from all distant stars = energy out
And so the heat flux is zero.
Needless to say, this local heater creates a perfect BB curve! And that’s what they are measuring.
Scientists are so amazed at this artificial perfection, that it could only be a result of Big Bang.
In reality it is just current star light. It says nothing about the past.
It would as stupid as saying that the Earth’s or the Sun’s Local BB radiation is a residual of some past event, but still persists.
-Z
zoe…”In reality it is just current star light. It says nothing about the past”.
That’s the way I see it. When you look out and see starlight you see it because that EM is striking your retina right now, not in the past.
I don’t believe in past or future. The mind creates past from memories and projects those memories as a future.
Who knows, maybe there is a way for the human mind to detect future events but thus far we are only guessing. Knowledge is a property of memory, which is old data. There is no way to have actual knowledge of the future, even though some climate scientists try to tell us otherwise.
‘The idiots can claim an object in space emits to 2.725K, but it also receives 2.725K from space, and therefore heat flux is zero.’
Very confused there Zoe.
We were talking about a warm object, say at 250 K, emitting to space. It will receive flux from space corresponding to 2.73 K, then the heat flux is not zero BECAUSE 250 K is much bigger than 2.73 K.
Wrong again, Nate.
The 250 K object will not “receive flux” from an object at 2.73 K.
You’re trying the “all photons will always be absorbed” scam, again.
Nothing new.
‘The 250 K object will not ‘receive flux’ from an object at 2.73 K.’
‘You’re trying the ‘all photons will always be absorbed’ scam, again’
You’ll have to argue it out with the other JD:
“Energy can transfer from the cold object to the hot object either by transfer of energetic particles or electromagnetic radiation, but the net transfer will be from the hot object to the cold object in any spontaneous process.”
JD ‘All that is correct, Norman.’
Nate, you are confusing “can transfer” with “will transfer”.
That’s just one of your deficiencies.
In a real world scenario, not all molecules of a hot object have the same energy. So it is entirely possible that a high energy photon from a colder object could be absorbed by a hotter object. But, it could not raise the average temperature.
However, in a perfect scenario, all the molecules in a homogeneous black body would have exactly the same energy. Consequently, the blue plate would not absorb the “downstream” energy from the green plate.
Not understanding that is just another of your deficiencies.
JD, you are confusing ‘can transfer’ with ‘can’t transfer’, not aware, it seems, that they are not compatible.
“However, in a perfect scenario, all the molecules in a homogeneous black body would have exactly the same energy.”
Sure! And they’re all doing the Macarena.
More of your never-ending stream of nonsensical made-up physics.
Yes Nate, to you reality is nonsensical.
Nothing new.
‘all the molecules in a homogeneous black body would have exactly the same energy.’
Complete and total failure to understand thermo and stat mech!
And that’s not all you don’t understand, Nate.
This was mentioned upthread, but since some clowns, like Nate, are still confused, it’s worth mentioning again.
Dividing by 4 is fine for an average energy, but the quotient is no longer the solar flux. And, it can not be used as such. It no longer has any valid connection to temperature.
The solar constant, after albedo, is typically 960 Watts/m^2. If that flux perpendicalarly impacted a black body flat surface, perfectly insulated on the back, the equilibrium temperature would be 361 K (87.6 °C, 190 °F).
But dividing by 4 yields 240 Watts/m^2, so the same surface corresponds to 255 K (-18 °C, -1 °F).
For clarity:
960 Watts/m^2 –> 361 K (87.6 °C, 190 °F).
240 Watts/m^2 –> 255 K (-18 °C, -1 °F).
Dividing the solar flux by 4 is pseudoscience. Such a procedure fools clowns like Norman, who then believes the Sun cannot warm the planet!
Nothing new.
JDHuffman
I just want to point out how totally ridiculous your example is.
The blackbody surface you have reach 361 K would not be exposed to a constant input energy of 960 W/m^2. It would only receive this amount of energy for a short time in a cyclic fashion. Half the time it would receive zero energy but continue to radiate. Other times is would receive only a fraction of the 960 W/m^2.
You are also wrong. I do accept the Sun will warm the planet Earth. The reality you and the Joseph Postma gang are not logical enough to understand is that the Solar energy is not enough to elevate the Earth’s average temperature to the level it is in reality. You do not even understand the argument but pretend you do (just as you pretend to understand physics. I did like that you have actually done a little math. That is more than you have done in 99% of your posts).
I have already clearly shown you that real world locations (such as Desert Rock, Nevada) do not have enough solar input energy to maintain the rate of surface energy loss. You make the phony claim I can’t understand the graphs or that the spectra is needed but you have yet to explain why either of your bogus statements has merits.
You won’t accept real world data. You need to hold on to your false made up physics so strongly that you need to reject facts and logic. Nothing new, you will do this as long as you are the blog troll.
norman…” The reality you and the Joseph Postma gang are not logical enough to understand is that the Solar energy is not enough to elevate the Earths average temperature to the level it is in reality”.
With an atmosphere and an ocean to store solar energy and delay its release, it appears there is more than enough incoming energy to keep the Earth at its current average temperature. There was no need to invent a theory like the GHE based on a misinterpretation of Stefan-Boltzmann.
Gordon Robertson
The location I chose is a desert. Your theories are invalid for this location. Why not show how the oceans warm the Earth above the temperature that the Solar energy is able to do. I will not say you are wrong. I want to see some evidence. Show some work. Do some math. If you took college level physics calculations like these would not be difficult for you.
“The location I chose is a desert.”
Poor Norman believes he can pick one spot to prove something about the entire planet.
He’s always good for a laugh.
JDHuffman
Actually you can pick multiple locations and get the same results. The energy input from the Sun is not enough to sustain the energy loss. Just not there. If you had the motivation you could find the information and do your own research. It is out there.
Are you the clownish fool, or the foolish clown?
How do we tell the difference?
When you finish arguing with yourself, you still haven’t identified even one time my physics has been wrong.
http://www.drroyspencer.com/2019/06/a-simple-no-greenhouse-effect-model-of-day-night-temperatures-at-different-latitudes/#comment-357662
norman…” Why not show how the oceans warm the Earth above the temperature that the Solar energy is able to do”.
We are talking about an alleged 33C warming between a planet with no atmosphere and oceans and one with both.
You do the math. If both the oceans and the atmosphere stores heat then the average temperature rises…right?
That’s what your GHE theory claims only they got it wrong. There’s no way that GHGs, making up 0.3% of the overall atmosphere could raise the average temperature by 33C.
If the ocean is responsible for the warming atmosphere then you’d expect the ocean to be cooling as it releases the stored energy.
backdoor guy, what do you mean by “if”?
Have you never heard of an El Niño?
bdg…”If the ocean is responsible for the warming atmosphere then you’d expect the ocean to be cooling as it releases the stored energy”.
So what? It gets warmed again real fast by the Sun.
Norman, you’re wrong again.
Nothing new.
My simple example had nothing to do with rotation. I was just demonstrating the pseudoscience of dividing flux by 4, and why such nonsense leads to bogus results.
As to the rest, I’ll let you argue with yourself:
“I do accept the Sun will warm the planet Earth.”
“I have already clearly shown you that real world locations do not have enough solar input energy to maintain the rate of surface energy loss.”
JDHuffman
You again show how ignorant you are. The two statements of mine you post, incorrectly thinking that they contradict is lame.
Yes the Sun warms the Earth to a certain temperature above the temperature it would reach without the Sun. Not hard to understand.
Your attempt to discredit a poster should be a little more advanced.
The Second one shows that the Sun could not warm the planet surface to the observed temperatures. It requires the GHE to achieve the elevated temperatures.
I have not seen you support your conclusion that the dividing by 4 in nonsense and leads to bogus results.
The amount of energy a surface emits by EMR must have received this much energy. The energy emission is continuous. The solar input is cyclic. The amount of energy the surface receives from solar can be calculated as can the energy emitted. The energy emitted is much greater than the energy received.
I have demonstrated this to you with actual empirical data. You still have to explain why spectra of the energy matters in these cases. You have yet to do this. The site does not provide spectra they provide total energy gains and losses. The data shows you are wrong. The data shows you don’t understand why you are wrong. The data demonstrates you need to pull a phony rabbit out of your magic hat (spectra) because the evidence is clear. You just can’t process it. If you had some math and logic skills maybe it would not be so daunting.
So who won in your debate with yourself–Norman the foolish clown, or Norman the clownish fool?
‘Dividing by 4 is fine for an average energy’
Then JD goes on and to ignore this and do something else.
Norman’s rotisserie meat is a good demo that the average energy received by the rotating meat from the fire over time is what determines its temperature and how cooked it is.
Just as when I set my microwave to 50% power. It only has one power output, 100%. So it simply turns it off for 50% of the time.
It works!
Nate is confused by a rotisserie and a microwave.
Is there anything that doesn’t confuse him?
JD, why bother to respond if you have nothing of substance to say?
Exactly….
norman…”With EMR there is NOT a potential difference between two plates at different temperatures that generates EMR. The rate of EMR emission is based upon the temperature of each surface. The hot surface emits more EMR than a colder one so the net energy flow is from hot to cold. The energy flows in both directions. More flows from hot to cold so you have a heat transfer from hot to cold”.
*********
I did not infer a potential difference although there is one if you insist on a two-way flow of EM.
As you say, a body radiates EM based on it’s temperature. You seem to think that temperature has nothing to do with an atom’s ability to absorb energy of a specific frequency. You continue to push your theory that any EM incident upon a body of any temperature will be absorbed.
You are confusing thermal equilibrium with bodies of different temperature. You are trying to apply Kircheoff’s Law where it doesn’t apply and applying blackbody theory incorrectly.
You posted a youtube link a while back with a video from the 1960s showing the relationship between absorp-tion and resonance. The video showed that energy of a certain frequency could be absorbed only by atoms that resonated with that frequency. They used mechanical springs as an example.
That’s true and it comes from quantum theory. The resonance in the atoms is about the electrons on the atoms. Depending on their energy level, which depends on the temperature of the atoms in the mass, the electrons have a specific frequency.
When those electrons emit EM, they impart that frequency to the EM as E = hf. When they absorb EM they do so based on the frequency of the incoming EM. An EM frequency lower than their resonant frequency has no effect on them. In order to excite an electron, the EM has to have a frequency equal to or greater than the electron frequency, which translates to a higher energy level.
And the energy has to be quantized. The available energy bands to which the electrons can jump are determined by E, which is the difference in eV between energy bands. If the frequency of the incoming EM is low, then E = hf is low and lack the intensity to raise the electron to a higher level.
EM from a cooler body lacks the energy and frequency to affect the electrons in a hotter body. Therein lies the explanation for why heat cannot be transferred by radiation from a cooler body to a hotter body and from a lower energy potential to a higher energy potential.
Electrons in an atom can spontaneously fall from a higher energy band to a lower energy band but they cannot rise to a higher energy band unless excited. EM from a cooler body cannot excite them.
That is known across all forms of energy. Energy of any kind cannot spontaneously move from a lower energy potential to a higher energy potential.
Nate,
“We were talking about a warm object, say at 250 K, emitting to space. It will receive flux from space corresponding to 2.73 K, then the heat flux is not zero BECAUSE 250 K is much bigger than 2.73 K.”
There is no emission to “space”. There is only emission to matter. Just like the plate receives 2.73K (due to all the SUPER HOT STARS and SUPER HOT GALAXIES), some other random part of space that HAS matter will only receive a minute nothing (due to inverse square law) from your effing plate AND 2.73K from the HOT STARS and HOT GALAXIES. Your plate is not sending 250K, it’s sending nothing.
The Earth sends 0.0004 W/m^2 to its Lagrange L2 point. That’s like 1.5mil KM away (6 moons away). There is no heat transfer from Earth to L2. And that’s why CMB mapper satellites go there!
Space can not be warmed. If you want to consider the 2.73K from distant matter, then you need to consider it also for your sun and your blue plate too, hypocrite. Make sure to complicate your “thought experiment” equally, don’t just leave it up to your green plate to do all the work.
Zoe,
‘There is no emission to space’
Then Im not sure where all the heat escaping from Earth is going?
“The Earth sends 0.0004 W/m^2 to its Lagrange L2 point. Thats like 1.5mil KM away (6 moons away). There is no heat transfer from Earth to L2.”
Huh? You just said it sends 0.0004 W/m^2. Is that nothing?!
You are a very strange little troll, zoe.
nate…”Then Im not sure where all the heat escaping from Earth is going?”
There is no heat escaping to space, heat is converted to electromagnetic energy and the heat is lost. The EM goes where EM goes until it meets a mass at a lower temperature than the Earth.
Mind you, it’s not as simple as that. The Sun can radiate EM to us at 93 million miles and the intensity is still 1369 W/m^2. Don’t know if radiation from the Earth would be absorbed anywhere but the Moon. It would be so diluted as to be useless, just like surface radiation more than a few feet above the surface.
Gordon,
“The Cesium-based atomic clock is irrelevant, its the second upon which we rely. The second is defined based on the rotation of the Earth.”
You must have travelled by time machine into the present. When are you from?
‘Since 1967, the second has been defined as exactly “the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom” (at a temperature of 0 K).’
zoe…”‘Since 1967, the second has been defined as exactly “the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom” (at a temperature of 0 K).’”
That’s cool, but the length of the second is pretty well the same. We need it to remain the same to suit our day, our calendars, etc. That second is based on the Earth’s period of rotation.
What really happened was that the transition impulses from the cesium atom were counted so they would approximate our current second. It’s the other way around, the cesium transitions were pegged to our second.
Once that was done, the cesium atom could be counted on as a super-accurate time base, unless someone put the atomic clock in an oven.
Gordon Robertson
In your post above: YOU: “norman…” Why not show how the oceans warm the Earth above the temperature that the Solar energy is able to do”.
We are talking about an alleged 33C warming between a planet with no atmosphere and oceans and one with both.
You do the math. If both the oceans and the atmosphere stores heat then the average temperature rises…right?
That’s what your GHE theory claims only they got it wrong. There’s no way that GHGs, making up 0.3% of the overall atmosphere could raise the average temperature by 33C.”
I guess since you won’t do the math I will. Your point actually works to demonstrate the GHE must exist to create the ocean water temperatures observed on Earth.
If you take Roy’s Solar energy graphs from the other thread.
http://www.drroyspencer.com/wp-content/uploads/Simple-no-GHE-model-solar-flux-550×344.jpg
Using his near equator graph, the maximum amount of energy 1 m^2 of surface water could absorb. You can assume a blackbody for this case. You could use an emission of 0.9 if you want to consider all the angles of emission (about 0.98 for direct emission) but the amount absorbed would be equivalent. Since the energy emitted and absorbed would have the same percent as a blackbody it will still work.
The amount of solar energy the 1 m^2 water receives in a 24 hour cycle comes to 39,456,000 joules. It won’t receive more than this (it can receive less depending upon what % you feel will not be absorbed by the water).
The water only receives energy for 12 hours but emits all 24. So the water surface (lower layers could get warmer like a solar pond if you prevented convective overturning as the surface gets colder and denser by IR emission it would sink and be replaced by the warmer water below) can only reach the temperature at which it will radiate away a total of 39,456,000 joules of energy. Even though the actual emission would cycle you can get the average of the cycle. 39,456,000 joules/24 hours = 1,644,000 joules/hour (also of course per meter squared as stated earlier).
1,644,000 joules/hour divided by 3600 seconds/hour = 456.67 Watts/m^2. When the surface water reaches the temperature that it radiates at this level it is all the warmer the surface can get from the Solar input.
If you use a blackbody, for this emission the surface temperature comes out at 299.57 K. So the surface temperature of water exposed to the maximum solar input could not get hotter than this. You would have to suppress evaporation to get this temperature. Evaporation is a big coolant of warmer water surface temperature.
In the real world with evaporation you have ocean temperatures above this maximum level.
https://www.holiday-weather.com/red_sea/averages/
The Red Sea averages 31 C in Summer. That puts it at 304.15 K. The next one I would need to find is evaporative cooling for this water and subtract this energy loss from the hypothetical value above. Then we could see how much energy the GHE provides the sea surface. Even with this, it shows the Solar flux is not enough to maintain the sea surface temperature observed in the real world. You won’t accept is but the evidence is actually there.
Gordon Robertson
I did find evaporative cooling of warm water. It can exceed 200 W/m^2.
https://books.google.com/books?id=Zi1coMyhlHoC&pg=PA113&lpg=PA113&dq=watt/m%5E2+evaporative+cooling+of+warm+tropical+water&source=bl&ots=_ViN8LD38l&sig=ACfU3U217EeHiHgCvqx-_5RscvMa0j5NsA&hl=en&sa=X&ved=2ahUKEwj59M73w-_iAhVHmK0KHRrgChYQ6AEwBXoECAkQAQ#v=onepage&q=watt%2Fm%5E2%20evaporative%20cooling%20of%20warm%20tropical%20water&f=false
If you go conservative and have this only at 100 W/m^2 for our hypothetical you would now have a surface that would lose 456.67 W/m^2 minus the evaporative cooling loss of 100 W/m^2 so the radiant energy loss would be at 356.67 W/m^2 allowing the surface to reach a maximum of 281.62 K. Now the hypothetical is 22.52 K below the actual observed value. If you use a higher number for the evaporative cooling (will try the 200 W/m^2 value). The surface water temperture in the hypothetical will now only reach 259.39 K or 44.7 K variation.
Since the real world has evaporative cooling and the solar energy can only get water surface to a maximum of 299.57 K you have to have the GHE to make up the difference. The book even explains it to you. Because of GHE (atmosphere radiating back to the surface) The heat loss by IR is small. Without the GHE the heat loss by IR would be the dominant form of heat loss.
Poor Norman. He’s so hopelessly lost.
Now he is trying to convert evaporative heat loss to a flux! Then, he subtracts that “flux” from the bogus, divided by 24, divided by 3600, ocean surface “flux”!
He makes one error after another.
He ends up with the ocean being too cold, due to his made-up calculations, so he then claims CO2 must be supplying the extra energy. But, the atmosphere is always colder than the surface. This morning my IR thermometer read -7.6 °F, directly overhead sky, and 65.3 °F, ground temperature.
Poor Norman.
Poor JDHuffman
A hopeless and stupid troll that can’t be helped. Can’t learn, can’t think but can annoy with empty troll comments by the hundred per thread.
Troll why not pretend you have physics knowledge. Pretend you went some University and studied physics. It is fun to pretend. You even have an imaginary friend that supports you when you need him to pop up. Pretending is fun, isn’t it? Much easier than reading a textbook on physics. I have not seen you look up some words lately to pretend you know the concepts involved. I have not seen you use words (you don’t have a clue what they mean) lately like Poynting Vectors or Enthalpy or Entropy. You have no clue what any of these mean and you certainly could not use them in any rational way. You can pretend you understand them, reality is you don’t. Fool away troll. See who else will believe your pretend world. So far a crackpot and your imaginary friend. Not doing too well in proving you are the “Physics Expert” on the blog.
Norman, I wouldn’t call myself a “physics expert”, but I appreciate you making the connection.
JDHuffman
I would certainly NOT call you a physics expert. You are finally attempting a bit of truth now. We all know that you have little physics knowledge but pretend that you have quite a bit.
You guys have a whole “Team”, a whole network of “imaginary friends”, any of whom just love to “pop up” whenever they’re needed. You should try looking at things from the other side, occasionally, Norman.
That’s a pretty interesting observation JD. Let’s the assume the emissivity setting on your IR device was set at 0.95 (the typical default value). That means the downwelling radiation would appear to be ~214 W/m^2 and the upwelling radiation would appear to be ~386 W/m^ assuming that the SB law provides a reasonable approximation. Of course, the emissivities of the ground and sky aren’t really 0.95 so there’s going to be some error here. That’s okay. That’s not important right now. The question…why do you think your IR device was able to provide a reasonable estimation of the ground temperature despite the significant amount of downwelling radiation your device also recorded that would have to be interacting with the ground? What do you think happened to that downwelling radiation when it hit the ground?
See what I mean?
norman…”I guess since you wont do the math I will”.
How the heck can anyone do the math? It’s far too complex a problem.
However, you alarmists rant about a heat budget, claiming heat in = heat out. With an alleged 15C average for the planet it means solar energy created that temperature, balanced between the oceans and the solid surface. Since the oceans make up 70% of the planet’s surface, I’d say they radiate most of the heat as IR and conduct most of it to the atmosphere.
This nutty notion that GHGs back-radiate enough EM to raise the temperature of the surface that warmed them contradicts the 2nd law and represents perpetual motion. You cannot recycle heat in such a manner as to raise the temperature of the source.
The other theory about GHGs slowing the rate of radiation from the surface is just plain wrong. The rate of heat dissipation is controlled by the temperature of the atmosphere which is in contact with the surface. That temperature is controlled by N2 and O2, which make up 99% of the atmosphere.
It doesn’t take a lot of science to figure out that 0.04% CO2 and 0.3% atmospheric WV cannot raise the Earth’s temperature by 33C. Another explanation is required.
I borrowed the ocean as a hot water bottle from Stephen Wilde:
http://www.newclimatemodel.com/the-hot-water-bottle-effect/
And the atmospheric heat retention theory from R. W. Wood. He was a highly respected authority on gases like CO2, which he felt could not warm the atmosphere.
Both together explain what the GHE can’t.
Of course, let’s not forget that we have been re-warming since the Little Ice Age and that warming accounts for most of the warming since.
Gordon,
‘This nutty notion that GHGs back-radiate enough EM to raise the temperature of the surface that warmed them contradicts the 2nd law and represents perpetual motion. You cannot recycle heat in such a manner as to raise the temperature of the source.’
I have insulation in the walls of my house. if its a cold day and Ive been awy with the heat off, the insulation in my walls has reached the temp of the outdoors.
If I come home and turn on the heat, the house warms and starts losing heat thru the walls. The insulation warms and reaches a temp between the inside and the outside.
The insulation was warmed by the heat from the house, but now it helps keep the house warm by limiting the heat flow through the walls.
No contradiction of 2nd law, no perpetual motion. Yet with the insulation, and with the same furnace heat input power, the house will be warmer.
Nate, set your furnace thermostat at 75 °F, on a cold day. Your insulation will not raise the temperature above 75 °F.
But, in your pseudoscience, the insulation would “add” more energy, raising the temperature above the set point.
That’s not possible, as Gordon pointed out.
JD,
The analogy is to allow your furnace to run without a thermostat. The same furnace will bring the home to a higher equilibrium temperature with the insulation than without.
The insulation isn’t adding energy. It is impeding the flow of heat and energy loss to the outside thus augmenting the furnaces ability to bring the home to a higher equilibrium temperature.
In the analogy the temperature of the home is like the global mean temperature of Earth, the insulation is like the GHGs, and the furnace is like the Sun. Neither violate the 2LOT.
Well backdoor guy, your analogy fails.
1) Earth has a set-point.
2) CO2 is NOT an insulator.
3) For the same solar constant, adding CO2 will NOT raise the temperature.
A much better analogy is the “dry sponge”. Does a dry sponge increase the moisture in your house?
(The backdoor is open, and you are cleared for departure. Before popping up next time, you might want to learn some physics.)
JD is a professional point misser. The sun, like our furnace, supplies a certain heat input rate to Earth.
Add insulation to the house or the Earth, with fixed heat input, they both warm.
Yes Nate, you missed one of the points:
“2) CO2 is NOT an insulator.”
Physics can be hard to understand, so think of the “dry sponge” analogy. A dry sponge does not add moisture to the system. So 2 dry sponges wouldn’t add moisture either.
Hope that helps.
‘Earth has a setpoint’ Evidence?
Why are some posters so fond of declaring things that they just made up?
Dry sponges are not very good to eat. Nor are they good at conversation. Your point is what?
Mostly they’re just good at soaking up water. But they need to be in contact wit the water, unlike radiating substances..
Wow Nate, you responded in 2 minutes.
Then, you responded to yourself!
You really are obsessed with this, huh?
It appears to just make you frustrated and hyper–probably not a good hobby.
Have you considered knitting?
2 min, no more like 20 min. I was responding to your previous one.
In any case Pot-kettle!
Wrong again, Nate.
It was 2 minutes, as I indicated. You can’t get anything right.
JDHuffman says:
June 17, 2019 at 2:59 PM
Nate says:
June 17, 2019 at 3:01 PM
See why you have no credibility?
What part of ‘responding to previous one’ dont you get?
I even quoted it, dumbass!
See why you have no credibility?
JD, what is Earth’s set-point temperature right now?
288 K +/- 1 K
bdg…”The insulation isnt adding energy. It is impeding the flow of heat and energy loss to the outside…”
Heat is the energy, R-rated insulation won’t impede radiated energy. So you are trying to slow down the heat dissipation in the atoms of air that have been heated by the furnace.
Normally, those atoms would butt up against the walls and the ceiling and transfer their energy via conduction to the walls and ceiling. Wood has a lower resistance to heat transfer than R-rated insulation. Normally, insulation is strands of spun fibreglass or mineral wool that has air spaces built in. Air does not transmit heat as well as wood, nor does the fibre/wool.
Radiation will go straight through that kind of insulation.
There is nothing in the atmosphere that can slow down EM either. The notion that CO2 and WV can trap heat or slow its dissipation from the surface is sheer pseudo-science. Heat cannot be trapped by a gas, it can be trapped by glass in a greenhouse since atoms of air won’t pass through the glass.
The mistake made in AGW theory is mistaking heat for something other than what it is. Normally, heat is confused with EM but they are totally different forms of energy. GHGs can trap a certain amount of EM, but EM is not heat.
According to the 2nd law, in order to transfer heat, EM has to flow from a hotter body and be absorbed by a cooler body.
Trapping a trivial amount of EM has no effect on the rate of heat dissipation at the surface. The rate of dissipation is related to the temperature difference between the surface and the layer of the atmosphere butting against it. The temperature of that layer is determined by nitrogen and oxygen which make up 99% of the gases.
If you want to increase surface heat dissipation you need convection. The surface warms the layer butted against it and that layer rises. Colder air rushes in and creates a higher temperature gradient between surface and atmosphere, increasing heat dissipation. When that layer warms, dissipation slows.
GR,
Conduction, convection, and radiation are different ways in which heat can be transferred. Just because a material is ineffective at impeding one pathway for thermal transmission does not mean that it is equally ineffective with the other two. It’s still a thermal barrier as long as it is effective a blocking the transmission of heat via at least 1 of the transmission pathways. And no, typical residential insulating materials are not completely ineffective at blocking the radiation pathway anyway. I’m not sure where you got that idea. It doesn’t really matter though because the analogy is not dependent on the heat transmission pathway anyway.
The point…a thermal barrier will result in a higher equilibrium temperature in the presence of a heat source than if the barrier had not been there. And if the barrier is placed into a system that was already in equilibrium then the temperature of the system will increase so that it achieves a new and higher equilibrium temperature. The increase in temperature we call warming. This does not violate the 2LOT. This experiment is ran by millions or even billions of people on a continual basis during the cool season.
nate…”The insulation was warmed by the heat from the house, but now it helps keep the house warm by limiting the heat flow through the walls.
No contradiction of 2nd law, no perpetual motion”.
R-rated insulation used in most homes is designed to slow the CONDUCTION of heat through walls and ceilings. Radiation goes straight through it and hardly affects the heat level in the home, if at all.
That process does not describe the AGW pseudo-science about a cooler gas being warmed by a hotter surface via radiation then having the cooler gas transfer heat back to the hotter surface to raise its temperature.
The 2nd law says that cannot happen and the perpetual motion described by such a process is verboten.
GR said…”That process does not describe the AGW pseudo-science about a cooler gas being warmed by a hotter surface via radiation then having the cooler gas transfer heat back to the hotter surface to raise its temperature.”
You are correct in the strictest sense. But you’re also building a strawman here because that is not how the process actually works.
In your home the furnace provides the energy source and the ability to transfer heat into the system. The insulation is impeding the flow of heat to the outside. It is slowing the rate at which energy within the home is lost to the outside. The insulation by itself isn’t adding energy or actually doing the heating. That’s the job of the furnace. The insulation just augments its warming ability by trapping more of the energy so that it doesn’t escape. The effect is that the equilibrium is higher than it would be otherwise.
In the atmosphere the sun provides the energy source and the ability to transfer heat into the system. The GHGs are impeding the flow of heat to space. They are slowing the rate at which energy within the geosphere is lost to space. The GHGs by themselves aren’t adding energy or actually doing the heating. That’s the job of the sun. The GHGs just augment its warming ability by trapping more the energy so that it doesn’t escape. The effect is that the equilibrium is higher than it would be otherwise.
The reason why you are incredulous is because you think the argument is that GHGs can warm the surface even in the absence of the Sun. But that’s not how it works. That’s why what you are presenting is a strawman.
bdg, you start out with only a hint of pseudoscience, as revealed by your choice of words.
The GHGs are impeding the flow of heat to space. They are slowing the rate at which energy within the geosphere is lost to space…The GHGs just augment its warming ability by trapping more the energy so that it doesn’t escape.
But in your last sentence, you make your move: “The effect is that the equilibrium is higher than it would be otherwise.”
Now you have implied that the system temperature will be higher.
Very tricky.
How does this “re-warming” work exactly? How do you know that it accounts for most of the warming since the LIA? Is there a “re-cooling” that we should consider as well?
backdoor guy, did you learn to ask stupid questions from DA, or from your indoctrination?
bdg…”How does this re-warming work exactly? How do you know that it accounts for most of the warming since the LIA?”
It was calculated by the geophysicist, Syun Akasofu.
GR,
I’ve read Akasofu’s argument. His argument is that because the Earth has been warming since 1800 and because he was able to draw a linear regression trendline then the warming has to be natural. Now, aside from the fact that “natural” is not an actual cause, but only a classification under which you categorize climate modulating agents, he makes really weak arguments to justify his idea. And his idea cannot explain past climate change events nor does it make predictions that can be tested so I question whether it is even a theory.
bdg…”Ive read Akasofus argument. His argument is that because the Earth has been warming since 1800 and because he was able to draw a linear regression trendline then the warming has to be natural. Now, aside from the fact that natural is not an actual cause, but only a classification under which you categorize climate modulating agents…”
Nothing to do with trend lines, he stated initially that the IPCC failed to account for re-warming from the Little Ice Age. He was inferring that they completely ignored a natural explanation for the current warming and went with a cockamamey explanation based on CO2 bubbles trapped in ice.
The IPCC has a mandate to find evidence for anthropogenic warming. They are so focused on that wild goose chase that they completely ignore natural evidence right in front of them.
Why would anyone invent a ridiculous argument that a trace gas in the atmosphere, making up 0.04% of the atmosphere, could create catastrophic warming/climate change when the beginning of the warming the claim began smack-dab in the middle of the LIA while global temps were 1C to 2C below normal?
That 0.04% is 95% from natural sources so why would the natural CO2 not have caused catastrophic warming/climate change centuries before?
I’ll tell you why. The IPCC was started by the United Nations after urgings by Tory PM Margaret Thatcher. She was looking for a way to make coal unpopular so she had a hammer when dealing with UK coal miners, who were a thorn in her side.
Meantime, the UN had been looking for a way to impose a universal tax to help them with their pet projects re funding poor nations. Thatcher handed them the solution and the UN road it the rest of the way.
AGW is political bs, nothing more. Well, maybe a bit more. It’s a gravy train for any scientist willing to sell out for the lucrative associated funding.
Hi there,
You dont seem to be answering my questions ??? If as your very wild claim without the GHG’s the Sun would only heat us to -18 So why on a summer day is it 20c in the UK and 40c in Dubai. If as you claim it’s only the GHG’S warming us, from -18c to 16c, then most people would say Dubia is so much warmer cos of its position on Earth. But if you are right, then the Sun would not heat different Countries like it does, it would be basically the same temp. all the time ??? You GHG’s idea is ok for a fun thing to think of, but not if your serious on science. I JUST PROVED YOU SO WRONG.
Wayne
wayne,
The 255K (-18C) and 289K (16C) figures are in reference to the global mean temperature. Temperatures at specific locations and specific times are highly variable and are the product of dozens of processes acting on the atmosphere.
I’ve claimed that GHGs are the only agent that affects temperatures at specific locations at specific times. And Dubai is typically warmer than the UK for a bunch of reasons; not just it’s position on Earth.
I could not reply in the other place ???
bdgwx wrote;
The 255K (-18C) and 289K (16C) figures are in reference to the global mean temperature. Temperatures at specific locations and specific times are highly variable and are the product of dozens of processes acting on the atmosphere.
Wayne wrote;
Roy stated and others that without the GHG’s it would be -18c averge temp. on Earth. So what are you saying the averge temp. would be on Earth, your changing what Roy and I thought all of you said now. So where on Earth would it be -18c, and where would it be 16c without the GHG’s ???
bdgwx wrote;
Ive claimed that GHGs are the only agent that affects temperatures at specific locations at specific times. And Dubai is typically warmer than the UK for a bunch of reasons; not just its position on Earth.
Wayne wrote;
NO, you, or Roy stated that it would only be -18c without the GHGs, with is ridiculous, I mean without the GHG’s it would be like the SUN, 120c plus in the day, and -40 in the night.
Dubai is basically so much hotter than the UK, primary cos of its location. BUT, if as Roy seems to claim, it would be just -18c without the GHGs, we would have basically the same temp. all over the Earth, BUT, we don’t.
“SO” questions, what temp do you think the averge would be over all the Earth without the GHG’s ??? Why did Roy say it would be only -18c ??? I mean it would be 120c without the GHG’s in the day ???
You started that dividing by 4 was to do with distance now, but you still have not explained what this distance is, and what different dividing by 4 means ???
Why not if you think your right comment on here after answering my questions.
https://www.youtube.com/watch?v=3Y9wV4cciXA&t=589s
Wayne
I only support the claim that the Earth would have a global mean temperature of 255K without an atmosphere and assuming the albedo were the same. I’m not saying that Dubai would be -18C without the GHE. It definitely wouldn’t be.
And no, I’ve never claimed that dividing the solar constant by 4 had anything to do with distance. It has nothing to do with distance. The reason you divide by 4 is because that is what the formula for energy mandates.
E-year = (S/4) * (365.24 * 24 * 3600) * A
…where E is energy in joules, S is the solar constant in W/m^2, and A is the surface area of a sphere.
The reason why S is divided by 4 is because S is relative to a 2D plane by definition while Earth is actually a 3D sphere so it has to be rescaled to the proper the shape.
Perhaps one way to clear up the confusion is to define Sp as the solar constant on a plane and Ss is the solar constant on a sphere. Because of canonical geometrical princiapls Ss = Sp/4 with the specific values for 1 AU being Sp = 1360 W/m^2 and Ss = 340 W/m^2.
bdgwx, please stop trolling.
Norman, you ca’t average flux that way. You are violating the laws of physics. That’s why you end up with bogus results.
See if you can understand this simple example:
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-358018
Nate,
“Then Im not sure where all the heat escaping from Earth is going?”
Radiation is not heat.
“Huh? You just said it sends 0.0004 W/m^2. Is that nothing?!”
This radiation came from the sun. The sun can heat L2 more than the Earth. From L2’s point of view, the heat only comes from the Sun. The Earth heats nothing.
Ok zoe, declare whatever strange things you want to.
http://www.drroyspencer.com/2019/06/a-simple-no-greenhouse-effect-model-of-day-night-temperatures-at-different-latitudes/#comment-358279
Space used as a heat sink
Nate
You have been mobbed by trolls and a crackpot. JDHuffman troll is really quite the idiot. I respond to him because he will never leave me alone. If I post to crackpot Gordon Robertson this idiot annoying troll jumps in with the stupidest comments you would not imagine a rational person of making. He does it only to annoy and try to get a knee-jerk reaction from some unsuspecting poster. I know he is a troll and that is what he does. I have ignored this stupid poster but he never leaves one alone. You can’t get anywhere with Crackpot Gordon Robertson. I am not sure about Zoe Phin. He seems to troll you with stupid semantics about what heat is. If you look at the definitions it has such a variety. Hard to pin down. In normal physics you are quite correct. It would not even matter if the space had no temperature. There is still a heat flow from a hot surface. Heat flow is about a transfer of energy. Heat will flow from a hot object to space. You are quite correct. You will not get too far with the trolls. I hope you can see that JDHuffman is not only a troll but he is really quite stupid. His posts are not even good trolling. They have horrible logic. It would be nice if I could post on this blog without that troll searching the threads and looking for my posts. It won’t happen. He loves to annoy and annoy he does. In this iteration of ger>an he is careful to avoid insulting Roy Spencer. Same stupid person, same stupid troll. Even Gordon Robertson might have a good point once in a long while. JDHuffman never has a good point. All stupid trolling crap over an over. Endless.
Poor Norman is in full meltdown.
I probably shouldn’t laugh….
As stated, the blog’s stupid troll just can’t leave it alone. The idiot has to post and he posts stupid things every time. I would hope this idiotic troll might post something of value just once. Sad but true this irrational troll does not possess enough intelligence to get lucky by accident and have a useful point.
A mindless troll out to annoy. Can’t be ignored. Seems more like a human version of a mosquito in a dark room that buzzes around the ears. Only value is to annoy. Don’t worry, this troll will keep up with the infantile empty posts. It is what the troll does. No luck with him. He will not even make a good point by mistake. Far too stupid to hope for that.
The meltdown continues, but I mustn’t gloat.
Schadenfreude is just too rude….
Just another stupid troll comment that you could have easily avoided. A troll addict. You have to troll regardless. I guess you think it is fun. Not sure who else but you enjoys your troll comments.
“The kid having a meltdown has no understanding of himself and of what he is feeling. He feels fragmented and out of control.”
https://www.psychalive.org/what-to-do-about-tantrums-and-emotional-meltdowns/
JDHuffman
Amazing, you actually had an entertaining post. Keep it up and you might not be so redundant.
Only a two sentence comment!
And Norman didn’t use the word “troll” even once.
The medicine must have finally kicked in….
norman…”You can’t get anywhere with Crackpot Gordon Robertson. I am not sure about Zoe Phin”.
Would help if you knew anything about physics and stopped being such a self-righteous goof who attacks when he doesn’t understand.
You should get a job with the National Enquirer, they are always looking for someone full of bs who specializes in propaganda.
bdg…”typical residential insulating materials are not completely ineffective at blocking the radiation pathway anyway. I’m not sure where you got that idea. It doesn’t really matter though because the analogy is not dependent on the heat transmission pathway anyway.
The point…a thermal barrier will result in a higher equilibrium temperature in the presence of a heat source than if the barrier had not been there. And if the barrier is placed into a system that was already in equilibrium then the temperature of the system will increase so that it achieves a new and higher equilibrium temperature. The increase in temperature we call warming. This does not violate the 2LOT”.
***********
I say R-rated insulation won’t block radiation because it has no metal in it. EM cannot be blocked by such a low density wool as R-rated insulation. If it could block EM, the insulation would also block radio waves and other forms of communication that flow straight through wood and other building materials other than concrete, especially concrete with rebar.
In order to block EM from leaving a home, you need a metal reflective surface like aluminum foil. The outer walls of a house, and the roof, will radiate so you need to slow down conduction through the walls.
I don’t know why you are bringing in equilibrium. This is a heat loss problem. If you can slow the loss of heat by conduction, you slow the temperature decline in the house. Slowing down radiation loss has a minimal effect at terrestrial temperatures.
I don’t know why you are referencing the 2nd law either. I did not claim the process of slowing heat dissipation via insulation contradicted the 2nd law. I merely said that heat cannot be recycled from a hotter body to a cooler body then back to the hotter body so as to raise its temperature. Or that radiation from a cooler body cannot raise the temperature of a hotter body, or even be absorbed by the hotter body.
So my post was deleted simply because I pointed out that “Zoe Phin” is actually Bob Phin. Sad.
Or maybe you just forgot where you made the comment, des?
http://www.drroyspencer.com/2019/06/a-simple-no-greenhouse-effect-model-of-day-night-temperatures-at-different-latitudes/#comment-358408
I think they have treatments for paranoia….
I can’t reply to your answers above ???
bdgwx says:
June 22, 2019 at 3:16 PM
I only support the claim that the Earth would have a global mean temperature of 255K without an atmosphere and assuming the albedo were the same. I’m not saying that Dubai would be -18C without the GHE. It definitely wouldn’t be.
So going on the Earth now have a mean temp. of ??? 16c. So do you think Dubia would be something like 8c ??? In all this your saying, or thinking, that on a very hot day its not that red ball in the sky that I feel huge great hitting my face and burning me so bad, but the GHG’s ??? Thats a wild claim ??? Your saying that the GHG’s don’t very slightly cool us like we are saying, and shelter us from the huge Suns eney/heat, but heat us from -18c to 16c ??? Odd thing is that your saying they heat us so much, but on a solar eclipse, its goes “SO” cold in seconds, thus proving there is “NO” heat there from the GHG’s !!!
Wayne
Wayne
wayne…”on a very hot day its not that red ball in the sky that I feel huge great hitting my face and burning me so bad, but the GHG’s ??? Thats a wild claim ??? Your saying that the GHG’s don’t very slightly cool us like we are saying, and shelter us from the huge Suns eney/heat, but heat us from -18c to 16c ???”
Good points Wayne. When we experience records of 30C+, what I notice is the intensity of the Sun. There’s no way that a trace gas could raise temperatures that much. Besides, those records were often set decades before and exceeded by a fraction of a degree C.
And there’s no way a trace gas could raise the atmospheric temperature by 33C. The increase has to be due to a combined effect of the atmosphere, including all gases, and the oceans.
Hi Gordon,
you just made another light bulb go of in my head. GOD, of course………I live in the UK, and at times in in the summer, one day might be cloudy, and like tomorrow its going to be an average of 17c, from 12 till 20.00. Now look at Friday, again most of the day is cloudy, no cloudish for about 5 hours, but from 12 to 20.00, the average temp. goes up to 24c. Thus its the damned Sun doing this, not the GHG’s as some claim. The Clouds and GHG’s when there are clouds, shades us from the Sun, they don’t warm us. Same as with a solar eclipse, why does it get so cold in seconds, if they were right, there should be so much more heat up there accumulated throughout the day, but there’s not.
Same in the summer here it could be 25c, but then in winder its 6c. Its the Sun driving all temp, as with the position the UK is to the Sun, and Uk’s position on Earth. This rise in summer and drop in winter, shows its the Sun, and also the heat that’s being held in the seas and earth and slowly being let out in the air we live in, and cycle goes on. Yes GHG’s shade us, and cooler not warmer.
Wayne
GHGs do not shade you from the Sun because they do not respond in the shortwave spectrum. That’s actually one aspect of them that makes them a GHG.
The reason why the temperature behaves the way it does in the UK is because there are dozens of processes acting on the atmosphere at any given time. The downwelling infrared radiation from GHGs is just one among many factors that are influencing the temperature in your backyard.
wayne said…”So do you think Dubia would be something like 8c ???”
I don’t actually know what Dubia’s mean temperature would be without GHGs. That would take some research to figure out.
What we can say is that individual temperature measurements at Dubai would have a higher standard deviation and larger range, but the mean temperature would be lower.
wayne said…”In all this your saying, or thinking, that on a very hot day its not that red ball in the sky that I feel huge great hitting my face and burning me so bad, but the GHGs ???”
I’m not saying that at all.
wayne said…”Your saying that the GHGs dont very slightly cool us like we are saying, and shelter us from the huge Suns eney/heat, but heat us from -18c to 16c ???”
GHGs do not shelter you from solar radiation. They are actually transparent (mostly anyway) to incoming radiation. That’s one defining characteristic that makes them GHGs. The fundamental reason behind this is explained by molecular physics and quantum mechanics.
wayne said…”Odd thing is that your saying they heat us so much, but on a solar eclipse, its goes SO cold in seconds, thus proving there is NO heat there from the GHGs !!!”
Well, actually the lag between peak eclipse and the drop in temperature is between 30 and 60 minutes (give or take) on average. Part of the reason why there is a lag is because of GHGs. Without GHGs this lag would be much shorter. The fact that the lag is so long is yet another line of evidence that proves the GHE is real.
bdgwx says:
Dr Roy Spencer’s graph has the Dubai max just under 50 F, 10 C.
It’s below freezing most of the time.
http://www.drroyspencer.com/wp-content/uploads/Simple-no-GHE-model-output.jpg
bdgwx, Svante, please stop trolling.
Hmm, I am not getting the below. Still dont get why you’re dividing by 4 ??? Could you try and explain it in layman’s terms please ??? solar constant is the quantity of radiant solar energy received at the outer layer of the earth’s atmosphere that has a mean value of 1370 watts per square meter. If per square meter, why divide by 4 ???
Geometry tells us that the surface area of a sphere is four times pi times the radius of the sphere squared. Has this something to do with it ??? But I would think the Suns energy/heat is hitting close to half the area of the Earth at one time, in 24 hours it hits it all, thus why divide by 4 ??? Just dont get you, i would have thought you would x by 2 ???
bdgwx says:
June 22, 2019 at 3:32 PM
And no, Ive never claimed that dividing the solar constant by 4 had anything to do with distance. It has nothing to do with distance. The reason you divide by 4 is because that is what the formula for energy mandates.
E-year = (S/4) * (365.24 * 24 * 3600) * A
where E is energy in joules, S is the solar constant in W/m^2, and A is the surface area of a sphere.
The reason why S is divided by 4 is because S is relative to a 2D plane by definition while Earth is actually a 3D sphere so it has to be rescaled to the proper the shape.
Perhaps one way to clear up the confusion is to define Sp as the solar constant on a plane and Ss is the solar constant on a sphere. Because of canonical geometrical princiapls Ss = Sp/4 with the specific values for 1 AU being Sp = 1360 W/m^2 and Ss = 340 W/m^2.
Wayne
wayne…”solar constant is the quantity of radiant solar energy received at the outer layer of the earths atmosphere that has a mean value of 1370 watts per square meter. If per square meter, why divide by 4 ???”
It’s based on a bad presumption, Wayne. That 1360 W/m^2 has to be summed over a
If we are considering a square metre of solar energy at TOA, we must extend that square metre to the surface and measure it there. In that case, we are concerned with dissipation due to the atmosphere between TOA and the surface.
If there was no atmosphere, and we measured at the equivalent altitude of TOA, why would we have to divide by 4 to get the equivalent solar radiation at the surface? That would make no sense since the 1360 W/m^2 would not be much different than at TOA. It the atmosphere that dissipates the 1360 W/m^2 not an arbitrary factor of 4.
Furthermore, since the broad spectrum solar energy heats all atoms/molecules on the surface, why would it not heat all of the molecules in the atmosphere? There is no focus issue, anything the solar energy encounters it should warm due to its broad frequency spectrum and high intensity.
The degree to which solar energy heats the atmosphere comes down to density. Near TOA, there is very little mass to heat but closer to the surface there is significant mass in the atmosphere.
Why has that not been taken into account?
Dividing by 4 has nothing to do with the atmosphere or the dissipation of the radiation caused by the atmosphere. That is a separate issue completely unrelated to top-of-atmosphere considerations. Dividing by 4 is entirely a geometrical principal.
bdg…”Dividing by 4 is entirely a geometrical principal.”
If you are dividing by 4 you are spreading all of the solar energy incident upon the Earth over and entire sphere during one instant of time. You cannot divide 1 m^2 of the total solar energy by 4 and claim that as an average over the entire sphere.
Then you must sum each instant over a 24 hour period while considering that half the planet is not receiving solar energy at any one time. This is a very complex problem that takes in the tilt of the Earth and its orbit as well as heat dissipated on both the dark side and the heated side.
Dividing by 4 does nothing. You are right, it is purely a geometrical conversion factor for converting the ratio of the surface area of a circle to the surface area of a sphere.
GR said…”You cannot divide 1 m^2 of the total solar energy by 4 and claim that as an average over the entire sphere.”
Not only can you do it, but you HAVE to do it to calculate the correct amount of energy. The formula for energy is…
E = (S/4) * T * A
…where S is the solar constant, T is time, and A is area.
GR said…”Then you must sum each instant over a 24 hour period while considering that half the planet is not receiving solar energy at any one time.”
Yep. We are 2 steps of ahead of you. Already done. You even a posted a link in which they did this very thing and showed that the result was equivalent to just dividing the solar constant by 4.
Well said. See my layman’s, make far more real World sence, do you thing Big and Gordon ???
Wayne
bdg…”Not only can you do it, but you HAVE to do it to calculate the correct amount of energy. The formula for energy is…
E = (S/4) * T * A”
***********
You fail to understand the immensity of this problem.
For one, the factor 4 converts the surface area of a sphere to the surface area of a circle. It has absolutely nothing to do with the distribution of solar energy over the entire surface of a rotating planet with a tilted axis and an elliptical orbit.
To get the total solar energy available for the Earth you must first sum the one metre square areas to get the total power incident upon the surface. That will give you the equivalent of all 1360 W/one metre square areas reaching the surface, some of them striking dead on and some striking at variable angles as the the energy strikes toward the poles.
There is only one half of the sphere receiving solar energy of a variable amount as the other half radiates, convects, and conducts the received energy away.
If you think you can reduce this problem to a division by 4, you are seriously deluded.
Your equation is rubbish. There is no time factor in it, no heat dissipation factor, no allowance for tilt, rotation, and orbit.
A better approach is the one I suggested. Focus on one square metre of energy at TOA being projected onto the surface after passing through the atmosphere. I am damned sure the value found will not be (1360 W)/4.
GR said…”For one, the factor 4 converts the surface area of a sphere to the surface area of a circle.”
Nope. That’s not right. The factor 4 in the formula converts the surface area of a circle to the surface of a sphere. The solar constant is defined as the flux over a circular area. But, the Earth is actually a sphere. That’s why the conversion must happen.
GR said…”Your equation is rubbish. There is no time factor in it, no heat dissipation factor, no allowance for tilt, rotation, and orbit.”
The formula I posted does have the time factor (T). The amount of energy received by Earth at TOA is not dependent on heat dissipation, tilt, or rotation. It is, however, dependent on the orbit though.
GR said…”A better approach is the one I suggested.”
Yep. It’s a valid approach. I suggested do just this WAY up in this thread long before you did. Later on you posted a link to somewhere did it as well. And guess what? Yep, it turns out to be EXACTLY equivalent to dividing the solar constant by 4.
bdgwx, please stop trolling.
wayne said…”Geometry tells us that the surface area of a sphere is four times pi times the radius of the sphere squared. Has this something to do with it ???”
YES! It has EVERYTHING to do with this.
wayne said…”But I would think the Suns energy/heat is hitting close to half the area of the Earth at one time, in 24 hours it hits it all, thus why divide by 4 ???”
When the Earth is 1 AU from the Sun…
1360 W/m^2 is hitting the square meter closest to zenith.
680 W/m^2 is hitting the lit hemisphere.
340 W/m^2 is hitting the entire Earth.
…at any given moment in time.
bdg…”340 W/m^2 is hitting the entire Earth.
at any given moment in time.”
***********
How does solar energy hit the unlit side?
On average Gordon.
680 W/m^2 over 205e12 m^2 and 0 W/m^2 over 205e12 m^2 is equivalent to 340 W/m^2 over 510e12 m^2. Note that the area of Earth is 510e12 m^2.
On average Gordon…ON AVERAGE.
And it’s important to note that the solar constant itself is an average!
bdg…”And its important to note that the solar constant itself is an average!”
That has nothing to do with division by 4.
As far as averages are concerned, an average is a statistic and as Mark Twain pointed out there are 3 kinds of lies: lies, damned lies, and statistics.
This is like the global average temperature, nothing more than a number. It has no meaning whatsoever.
Same with this S/4 number, absolutely no meaning.
GR, do you agree that the energy hitting the Earth is E = (S/4) * T * A where S is the solar constant, T is the time period in question, and A is the area of Earth?
bdgwx, please stop trolling.
Your πr^2 and 4πr^2, areas of sphere dont work out in real World time. Also the light hitting the Earth, is basically half of the Earth. This is one of the problems I had with a few physicists working out power and force, in static lifts, but that’s another story. Give you another lay mans and thought experiment for you, showing you how wrong dividing by 4 is. Some physicists, and I am not saying you are like that, but some dont have common scene.
I wrote this a few weeks ago on Roy Spencer s site above.
They divide the energy by 4 as over a 24 hour period this would be the average amount of energy the entire surface receives. Its erroneous because in real time no area of the planet is actually receiving that specific amount of energy. The Poles are always receiving less than this, the equator and tropics receive much more than this for most of the day. Nothing on the night time side of the planet receives anything from the sun, so the entire exercise of averaging energy is meaningless.
If you put a chicken over a roasting heater at 120c and slowly rotated it for 24 min. you would cook the turkey. If you put 4 roasting heaters around the turkey at 30c and slowly rotated it for 24 min. the entire chicken received the same energy but averaged all at once over the entire bird, no part of the chicken would ever get warm, much less cook.
So, lets just say the heat/energy from the Sun is 16c, and it hits half of the Earth, and in 24 hours it hits all the Earth, how would we average the temp. Out ??? Hmm, we could say that half the Earths not being hit, so we could say the average = 8c, but that would not tell us much about the side being hit 16c to the side not being hit 0c ??? We NEED the FULL energy/heat hitting the Sun for 24 hours, NOT the average, that’s not what hitting the Earth in 24 hours, its the TOTAL energy/heat, and what would this be ??? Lets just say the heat moved around the Earth like a ticking watch, and gave out 16c of heat/energy every second, 16c x 24 = 384c, now of course you would have to average that out to real time heat/energy hitting each part of the Earth, 384c divided by 24 = 16c. As the Earths rotation is 24 hours, and the Suns energy/heat is constantly 16c and hitting the Earth. Yes I know the heat/energy would be less the further out you go, but you get my jest.
Wayne
Stole this from above. THE SUN HITS WILL FULL FORCE OVER THE WHOLE EARTH IN 24 HOURS, YOU CANT AVENGE THE FORCE NOT HITTING THE EARTH, THATS NOT WHATS HAPPENING IN 24 HOURS. WE NEED THE TOTAL FORCE IN 24 HOURS.
Dividing by 4 is fine for an average energy, but dividing the 960W by 4 = 240W, is then NO longer the solar flux. And, it can not be used as such. It no longer has any valid connection to temperature. Yes ??? Light bulbs ???
The solar constant, after albedo, is typically 960 Watts/m^2. If that flux perpendicularity impacted the black body flat surface, perfectly insulated on the back of it, the equilibrium temperature would be 361 K (87.6 C, 190 F).
But dividing by 4 yields 240 Watts/m^2, so the same surface corresponds to 255 K {-18 C, -1 F.} Total madness.
For you too see where I am coming from, but I am sure you see.
960 Watts/m^2 > 361 K {87.6 C, 190 F.}
240 Watts/m^2 > 255 K {-18 C, -1 F}
Dividing the solar flux by 4 is pseudoscience. Such a procedure fools some people, who then believes that the Sun cannot warm the planet !!!
Wayne
wayne said…”Dividing by 4 is fine for an average energy”
Yeah…so what’s the problem? These simple 3-layer box models are intended to be used for energy budget analysis. Nothing more.
wayne said…”but dividing the 960W by 4 = 240W, is then NO longer the solar flux.”
It may not be the instantaneous solar flux received at a specific location and time, but it is the global mean solar flux.
wayne said…”It no longer has any valid connection to temperature.”
It may not be valid to infer the instantaneous temperature at a specific location and time, but it is valid to infer the global mean temperature.
Here’s a question for you…do you believe a guy who tells you the Earth is 2x further away than it really is or do you believe the thousands of experts who correctly compute the orbital distance of Earth?
wayne said…”For you too see where I am coming from, but I am sure you see.
960 Watts/m^2 > 361 K {87.6 C, 190 F.}
240 Watts/m^2 > 255 K {-18 C, -1 F}”
Interesting…so if you don’t divide by 4 then you wrongly conclude that the Earth is near the boiling point. Does that make any sense to you? Do you want to reconsider whether dividing 4 sounds logical now?
And by the way, what do you think the 240 -> 255K connection is telling you? Why does the solar-only flux yield a temperature that is lower than it really is? Hmm…what could we be missing?
bdg…”wayne saidDividing by 4 is fine for an average energy
Yeahso whats the problem?”
**********
As I stated in another post, averaging is a statistic that can be seriously abused and wrongly applied. Unless the person doing the averaging understands the context in which the average is being applied, it is likely the average will mean nothing.
The UAH 0.13C/decade warming trend is a good example. Statistically-speaking, it is correct over the range from 1979 till present. However, it tells you nothing about the context, which is everything.
If you cool a room to 15C from an ambient temperature of 20C using an air-conditioner, then you turn on the furnace and heat it to 25C, you cannot claim the room has heated till you reach the 20C temperature.
Over the range of the UAH record, that’s essentially what has happened. Till nearly 1997, the planet was re-warming from a cooling period then an abnormal El Nino in late 1997 drove the global average up by 0.8C above the baseline. That was nearly a full degree C above the 1979 temperature.
After the EN receded, the global average flattened out for 15 years till another major EN drove the global average up nearly 1C. We are still recovering from that EN.
Alarmists have taken the UAH average seriously and claimed there has been 4 decades x 0.13C/decade warming DUE TO ANTHROPOGENIC CO2.
That is a blatant lie. No one understands the context behind this data or the long term effect of two major EN’s in 17 years. The trend has NOT actually been 0.13C per decade of TRUE warming in reality but it is on paper. Furthermore, there is no proof associating any warming with CO2.
It’s the same with your divide by 4 factor. This is a statistical average that denies and ignores the actual problem which is exceedingly complex. The energy budget is a fictitious work by statisticians who fail to understand the underlying physics.
bdgwx wrote; Here’s a question for you…do you believe a guy who tells you the Earth is 2x further away than it really is or do you believe the thousands of experts who correctly compute the orbital distance of Earth?
Not sure, no one has said what distance the solar flux is. its not whats hitting the Earth, its far higher up.
wayne said…”For you too see where I am coming from, but I am sure you see.
960 Watts/m^2 > 361 K {87.6 C, 190 F.}
240 Watts/m^2 > 255 K {-18 C, -1 F}”
bdgwx wrote; Interesting…so if you don’t divide by 4 then you wrongly conclude that the Earth is near the boiling point. Does that make any sense to you? Do you want to reconsider whether dividing 4 sounds logical now?
Wayne wrote; I did not say you don’t divided. As stated above, the solar flux does NOT hit the Earth like they are saying it does, hitting a flat disc, and cos a flat disc is four times the surface area of a globe, you then divide by four, that’s NOT what happens in the real World. The solar flux hits the globe on bascaly one side of it, yes the solar flux will be less the more its spread out, but as they don’t take this into account on their solar flux hitting a flat disc then dividing it by four, we don’t need to here. So we divided by two, and as the solar flux is at a higher temp. than what’s actually is further down on the Earth where we live, if the solar flux temp. was taken down at the surface, and then divided by two, I bet you my car against your, that the temp. would be not far off the average that we get for the whole Earth, whats that about 16c ???
Still dont get how your dividing the above ??? If you divide 87.6c by 4 to me you get 21.9c, not far off the average temp. of the Earth.
bdgwx saysAnd by the way, what do you think the 240 -> 255K connection is telling you? Why does the solar-only flux yield a temperature that is lower than it really is? Hmm…what could we be missing?
Wayne wrote; Cos you are wrongly working out the solar flux for a non rotating Earth, and your seeing the solar flux hitting a flat disc, then you divide by 4 ??? I mean that’s not science at all ???
Thank you for the debate, and friendly one. UK, Wales calling, but then you might have looked at our Web-Site.
Wayne
I “DONT” get how your devieing by 4 ??? 87.5c divided by 4 = 21.9c ??? That’s not far off, how do you get to -18c ???
YOU, divided by two, not by 4. And the solar flux would not be the real heat, is not the solar flux much further up in the air, thus lower down its going to be less ???
The solar flux, and constant. To determine the average amount of solar energy that reaches the Earth, we must consider what the Earth looks like to the Sun. When looking at Earth from the Sun, only one half of the Earth can be seen. Thus to make an appropriate estimate of the average amount of solar energy over the entire surface area of the Earth the value for Isc, must be divided by 2. And then as it hits only half the earth at once, you divide by two again. YES, I am with you there.
BUT, the Earth is rotating, so the averge of divide by 4 is NOT telling thus what we are needing to know, the TOTAl energy/heat hitting the Earth in say a full rotation, and in a full rotation the solar flux is hitting the whole Earth, right ???
WHAT your doing is averaging an Earth that does not rotate ??? That makes no sense when the solar flux will hit the whole Earth in 24 hours. Thats like saying, ho, I was laying in the Suns heat 20c for 24 hours, but it then got very cold for the other 12 hours, thus the averge heat was just 10c. But still in the 12 hours at 20c I got burn very bad. But if it was 10c for 12 or 24 hours you would not even get a tan. BUT, if you were in the Sun, with the whole Earth is for 24 hours, you would be so much more burnt than being in it for 12 hours at 20c.
1,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun just hit your all your face at 10c for 24 hours. This is what your averaging seems to doish.
{a very light tan if any}
2,
Lets say the Earth was your face, and your face did not rotate around the Sun. And your half your face had the heat of the sun at at 20c for 24 hours. This is what your averaging seems to doish.
{a very deep tan on half your face}
3,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun hit all your face at 20c for 24 hours. This is not your averaging and what happens in the real World.
{you would get a very deep tan all the way around your face}
IF, you agree, at that is the way they work out things, and why they say without the GHG’s it would be -18c, THERE IS SOMETHING VERY WRONG WITH THE WAY THEY WORK PHYSICS OUT FOR THIS. Lots of physicians have slipped up and overlooked this big time ???
Wayne
wayne said…”YOU, divided by two, not by 4. And the solar flux would not be the real heat, is not the solar flux much further up in the air, thus lower down its going to be less ???”
I divided by 2 to get the average flux over the lit hemisphere.
I divided by 4 to get the average flux over the full sphere.
wayne said…”WHAT your doing is averaging an Earth that does not rotate ???”
It doesn’t matter if the Earth is rotating or not. The amount of energy hitting Earth is always E = (S/4) * T * A. It doesn’t matter if T is 1 second, 24 hours, or 1 year. The amount of energy is ALWAYS defined by the formula above which divides S by 4.
bdgwx, please stop trolling.
Hadd to edit number two, cant seem to do it without posting again. AND ON NUMBER TWO THE WAY SOME PHYSICISTS ARE DOING IT, THE EARTH WOULD NOT WARM UP LIKE IT DOES, WITH THE SEAS AND THE EARTH ITSELF WARMING. YOU HAVE TO WORK OUT THE SOLAR FLUX ON A ROTATING EARTH, WHY THEY DO IT ON A FLAT DIEC AND NON ROTATING EARTH IS BEYOND ME.
WHAT your doing is averaging an Earth that does not rotate ??? That makes no sense when the solar flux will hit the whole Earth in 24 hours. Thats like saying, ho, I was laying in the Suns heat 20c for 24 hours, but it then got very cold for the other 12 hours, thus the averge heat was just 10c. But still in the 12 hours at 20c I got burn very bad. But if it was 10c for 12 or 24 hours you would not even get a tan. BUT, if you were in the Sun, with the whole Earth is for 24 hours, you would be so much more burnt than being in it for 12 hours at 20c.
1,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun just hit your all your face at 10c for 24 hours. This is what your averaging seems to doish.
{a very light tan if any}
2,
Lets say the Earth was your face, and your face did not rotate around the Sun. And your half your face had the heat of the sun at at 20c for 24 hours. This is what your averaging seems to doish.
{a very very deep tan on half your face}
3,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun hit all your face at 20c for 24 hours. This is not your averaging and what happens in the real World.
{you would get a very deep tan all the way around your face, and whole of your face would warm up and we would be able to live on it}
IF, you agree, at that is the way they work out things, and why they say without the GHG’s it would be -18c, THERE IS SOMETHING VERY WRONG WITH THE WAY THEY WORK PHYSICS OUT FOR THIS. Lots of physicians have slipped up and overlooked this big time ???
Wayne
wayne…”WHY THEY DO IT ON A FLAT DIEC AND NON ROTATING EARTH IS BEYOND ME”.
The answer is simple, according to Gerlich and Tscheuschner, two experts on thermodynamics. It cannot be done with state of the art computer power. The problem is far too complex to model, let alone calculate it.
The calculations would involve complex arrays of sine and cosine functions that would choke any known computer.
Current climate modeling is a sham based on invented physics and pulling the warming effect of CO2 out of a hat. Mainstream physics and other science has been abandoned, either because the climate scientists lack the ability to apply physics properly, or because they prefer consensus to the scientific method.
wayne said…”WHAT your doing is averaging an Earth that does not rotate ???”
The 3 layer box model described in this blog post is based on averages obtained over 1 sidereal year period. The Earth rotates on it’s axis 366.24 during this period. The averages represented by this simple model are actually more meaningful because the Earth rotates.
wayne said…”Thats like saying, ho, I was laying in the Suns heat 20c for 24 hours, but it then got very cold for the other 12 hours, thus the averge heat was just 10c. But still in the 12 hours at 20c I got burn very bad. But if it was 10c for 12 or 24 hours you would not even get a tan. BUT, if you were in the Sun, with the whole Earth is for 24 hours, you would be so much more burnt than being in it for 12 hours at 20c.”
Remember, we are discussing a simple 3 layer box model that is meant to represent the average flux over a 1 year period. If you want to analyze the diurnal cycle then this is not the analysis tool for you. The problem with a model that can accurately represent the diurnal cycle is that they are VERY complicated. The models that best describe the diurnal cycle are global circulation models. Are you certainly free to use one if you wish, but I think you’re going to find that they are overly complex for answering the questions that the 3 layer box model is designed to answer. You’ll bet the same answers either way though.
wayne said…”IF, you agree, at that is the way they work out things, and why they say without the GHG’s it would be -18c,”
What do you think the mean temperature of the Earth would be without an atmosphere? What about an nitrogen/oxygen atmosphere sans GHGs?
wayne said…”WHY THEY DO IT ON A FLAT DIEC AND NON ROTATING EARTH IS BEYOND ME”
These 3 layer box models assume a spherical planetary body that rotates.
Postma is the one that misrepresented the model. That’s why he gets the wrong answer for the average orbital distance of the Earth. That’s all his doing and his alone.
bdgwx, please stop trolling.
Hi there big.
Late here not much time. Your for some reason working out the Suns heat hitting a flat disc ??? The Sun does not hit a flat disc, it hits basically half the Earth at once ??? Your working out a flat disc, and cos its 4 flat discs to one round Earth you divide by 4. So you know the heat that’s hitting this flat disc, how do you work that out ??? As I thought the solar flux was the amount that hit one square meter several miles up in the air ???
I DONT get how your devieing by 4 ??? 87.5c divided by 4 = 21.9c ??? Thats not far off, how do you get to -18c ???
YOU, divide by two, not by 4. And the solar flux would not be the real heat, is not the solar flux much further up in the air, thus lower down its going to be less ???
The solar flux, and constant. To determine the average amount of solar energy that reaches the Earth, we must consider what the Earth looks like to the Sun. When looking at Earth from the Sun, only one half of the Earth can be seen. Thus to make an appropriate estimate of the average amount of solar energy over the entire surface area of the Earth the value for Isc, must be divided by 2. And then as it hits only half the earth at once, you divide by two again. YES, I am with you there.
ALSO, your version of the -18c, you have NOT counted in the heat that’s heating the seas and the land itself over the thousands of years have you ??? And other things, I mean even if the -18c was right, you HAVE to add this in, and that heat would be very high.
Wayne
http://www.drroyspencer.com/2019/06/on-the-flat-earth-rants-of-joe-postma/#comment-355878
Wayne you asked this first on June 4. Then many times since.
I see that it has been answered many times as well.
I see plenty of clear answers. One or two of mine.
If you havent understood by now…maybe you need to go off and read up on it.
Hi Nate,
Yes I did ask this before. Where did you answer me please ???
Right, Big wrote; “wayne said…”But I would think the Suns energy/heat is hitting close to half the area of the Earth at one time, in 24 hours it hits it all, thus why divide by 4 ???”
When the Earth is 1 AU from the Sun…
1360 W/m^2 is hitting the square meter closest to zenith.
680 W/m^2 is hitting the lit hemisphere.
340 W/m^2 is hitting the entire Earth.
…at any given moment in time.”
Wayne wrote; RIGHT, I get you. BUT, I think this is where your going wrong, and your average numbers just can’t work this physics equation out in the real World this doesn’t happen.
1,
Closest to zenith, where, what distance is this closest to zenith.
2,
has this number you have come to by averaging, taken into account the HUGE warming of the seas and land that the Sun has warmed up over the years ??? If not why ??? Your -18c has to add on to it this huge warming, yes ???
3,
340 W/m^2 is NEVER hitting the entire Earth at any moment. THUS, why even try to average this out. COS averaging this out can not work, all the other equeston are mute. 4 men in 4 rooms, 1 has £1000,000, one has £2000,000, one has £3000,000, and one has £0. The averge man in those rooms has £150,000 ??? But in reality one man have nothing ??? Averaging things in the real World does “NOT” work out. DO YOU ALL AGREE ???
4,
K, get you, this number is on a rotating Earth ??? 680 W/m^2 is hitting the lit hemisphere. You then divide by 2, this is where I think your going wrong. When 680 W/m^2 is hitting the lit hemisphere, no heat from the Sun is hitting the unlit side, so you must not divide. As we just want the heat hitting the Earth, and at any given time 680 W/m^2 is hitting the lit hemisphere. So just say I am right, we add 18c to -18c = 0c, add in the heat from the seas and land that’s got heated up over the years ??? BUT STILL HOLD ON, YOU GUYS ARE SAYING WITHOUT THE GHG’s it would be -18c RIGHT ??? SO ARE YOU TELLING ME ON A SUMMER’S DAYS AT SAY 25c, that, that heat I feel from that red ball in the sky is not from it ??? Please got to know your answer to this ???
5,Big wrote; What do you think the mean temperature of the Earth would be without an atmosphere. Just look at the Moon in daytime to answer that, the Earth would be boiling, as the atmosphere shades is so much from the Sun.
6,
“EDITED”.
a,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun just hit your all your face at 10c for 24 hours. This is what your averaging seems to doish.
{a very light tan if any}
b,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun hit all your face at 20c for 24 hours. This is not your averaging and what happens in the real World, “YES” ??? Actually this is what happens, that’s why averaging does not work in the real World, DO YOU ALL AGREE ???
{you would get a very deep tan all the way around your face, and whole of your face would warm up over time, and we would be able to live on it}
You can’t average, as 680 W/m^2 is hitting the Earth “ALL” the time.
Wayne
a,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun just hit your all your face at 10c for 24 hours. This is what your averaging seems to doish.
{a very light tan if any}
b,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun hit all your face at 20c for 24 hours. This is not your averaging and what happens in the real World, “YES” ??? Actually this is what happens, that’s why averaging does not work in the real World, DO YOU ALL AGREE ???
{you would get a very deep tan all the way around your face, and whole of your face would warm up over time, and we would be able to live on it}
You can’t average, as 680 W/m^2 is hitting the Earth “ALL” the time.
Wayne
1. It changes throughout the year. On the summer solstice in the NH (June 21st) it is the Tropic of Cancer (~23.5N).
2. No. The amount of energy received at TOA (top of atmosphere) has absolutely nothing to do with how the energy is distributed. And the no-GHG temperature of 255K is assuming that Earth’s albedo remains the same thus allowing only 240 W/m^2 of the 340 W/m^2 to be used for heating. Obviously being only a hypothetical scenario it assumes that the Earth never had a GHG effect and thus 255K was the final equilibrium. Though I’m not sure it matters either way because if you start with 289K and then remove the GHE the equilibrium would still establish itself at 255K after the Earth cooled due to the removal of GHGs.
3. Sure it works out. If I know the average and I know the number of men I can compute the total wealth. It is 4 * 150000 = 600000. Likewise, if I know the average flux and I know the total number of square meters and amount of time I can compute the total energy. It is E = (S/4) * A * T. You have to remember that S is already a temporal average so it’s not like extending it to a spatial average should be any more or less offensive especially considering that it is mathematically correct regardless how much offense someone takes. All we’re doing is converting it from a 2D circular plane representation to a 3D sphere representation. That’s necessary since the Earth really is a 3D sphere.
4. Yes. We are all well aware that the lit hemisphere receives 680 W/mi^2 on average and the unlit hemisphere receives 0 W/m^2 at any instantaneous moment when the Earth is 1 AU from the Sun. That’s what gives Earth its diurnal effect. That’s not being questioned by these 3 layer box models. We aren’t saying that every location on Earth will be 255K without GHGs. We are saying that the global mean temperature would be 255K. That’s a completely different statement. Interestingly the diurnal temperature range without GHGs would be much higher. This means daily highs would be higher. But daily lows would be lower as well. And overall the daily mean temperature would be lower. Look no further than the Moon for how this effect plays out in real life under 340 W/m^2 of average flux. And the heat you feel on a clear day is from the Sun. None of that changes because of GHGs.
5. Yes. The lit side would be boiling. But the unlit side would be insanely cold. The mean over the entire surface is still lower overall. Remember, the statements about GHGs making a planet warmer are in reference to the mean temperature only.
6. Of course the side actually facing the Sun will be warmer than the side facing away. We are all very aware of that fact. But a global mean temperature does not preferentially select just the lit side or the unlit side. A global mean temperature is…well…global. And I would never say that 680 W/m^2 is averaged over the entire planet because it’s not. It’s 340 W/m^2 that’s average over the entire planet. That’s the value you use if you want to do energy budget analysis over the course of a full sidereal year which involves 366.24 rotations (or 365.24 relative to the Sun). Remember, these simple 3 layer box models are only meant for “big picture” average yearly energy budget analysis. You can’t use them to analyze diurnal or local temperature variations. For that kind of analysis you need more complex models which do exist.
Hi Big, I really appreciate your answers here, thank you for everything.
OK, lets get back to the basics. Lets try and work this one out in more real World ways, on what actually happens. If you think this scenario is not right and you have another one please say. I think this is right, but will think it over and see if its not right in some way. We will concentrate on 24 hours, seems fair ???
a,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun just hit your all your face at 10c for 24 hours. This is what your averaging seems to doish. BUT, the average does NOT apply here, as the FULL force of 680 W/m^2 is hitting all parts of the Earth in 24 hours.
{a very light tan if any}
b,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun hit all your face at 20c for 24 hours. This is not your averaging and what happens in the real World, YES ??? Actually this is what happens, thats why averaging does not work in the real World, DO YOU ALL AGREE ???
{you would get a very deep tan all the way around your face, and whole of your face would warm up over time, and we would be able to live on it}
You cant average, as 680 W/m^2 is hitting the Earth ALL the time. Do “YOU” agree averaging does not work in my example, thus how can it work in yours ???
c,
HAS these three modals taken into account that the Sun has warmed up the seas and land by a huge amount through the years ???
d,
has your modals taken into account the heat rising from inside the Earth ??? Also the heat of volcanoes and such ???
e,
THIS IS WHERE I SAY YOUR GOING WRONG, THE EQUATION AVERAGE, CANT BE USED IN THIS SITUATION, YOU NEED ANOTHER ONE. Big wrote; Sure it works out. If I know the average and I know the number of men I can compute the total wealth. It is 4 * 150000 = 600000. BUT THIS IS WHAT I AM TRYING TO SAY, in the real World one man have 0, but your your World this man who has nothing has a 150,000, BUT, thats not true is it ??? And that man does not have 150,000, and never has 150,000, he has 0. THUS WHAT I AM SAYING, THIS AVERAGING DOES not WORK IN REAL wORLD SITUATIONS. Do you agree that one man will always have 0 ???
Wayne
Of course a global average is inappropriate to use for analyzing a specific location at a specific time. No one (except maybe Postma) is suggesting that we take the global mean flux of 340 W/m^2 or the global mean temperature of 15C and pretend like every spot on Earth at every moment in time would observe those exact values. No reputable climate scientist has or ever would suggest otherwise. If you want to know what effect the Sun has at specific location at a specific time then a 3 layer box model isn’t going to help you at all in answering that question.
What scientists are doing with these simple 3 layer box models is mainly to help understand the movement of energy in the climate system on a *global* scale and over long periods of time. Any link between the average fluxes and average temperatures are only on a *global* scale. Scientists aren’t suggesting that there isn’t a diurnal cycle nor are they suggesting that atmospheric properties (temperature, WV mixing ratios, wind speeds, etc.) are the same everywhere.
Regarding a and b above…is your face rotating such that it only receives sunlight 50% of the time. It wasn’t clear which scenario had rotation about and which didn’t. When you say “rotated around the Sun” I interpret that as orbiting around the Sun which is different than rotating about an axis.
Hi Big, thank you again for your answers.
I never thought you was suggesting that we take the global mean flux of 340 W/m^2 or the global mean temperature of 15C and pretend like every spot on Earth at every moment in time would observe those exact values.
What I was suggesting, for too see what is actually happening in the real World, averages don’t seem to work in lots of cases, as in my one man has a Million, other man has a Million, other man has a Million, other man has nothing, averge this out all men have 1/4 of a Million, but in the real World one man still has nothing.
I thought you said without the atmosphere, the Earth would be -18c, so you’re suggesting the atmosphere puts the temp. up from -18c to 15c, up by 33c ??? And your suggesting is the atmosphere has put the temp. up by 33c, and that heat I feel on a summer’s day when its 30c, the heat I feel on my face from the red ball in the sky is not actually the Sun, but the atmosphere ??? That sounds mad, as when the clouds go over the Sun, its instantly far less burning on my face ??? But your saying without the atsmohre it would only be -18c, thus the heat on my face is not the Sun but the atmosphere ???
Lets say you are right for a minute, and that’s without the atmosphere its -18c, and that the atmosphere has warmed it up 33c, BUT, have you added as I asked the huge amount of heat the seas and land of the Earth gives out, after the Sun has warmed it up all these years, and the heat coming out from the Earth itself from the inside ??? IF YOU HAVE NOT ADDED THESE IN, your numbers are definitely NOT right, as you know there’s huge amount of heat in the seas and land.
SO I QUESTION YOU ON THIS AGAIN.
Regarding a and b above, is my face rotating such that it only receives sunlight 50% of the time. “NO”, My face is rotating and orbiting exactly as the Earth does the Sun.
THIS IS WHY I SAY, IN THIS CASE, AVERAGING CAN NOT WORK.
a,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun just hit your all your face at 10c for 24 hours. This is what your averaging seems to doish. BUT, the average does NOT apply here, as the FULL force of 680 W/m^2 is hitting all parts of the Earth in 24 hours.
{a very light tan if any}
b,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun hit all your face at 20c for 24 hours. This is not your averaging and what happens in the real World, YES ??? Actually this is what happens, thats why averaging does not work in the real World, DO YOU ALL AGREE ???
{you would get a very deep tan all the way around your face, and whole of your face would warm up over time, and we would be able to live on it}
You cant average, as 680 W/m^2 is hitting the Earth ALL the time. Do YOU agree averaging does not work in my example, thus how can it work in yours ???
Wayne
I’m still unsure about scenarios A and B. I get that your face is orbiting the Sun, but is your face rotating such that 50% of the time it is lit and 50% the time it is unlit? Which scenario is it rotating? A or B?
And by the way, I’m still not saying that 680 W/m^2 is hitting the Earth on average. I’m saying 340 W/m^2 is hitting the Earth on average.
Your saying the atmosphere heats us up to 33c more as otherwise you think it would be -18c, BUT, I say the atmosphere shaded us from the Sun, look at the Moon in day, its 140c ???
Lets say im right, and its 680 W/m^2 hitting the Earth, but the atmosphere shades us a lot from this heat. And that the Sun as we know have heated up the seas and land so much over the years. This seem more like whats happened in the real World ??? As in your example, the heat should be there all the time from the GHG’s the atmosphere, BUT, as soon as you get cloud cover it goes so much cooler, same on a solar eclipse, it goes very cold so fast ???
And we have a problem on your example, as years ago the Co2 levels have been so much higher, but temp. lower ???
But lets keep to the top part for now. How averaging does not seem work out.
Wayne
wayne said…”Your saying the atmosphere heats us up to 33c more as otherwise you think it would be -18c, BUT, I say the atmosphere shaded us from the Sun, look at the Moon in day, its 140c ???”
I’m only saying that the global mean temperature would be ~255K if you removed GHGs and the albedo stayed the same. In reality removing GHGs would almost certainly change the albedo. Using estimates from Wild et al. 2013 you might say the removal of H2O might add 76 W/m^2 to the downward flux from the Sun and remove the 84 W/m^2 of evaporation loss. This would bring the incoming surface flux up to 320 W/m^2 and if you subtract of sensible heat loss you’re left with about 300 W/m^2. That would yield a BB temperature 270K (-3C). To summarize…no GHG with the same albedo would be 255K (-18C) and no GHG with more realistic albedo would be 270 (-3C).
The atmosphere DOES shade the surface from the Sun a bit. Again using the estimates from Wild et al. you can conclude there is an -80 W/m^2 shading effect. This is part of the reason why maximum temperatures on Earth are less than on the Moon. However, that’s a double edged sword as the atmosphere also redistributes the heat and insulates the Earth leading to significantly higher minimum temperatures. The insulating and redistribution effect is much greater than the shading effect. That’s why the *mean* temperature is higher overall here on Earth as compared to the Moon.
wayne said…”Lets say im right, and its 680 W/m^2 hitting the Earth,”
Careful…680 W/m^2 is hitting the lit hemisphere. 0 W/m^2 is hitting the unlit hemisphere. 340 W/m^2 is hitting the Earth.
wayne said…”And that the Sun as we know have heated up the seas and land so much over the years. This seem more like whats happened in the real World ???”
Yes. Of course!
wayne said…”As in your example, the heat should be there all the time from the GHG’s the atmosphere, BUT, as soon as you get cloud cover it goes so much cooler, same on a solar eclipse, it goes very cold so fast ???”
Clouds are tricky. They can both shade the Earth and insulate it depending on the height and time of day in which they occur. Clouds by day put a downward pressure on mean temperatures. Clouds by night put an upward pressure on mean temperatures. Low clouds have a net downward pressure on mean temperatures. High clouds have a net upward pressure on mean temperatures.
Regarding solar eclipses…I happen to be lucky enough to live in the path of totality during the 2017 eclipse. It was a great show! Anyway, as is typical of all solar eclipses temperatures on a clear day don’t bottom out until 30-60 minutes AFTER solar radiation has essentially gone to 0 W/m^2. Why the delay? There are many reasons but a significant factor is the GHE! Land and bodies of water heat during the lead up to peak eclipse and they continue to radiate throughout the event and warm the atmosphere just above the surface. GHGs return a large portion of this energy back to the surface and maintain the temperature for a much longer period of time. Without GHGs the lag between peak eclipse and temperature drop would be shortened dramatically; perhaps to as little as a few minutes or just long enough to get the surface in equilibrium with the air temperature 1 m off the ground.
wayne said…”And we have a problem on your example, as years ago the Co2 levels have been so much higher, but temp. lower ???”
We have to go back millions of years to find an era when CO2 levels have been higher. But this is an interesting line of discussion nonetheless. Yes, CO2 levels have been higher in the distant past with lower temperatures. The biggest reason why is because the Sun was dimmer in the distant past. The Sun, like all main sequence stars, brightens as it ages. The rate at which it brightens is about 1% ever 120 million years. 600 MYA ago the Earth received -12 W/m^2 less radiation than today. To make up for that shortfall CO2 concentrations would have had to produce a +12 W/m^2 offset which is equivalent to about 2500 ppm to maintain an equivalent radiative balance we have today.
wayne said…”How averaging does not seem work out.”
Averaging certainly isn’t going to help answer all questions releated to the climate, but it is extremely helpful in answering some questions easily especially concerning global scale questions like the above related to the faint young Sun paradox.
Hi and thank you again.
Just a quick one, back with more later.
you dont seem to believe in that averages dont work like I do ???
You seem to be saying that 680 W/m^2 is hitting the Earth, then you just divide by 2 ??? BUT, all the heat that hits this one side of the earth, lots of this heat stays there when the Sun moves on, stays in the air, and slowly cools down throughout the night. Lots of this heat, heats up the seas and Earth as well. Then we have to add in the heat from deep inside the Earth, as in molten, and the volcanos as such.
THIS IS WHY I SAY, IN THIS CASE, AVERAGING CAN NOT WORK.
a,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun just hit your all your face at 10c for 24 hours. This is what your averaging seems to doish.
BUT, the average does NOT apply here, as the FULL force of 680 W/m^2 is hitting all parts of the Earth in 24 hours.
{a very light tan if any}
b,
Lets say the Earth was your face, and your face rotated around the Sun. And your face rotated all the way around the sun, and the Sun hit all your face at 20c for 24 hours. This is not your averaging and what happens in the real World, YES ??? Actually this is what happens, thats why averaging does not work in the real World, DO YOU ALL AGREE ???
{you would get a very deep tan all the way around your face, and whole of your face would warm up over time, and we would be able to live on it}
You cant average, as 680 W/m^2 is hitting the Earth ALL the time. Do YOU agree averaging does not work in my example, thus how can it work in yours ???
THIS IS WHERE I THINK YOU GOING WRONG.
big wrote;
Im only saying that the global mean temperature would be ~255K if you removed GHGs and the albedo stayed the same. In reality removing GHGs would almost certainly change the albedo. Using estimates from Wild et al. 2013 you might say the removal of H2O might add 76 W/m^2 to the downward flux from the Sun and remove the 84 W/m^2 of evaporation loss. This would bring the incoming surface flux up to 320 W/m^2 and if you subtract of sensible heat loss youre left with about 300 W/m^2. That would yield a BB temperature 270K (-3C). To summarizeno GHG with the same albedo would be 255K (-18C) and no GHG with more realistic albedo would be 270 (-3C).
Your averaging again ??? But averaging cant work in this situation, your saying the Earth is on average 340 W/m^2, that’s fine, BUT, is does not work like this in the real World. Try this one. The suns heat is highest on the equator. Lets just say the heat coming from the Sun on the equator is like a laser beam, and carves a small line into the Earth, as this laser beam light is 680 W/m^2 it carves a line deep right around the Earth in 24 hours, as the total heat = 680 W/m^2.
Your divideing this number and getting 340 W/m^2, if you divide this 680 W/m^2 by 2, you will NOT get line as deep. I dont think your seeing this ??? You will not get a line as deep, thus your equation as in averaging, does not work in the real World. WHAT we want is the total average heat, temp. thats hitting the Earth all the time, and thats 680 W/m^2. You have to forget about averge completely, as its throwing everyone. Two rooms, one is 30c and the other is 0c, your saying the average temp. in both rooms = 15c, BUT, its never 15c in one room, and never 15c in the other room. The 680 W/m^2 is moving over the Earth all the time, and its warming the Earth up over time, and some of this heat stays in the air. Also there is heat coming out of the Earth from the molten rocks and things.
WHAT WOULD HAPPEN IF YOU DIVIDED, night time temps. ??? That would be the wrong answer right ??? As its not always night, its day.
We could say one side of the Earth = 680 W/m^2 and the other side is 340 W/m^2, {ACTUALLY THIS MIGHT WORK, and average this number up, as this is more like what actually happens in the real world} divide 680 W/m^2 and 340 W/m^2 = 510 W/m^2, add in the Sun has heated up the seas and land over time, and the heat coming out of the Earth from molten rock ???
Wayne
Wayne
Wayne saidAnd that the Sun as we know have heated up the seas and land so much over the years. And the heat coming out of the Earth as in molten rock. HAVE you added these into you equations, it seem you have not. But I think you will agree, you have to add these in ???
Wayne
Not sure how in another way I can get over the average does not work. As its never 340 W/m^2 over all the Earth, ever, its just about always daylight = warm, night = coldish.
Wayne
It’s not that much different than any other average. For example, the average life expectancy is 72 years. That doesn’t mean you’re going to die at 72. Each person has different risk factors for death. That doesn’t make the average any less useful.
Hi there,
I “THINK” I said this one before ??? Whats the averge temp. of the half of Earth the Sun is hitting in c ??? And whats the whats the temp. of the half the Earth the suns not hitting in c ??? Add those together, and this “WOULD” be the averge temp. of the Earth yes ???
So it would be 15c of the lite half, and say ??? 6c for the unlite half, thus averge temp. on Earth would be more like 10.5c ??? Sounds about right ???
Add in the heat that’s in the seas and land, and the heat coming out of the Earth as molten rock ??? Not sure how much heat this is, in c ???
Wayne
The diurnal temperature range (DTR) over land is about 11C. It’s quite a bit less over the ocean. If the mean on the lit side were say 18C and the mean on the unlit side were say 12C then the global mean would be 15C. The math works out to 15C for 19/11, 20/10, 16/14, etc. combinations as well. The global mean can increase even if daily highs decrease. This would happen if the daily lows increased more than the daily highs decreased.
RIGHT, BEEN OUT FOR A DRIVE AND THINK I GOT IT. Sameish as above.
Wayne saidLets say im right, and its 680 W/m^2 hitting the Earth,
Big wrote; Careful680 W/m^2 is hitting the lit hemisphere. 0 W/m^2 is hitting the unlit hemisphere. 340 W/m^2 is hitting the Earth.
1,
680 W/m^2 is hitting the lit hemisphere. 0 W/m^2 is hitting the unlit hemisphere. 340 W/m^2 is hitting the Earth.
“HOWEVER”, we have to add on heat here, we can’t just say 0 W/m^2 is hitting the unlit hemisphere. As there is still heat in the unlit hemisphere, from the Suns heat in the day, as in just the air, this heat is slowly cooling and rising. HOW much heat is left in the unlit hemisphere ??? We “HAVE” to add this in, do you agree ???
2,
We also have to add in heat from the seas and land that’s being warmed over the years. How much more heat should we add in for that ??? Then the heat we must add in from the heat thats coming from the molten rock, and volcanoes all over the Earth and in the Seas, “HAVE” you included this heat ???
So if you can add in these heats, and add them all up, what c would the average Earth be then ???scientifically
3,
As do you agree that if we don’t add in the heats from 1 and 2 above, then your 680 W/m^2 is hitting the lit hemisphere, and dividing by 2 “IS SCIENTIFICALLY VERY WRONG”, do you agree ??? As your only adding up the heat/energy of 680 W/m^2 that is hitting the lit hemisphere, BUT there is more heat to add on here, do you agree ???
Wayne
1. We were discussing incoming solar radiation at TOA. It is 0 W/m^2 on the unlit side. Review Wild et al. 2013 for estimates on the average of various outbound fluxes if you are curious.
2. Geothermal is about 0.1 W/m^2. Tidal dissipation of the Earth/Moon system is about 0.01 W/m^2. Obviously these are small fractions of the 340 W/m^2 of solar at TOA (or 240 W/m^2 at the surface). Again, review Wild et al. 2013 for estimates on all of the various inbound and outbound fluxes.
3. Geothermal and tidal energy sources are so small compared to solar radiation that they can be largely ignored. Dividing the solar constant by 2 to get the average over the lit hemisphere or 4 to get the average over the whole Earth is scientifically compulsory. You’ll get the wrong answers to energy budget questions if you don’t divide by 4.
Hi there big. You know a lot about this subject, is this part or your job, is so what is your job please, or if not is it just an interest ???
But you did not answer a few of my questions. I am trying to explain toy you that the equation averge, can not work in this example. One man on the lit side has £100,000, other man on the unlit side has £0. if you averge these out, both men have £150,000. BUT, never in the real World does both men have £500,000 each ??? In the real World one man only has £100,000, other man has £0.
Lets try another way. AS WE NEED THE TOTAL HEAT/ENERGY. You know your 680 W/m^2 that is hitting the lit hemisphere, and you dividing this in two, your talking of the heat/energy for 24 hours, right ??? So you divide 680 W/m^2 that is hitting the lit hemisphere, by nothing thats hitting the unlit, yes ??? And you get 340 W/m^2 in 24 hours am I right so far ???
If I am right, I dont get you, as the solar flux goes all the way around the Earth in 24 hours, constantly giving out 680 W/m^2. So this 680 W/m^2 is the total heat/energy been giving out in 24 hours, so why divide this number ???
Try this. I have a plane, and go around the World in 24 hours, I shine a beam of light out constantly. To shine a beam of light on one side of the Earth for 12 hours takes a 100 Watts of energy, I move to the other side of the Earth, and shine a beam of light on other side of the Earth for 12 hours takes a 100 Watts of energy. You cant HALF the energy used, 200 Watts is the total energy. What your doing is taking the energy used for 12 hours it seems
Its the same as 680 W/m^2, that’s hitting half the Earth in 12 hours, it then stays at 680 W/m^2, and hits the other side of the Earth. At “NO” time it the power “EVER” 340 W/m^2 on both sides of the Earth. Its “ALWAY” 680 W/m^2 going all around the Earth in 24 hours.
I drive my car around the Earth, it takes 1000 gallons to drive it half the way around the Earth in 12 hours. You don’t then divide this number by 2 to say the average gallons I used was 500, cos that does not make sense, just like dividing the 680 W/m^2 by 2 does not make any sense in the real World.
Your just adding the solar flux for 12 hours hitting one side of the Earth, then dividing it by 2 ??? You may as well take the 0 W/m^2 hitting the unlit side and divide this by 2, = 0 W/m^2
Whats the solar flux hitting one square meter of the Earth ??? Times this by how many square meters on the Earth, and you get your total W/m^2.
I just dont get how you dont use the total flux hitting the Earth for 24 hours, but you use half of this ???
Wayne
Watch what happens here. We know 680 W/m^2 is continuously hitting the Earth; but only 1/2 of the Earth at a time. It’s always this way regardless of how much time we are considering. It can be 1 second, 1 hour, 24 hours, or 365 days. No matter what the Earth is constantly exposed to that 680 W/m^2 (on average) beam. But again, only 1/2 of the Earth gets it at a time. And doesn’t matter in the slightly if the Earth is rotating or not.
So the total amount of energy received in a specific amount of time is:
1 second = 680 W/m^2 * 205e12 m^2 * 1 s = 1.4e17 joules
1 hour = 680 W/m^2 * 205e12 m^2 * 3600 s = 5.0e20 joules
1 day = 680 W/m^2 * 205e12 m^2 * 3600 * 24 = 1.2e22 joules
…which is exactly the same as…
1 second = 340 W/m^2 * 510e12 m^2 * 1 s = 1.4e17 joules
1 hour = 340 W/m^2 * 510e12 m^2 * 3600 s = 5.0e20 joules
1 day = 340 W/m^2 * 510e12 m^2 * 3600 * 24 = 1.2e22 joules
Notice that when using 680 W/m^2 I use 1/2 of the Earth or 205e12 m^2, but when using 340 W/m^2 I use the whole Earth or 510e12 m^2. It works out exactly the same either way.
Regarding your plane and light thought experiment…
1st side: 100 W * (12 * 3600 s) = 4320000 joules
2nd side: 100 W * (12 * 3600 s) = 4320000 joules
Total: 4320000 + 4320000 = 8640000 joules
And the average power over 24 hours is 8640000 / (24 * 3600) = 100 W!
Watts is a unit of power or j*s-1 (joules/second)
Joules is a unit of energy or kg*m2*s-1.
Make sure you’re not conflating power with energy. The statement “100 W of energy” and the like is technically incorrect terminology.
However, we can use units that are common electricity billing…the watt-hour. When using these units the energy works out to…
1st side: 100 W * 12 h = 1200 W-hours
2nd side: 100 W * 12 h = 1200 W-hours
Total: 1200 W-hours + 1200 W-hours = 2400 W-hours.
Hi Big, is this part or your job, or just an interest ???
So you seem to agree with me here ??? First side 4320000 joules, second side 4320000 joules, total energy used by lets say the Sun = 8640000 joules. THIS IS WHAT I HAVE BEEN TRYING TO GET TO YOU. We need the total energy thus heat hitting the whole of Earth, not the averge.
{I don’t get, or was it type error, where you said the average was 8640000 joules ??? Average would be; 4320000 joules ???}
Big wrote; Regarding your plane and light thought experiment
1st side: 100 W * (12 * 3600 s) = 4320000 joules
2nd side: 100 W * (12 * 3600 s) = 4320000 joules
Total: 4320000 + 4320000 = 8640000 joules
And the average power over 24 hours is 8640000 / (24 * 3600) = 100 W!
Watts is a unit of power or j*s-1 (joules/second)
Joules is a unit of energy or kg*m2*s-1.
Make sure youre not conflating power with energy. The statement 100 W of energy and the like is technically incorrect terminology.
However, we can use units that are common electricity billingthe watt-hour. When using these units the energy works out to
1st side: 100 W * 12 h = 1200 W-hours
2nd side: 100 W * 12 h = 1200 W-hours
Total: 1200 W-hours + 1200 W-hours = 2400 W-hours.
Wayne wrote; I don’t think you should be dividing in the below. I have a torch shining a light on half the Earth, it uses a 100E to put out a 100C, I have an energy bill of 100E from the torch/Sun, and I have a thermometer to tell the people of the Earth the temp. with is 100c.
I cant tell the torch/Sun I am averaging out your energy and heat as your only hitting one side of the Earth, I have to give the Sun the bill for a 100E, as my thermometer tells me the Suns heat was 100c
I do work for a firm and they pay me by hole much heat I produce, I produce 100E, and dig 100 holes producing 100C, in 12 hours, in the night I do nothing. The firm and myself, don’t want the averge I work in 24 hours 50E or 50 holes and 50C. they want the total energy I have used, and the total heat I have produced.
Sun produces 100E, and in turn hits the Earth with a 100C, you should not be interested whats happening on the other side of the Earth ??? I have a room using 100E, and its heated up to 100C, you have a room using 0E and heated up to 0C. averaging both rooms out, does “NOT” tell us what is happening in my room, its not what is happening in the real World.
I know its hard to say your wrong, maybe your not wrong, and maybe I am. but please could you spent several hours thinking my points over, and then if you still dont see my point, try and explain why you think you need to half energy bills from the Sun, and the heat they produced for you in 12 hours ???
I run for 12 hours, produce 100E, and this in turn warms half the Earth to 100C, I then do nothing for 12 hours. Still I have used 100E and produced 100C in these 24 hours. “NOT” ever 50E or 50C ??? WE NEED THE TOTAL E FROM THE SUN, AND THE TOTAL H, NOT HALF OF IT.
Wayne
wayne said…”At NO time it the power EVER 340 W/m^2″
To be pedantic there really are locations that are receiving 340 W/m^2. It’s just not happening everywhere. The same can be said for 680 W/m^2 as well. Except there are fewer square meters receiving 680 as compared to 340. We can extend this to 1360 W/m^2 as well except now there is only a single square meter that is receiving this amount.
But I do understand the intent of your statement here. The crux of the matter has to do with the average beam flux. The beam flux is 680 W/m^2 relative to a half-sphere. Only half of the Earthly sphere is pointed into the beam. The other half receives zero. You are focused only on the lit half. I totally get that. That’s completely okay as long as you are aware of the caveats; the main caveat being that is valid only for 1/2 of the Earth at any instant of time.
wayne said…”I drive my car around the Earth, it takes 1000 gallons to drive it half the way around the Earth in 12 hours.”
Let’s do some calculations with this thought experiment. The distance around the equator is 25000 miles. If you drive 12500 miles on 1000 gallons then you are averaging 12.5 miles per gallon. If this average is maintained you will need a total of 25000 / 12.5 = 2000 gallons to go all the way. In this case you would multiple by 2 to get the correct answer.
So what’s the difference? Why multiple in this case? The reason is because you are repeating an effect. Why divide in the solar flux case? The reason is because you are spreading out the effect. Or said another way 1000 gallons represents 1/2 of a whole while the 1360 W/m^2 solar constant is 4/1 of the whole.
I think the crux of the confusion is in how scientists decided to define the solar constant. Here are some proposals for alternate definitions that might help clear things up…I don’t know.
Ss = solar constant on a sphere = 340 W/m^2
Sd = solar constant on a dome = 680 W/m^2
Sp = solar constant on a plane = 1360 W/m^2
Some shapes, like a cylinder or a pyramid, have different cross sections depending on how they are oriented so I can’t easily give you the solar constant for these shapes. I’d have to know how the shape was oriented and even then cross section might be difficult to calculate.
Hi Big. Have to say thank you again for all this.
Ok try this one. Its not a trick question, or maybe it it. Simple physics question.
I have a energy heat source, that shines light to heat things up. You ask me to heat your garden up. I use 100E in 12 hours, in heating your garden to 100C, for 12 hours.
How much E have I used in 12 hours, and how much C have I heated up your garden to in 12 hours ???
Wayne
You have used 100E and the temperature increased by 100C – X in 12 hours where X is the unknown initial temperature.
This works out to an average rate of change of 8.3 E/hour and (100-X)/12 C/hour.
Wayne wrote; I have a energy heat source, that shines light to heat things up. You ask me to heat your garden up. I use 100E in 12 hours, in heating your garden to 100C, for 12 hours.
How much E have I used in 12 hours, and how much C have I heated up your garden to in 12 hours ???
Big wrote; You have used 100E and the temperature increased by 100C X in 12 hours where X is the unknown initial temperature.
This works out to an average rate of change of 8.3 E/hour and (100-X)/12 C/hour.
Wayne wrote;
Now we are getting somewhere. So what if I said your garden was half a globe, I ask the same question; I have a energy heat source, that shines light to heat things up. You ask me to heat your garden up, its half the Earth, I only shine on half the Earth. I use 100E in 12 hours, in heating your garden to 100C, for 12 hours.
How much E have I used in 12 hours ???
Wayne
Hi Big,
Also try this one.
A 100 watt Sun uses 0.1 kilowatts each hour to shine energy/heat on a “flat” disc, to heat it to 100c. It would take 12 hours for the Sun to consume 1 kWh of energy, to heat the flat disc to 100c for 12 hours. If your Sun was a company, and charges you a rate of 1, for each kWh, you would pay 12 to use a light every 12 hours.
Questions,
1, How much energy/heat does the sun use in 12 hours, to heat the flat disc ???
2, How much c, does the Sun heat the flat disc up to for the 12 hours ???
3, How much do I pay the Sun for its service, for 12 hours ???
Wayne
1. 0.1 kW * 12 h = 1.2 kWh
2. 100C
3. 1.2 kWh * 1 $/kWh = 1.2 $
wayne said…”If your Sun was a company, and charges you a rate of 1, for each kWh, you would pay 12 to use a light every 12 hours.”
As shown above the actual amount you pay during a 12 hour period is 1.2; not 12.
Hi Big,
Yes your right, sorry typo.
A 100 watt Sun uses 0.1 kilowatts each hour to shine energy/heat on a flat disc, to heat it to 100c. It would take 12 hours for the Sun to consume 1 kWh of energy, to heat the flat disc to 100c for 12 hours. If your Sun was a company, and charges you a rate of 1, for each kWh, you would pay 12 to use a light every 12 hours.
Questions,
1, How much energy/heat does the sun use in 12 hours, to heat the flat disc ???
2, How much c, does the Sun heat the flat disc up to for the 12 hours ???
3, How much do I pay the Sun for its service, for 12 hours ???
Wayne
Big wrote;
1. 0.1 kW * 12 h = 1.2 kWh
2. 100C
3. 1.2 kWh * 1 $/kWh = 1.2 $
wayne saidIf your Sun was a company, and charges you a rate of 1, for each kWh, you would pay 12 to use a light every 12 hours.
As shown above the actual amount you pay during a 12 hour period is 1.2; not 12.
Wayne wrote;
So you agree that the energy used was 1.2KWh ???
BUT, I thought you would average his out ??? As in reality, in the real World, this flat disc must have three sides like a coin. So if these sides are equal in area, You, not me would suggest the whole coin has an average of 0.4KWh. BUT, the energy/heat is 1.2KWh on just one side for 12 hours.
I understand you averaging it out, but in reality the energy.heat is 1.2KWh, “NOT” 0.4KWh……….YES ???
Wayne
wayne saidI drive my car around the Earth, it takes 1000 gallons to drive it half the way around the Earth in 12 hours.
Big wrote; Lets do some calculations with this thought experiment. The distance around the equator is 25000 miles. If you drive 12500 miles on 1000 gallons then you are averaging 12.5 miles per gallon. If this average is maintained you will need a total of 25000 / 12.5 = 2000 gallons to go all the way. In this case you would multiple by 2 to get the correct answer.
So whats the difference? Why multiple in this case? The reason is because you are repeating an effect. Why divide in the solar flux case? The reason is because you are spreading out the effect. Or said another way 1000 gallons represents 1/2 of a whole while the 1360 W/m^2 solar constant is 4/1 of the whole.
This is where think your going wrong.
But with the solar constant you’re NOT spreading out the effect ??? The solar constant is only on one side of the Earth, its not spread out over the Earth ??? If you freeze time, the solar constant is just over one side of the Earth, and its putting out 680 W/m^2 is continuously hitting half the Earth. Its not hitting the other side of the Earth at all, so why half the 680 W/m^2 ???
As I keep saying, one man shines a light on one side of the Earth and gets 100,000 for 12 hours. Another man does not shine light on the other side of the Earth for 12 hours and gets 0. But you stated in averaging, that both men shine half light on each side of the Earth and get 150,000 each ??? BUT, in the real World only one man shines the light and gets 100,000 ???
Am I right in that the real World, only one man shines the light and gets 100,000 ???
Wayne
wayne said…”The solar constant is only on one side of the Earth”
No it’s not. By definition the solar constant is referring to a 2 dimensional plane. This is not debatable. It is most definitely NOT what is hitting one side of Earth.
wayne saidThe solar constant is only on one side of the Earth
No its not. By definition the solar constant is referring to a 2 dimensional plane. This is not debatable. It is most definitely NOT what is hitting one side of Earth.
solar constant on a sphere = 340 W/m^2
Wayne wrote; “BUT”, the Sun is not shining on the other side of the Earth, the solar constant = 680 W/m^2
Ok, try this one, I don’t think you answer this one.
The Sun shines its solar constant on a flat disc, the solar constant = 680 W/m^2 do you agree ???
1,
Do you agree on a flat disc the solar constant = 680 W/m^2 ???
2;
Do you agree the solar constant only hits one side of our Earth at one time ??? Yes I know it moves around the Earth, but its basically hitting one side of the Earth, and that solar constant = 680 W/m^2 ???
3,
Two men, one man has a light beam, shines a light on one side of the Earth, he gets paid 100,000, another man has a light beam, but does not shine a light on the other side of the Earth, he gets paid nothing.
a;
If you average this out both men get 150,000.
b;
In the real World, just one man has 100,000.
c;
Which is the right answer in the real World, a or b ???
Wayne saidThe solar constant is only on one side of the Earth
No its not. By definition the solar constant is referring to a 2 dimensional plane. This is not debatable. It is most definitely NOT what is hitting one side of Earth.
But its “NOT” two dimensional ??? How do you work a flat round Earth = two dimensional ???
d;
Half the Earth is in daylight, half the Earth is in dark, in your World as you average, half the World is half light, and half the World is half light. BUT, in the real World its NOT like that.
We all dont understand why you think you should half the power of the Suns 680 W/m^2 ??? The Suns power is NEVER 340 W/m^2, its alway 680 W/m^2 ??? Its only hitting one side of the Earth, so why devied ???
e;
Two men, one man has a light beam, shines a light on full power on one side of the Earth, another man has a light beam, but does not shine a light on the other side of the Earth.
QUESTION, does one man use half his light bean and shine it on one side of the Earth, and does the other man use half his light beam and shine it on the other side of the earth ??? Or does just the one man shine his light beam full on, on one side of the Earth ???
Your dividing dose not make sense in what happens in the real World, do you agree, if not please explain how when the suns constant = 680 W/m^2, but you can magically half it, when its never half. We all dont get why you have something that’s never half, and its only hitting one side or the Earth at a time ???
Wayne
Hi bdgwx,
I just asked Roy to reopen this and he has. Hope we can carry on our debate, as I want to get to the bottom of this.
Wayne
The explanations provided by bdgwx make sense to me. I’m convinced.
The Joe guy makes one valid observation but it goes nowhere because it concerns a different part of the problem that no one knows how to solve. The observation is that the intensity and variation of solar heat hitting the sunny side of Earth has tremendous influence on the climate.
Yet this observation says nothing about the question of energy balance, except for the issue of clouds and how they impact the amount of energy that reaches the Earth’s surface.
We also know that water vapor acts as a damper – slowing the temperature increase during the day and slowing the temperature decrease during the night. Does the peak intensity of solar energy, not incorporated in the “flat” model, change how water vapor should be modeled?
You have a chicken, you have what happens in the real World. You have a energy/heat bean heating one half of the chicken, at 680 W/m^2, half of the chick gets cooked well done.
In your, what seems like a fantasy, you macicaly take the energy/heat beam and half it, some that it chicken gets cooked all over by 340 W/m^2, the chicken gets slightly cooked all over. But this is not whats happening in the real World.
Reality tell us that in 24 hours the chicken gets cooked at 680 W/m^2 all over well done.
You averages is “NOT” what happens in the real World.
Unless you can prove it does ???
Wayne
Bobdesbond,
I already replied to your conspiracy theory:
“Bobdesbond,
Bob seems to have disappeared from YouTube. Why is that Zoe ?
He went back to work. Doesnt have time. Now he just advises me.
What do you think his avatar represents?
Me looking at his name. Me looking at him. Me being trained.
The avatar was specifically chosen to tell a story. You psycho freak got the story wrong, and now peddle conspiracy theories. Now why would you do that if you didnt feel like a loser?
And what the eff is up with the Asian thing? Do I look asian to you? Youre nuts.”
bdgwx,
https://media.giphy.com/media/GlX6FnsFFAnEA/source.gif
If the input was 46 W/m^2, your cult would claim emission of 5 W/m^2 on all sides. So why do we see 46 W/m^2 on all sides?
Wayne is correct.
https://www.researchgate.net/profile/Yifan_Fan/publication/319041915/figure/fig5/AS:625332176556033@1526102217918/Diurnal-variation-of-ambient-air-temperature-and-solar-radiation-for-the-period-of.png
The Spencer method is to take the flux and divide by 2, in this case we get ~380 W/m^2, but actual average temperature is ~26C (454 W/m^2).
Spencer’s way doesn’t work anywhere at any time of the year, why would it work on average? It doesn’t.
Current warming: “What else could be causing it?”
Perhaps the same natural forces that caused the MWP.
https://principia-scientific.org/empirical-evidence-refutes-greenhouse-gas-theory/
All this rancorous bickering and scientific handwavium revolves around explanations for, mechanisms behind, evidence of a greenhouse style atmospheric warming of the earth and mankinds carbon dioxide warming that atmosphere.
I have presented what I consider a rather solid case that the atmosphere cools the earth like the reflective panel sitting on the dash which renders all of this RGHE debate and discussion moot.
All this surely proves that the IPCCs contention that 97% of scientists agree with AGW is total rubbish. The science certainly is not settled.
Really this is fantastic Animal illustration
The surface area of the earth in sunlight is one half of the total surface at all times, even slightly more than half due to the sun’s larger diameter. To imagine the earth as a disc that represents the ‘shadow’ of the earth and is one quarter it’s total area ignores the fact that this energy is spread over one half of the surface not one quarter.
the earth must receive this 1370 w/m2 over an area the size of the ‘disc’ (1/4 earth surface area) but it is spread over the area of a hemi-sphere. i would calculate the total solar energy hitting earth as:
TSI X(area of earth ‘disc’ shadow)=
1370 x (6371000*6371000)*3.142)= 1.75 e17 W
is that wrong or right? if wrong, where have i messed up?
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