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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.
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