I saw a Space.com article today entitled, Can Venus teach us to take climate change seriously? While Space.com writers should know quite a bit about the other planets, the article was a fount of misinformation and gross exaggeration.
The obvious purpose of the article was scare us into taking increasing carbon dioxide levels seriously, following on the Fourth National Climate Assessment (NC4) report (which I’m still trying to digest). After repeating the NC4 claim that “10 percent of the U.S. economy could evaporate by 2100” due to climate change, the Venusian silliness begins:
…a nearby world has an even hotter climate problem than ours, and scientists say we could learn some valuable lessons from it. That world is Venus, Earth’s “evil twin,” which was once nice enough until something went wrong and the atmosphere began trapping a little too much heat.
First of all, it is extremely speculative to claim that the climate of Venus “was once nice enough“. Then to further claim, “something went wrong and the atmosphere began trapping a little too much heat.” This makes it sound like Venus was just another hospitable planet until it reached some tipping point and a runaway greenhouse effect ensued.
This kind of prose might be good for science fiction, but it does a disservice to the facts.
While it is true that the surface of Venus has a temperature around 860 deg. F., the fact is that Venus has over 220,000 times as much CO2 as Earth’s atmosphere!
By way of comparison, since the Industrial Revolution started, it is widely believed we are now approaching 1.5 times as much atmospheric CO2 as we had hundreds of years ago.
220,000x versus 1.5x. Do you see the grotesqueness of comparing Venus to Earth?
If we used distances to make the comparison, and 1.5x was 1.5 inches, then 220,000x would be 2.3 miles! Imagine that the water level in the ocean increased by 0.5 inches, and people were trying to scare you about what might happen if it goes up another half an inch in the next 50 years by comparing it to 2.3 miles of water depth.
To make matters worse, the article then builds upon this Venus-is-so-much-like-Earth theme with the supposed reason why Venus is hotter than the Earth:
Much of the appeal of Venus comes from the fact that despite its horrifying modern appearance, it’s actually really similar to Earth. “Picture a planet that’s just like Earth but it’s a little hotter because it’s a little closer to the sun and that would be Venus,” David Grinspoon, a planetary scientist and astrobiologist at the Planetary Science Institute, told Space.com.
Wrong again, Dr. Planetary Scientist.
Because Venus is perpetually cloud-covered, it actually absorbs less sunlight than the Earth, even though it is closer to the sun. Were it not for its extreme greenhouse effect, its surface temperature would actually average colder than the Earth.
Using Venus to scare people about increasing CO2 levels on Earth is nothing new. But I’m surprised that a reputable website like Space.com would stoop to such a level of pseudo-science to support reductions in greenhouse gas emissions.
Venus’ atmosphere is 96% CO2, 3.5% Nitrogen, and 0.5% other gases, compared with earth’s which is 0.04% CO2.
I don’t think we are in danger of becoming Venus. We will probably run out of fossil fuels before we hit 0.06% CO2, assuming the we continue our current usage patterns.
It will never run out. It may be priced out, but new technology lowers the cost. Think fracking.
Fracking isn’t cheaper. It only lowers the cost in the US by increasing supply. But, it is more expensive. Fracking is the weak hand in the market place and the market will continually try to wipe out fracking.
Yes, but improved technology is driving the break-even cost down.
It’s the same story for deep water drilling:
https://tinyurl.com/ycvqnlaw
I’ll say it again: fossil fuel will not run out.
Not sure if I understand that graph. Fracked oil is more expensive than conventional drilled oil.
You are right, the market will wipe out fracking if the price is less than $60 a barrel. My only point is that this break-even price is dropping. Improved technology makes new resources available.
–Paul says:
December 12, 2018 at 11:08 PM
Not sure if I understand that graph. Fracked oil is more expensive than conventional drilled oil.–
The graph indicates break even of tight oil is below $60 and conventional is below $40
And few year ago tight oil was just below $100 per barrel.
Break even must be average, or some fracked well make more than others.
Also price of crude oil will control costs, if price is low, one will attempt to not to lose money by drilling if cost seem higher than expected yield. Or if price is higher it allows drill well which are considered to have higher cost.
It also has risks or it’s like gambling- some time you make money and some time the lose money, and trick is make more money than you lose. Or if you have more skill plus some luck you can make a lot of money.
Anyhow the graph indicate a huge improvement in cost of fracking- this could be largely improvement in technology, but also lowering of crude oil will also part of the lowered cost.
Or it could be if crude oil lowers more, you could see further improvement, though probably a fair risk of lower production.
You would have to know the business fairly well to make a good guess in this regard.
Yes, and short term they may have to sell below the break-even price when the equipment is already there.
Better to sell at a loss and get something towards the fixed cost, as long as the variable cost is covered.
Vice versa, you may be able to make gasoline from coal for less than a $100/barrel, but no-one will invest if the price is short term.
Not sure if that is true about the break even price dropping. I work for a supplier to the fracking industry, several customers, and they claim their costs are going up.
Yeah gbaikie,
The only costs that have been dropping are energy costs. But, according to them all other costs are going up, exploration, labor, equipment, etc. etc. I’ve not heard any claims from them that their break even costs are dropping-maybe not quite as much upward pressure.
Paul says:
“I work for a supplier to the fracking industry, several customers, and they claim their costs are going up.”
If that is long term then fracking is doomed, and something else will be invented.
We will never run out because all the carbon ever burned as a fuel is still here on earth and is still carbon. We can even change it back into fuel. All we need is energy and know-how. In the future, once we get beyond our nuclear energy phobia, we will have enough energy. We wont need to burn carbon fuels as much, and we will have as much carbon as we need.
“We can even change it back into fuel. All we need is energy and know-how.”
Fortunately trees already have the know-how. We just have to supply enough CO2 for them.
Trees are a very temporary CSS system. When they die and decompose they release any CO2 they have stored.
Canada’s vast forests are net emitters of CO2.
https://www.nrcan.gc.ca/forests/report/disturbance/16552
Not if they are chopped down and used for fuel.
But you knew that. Surely.
If a tree is chopped down and burned it emits CO2. It does so in forest fires as well.
When burning, wood reacts with Oxygen, which is contained in air:
C6H12O6 + 6 O2 = 6 CO2 +6 H2O
http://www.whatischemistry.unina.it/en/burn.html
But you knew that. Surely.
Jack, it appears you have an interest in wood. Maybe I can help.
C6H12O6 is actually the formula for glucose. It’s a simple sugar found in many plants. But wood is primarily cellulose, hemicellulose, and lignin. The simplest of the three is cellulose, C6H12O5. Just dropping one oxygen atom from glucose makes it indigestible for humans. In dietetics, it’s called “fiber”.
Hemicellulose is a hodgepodge of pentoses.
Lignin is the most complex, having the chemical formula C31H34O11.
A simple tree is not so simple. It is more complicated than an automobile factory.
Another
The main products of wood combustion, including cellulose, are CO2 and H2O with some CO.
http://virtual.vtt.fi/virtual/innofirewood/stateoftheart/database/burning/burning.html
JD,
You have the chemical formula for cellulose wrong.
It is (C6H12O5)n
You are probably just as short of a minor in chemistry as you are in physics.
You are not even in the ballpark much less in the hole.
Excuse me,
make that (C6H10O5)n
Yes bob, typos are easy, as you so aptly demonstrated.
So … Bob makes a mistake. A day later he notices, acknowledges, and corrects the mistake.
Meanwhile … JD makes a mistake. Doesn’t notice it. Doesn’t acknowledge it. Doesn’t correct even when pointed out to him. Instead he makes snide comments (presumable to distract from his own original shortcoming).
I know which of the two gets my respect.
*****************************************
Also, JD makes typos WITHIN a post making snide comments about typos. “Tim, you don’t even knew the difference between “to” and “two”.” The least he should do is proofread his own comment about typos before submitting.
JDHuffman
I gather you neglected to read the link.
The chemical nature of wood is closely related to sugars. To make things simpler, let’s consider wood to be composed just of Sugar, whose formula is C6H12O6.
Meanwhile
Read this to learn about wood combustion.
https://is.muni.cz/el/1423/podzim2013/MEB423/um/Wood_Lesson_02.pdf
Slide 13 is all you need.
Jack, you remind me of someone, but I can’t remember who.
He was also inconsequential.
We won’t ever run out of fossil fuel either. If the world ever gets to the point in a few hundred or thousand years of running out of fossil fuel increased effort will be made to develop alternative energy and also by that time nuclear fusion power should be pretty well evolved so that less and less fossil fuel is consumed and we never run out. It will just eventually become the less preferred energy source.
That would also be assuming the rise is primarily from burning fossil fuels. Several scientists have calculated that fossil fuels contribute less than 5% and the natural contribution is greater than 95%. If we shift into a cooling phase and the level of CO2 in the atmosphere is primarily natural, the level will go back down. Stay tuned.
Using carbon isotope analysis the 43% increase in atmospheric CO2 over the past couple of centuries can be directly attributed to the burning of fossil fuels.
http://www.jamespowell.org/Stuff/Ourfault/Ourfault.html
No, isotopic analysis does not establish attribution on its own. It merely fails to contradict the hypothesis.
Jack Dale, I think it is important to note that the article you cite starts with a metaphor on Lance Armstrong before stating it’s case for AGW. Attempting to make a weak argument seem stronger with the “If this is right then this must also be right” strategy. May be convincing to some but speaks more to the shortcomings of the argument to those that adhere to the scientific method. Just more pseudoscience claiming more than can be warranted, as Bart points out.
Do you have a hard time thinking metaphorically?
BTW – who are your several scientists?
jack…”Do you have a hard time thinking metaphorically?”
Why would anyone interested in science want to think metaphorically? Unless, of course, they believe in Big Bangs, black holes, the natural selection of evolution, space-time, and alarmist climate science.
Science is about evidence, not metaphors.
Jack Dale, do you have a hard time with the scientific method?
Authors who conclude human emissions cause only a minor increase in the level of atmospheric CO2 include Revelle and Suess (1957), Starr (1992), Segalstad (1992, 1996, 1998), Rorsch et al. (2005), Courtney (2008), Siddons and D’Aleo (2007), Quirk (2009), Spencer (2009), MacRae (2010, 2015), Essenhigh (2009), Glassman (2010), Wilde (2012), Caryl (2013), Humlum et al. (2013), Salby (2012, 2014, 2016), Harde (2017a), and Berry (2018). Note Dr. Roy Spencer’s article with comments “The Sorry State of Climate Science Peer Review” November 14th 2018 regarding Harde. Harde vs Kohler speaks volumes to which side of the AGW fence has any interest in the truth.
Revelle and Suess, 1957
“The increase of atmospheric CO2 from this cause is at present small but may become significant during future decades if industrial fuel combustion continues to rise exponentially.”
And it did rise exponentially
1957 8,324 tonnes of anthropogenic CO2 per annum
2014 36,138 tonnes of anthropogenic CO2 per annum
https://cdiac.ess-dive.lbl.gov/ftp/ndp030/global.1751_2014.ems
BTW – Since 1751 we have dumped 1.5 trillion tonnes of anthropogenic CO2 into the atmospheric.
As you know CO2 is a persistent non-condensing GHG.
Emissions rose rapidly, but concentration has not. Currently, emissions are still accelerating, and the rate of concentration is not.
http://oi63.tinypic.com/11gniqg.jpg
In the years 2002-2014, emissions rose 40%, but the rate of concentration flatlined, with no indication of any acceleration whatsoever. I could not find emissions data beyond 2014, but you may be assured China and India kept them increasing. Concentration, after the 2016 El Nino induced blip, is back to the same rate as prevailed during most of the past two decades.
The better fit is with temperature anomaly:
https://tinyurl.com/muo5shh
AGW theory is largely based on CO2 being a persistent gas in the atmosphere. However, the data shows it is not. Several of the authors I show above find the residence time of CO2 in the atmosphere to be between 2 and 7 yrs. The IPCC’s own data suggest 4.5 yrs. AGW theory and IPCC models suggest that temperature will rise with the rise of CO2. It hasn’t. Atmospheric CO2 follows the integral of temperature in the data and in proxy records, showing temperature drives CO2 levels. Alarmists believe the tail is wagging the dog. Metaphorically speaking. AGW theory sounds as simple and convincing as an extra blanket on a bed, but it doesn’t match the data. So it’s wrong!
“It merely fails to contradict the hypothesis.’
Ignore Bart.
He dismisses all data that contradicts his beliefs, no matter how compelling.
It’s not dismissing anything. It is merely recognizing that this is an argument along the lines of: “Dogs have four legs. This animal has four legs, therefore it is a dog.”
It is the same deal here. The argument is: “Our theory says the isotopic composition should change. The isotopic composition has changed, therefore our theory is correct.” It is a syllogistic fallacy.
Please do not protest that the isotopic composition is predicted and verified. It isn’t. The prediction is actually a posteriori based upon modeling which eventually runs in a circle back to the measured values. It’s logical fallacies all the way down.
The isotopic evidence is really four or five independent pieces, C14 in tree rings prior to 1950, C14 in ocean vs depth, C13/C12 ratio in atmosphere, in the ocean, in biota, and in ice cores going back hundreds of years.
All contradict the natural hypothesis.
‘The prediction is actually a posteriori based upon modeling which eventually runs in a circle back to the measured values. Its logical fallacies all the way down.’
This is normal science in operation. A hypothesis is tested with measurements. All measurements have to be analyzed, and modeled with standard methods.
The claim of ‘runs in a circle’ and ‘logical fallacies’, are just assertions, wishes, hopes.
The idea that large numbers of professional scientists fail to understand proper hypothesis-testing is not plausible. Nor is there any evidence of it.
Very impressive CO2 and C13 isotope effect in ice cores going back 1000 years.
https://wol-prod-cdn.literatumonline.com/cms/attachment/5bfc0b16-a0d7-4b55-8267-21c4069d303b/jgrd50668-fig-0005-m.jpg
“The new records between 1000 and 2012 A.D.: (a) CO2 concentration (black circles) and the δ13C (brown circles); solid lines are results of the double deconvolution for, respectively, (b) the atmosphere‐terrestrial biosphere and (c) the atmosphere‐ocean CO2 flux. Error bars are analytical uncertainties when the analytical uncertainty was larger than the minimum uncertainty (1.2 ppm and 0.05‰) as described in the text (section 3.1). Dashed lines show the 1σ uncertainty associated to the KFDD.”
It is abundantly clear the the isotope effect is only occurring during the modern period of high human emissions.
Natural fluxes are calculated.
Would CO2 level have been the same whether fossil fuel burning or not?
Pau,
The overwhelming evidence says no.
It would not have been the same, but the overwhelming evidence says the difference would have been insignificant. The “evidence” to which Nate refers is merely post hoc ergo propter hoc. But, the consistency between the rate of change of atmospheric CO2 concentration and the temperature anomaly establishes that the former is largely driven by the latter.
Bart again fails to apply his logic to his own ideas.
‘the consistency between the rate of change of atmospheric CO2 concentration and the temperature anomaly establishes that the former is largely driven by the latter.’
is a perfect example of ‘post hoc ergo propter hoc’.
It is a correlation according to Bart’s eyeball, with no evidence of causation.
With all the measurements being made on the Earth, isotopes and carbon fluxes, ocean carbon content, and ice cores, Bart can’t see to find one piece of corroborating evidence of causation.
“…isotopes and carbon fluxes, ocean carbon content, and ice cores…”
Not convincing. The data supporting these are spotty at best, and the correlations are vague. Nothing like the close fit between temperatures and CO2 rate of change.
‘Not convincing. The data supporting these are spotty at best’.
The standard disclaimer for all non-supportive data–no longer credible.
‘Nothing like the close fit’
The closeness of fit is subjective at best.
Still a ‘post hoc propter hoc’ with no evidence of causation.
Will you ever face up to these realities?
These are not realities. They are excuses to avoid accepting what your eyes can plainly see.
My eyes are capable of seeing data like this and incorporating it into reality:
https://wol-prod-cdn.literatumonline.com/cms/attachment/5bfc0b16-a0d7-4b55-8267-21c4069d303b/jgrd50668-fig-0005-m.jpg
Why aren’t yours?
These plots are meaningless. They are not data. They are speculation based upon unverified and unverifiable models using closely correlated observables.
You’re a hard nut to crack.
Well, either I’m a nut, or I speak the truth. It’s easy to rationalize something you want to be true. But, when you have conflicting evidence, you have to go with that which is most assured.
The best, most modern, most reliable data tell us unequivocally that the rate of change of atmospheric CO2 is proportional to appropriately baselined temperature anomaly. That simply would not be true if, as is alleged, humans were the driving force behind concentration, and the temperature anomaly responded to concentration.
‘These plots are meaningless.’ to a religious zealot yes.
These are data (top two).
If data like this is meaningless to you, then stop pretending you are doing science.
You are doing something else.
best data
“tell us unequivocally that the rate of change of atmospheric CO2 is proportional to appropriately baselined temperature anomaly. That simply would not be true if, as is alleged, humans were the driving force behind concentration, and the temperature anomaly responded to concentration.”
More logic fails.
That simply would not be true, if and only if, one ignores know mechanisms for this effect. Such as: ENSO’s effect on land carbon uptake in the tropics, AND the known correlation of ENSO with global temperature. Together, this means CO2 rate is correlated to temperature on ENSO time scales.
On longer time-scales you have CO2 rate-of-change noisily increasing, CO2 emissions increasing, and temperature noisily increasing.
Without other data on causal relationships, we simply have three variables increasing. Not evidence of anything.
Nate says:
“Without other data on causal relationships, we simply have three variables increasing. Not evidence of anything.”
Isn’t his correlation pretty crazy in itself?
His derivative is always positive, i.e. CO2 goes up when:
1) Temperature goes up.
2) Temperature stays the same.
3) Temperature goes down.
It’s a runaway situation. If things are so unstable, why would it strike now and not before?
The 2nd derivative is also positive, so it gets worse.
Conversely, CO2 goes down when:
1) Temperature goes down.
2) Temperature stays the same.
3) Temperature goes up.
It’s quite symmetric in that regard. What determines the current regime is the temperature relative to a threshold value, which depends upon the present climate state, much in the way the gain of a transistor circuit depends upon the current bias voltage. And, what determines the temperature is a complex forced feedback network, also state dependent. The entire feedback loop is evidently stable, as we would not be here otherwise.
This is a local-in-time, linearized model. Linearization is a standard engineering technique. We use it, e.g., to design transitor circuits, where the gain of the circuit is determined by the linear approximation of the transistor response at the current bias voltage.
Thus, it is not guaranteed to hold at all times in all climate states. But, it evidently holds in the present climate state. The only way you’ll get instability is if the CO2 level feeds back into greater temperature, so as to overwhelm other stabilizing feedbacks, producing positive feedback in which temperature rises, producing a rise in CO2, producing a rise in temperature, producing a rise in CO2, and so on ad infinitum.
That is, in fact, how we know CO2 level cannot have a significant aggregate impact on temperature in the present climate state – if it did, we would have a runaway condition, and we wouldn’t be here. The closed loop CO2 level is stabilized by feedbacks that operate on very long timescales. Hence, the aggregate temperature-raising potential of increasing CO2 must be negligible within that timeline.
This may seem like a bunch of gobbledygook if you are not intimately familiar with feedback systems. But, if you were, it would be very ho-hum.
There is such a feedback: the CO2 thermostat.
It will fix the current CO2 spike within a million years.
That is a positive feedback. Unimpeded, it would lead to a saturation condition. As we are not in a saturation condition, and there have been many eons for it to evolve, it follows that this is not a dominant term.
Earth is presently too far from the Sun to become Venus, but only by about 0.05 AU. But it could still get very hot here, if we put all fossil fuel carbon into the atmosphere (~13,000 Gt C from Swart and Weaver, Nature Climate Change, 2012).
Surely we aren’t that stupid, I hope.
Eventually, in ~1 Byrs, the Sun’s output will increase enough to cause a runaway greenhouse effect on Earth.
As far as I remember, there is another phenomenon which explains the I temperature at the surface of Venus: its atmosphere is 100 times denser than Earth’s.
From the high altitude sulfuric clouds where radiative equilibrium takes place, the ratio of temperature gradients is proportional to the ratio of atmospheric densities (the surface gravity being comparable).
This recent study suggests that atmospheric pressure and solar irradiance play a much larger role in planetary temperature than does atmospheric composition.
https://www.omicsonline.org/open-access/new-insights-on-the-physical-nature-of-the-atmospheric-greenhouse-effect-deduced-from-an-empirical-planetary-temperature-model.php?aid=88574
https://tinyurl.com/yafw3yxu
So what’s your point Svante?
I guess this means you can’t refute any of their analysis or the conclusions in their paper and neither does the blog you attached a link to. You just rely on some website that is dedicated to smearing the reputation of scientists that don’t believe in man-made catastrophic climate change? What a disgustingly lame reply.
It says:
“In a recent study Volokin et al. [1] demonstrated that the strength of Earth’s atmospheric Greenhouse Effect (GE) is about 90 K instead of 33 K as presently assumed by most researchers”.
It’s their own study with names reversed.
Yes, Svante. That’s a part of their point:
“A recent study has revealed that the Earth’s natural atmospheric greenhouse effect is around 90 K or about 2.7 times stronger than assumed for the past 40 years. A thermal enhancement of such a magnitude cannot be explained with the observed amount of outgoing infrared long-wave radiation absorbed by the atmosphere (i.e. ≈ 158 W m-2), thus requiring a re-examination of the underlying Greenhouse theory.”
No need to keep replying to me until this article closes for comments.
‘we decided to try an empirical approach not constrained by a particular physical theory. ‘
IOW there model need not agree with established physics…
Thanks for that link.
Hi Paul. Yes Venus flux balances mainly from the high altitude cloud decks where liquid droplets and solid ice emit the same energy back to space as the absorbed solar flux. Within the troposphere of Venus from this level and to the surface a stable thermal gradient runs that is independent of solar heating rate, independent of convection or radiation dominance and independent of gaseous opacity. The gradient is a reversible adiabatic which follows an isentropic profile or is simply described by the change in gravitational potential energy expressed through the isobaric specific heat capacity.
This gradient is also found in Earths troposphere and in modified by the specific latent heat of water as a condensing gas greatly adding to the specific heat capacity and thus reducing the thermal response to energy changes.
Little to do with the composition of the atmosphere and the inaptly named “greenhouse” effect and everything to do with the fact that the atmosphere is 93 bar at the surface.
Furthermore, at 450 C at 90 bar, CO2 is a supercritical fluid, quite a little bit more opaque than gases don’t you think?
The basic physics indicates CO2 is NOT a “heat source”. So CO2 is NOT heating planet Venus. They need to look at the other REAL heat sources, such as vulcanization and solar. Venus gets about twice the solar flux as Earth. Although the lower energy photons may not be able to penetrate the atmosphere, the higher energy photons would not be stopped.
But why consider anything that doesn’t “feed the kitty”?
“The basic physics indicates CO2 is NOT a “heat source”.”
Of course. The sun is the heat source for the surface (along with small amounts of geothermal energy).
“They need to look at the other REAL heat sources”
… and they need to look at the real heat sinks for the surface. After all, the heat sources are only half the equation. Once we understand how the energy LEAVES (thermal IR) as well as how it ARRIVES (solar and geothermal), then we can start to calculate the surface temperature.
tim…”The sun is the heat source for the surface (along with small amounts of geothermal energy).”
Small amounts? The surface temperature is 450C. Astronomer Andrew Ingersoll claimed such a surface temperature contradicts the 2nd law of thermodynamics if it is claimed to come from atmospheric feedback.
There must be another source for the heat that escapes us.
Gordon, Perhaps you are referring to this:
”
Venus lower atmosphere heat balance, Andrew P. Ingersoll Judith B. Pechmann
Abstract: Pioneer Venus observations of temperatures and radiative fluxes are examined in an attempt to understand the thermal balance of the lower atmosphere. If all observations are correct and the probe sites are typical of the planet, the second law of thermodynamics requires that the bulk of the lower atmosphere heating must come from a source other than direct sunlight or a thermally driven atmospheric circulation. Neither the so‐called greenhouse models nor the mechanical heating models are consistent with this interpretation of the observations. One possible interpretation is that two out of the three probe sites are atypical of the planet. Additional lower atmosphere heat sources provide another possible interpretation. These include a planetary heat flux that is 250 times the earth’s, a secular cooling of the atmosphere, and a chemically energetic rain carrying solar energy from the clouds to the surface. Other data make these interpretations seem unlikely, so measurement error remains a serious possibility.”
https://agupubs.onlinelibrary.wiley.com/doi/10.1029/JA085iA13p08219
This nearly 40 year old paper leaves open many possibilities, including faulty models, exotic heat sources, measurement error, and even random bad luck!
No additional heat is required. The flux balancing temperature exists at a level in a gravity field where gravitational potential energy is completely compatible with kinetic energy whose mean is temperature in a gas.
Yes, earth reflects about 30% of it’s TSI, venus reflects 70%, so it should be colder than earth without the greenhouse effect.
It’s essential that rising water vapor condenses before it is knocked out by solar radiation.
Yet another thing that makes our planet just right.
When the altitude temperature profile does not cause condensation, oceans boil off. More water vapor causes more warming until all water is lost to space.
https://static.cambridge.org/resource/id/urn:cambridge.org:id:binary:20170503120102304-0104:9781139020558:84412fig13_8.png?pub-status=live
svante…”earth reflects about 30% of its TSI, venus reflects 70%, so it should be colder than earth without the greenhouse effect”.
What greenhouse effect?
It’s not just me claiming that, astronomer Andrew Ingersol:
From an abstract to a paper behind a paywall…
https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/JA085iA13p08219?deniedAccessCustomisedMessage=&userIsAuthenticated=false
“If all observations are correct and the probe sites are typical of the planet, the second law of thermodynamics requires that the bulk of the lower atmosphere heating must come from a source other than direct sunlight or a thermally driven atmospheric circulation. Neither the so‐called greenhouse models nor the mechanical heating models are consistent with this interpretation of the observations”.
From 1980, quite recent for you, any advances since then?
The surface exists at an equilibrium temperature with the flux balancing temperature at around 60Km in a gravity field. No addition energy, heat or mechanism required or implied.
Meghan earned an MA in science journalism from New York University and a BA in classics from Georgetown University.
Do we even know the surface temperature on the night side of Venus?
Good question. Days and nights are very long. I suspect it is still very hot due to the density of the atmosphere and the strong GHE.
Strong GHE: Is that because of the density and the composition of the atmosphere?
With a pressure 93 times that of the Earth it must also stretch far higher than at the Earth.
I have heard that at 1000hPa the temperature is similar to the Earth.
That mekes you think about how much the atmosphere and its density influences the surface temperature an how it does that.
I’ve discussed this many times before. The GHE depends upon how much greenhouse gas exists in the atmosphere, and the higher the pressure the more GHG there is. Pressure, by itself, can only explain temperature CHANGES with height in a convecting atmosphere. (The stratosphere covers a pressure range of 100X [200 to 2 hPa], yet is nearly isothermal). Pressure does not give you the baseline temperature from which to extrapolate to higher temperatures at higher pressures. ABSOLUTE temperature is a function of energy gain and energy loss. If the atmosphere did not absorb and emit IR radiation at all (and did not absorb sunlight directly), the surface temperature would be the same as if the atmosphere did not exist, even with a 93 bar atmosphere. Greenhouse gases are what help destabilize the atmosphere and make it convect, so are necessary to explain why convecting atmospheres have approximately adiabatic lapse rates.
“the surface temperature would be the same as if the atmosphere did not exist…”
The atmosphere would still make two important contributions.
1) The atmosphere would add thermal mass, making the surface warm up slower and cool down slower.
2) The atmosphere would carry energy from warm daytime areas to cool, nighttime areas.
Both of these would tend to even out temperature extremes. In turn, a more uniform planet would have a higher average temperature. (but never above the effective temperature for the sunlight averaged over the whole surface.)
Dear Dr. Roy,
While I am a fervent supporter who loves what you do there are times when I must respectfully disagree.
For example you (and Richard Lindzen too) talk about climate sensitivity in terms of degrees Centigrade per doubling of CO2 concentration. This suggests that you think the Arrhenius (1896) hypothesis is valid.
Now you say:
” If the atmosphere did not absorb and emit IR radiation at all (and did not absorb sunlight directly), the surface temperature would be the same as if the atmosphere did not exist, even with a 93 bar atmosphere.”
This statement is false because radiation is not the only mechanism for transferring heat to planetary atmospheres. Imagine a planet that has an atmosphere of pure helium that will not absorb any IR radiation emitted from the surface. Let us also stipulate that there are no liquids present to affect heat transfer via phase changes.
There is still one process (other than radiation) left to transfer heat from the the surface of a rocky planet to the atmosphere and it is a process that is dominant whenever the absolute pressure at the surface exceeds 0.1 bar. I refer to convection and it is all you need to calculate the temperature gradient in a planet’s troposphere from first principles using thermodynamics as my high school physics teacher (Dr. John Pallister) taught me sixty years ago.
Dr. Pallister’s used a slide rule but even with modern computers the picture has changed very little. Here is a paper that I have linked before on your blog:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5820795/
Roy W. Spencer, Ph. D. says, December 3, 2018 at 1:22 PM:
No, it doesn’t. An “atmospheric thermal effect” on the surface only needs for the atmosphere to be IR active. It doesn’t matter to what extent it is IR active. As soon as a stable convective circulation is established, changing the chemical makeup of the atmosphere in some direction or other won’t do a thing, thermodynamically speaking. At that point, at the sfc-atm interface, the working mechanisms of fluid dynamics completely override those of radiation physics, the sfc heat budget and mean temperature is fully controlled by convection-related processes, which will easily and readily negate any potential radiative effects from an increase in atmospheric IR opacity alone to take hold.
Kristian, you seem to have overlooked the atmosphere-space interface, This WILL change with the addition of more GHGs, which will in turn change everything below (including the surface temperature).
Tim Folkerts says, December 4, 2018 at 6:48 AM:
Tim, you know I haven’t. We’ve been through this a hundred times before. Please stop it with this feigned memory lapse tactic of yours.
On what observational evidence do you base your assumption that a lower OLR at the ToA will necessarily translate into an increase in the average temperature at the surface far below? Your assumption completely disregards the various (mostly NON-radiative) energy/heat transport mechanisms operating BETWEEN the surface and the tropopause …!
https://okulaer.wordpress.com/2017/04/15/the-congo-vs-sahara-sahel-once-more/
https://okulaer.wordpress.com/2014/11/16/the-greenhouse-effect-that-wasnt-part-2/
Also, why do you assume that simply making the atmosphere nominally more opaque to IR will necessarily (MUST) result in a lowering of the total All-Sky OLR at the ToA (at equal T_s)? We do not see this in the observations, Tim. At all. Again you’re ignoring the power and/or independence of convective processes within the troposphere.’ What happens to the energy/heat between being transferred from the sfc to the atm and finally escaping to space via radiation?
Why would simply making it harder for energy/heat to move through a fluid, dynamic atmosphere specifically via RADIATION (Beer-Lambert Law-wise) necessarily make it harder for that energy/heat to move through that same atmosphere in general, overall? What actual real-world observations do you base this one-dimensional idea on?
The physical phenomenon that makes the ENTIRE difference in this respect between a fully radiatively inert steady-state atmosphere and one which is IR active (at ANY level) is … convection. Once, and as long as, we have stable (small- and large-scale) overturning of air masses within the atmosphere (troposphere, really – it’s in the name, after all), no potential radiative thermal effects will be able to take hold – not in the troposphere itself, and NOT at the surface below.
You need to start addressing what is actually happening in the real world (like, the relevant DATA) rather than just constantly keep referring to “The Theory” (‘This is what SHOULD or COULD happen!’ (IF so and so; because there’s always an ‘if’, isn’t there? The always implied ‘All Else Being Equal’ stipulation.)).
Kristian, you claimed “mean temperature is fully controlled by convection-related processes”.
Based on this claim …
* Convection-related processes lead to a fixed, stable lapse rate.
* A fixed lapse rate leads to lower temperatures higher in the atmosphere.
* Lower temperatures higher in the atmosphere lead to lower levels of radiation to space.
* Lower levels of radiation to space lead to more energy staying in the earth system.
* more energy staying means rising temperatures.
Now, convection does not FULLY control temperatures (as we both know). and the lapse rate will adjust a bit. But the theoretical connection is clear — especially with your assumption that convection is in control!
Tim Folkerts says, December 4, 2018 at 12:43 PM:
To be more precise, though, CHANGES in sfc mean temp are fully controlled by solar input AND convection-related processes.
Yes.
Yes.
Er, yes. If the radiation in question is emitted from those lower-temperature atmospheric levels.
Sure. However, the lower level of radiation to space would be a result of lower tropospheric temperatures, not of a higher tropospheric concentration of IR-active gases. A more radiatively active troposphere could never end up WARMER than a less radiatively active troposphere. Because the purely radiative processes operating in the troposphere on balance provide a strongly COOLING contribution to the troposphere’s thermal budget, NOT a warming one …
https://okulaer.wordpress.com/2018/02/25/an-atmospheres-ir-activity-wont-make-it-warmer-and-so-cannot-be-the-cause-of-surface-warming-either/
Sure enough. As pointed out above. It is a tightly coupled radiative-convective collaboration effort. But Earth’s own thermal radiation does NOT contribute towards making the lapse rate LESS steep, thus reducing the heat transport rate up and away from the surface. Quite the opposite. Don’t you agree?
“However, the lower level of radiation to space would be a result of lower tropospheric temperatures, not of a higher tropospheric concentration of IR-active gases. “
I would suggest it is a result of *BOTH* higher GHG concentration *AND* lower temperatures.
In particular, the radiation to space comes from the ‘top layer’ of GHGs. (Radiaion from ‘lower layers’ is blocked by the higher layers.) If you add more GHGs, the ‘top layer’ will be higher. And higher = cooler in the atmosphere.
“Greenhouse gases are what help destabilize the atmosphere and make it convect, so are necessary to explain why convecting atmospheres have approximately adiabatic lapse rates.”
It gives good meaning. Without radiating gasses the upper parts of the atmosphere would never loose heat.
I imagine the atmosphere would have the same temperature as the ground (maybe higher) and hardly any convection.
Hi Dr. Spencer,
This recent study suggests that atmospheric pressure and solar irradiance play a much larger role in planetary temperature than does atmospheric composition.
https://www.omicsonline.org/open-access/new-insights-on-the-physical-nature-of-the-atmospheric-greenhouse-effect-deduced-from-an-empirical-planetary-temperature-model.php?aid=88574
Tim, you write:
Tim, Tim, Tim. Without these gases, the energy wouldn’t escape the atmosphere at all.
Look, as soon as energy is being transferred from the surface to the atmosphere above in the form of heat, most all of this energy will be unable to escape the Earth system, if it weren’t for thermal radiation. This physical circumstance alone necessitates a radiatively active atmosphere. The atmosphere simply needs to be radiatively active in order to be able to adequately COOL. Not in order to be able to warm. To COOL.
Simply increasing the level og IR activity of an atmosphere will not make it warmer. Which means it will not be able to make the SURFACE underneath any warmer either. Because the temperature of the surface, once thermally and convectively connected to the atmosphere above, is determined by the balance between solar heat input and convective power, the latter to a large extent controlled by the temperature gradient away from the surface and up through the tropospheric column. The only reason why a radiatively active atmosphere makes the surface underneath warmer, is because the atmosphere is warmer than space.
But, crucially, it is NOT warmer than space because of its RADIATIVE properties specifically. Quite the contrary. It is warmer because it is able to HOLD energy tranferred to it in the form of heat as internal energy. The radiative properties themselves don’t help in this regard. So when you say that the atmosphere is forced to release its thermal radiation to space from high up as a result of its content of IR-active gases, you basically misrepresent reality. Most all of the energy transferred to the atmosphere in the form of heat would ALWAYS remain “trapped” within the atmosphere as internal energy if it weren’t for the fact that it is ABLE to escape to space via radiation. And thus it is ultimately only BECAUSE of its IR-active constituents that the atmosphere is able to balance its heat input. Not in spite of them.
Emitted IR escapes from Earth to space all the way from the surface and up through all layers of atmosphere to the top, but as the bulk air becomes less and less dense as you move upwards, the RATE of escape will naturally increase. In the end, as you move towards the upper convective limit around 100-200mb (you will find this general limit on virtually all planets and moons with substantial atmospheres/tropospheres in our solar system), the radiative loss from the moving air masses becomes so great that their upward momentum is finally drained, and so they halt and start moving outwards instead, before they, as a consequence of continued radiative cooling, slowly subside.
Earth’s thermal radiation is but an EFFECT of the fluid-dynamical processes that drive the tropospheric circulation. It doesn’t CAUSE temperatures. It is an EFFECT of temperatures and temperature distribution.
Yes, in order for the surface to warm to an average temperature higher than the Moon’s, the atmosphere definitely needs to be IR active. But the IR-activity is not the CAUSE of the elevated average temp. It is just a necessary tool or device to enable the REAL cause (tropospheric temperatures) to create the effect …
No. That’s once again only what the “THEORY” says. It is not what we OBSERVE. And I just told you above why you can’t simply assume this as fact.
Again, you need to address the OBSERVATIONS, Tim, not just your “THEORY” about what SHOULD happen. Simply restating your theory isn’t an argument. You need to back it up …
Wrong again Dr Scientis Roy. How can you not understand that there is no radiative greenhouse effect anywhere in the universe, not even in a real greenhouse? How can you not understand that a cooler source of energy can never add energy to an already warmer source of energy? How can you not understand that two locomotives can only work to complement each other if they are both exactly the same strength; if not, the weaker one will detract from the stronger one.
Hans…”How can you not understand that a cooler source of energy can never add energy to an already warmer source of energy?”
I agree with Roy on much of his science and I applaud him for his integrity in climate science. On this part, however, I fail to understand the basis of the net energy balance upon which it is based. IMHO, it comes from a misinterpretation of the Stefan-Boltzmann equation.
In fact, I find it somewhat ironic that Roy’s data is proving GHGs have little or no effect in the atmosphere.
The original Stefan equation was R = (fi).T^4. Stefan derived that equation from data provided by Tyndall in which he electrically heated a platinum filament wire till it glowed colours related to the current. Another scientist correlated the colours over a range of currents to the real temperatures that would produce those colours, and Stefan found the T^4 relationship between those temperature and EM emission.
Later, Stefan’s first Ph.D student, Boltzmann, tried to replicate Stefan’s equation statistically and arrived at the same equation with a temperature gradient as:
R = e.b.A(T^4 – To^4)
where e = emissivity, b = S-B constant, A = area of emission, and the T’s represent a temperature gradient between the emitting body and its environment.
Both equations were supposed to represent blackbodies emitting to their environments and there is no doubt they represented a transfer of heat from a hotter body to a cooler environment. The 2nd law was preserved.
There has never, to my knowledge, been a situation in which a cooler blackbody radiated to a hotter blackbody.
Later, someone seemed to reason that if two blackbodies were emitting in the same area, that the S-B equation could be modified to include the effect of the 2nd BB on the first.
You cannot do that, the math does not represent the reality other than at thermal equilibrium, where no heat can be transferred. The reality came later via Bohr and Schrodinger who described electrons in atoms converting heat to EM in a relationship of E = hf. E represents the difference between two energy orbitals in which an electron can reside and the electron frequency is related to the temperature of the body.
Bohr made a stipulation, however. In order for an electron to absorb EM, the EM had to match exactly the frequency of the electron. It’s a resonant system like any other resonant system, like a bandpass filter, in which only certain frequencies are passed with the others being attenuated or outright rejected.
The frequency of the electron, f, is related to the temperature of the body of which it is a part. The hotter a body, the more energy it has…to a degree. Planck discovered that really high energy bodies are not as likely as mid energy bodies in the visible range and he built an exponential factor in his equation to allow for that.
Planck’s equation produced a probability-based bell-shaped curve for EM over its range whereas E = hf would rise indefinitely toward infinity. That was called the ultraviolet catastrophe.
The frequency and strength of EM radiation for cooler bodies, therefore, does not match the requirement of electrons in hotter bodies. That preserves the 2nd law since EM from a cooler body cannot be absorbed by a hotter body.
That fact bypasses the notion that blackbodies must absorb all incident EM upon them. If anything, that definition applies only at thermal equilibrium.
The notion that energy can be transferred both ways at the same time contradicts the understanding in physics that energy flows from a region of higher potential to a region of lower potential. The understanding applies to all forms of energy. Water flows downhill, masses fall from higher elevations to lower elevation, and electrons flow from higher electrical potentials to lower electrical potentials.
Why should EM behave differently? Hotter bodies represent higher potentials and cooler bodies lower potentials. It stands to reason that in a thermal context, EM can only flow hot to cold, unless external energy and systems are in place to compensate.
“There has never, to my knowledge, been a situation in which a cooler blackbody radiated to a hotter blackbody.”
Of course not Gordon since blackbodies do not exist. However, Planck’s referenced experiments in the late 1890’s determined that cooler blackbody radiation was absorbed by a hotter real body. This escapes Gordon’s knowledge because Gordon determines the library is too far away to learn about these experiments first hand.
“I find it somewhat ironic that Roy’s data is proving GHGs have little or no effect in the atmosphere.”
Roy’s data is proving GHGs have little efffect in the atmosphere. His various data are all in line with added ppm CO2 over 75 years to 2013 only adding a little ~0.7C temperature to earth surface global climate.
What a discussion i have started.
“The notion that energy can be transferred both ways at the same time contradicts the understanding in physics that energy flows from a region of higher potential to a region of lower potential.”
The net energy flow follows your statement, but should the radiation from a surface change because you put an other surface some where in the vicinity. Or you could place a mirror. If it do not reflect back to the surface, the surface would radiate according to SB. If you turn the mirror so that some of the reflected radiation hits the surface, then it should suddenly emmit a lower radiation.
It is in my opinion a fact that a surface at a given temperature emmits according to SB.If a hot surface is hot because of a constant powersource inside it will turn a bit warmer by placing a colder body nearby, a body that is a bit hotter than the other surrounding.
If the body is a bit colder than the surrounding the suface will cool a bit. It is easy calculated by the simple use of SB for each body and surface by itself.
I investigated if surfaces with very selective radiation and receiving bands could do the trick. They could not. A colder body have less radiation in any band than a hotter body. The net flow goes always the right way, and the result is the same, but easier to calculate as if every body radiates as if was alone.
ball4…”Plancks referenced experiments in the late 1890s determined that cooler blackbody radiation was absorbed by a hotter real body”.
1)Planck did no experiments other than thought-experiments related to statistical analysis. Others did the experiments to corroborate his equation.
2)where in that equation is there any reference to radiation from a cooler body raising the temperature of a hotter body that intercepted the radiation?
Each Planck curve shows the EM spectrum for one temperature.
3)Planck’s equation related EM radiation intensity to its frequency. It produces a bell-shaped curve with a frequency related to the colour green at its apex. Then it drops off toward the UV frequency range.
That was unexpected, since the relationship between EM energy and frequency had been established at E = hf, meaning energy should increase toward infinity with frequency.
By fudging the math, Planck was able to add an exponential function to attenuate the energy at higher frequencies to better fit the reality.
4)Roy’s data is not in line with the previous 75 years, which shows a steadily increasing trend. The UAH data, in its range, shows 17 years of temperatures below the average from 1980 – 2010, only exceeded the baseline following the extreme El Nino of late 1997.
Following the EN, a mysterious jump of about 0.2C appeared in the global average and the overall trend from 1998 – 2015 was flat. It was a second strong EN in early 2016 that drove the global average away from the flat trend level. The average has since dropped back to nearly the flat trend level.
“2)where in that equation is there any reference to radiation from a cooler body raising the temperature of a hotter body that intercepted the radiation?”
This question makes no sense Gordon. Walk or drive over to the library or use google to learn from the Planck ref.d experiments that developed Planck distribution of radiance at each frequency & each temperature of a real body emitting blackbody radiation demonstrating that “cooler blackbody radiation was indeed absorbed by a hotter real body” or Planck’s eqn. would have been much different.
“3)Planck’s equation related EM radiation intensity to its frequency.”
No, Planck’s ideal distribution shows BB radiance (sometimes depicted in units of intensity) for each frequency and each temperature.
“By fudging the math”
There was no fudging in the end, the experimental curve was later developed also from first principles.
“The average has since dropped back to nearly the flat trend level”
Right UAH data in the satellite era is in line with “nearly the flat trend” of about +0.7C per 75 years from additional CO2 ppm as predicted long ago.
Gordy, May I remind you that you still haven’t provided a logical, physics based explanation for my Green Plate Demo results. The energy flows from the high intesnity light source to the Blue Plate and then to the Green Plate, then to the bell jar. It’s all downhill. You are confusing temperature with energy flow. Temperature represents the state of energy stored in the solid body, not a flow.
svend…”should the radiation from a surface change because you put an other surface some where in the vicinity…”
That’s a good question.
The problem as I see it is this: we tend to think in terms of radiation only. Even in a vacuum, where a heat source is contained within the vacuum and radiating, say to glass walls, those walls have cooler air on the outside of them, and or there is the absolute zero of space out there.
Therefore, heat conduction and convection are always affecting the temperature of a radiating body and that’s where the temperature of another body comes into play.
A heated body radiating freely to air will arrive at an equilibrium temperature with the cooler air, and that temperature will be lower than the temperature produced by the electrical heating. That’s because the moment you begin heating the body electrically, the body will begin transferring heat to the air as natural dissipation.
The dissipation under a non-vacuum will be caused by direct conduction to the air, heat removal by convection of the air, and by radiation. If you interfere with any one of those three parameters, the heated body’s temperature will rise back toward the temperature at which the electrical current heats the body by itself.
That has nothing to do with energy gained by a back-radiation from the environment. The rise in temperature is due to a restriction in the body’s ability to dissipate the heat caused by the electrical current.
In a solid conductor, it is well known that the transfer of heat is due to the temperature gradient in the conductor. The heat always flows from a high temperature level to a lower temperature level. That is also the 2nd law of thermodynamics.
In a conductor, valence electrons, which are outer shell electrons and free to move atom to atom, transfer electric charge through a conductor. The electron drift is relatively slow through the conductor but the charge travels at the speed of light. That’s why you get light instantly when you turn on an incandescent bulb.
Those same valence electrons also transfer the energy we call heat. I presume that’s why heat travels so quickly through a conductor as well.
Suppose, rather than a straight conductor, you have a solid metal sphere with a heater element in its core. The heat transfers from the core out to the surface, presuming the aurface is in air at a much lower temperature. The electrons transfer the heat atom to atom to the surface but what happens with the surface electrons?
Something like that has to happen to cause the surface electrons to emit EM en masse. One would think they interact with the electrons in the air molecules adjacent to the surface.
I think you arr wondering whether they can sense a nearby surface with a different temperature. I think you first have to answer the question as to how electrons on the cooler end of a conductor with a temperature differential know the other end is hotter. That would be easier to explain with electron to electron contact throughout the length of the conductor.
Biologist, Rupert Murdoch put forward a similar explanation for why humans can sense someone staring at them from behind. He has hypothesized that both parties are emitting some sort of field of energy and that the energies are interacting. It would not surprise me in the least if an electron is able to sense its environment via the fields it emits and receives.
I am not claiming that in a human sense of sensing, I am talking more a servo-mechanism in which fields affect the behavior of the electron, causing it to change orbital levels.
Of course, we must remember the three modes of heat transfer. The electrons at the surface will be transferring energy as heat to the electrons in the air molecules. As those molecules heat, they will rise away from the surface and cooler molecules will replace them. There will be an ongoing cycle of heat dissipation due to conduction, convection, and radiation.
But what happens in an evacuated chamber with a heated body at its core and conductive walls surrounding it in a sphere. How does the heater know about the temperature of the conductive sphere with a vacuum between heated body and the cooler sphere?
I don’t know.
I can conjecture that surface electrons have a dense electric field around them due to their electric charges, and as they move, the charges produce a magnetic field. I am sure therein lies the root explanation for electromagnetic energy.
Can those fields travel through the vacuum to the conductive wall, get reflected back, and interfere with the surface radiation to tell the surface electrons there is a cooler body surrounding them?
Or how about the situation where the surrounding sphere is at a higher temperature than the inner heated body? I don’t pretend to understand the process to that degree although it interests me deeply.
I have reasoned in the past, based on an experiment done in a vacuum by another poster (swannie), that if you have a heated body radiating isotropically in an evacuated chamber that the body will reach an equilibrium temperature between the energy supplied to it and the energy it can radiate.
In other words, if all that radiation causing the dissipation is blocked, the heated body’s temperature will rise to the level it was heated by the incoming energy.
If you now bring a cooler metallic plate near the heated body, in vacuum, so that it blocks one side of the radiation from the body, the body’s temperature will rise. Metal blocks EM radiation completely. By moving the plate next to the radiating body you have effectively suppressed half of its ability to dissipate heat, so it’s temperature increases FROM WITHIN.
Based on arguments proposed here, the body is warming due to radiation from a cooler body. That contradicts the 2nd law and is justified based on a mysterious ‘net balance of energy’. There is no such phenomenon in physics, of which I am aware, as applied in this case.
The body is warming only because its means of dissipation has been lowered.
With regard to the Earth surface to atmosphere interface, I don’t think it’s conceivable that CO2, at 0.04% of the atmosphere could possibly interfere with surface radiation to the same extent given the vast amount of surface radiation in comparison. Based on it’s mass percent in our atmosphere, CO2 should be able to warm the atmosphere an equivalent amount, a few hundredths C.
Besides that, radiation from the surface drops off rapidly due to the inverse square law, and is unlikely to be a major factor in warming GHGs. Radiation as terrestrial temperatures is simply not a good means of heat dissipation.
Gordo gives us another of his long winded but incorrect descriptions of physics. His claim that the imposition of a second body between a heated body and the surroundings in an evacuated enclosure, will “block” the EM energy leaving the heated body is pure nonsense.
The S-B equation gives the correct result, which is captured in the second term of the equation Gordo presented above. That is, the IR EM radiation from the second body is intercepted by the first body, the result being that the temperature of the first body must increase to achieve a new steady state. He has never described how this supposed “blocking” process might occur, instead relying on an image of conduction heat transfer in which the flow of thermal energy is directly related to the difference in temperature.
In engineering, models of heat transfer via conduction are constructed using electrical analogs with voltages and restive networks. Voltage is analogous to temperature and the network of resistive elements represent the physical properties of the various materials being analyzed. The flow of thermal energy is represented by the various current flows thru the network.
In my Green Plate demo, the system is powered with a constant energy flow, which would correspond to a constant current power supply in the electrical analogy. The voltages, i.e., the temperatures, at various points within the network are the result of the net flow of energy thru the network. But, radiation energy flows are a function of T^4 and energy is emitted by each body within the system, which complicates things and the simple resistive network analog no longer applies. Since all the elements of the problem emit IR EM radiation, Gordo must account for all the flows and there’s no “blocking” of the EM from the cooler body in the direction fo the warmer body. If the warmer body will emit EM radiation at the wavelength of that from the cooler body, as captured in the emissivity value, the warmer body will absorb that radiation as well.
Likely the planet Venus never cooled, the greenhouse gasses like a blanket that never let it cool. Earth cooled from the super warm / molten state. About 600 million years ago the planet Venus cooled enough to form it’s crust but since then there has been almost no erosion only some vulcanism. Winds are in the km speeds on the surface. It hardly rotates, a day is a year. 90% of the light reaching Venus is reflected, it’s not greenhouse gas that trapping sun energy, it’s trapping heat from genesis. 600 million years ago the Earth was cooling enough that water formed, this wasn’t the case for Venus who maintained the heat. Venus was never “Earth-like”.
Venus was never Earth-like.
Right on! Don’t buy the “Space.com” nonsense that claims Venus once had oceans of water. If that was true where did the water go?
Let’s suspend disbelief and assume that the water dissociated in the upper atmosphere so that the hydrogen escaped into space. Maybe that could have happened but given the “Escape Velocity” of Venus, oxygen molecules would not have escaped into space. So where are they?
My (highly speculative) scenario is that Earth and Venus had atmospheres that initially were rich in CO2. Earth had enough water for life to develop and that life converted most of the CO2 into oxygen.
On Venus life did not develop so CO2 is still 97% of the atmosphere. Color me skeptical but I will listen to reason!
Where did the oxygen go?
It oxidised the surface rocks.
Oxygen only persists in an atmosphere if it is constantly replenished by photosynthesis.
I would wonder if there was any evidence for this
“Lets suspend disbelief and assume that the water dissociated in the upper atmosphere so that the hydrogen escaped into space.”
You would think that since hydrogen occurs in three different isotopes of different weights, maybe the heavier hydrogen is more abundant in the atmosphere of Venus.
Maybe someone has looked into this.
Hmmm
The reason for the slow rotation is Venus does not have a moon. If everything else between earth and Venus were equal except no moon. Venus with it slow orbit synced motion would never have developed an ocean to remove co2 from its atmosphere.
Do not know why this is not obvious to everyone.
The earth with no moon would look like Venus
Terry, any equations to go with your imaginative opinions?
Mr Spencer, by how much do the thick sulphuric acid clouds on Venus mitigate the warming?
The sulphuric acid clouds both reduce the flux balancing temperature by being highly reflective but negate the effects of ghgs by having significant broadband long wave opacity. The CO2 present, even at pressure still only absorbs around available bands which would leave windows for the surface to radiate through. The clouds close this windows and cap surface to space radiance.
The cloud deck is the effective emission height which sets the essential flux balancing temperature irrespective of the presence of CO2.
The cloud altitude sets the surface temperature gain as the surface is in equilibrium with the total energy of the effective emission surface.
“which would leave windows for the surface to radiate through.”
No IR windows since Venus near surface atm. opacity is totally opaque in the IR bands due mainly the high surface pressure, density acting with each of the absorbers in the atm. due to collision induced absorp_tion in ~hydrostatic equilibrium. No temperature equilibrium is possible until the emission spectrum shifts to regions for which the emissivity is not so near zero.
“which would leave windows for the surface to radiate through.”
No IR windows since Venus near surface atm. opacity is totally opaque in the IR bands due mainly the high surface pressure, density acting with each of the absorbers in the atm. due to collision induced absorp_tion in ~hydrostatic equilibrium. No temperature equilibrium is possible until the emission spectrum shifts to regions for which the emissivity is not so near zero.
There are some IR windows BF even through the clouds in the near IR where solar reflected overlaps with near surface emissions. Taylor et al 1997.
Sorry but your last sentence makes no sense to me at all.
Emiision spectrum of what?
Emissivity of what?
Why not so near zero when you have been speaking of total opacity?
Taylor et al 1997 is about dwarf star X-rays. Need more specifics.
“which would leave windows for the surface to radiate through” means the emission spectrum of the surface. The surface pressure of the absorbers in the atm. means the emissivity of the atm. Again, for Venus’ near surface atm. there are no true atmospheric windows at IR wavelengths greater than 3 micron. Venus high clouds scatter & absorb (about half & half) radiation at solar wavelengths but are strongly absorbing in the near IR (H2SO4 and CO2) for a net warming in the long term equilibrium balances.
https://trs.jpl.nasa.gov/bitstream/handle/2014/21811/97-0211.pdf?sequence=1
Taller et al 1997.
You do state correctly ball4 that no significant windows exist below 3μm which I do uphold. However the surface of Venus at 737K has a calculated in vacuo emission peak at 3.9μm and so significant band radiance still exists below 3μm where, even with clouds present which pose additional broadband opacity and scattering potential, observations deep into the atmosphere are possible.
What remains impressively relevant is that the in vacuo radiative potential of the surface at 737K is completely killed by the presence of the atmosphere such that it loses next to nothing by radiation. The 16,500W/m2 radiative potential is reduced to a few W/m2 such that the surface exists at a temperature calculable from a single measurement at altitude within the troposphere without thought of changes in opacity or what stops the surface radiating. The internal long wave interactions do not alter the gradient to the surface at all.
“You do state correctly ball4 that no significant windows exist below 3μm which I do uphold.”
The statement is “Again, for Venus’ near surface atm. there are no true atmospheric windows at IR wavelengths greater than 3 micron.”
“without thought of changes in opacity or what stops the surface radiating.”
Sure, Venus near surface T is at long term equilibrium. Change the atm. pressure or the grey absorber gas mixing ratios in the troposphere enough and the global surface temperature will change in proportion just like on earth.
Slapped myself around the head ball4.
Now, no windows exist at wavelengths greater than 3μm. Some windows exist at wavelengths smaller than 3μm and this is supported by the fact that the full atmospheric column does not constitute one totally opaque layer to IR.
To counter your second statement. Despite having completely different grey body mixing ratios the T1bar temperatures of Earth, 288K and Venus 339K exist in line with the temperature ratio predicted from the solar flux ratio alone. That is, there is no sign of an enhanced greenhouse effect at one bar pressure on Venus despite the 96.5 percent CO2 atmosphere and differing grey body mixing ratio.
Geoff, there are obviously atm. windows at “smaller than 3 microns” as pictures from Venus surface show visible light scattered, emitted and transmitted from Venus clouds incident on surface.
“Despite having completely different grey body mixing ratios the T1bar temperatures of Earth, 288K and Venus 339K exist in line with the temperature ratio predicted from the solar flux ratio alone.”
Generally accepted P=density*R*T works on Venus too as that is the formula NASA used to obtain Venus’ T(z) from their measured atm. density(z) and known P(z). Setting P=1bar you can only write something fundamental about gas T at 288K and gas T at 339K if you also write about density which you have not mentioned. NASA found the 339K at 1 bar on Venus by measuring the atm. density there at 1 bar. With that in mind, consider your comment attempting to counter mine:
“there is no sign of an enhanced greenhouse effect at one bar pressure on Venus despite the 96.5 percent CO2 atmosphere and differing grey body mixing ratio.”
The atm. density at 1 bar on Venus was found to compute its T(1 bar)=339K precisely because of the giant GHE on Venus.
For example, consider the Venus open IR window for “smaller” & then opening the window greater than 3 micron also. This means a new atm. for Venus with nil grey absorbers and thus nil collision-induced absorp_tion with nil pressure dependence.
Already have the Venus measured temperature data for that GHE eliminated case being measured at 232K as well as earth’s GHE eliminated data measured at 255K.
So let’s substitute that measured data into your comment which should work if your comment is fundamentally correct: “Despite having same grey body mixing ratios (nil) the T1bar temperatures of Earth, 255K and Venus 232K exist in line with the temperature ratio predicted from the solar flux ratio alone.”
Using a 1:1.176 for the Earth to Venus fourth root ratio of solar radiation from orbital data, p = 1 bar T Earth 255K, T Venus 232K for 1 bar with appropriate density at Venus higher pressure, T Venus/1.176 = 232K/1.176 = 197K an error of (255 – 197) = 57K !!
So big error, no fundamental agreement.
Now put back all the grey absorbers into Venus atm. thus close its IR window above 3 micron from surface and do the same calculation:
p = 1 bar T Earth 288K, T Venus 339K 1 bar with NASA measured density at Venus much higher pressure, T Venus/1.176 = 339K/1.176 = 288K an error of (288K-288K) = 0K !!
Your comment then does work BECAUSE of Venus enormous GHE and proves the NASA measured Venus atm. density is from a giant GHE at 1 bar and NOT “no sign of an enhanced greenhouse effect at one bar pressure on Venus”.
Geoff actually has shown there definitely is a sign of an enhanced greenhouse effect at one bar pressure on Venus right in Geoff’s comment.
At least we both agree that the published T1bar temperature of Venus by NASA was and is 339K which is exactly what one would predict from the flux ratio and shows no enhanced greenhouse effect as a result.
Yes NASA measured density from the Radio Occultation Studies but this also retrieved pressure without using the equation of state.
From the boundary condition that the first Doppler detection of the signal is the start of the atmosphere, integration of the density as mass per unit volume gives the mass of the column per square metre and therefore the pressure to the tangent if you know the gravity, which of course you do.
So, knowing the mean molar mass, T is retrieved at all points of the sweep with no other variables from T = (M,P)/(ρ.R)
The other temperatures you have produced are bogus black body temperatures given that you openly protest that black bodies dont exist and then quote ε=1 temperatures.
The rest is all waffle that does not detract from the fact that we agree that T1bar on Venus is nothing other than in line with the flux ratio magically cancelling all other proposed effects, as the predicted temperature remains unaltered.
Knowing that the surface pressure is 92bar rather than 1bar we can then calculate the surface temperature from an isentropic equation without thinking about grey body mixing ratios and the temperature calculated leaves no room for any other effects.
So the data from Venus shows no enhancement from a heat trapping greenhouse effect at 1bar and it shows no sign of enhancement at the surface that isnt explained by the clouds as liquid and solids in suspension (with much higher long wave emissivity than low pressure gases) flux balancing with space at around 60Km in a gravity field.
Starting from planetary greenhouse 1bar of Earth (288K) after applying the flux ratio shows the enhanced greenhouse effect 1bar of Venus (339K) considering the equation of state; starting from no greenhouse on Earth I demonstrated the ratio does not work and you have found no fault.
“The other temperatures you have produced are bogus black body temperatures given that you openly protest that black bodies don’t exist and then quote ε=1 temperatures.”
Black body radiation does exist though and these temperature (255K) and (232K) are from measurements of such radiation as I wrote so they are not bogus at all.
“flux ratio magically cancelling all other proposed effects, as the predicted temperature remains unaltered.”
There is no magic needed Geoff, the equation of state & NASA measurements shows there is a huge planetary GHE at Venus at 1bar as ratioed from Earth’s GHE on the orbital difference.
“Knowing that the surface pressure is 92bar rather than 1bar we can then calculate the surface temperature from an isentropic equation without thinking about grey body mixing ratios and the temperature calculated leaves no room for any other effects.”
Then do so. You will need measurements with the grey body absorbers in place or calculations with grey body absorbers in place without these measurments or calculations the isentropic formula will not work.
The data from Venus shows a GHE simply because the surface is measured around 730K (from orbit and from landers) and the planetary disk radiation is around 232K (from orbit and from earth).
Geoff…”The CO2 present, even at pressure still only absorbs around available bands which would leave windows for the surface to radiate through. The clouds close this windows and cap surface to space radiance”.
Why do most posters talk only about radiative effects while ignoring the effect of direct conduction between the 450C surface and subsequent convection?
Whereas radiation from a 450C surface would have more impact than radiation from Earth’s surface, conduction and convection are equally important.
Hi Gordon. I talk of radiation where others are talking about radiation as part of the discussion. The point about the thermal gradient that exists in both tropospheres is that ithey are totally independent of radiation, conduction or convection.
There are regions in the Venus troposphere where radiation dominates and convection is suppressed and vice versa. The gradient however runs smoothly through these regions without indicating a change in heat transfer mechanism.
The thermal gradients in the vertical are strongly suppressed in the vertical to heat transfer by conduction and convection as diffusion and convective overturning both perpetually recreate the adiabatic lapse rate. What is more difficult to comprehend is that long wave radiation acting over respective optical depths also fails to modify the lapse rate. Long wave radiative exchanges within the atmosphere should drive the column towards being isothermal, but these same fluxes are held responsible for the existence of the lapse rate in the first place. The result as shown by the data is that long wave fluxes do not modify the gradient at all from the gradient proposed in their absence. This is a profound outcome given all of the discussions about long wave forcings and feedbacks.
“The result as shown by the data is that long wave fluxes do not modify the gradient at all from the gradient proposed in their absence.”
This is incorrect as shown by many papers on radiative-convective equilibrium since the early 1960s. When a well mixed grey absorber is increased in the troposphere, then the lapse curve adjusts as it rotates about a ~midway point to keep thermodynamic internal energy conserved (constant) in the troposphere by increasing the global planetary median T at the surface and decreasing global planetary median T at higher levels in the atm. This is what is meant by atm. optical depth increasing.
Ball4,
Very interesting, do you have links to these papers?
phi, these are the foundational papers for modern meteorology. Any in depth study of the field will surely lead you to them. Better for you to plot your own course through the educational process.
Ball4,
You do not answer my question: do you have links to these papers?
Well, I think you know that these papers say the opposite of what you claim.
Ball4.
Papers cannot disprove what the data shows.
From altitude on both planets the surface exists at a temperature predicted without thinking about grey body opacity or mixing ratios.
The notion that an isothermal column (diffusive, non radiative) is pulled super adiabatic by ghgs and then truncated to the observed lapse by convection is embarrassingly complicated given that we can calculate the reversible adiabatic gradient observed in a much simpler manner. These competing processes obviously do not compete at all but act together to run the system to maximum entropy.
You have stated that increasing ghgs increases the gradient by warming the lower atmosphere and cooling the upper,
Quote,
it rotates about a ~midway point to keep thermodynamic internal energy conserved (constant) in the troposphere by increasing the global planetary median T at the surface and decreasing global planetary median T at higher levels in the atm.
Unquote.
But this notion is not upheld by data. It remains theory that is not supported by the data of both planetary tropospheres.
The gradients running to the surface of both planets is a reversible adiabatic profile unaffected by grey body mixing ratios.
That is what the data from both planets show.
Geoff Wood,
The gradients running to the surface of both planets is a reversible adiabatic profile unaffected by grey body mixing ratios.
That is what the data from both planets show.
No. Observations show that the gradient in the Earth’s atmosphere is much lower than the adiabatic lapse rate. Observed: about 6.5 K / km. adiabatic : a little less than 10 K / km (given the asymmetry between volumes in ascent and subsidence).
Geoff: “Papers cannot disprove what the data shows.”
Proper papers can prove what the data shows, a good example for radiation is Planck’s 1912 treatise.
Improper papers (AND comments!) do not show measured data so cannot disprove what the data shows.
Geoff: “But this notion is not upheld by data. It remains theory that is not supported by the data of both planetary tropospheres.”
Improper papers (AND Geoff’s comment!) do not show measured data so cannot disprove what the data shows.
Hi Phi.
Quote
Observations show that the gradient in the Earths atmosphere is much lower than the adiabatic lapse rate. Observed: about 6.5 K / km. adiabatic : a little less than 10 K / km
Unquote
You have compared a dry adiabatic lapse with an atmosphere that has another energy storage mechanism. Include a term for latent heat and the gravitational potential energy change with altitude is then stored as thermal energy or specific humidity. No additional energy is required or heat exchanged to describe the adiabatic relationship between the upper troposphere and the surface.
Geoff Wood,
You’re right but for air masses in subsidence (the largest part of the troposphere), the dry adiabatic is relevant. Observations show a much lower lapse rate. This is not surprising since the troposphere radiatively cools throughout the column.
Phi, global means indicate that the dry adiabatic lapse rate and the resultant potential temperature is a conserved quantity in the vertical. At the surface the actual temperature is reduced from the potential temperature by the quantity of water vaporised.
Observations show a much lower lapse rate than the dry adiabatic because the surface is cooled and the atmosphere heated by latent heat transfer which reduces the lapse rate to the observed gradient. This is quantifiable by realising that at the surface the energy is in the form of thermal states and chemical potential of the vapour state of water with no gravitational potential energy at z=0.
At the top of the troposphere (z~10km) the energy is stored as thermal energy and gravitational potential energy with no chemical potential as the specific humidity here is effectively zero. But the column can be specified as a 310K isentrope by using potential temperature as a conserved quantity and expressing the other energy storage mechanisms as thermal equivalents.
Quote
This is not surprising since the troposphere radiatively cools throughout the column.
Unquote
Radiative cooling is not necessary to explain the observed thermal gradient in the troposphere as over the diurnal cycle the net cooling and heating rates are effectively zero. Also, the basis of the lapse rate is that we set dQ=0 such that changes in gravitational potential energy oppose thermal energy storage. ie we predicted the observed gradient in the absence of radiative cooling or heating.
Geoff Wood,
From the laws of thermodynamics, we know that thermal gradient and heat flow are linked.
Assuming an atmosphere totally inactive in IR frequencies, what would be, according to you, its thermal profile?
From a simple, low number count of non-interacting bodies, contained by and moving within a gravitational field it is equally simple to understand that changes in gravitational potential energy must be accompanied by equal but opposite changes in kinetic energy.
As gases are composed of atoms or molecules that have mass and therefore feel gravity we can expand upon this idea, and extend the model to be an atmosphere around a containment mass.
A non radiative model is simply one in which the collisions of the atoms or molecules are perfectly elastic and as such conserve energy.
Our gas can store energy as thermal energy, which is the sum of independent kinetic states and as gravitational potential energy due to its position in the gravitational potential energy field.
We can define the gas’ energy Q, as
Q = m . Cp . T + m . g . h
Where m is the mass
Cp is the isobaric specific heat capacity
T is the temperature
g is acceleration due to gravity
h is height as defined by g
We can also write from above,
dQ = Cp . dT + g . dh
Which expresses that changes in total energy result in changes in temperature and/or changes in height per unit mass.
In the absence of heat flows dQ=>0 which is effectively the adiabatic condition of no heat enters or leaves which suits our non-radiative example.
Thus we can write,
Cp . dT + g . dh = 0 for adiabatic, which can be rearranged to give,
Cp . dT = -g . dh
Which mathematically expresses that even for our non-radiative gas, changes in positive vertical position are accompanied by negative changes in temperature per unit mass.
We can further rearrange the above to give,
dT/dh = -g/Cp
Which is the adiabatic thermal gradient derived for a gas conserving total energy in a gravitational containment field.
For dry air the rate of change of temperature with height is
-9.81/1,006= 0.00975K/m or 9.75K/km
On Earth this gradient is reduced by the availability of chemical potential energy in the form of latent heat adding to the total available energy storage.
Geoff Wood,
Your explanation poses many problems, both theoretical and experimental.
1. Any closed system tends towards thermodynamic equilibrium by maximizing entropy, thus, tending towards isothermal.
2. Your relationships express a dynamic behavior of volumes in vertical motion but say nothing about the static state of the air masses.
3. When two air masses meet, the coldest slips under the hottest. The resulting gradient is inverted and corresponds to a stable state preventing vertical movements. The system will tend towards a thermodynamic equilibrium by conduction, thus towards isothermal. Isothermal is also a stable configuration that prevents vertical movements.
4. Convection, therefore vertical motions in the troposphere, occur spontaneously when the thermal gradient exceeds the adiabatic. When heat flows are low (eg at night or in winter or in high latitudes), the thermal gradient tends to reverse.
5. Free convection only appears when it increases the entropy of the atmosphere. Forced convection can only exist by providing work.
Thanks for the reply phi.
1 Any closed system tends towards thermodynamic equilibrium by maximizing entropy, thus, tending towards isothermal.
A reversible adiabatic gradient is a maximum entropy profile. By definition a reversible adiabatic is isentropic. The isothermal column is lower entropy as shown by Bohren and Albrecht, Atmospheric Thermodynamics, 1998.
The isothermal column requires a mechanism to explain why it systematically gains total energy with altitude and how it remains contained because of this.
2.Your relationships express a dynamic behavior of volumes in vertical motion but say nothing about the static state of the air masses.
No phi, I opened by stating that it is easy to explain that for individual particles moving in a gravity field without interaction, the trajectory of each trades kinetic energy for gravitational potential energy. You accept that volumes of air moving vertically map out a reversible adiabatic and particles moving individually in the tenuous lunar atmosphere will map out the same containment profile. There is no step change to volumes from individual motion. The bulk motion is the integration of microscopic events where small numbers maximising entropy work together en mass to maximise entropy more successfully.
3 When two air masses meet, the coldest slips under the hottest. The resulting gradient is inverted and corresponds to a stable state preventing vertical movements. The system will tend towards a thermodynamic equilibrium by conduction, thus towards isothermal. Isothermal is also a stable configuration that prevents vertical movements.
When two air masses meet it is the result of tiny teleconnected pressure differences in the horizontal that are overturning and mixing cells weighing billions of tonnes. That information is carried by collisional information of individual particles interacting. The motion and total mixing of the cells globally does not detract at all from the gradient set by gravity. In fact it is an example of how the system maximises entropy production in transferring thermal energy poleward such that the system flux balances by radiating to space at the lowest possible temperature.
4 Convection, therefore vertical motions in the troposphere, occur spontaneously when the thermal gradient exceeds the adiabatic. When heat flows are low (eg at night or in winter or in high latitudes), the thermal gradient tends to reverse.
Convection is the sum of advective flow and diffusion by definition. You are separating bulk motion from diffusion as if they are different things when collectively the momentum interaction of individual particles acting in a diffusion process produces the netted pressure differences that cause bulk motion to occur. There is no step change as both operate to maximise entropy which ultimately is what we find in the global atmospheric gradients.
5 . Free convection only appears when it increases the entropy of the atmosphere. Forced convection can only exist by providing work.
The gradients found within the atmosphere of all measured bound gaseous envelopes are adiabatic as global means and are not pulled away from this by any individual process.
The notion that a static atmosphere would be isothermal but is destabilised by radiation and truncated to the environmental lapse by convection is messy and is undermined by the fact that the observed global mean is the mean calculated in the absence of heat entering or leaving or heat transferred internally. That is what the data shows.
Geoff Wood,
Interesting argumentation but I disagree with just about everything you say. I will not take everything because that would lead to a difficult discussion about the concept of entropy. I will content myself with the following points:
1. The observations invalidate any decisive role of the adiabatic gradient. It appears locally and partially only in very strong ascent. I repeat, the troposphere is mostly in subsidence, the dry adiabatic should apply but the effective thermal gradient is much lower.
2. In an atmosphere with GHG, a thermal gradient that would be fixed to the adiabatic (what value exactly?) Would cause a fixed IR flow from bottom to top. This is obviously in contradiction with the definition of thermodynamic equilibrium and contravenes the first principle of thermodynamics.
1 The observations invalidate any decisive role of the adiabatic gradient.
Again. And this is becoming amusing, you conflate adiabatic with dry adiabatic. Adiabatic can apply to an atmosphere with moisture which you seem unable to grasp conceptually.
Quote
but the effective thermal gradient is much lower.
Unquote
Yes because as I keep telling you chemical phase change adds to the atmospheres heat capacity and lowers the thermal response accordingly. Add another term to account for this as I have already said and the gradient is easily shown to be conservant of total energy but not of thermal.
2 In an atmosphere with GHG, a thermal gradient that would be fixed to the adiabatic would cause a fixed IR flow from bottom to top
And there lies your problem.
You require radiation to produce the gradient by destabilising the isothermal column without it and then declare that radiation would act in a manner to drive it isothermal anyway.
So what exactly, do you think, produces the gradient (given that you are out on a limb because the gradient is demonstrably, extremely adiabatic)?
Phi,
Quote
I repeat, the troposphere is mostly in subsidence,
Unquote
Is obviously not true as mass is the appropriate quantity and the atmosphere is not in collapse being totally supported by the solar flux. The rising air of exact mass to that subsiding is water vapour laden and results in near isothermal steps from the lower condensation level to the free convection zone.
Regards
Geoff,
So what exactly, do you think, produces the gradient…
As in any thermodynamic system, the thremal gradient comes either from a work or from a heat flux between a hot source and a cold source. In our troposphere, these sources are the surface heated by the sun and GHG.
1. Subsidences can not generate a moist adiabatic gradient since they are not (or only briefly) supplied with water. It is therefore the dry adiabatic that should apply to the subsidences.
2. You consider that thermodynamic equilibrium correspond to the adiabatic thermal profile. I point out that this profile causes a radiative flux from bottom to top in an atmosphere with GHG. Therefore, there can not be thermodynamic equilibrium if there is a thermal gradient.
Ascent are fast and subsidences slow. There is therefore, whether in mass or volume, a clear predominance of subsidences.
I have seen more than a few posts around the Internet that use Space.com as a source of information to claim that the cooling of the thermosphere is a sign of things to come in the troposphere.
Dont tell that to you know who.
On the relation between rotation and atmospheric composition is this study https://www.forbes.com/sites/brucedorminey/2018/09/23/early-suns-goldilocks-rotation-rate-may-be-why-were-here/#46d5dfc0794b
Our early Suns rate of rotation may be one reason were here to talk about it, astrobiologists now say. The key likely lies in the fact that between the first hundred million to the first billion years of its life, our G-dwarf star likely had a Goldilocks rotation rate; neither too slow nor too fast.
Instead, its hypothetical intermediate few days rate of rotation guaranteed our Sun was active enough to rid our newly-formed Earth of its inhospitable, hydrogen-rich primary atmosphere. This would have enabled a more habitable, secondary atmosphere composed of nitrogen, carbon dioxide, hydrogen and oxygen to eventually form.
If it had been a fast (less than one day rotator), our Sun might have continually stripped our young planet of its secondary atmosphere as well. However, if it took more than 10 days to rotate, it might not have been active enough to strip Earth of its hypothetical primary atmosphere.
My snopsis is https://rclutz.wordpress.com/2018/09/24/tweak-the-suns-rotation-and-were-not-here/
To be clear, they are talking about the sun’s rotation rather than the earth’s.
I see an author who needs to read (or watch) The Expanse. But only just because it’s so awesome and there’s a bit in the story with Venus.
I’m pretty sure that when you look into Venus it isn’t the Greenhouse Gas Effect causing the warming. The thick atmosphere absorbs the incoming radiation as the source of warming. Trapping outgoing LWIR has nothing to do with it.
The high temperature on the surface of Venus was explained by Sagan & Pollock in 1967. While hundreds of people have piled on nothing much has changed since then.
https://tallbloke.wordpress.com/2014/06/28/venus-surface-temp-correctly-predicted-from-lapse-rate-in-1967-but-is-it-the-whole-story/
Venus High Temperatures Arent Due to the Green House Gas Effect; More Climate Sophistry
https://co2islife.wordpress.com/2018/03/06/venus-high-temperatures-arent-due-to-the-green-house-gas-effect-more-climate-sophistry/
The data for our entire solar system (Nasa etc) show all atmospheres to have temperature regimes dependent on solar distance and gaseous Mass alone. If minimum pressure is above 0.1bar.
Gas specie is irrelevant, and so any Scientist with both eyes open should grasp that the Gas Laws of the Poisson Relationship and Thermodynamics as in Maxwell’s Theory of Heat pages 330-350 etc., decide surface Temperature solely. Zero GHE! Brett
“Pressure Rools!”
How does pressure prevent IR from escaping from the surface to space?
Pressure controls IR opacity. Without pressure the natural line width is too narrow to block radiation even with 500,000 spectral lines. Pressure (and temperature) broadening smears these lines towards a continuum. At the point of collision a diatomic gas is a quadrupole with an effective dipole moment and is capable of absorbing and re- radiation IR. As the pressure increases (and temperature) weak continuum radiation becomes more apparent as all interacting matter producesradiation to some extent.
On Venus the gaps in CO2 opacity are rendered insignificant because of the presence of a blenket condensing gas (cloud) which caps radiance from below rendering the lower atmospheric opacity of ghgs impotent.
Geoff, in the lower atm. Earth and Venus, radiative transfer is dominated by collision-induced absorp_tion where grey opacity is proportional to pressure. It is in the higher levels of each atm. where Doppler broadening dominates and for that process the atm. aborbers grey opacity is independant of pressure (ref.s are Pollack Icarus 10, 1969, pp. 301-13 and McKay Icarus 137, 1991 pp 56-61).
Thank you for the input BF.
Dominated by does not negate some effect. Thank you for not detracting from the statement that pressure controls IR opacity.
Yes, and atm. IR opacity arises from total atm. pressure collision-induced absorp-tion not the partial pressures of the gaseous grey aborbers in each atm. as some comment around here.
Ball4 says:
Yes, and atm. IR opacity arises from total atm. pressure collision-induced absorp-tion
How do N2 and O2 absorb IR?
David, as I wrote, collision-induced absorp-tion (grey opacity proportional to pressure) leads to weak absorp-tion features of both N2 and O2 in the infrared [e.g., Hartmann et al., 2008]. Due to the abundance of N2, O2, even the weak infrared absorp-tion of these constituents is radiatively measurable in earth atm.
N2 has two major bands influencing Earth’s outgoing infrared radiation: the collision-induced roto-vibrational fundamental band at 2400 cm^-1 and the collision-induced roto-translational band at 100 cm^-1. For O2, the collision-induced fundamental vibration-rotation band at 6.4 microns is the major absorp-tion signature of O2 in earth’s atm. thermodynamic internal infrared.
On global average under clear-sky conditions Earth OLR is reduced due to O2 by around 0.11 Wm^-2 and due to N2 by 0.17 Wm^-2. For perspective, together this amounts to about 15% of the OLR reduction caused by CH4 at present atmospheric concentrations.
For you atmospherics pressure adherents
http://www.drroyspencer.com/2016/07/the-warm-earth-greenhouse-effect-or-atmospheric-pressure/
Dr. Roy is right! Most of the time.
Prove him wrong.
From the flux balancing temperature at altitude, the surface equilibrium temperature can be retrieved from the isentropic flow equation which for a given species relates temperature to pressure with no other variables. The gradient is a maximum entropy profile which by definition, all other profiles will evolve towards.
The system is purely radiative across the vacuum of space which sets the flux balancing temperature at the effective radiative height in a gravity field. Pressure is a convenient measure of this position at it includes total mass and mass above. As such it is closely related to geopotential which is gravitational potential energy per unit mass.
Within the atmosphere there is no sign of enhancement of temperature by long wave opacity effects. Excess energy is transferred and distributed by all available mechanisms to control the gradient to the stability gradient set by gravity.
Geoff Wood says:
Within the atmosphere there is no sign of enhancement of temperature by long wave opacity effects.
The troposphere gains more energy from GHG ab.sorp.tion.
Since when does more energy not mean a higher temperature?
“Since when does more energy not mean a higher temperature?”
Since detailed measurements of ice melting in a container of 32F water were carried out.
Good one, Ball4.
Svante,
. “More water vapor causes more warming until all water is lost to space.”
Hogwash. Water COOLS the planet…
Loss of water decreases the efficiency at which a planet cools, thus temps rise…
Venus has been cooking off its water for hundreds of millions of years and the temp there has risen accordingly…
PhilJ
Liquid Water would cool the planet as it takes energy to evaporate it to vapor state. Water vapor does not cool the planet, in either process it increases the temperature. When it condenses the energy that it removed in evaporation is returned higher up in the atmosphere (warming it). With the radiant energy, water vapor with temperature radiates up and down, some out to space (which is a measured value) and some back to Earth (which is also a measured value). The energy reaching the surface will add to the energy balance causing warming.
No Norman, the evaporation/condensation process is “heat transfer”. No new energy is created. There is no temperature increase to the system.
And the “back-radiation” from water vapor will NOT “add to the energy balance causing warming”.
You’re still making the same mistakes.
Nothing new.
JDHuffman (g.e.r.a.n)
You make your same unsupported declarations. I tire of your droning on and about things you don’t understand. Learn some real physics. I might have more interest in your comments. As it is now you act like a child that want attention when adults talk.
Not many care what you post, agree with your stupid declarations, or are the least bit interested in your comments.
You have shown many times you do not understand any real physics. You pretend to just like you pretend you are not the same person that posts as g.e.r.a.n. Just a big phony!
JDHuffman
Also your reading comprehension is nearly zero! You just make a comment about my post without even reading the material, or attempting to understand what is written.
YOU: “No Norman, the evaporation/condensation process is heat transfer. No new energy is created. There is no temperature increase to the system. ”
When did I claim new energy was created by evaporation/condensation process???
Evaporation cools the surface but warms upper layers of atmosphere when it condenses. Heat transfer, it removes it from the surface and moves it elsewhere, the surface is cooled in the process. I guess you are too stupid to interpret that when I put “planet” in the context it was referring to the surface. Not very intelligent or able to interpret anything. Please go off an play with your toy Ferris Wheel. You don’t have adult reading ability. People need to spoon feed you everything or else you are not able to figure out the meaning using logic or rational thought process.
Norman, after I pointed out your mistakes, which you labeled “unsupported declarations”, you sure made the necessary corrections pretty fast!
Now, remember next time, so you don’t make the same mistakes again. I’m trying to teach you enough so you won’t look so stupid.
But you always find some way to hinder my efforts….
These might be useful.
An energy budget for Venus.
https://davidappell.blogspot.com/2016/05/energy-balance-diagrams-from-around.html
And for Earth.
https://scied.ucar.edu/longcontent/energy-budget
That link to our resident troll (David Appell) is interesting so why does he have so few comments on his blog?
Could it be that he censors dissident ideas?
I understand that you don’t like to hear any views other than your own, but why does that make me a troll?
It might be your many stupid questions, like this one.
So do you censor dissident or opposing ideas on your site or not?
Come find out.
You are a one-liner type of troll, David. You can get into any deep discussion of an issue.
gallopingcamel says:
Could it be that he censors dissident ideas?
Since you never comment except on safe, denier blogs, you’ll never know, will you?
Which undenier blogs did you have in mind?
The relationship is 1st order on atm. pressure and inconsisted on atm. composition, yet the GHE lives on.
Why is the atmosphere so much thicker, and denser than earth’s? Gravity on Venus is a little less than earth’s, correct?
IIRC
Water disappeared early, dissociated by more intense UV than Earth gets.
No water, so no weathering to remove CO2 from atmosphere into rocks.
CO2 from a billion years of volcanic activity accumulated to produce the present thick atmosphere.
Is any of that science, as in FACT?
It looks like the typical guesses, assumptions, and estimates found in all pseudoscience.
Why don’t you the science of all this…?
Interesting that a “science writer” doesn’t know any science and doesn’t know how to compose a sentence either.
Entropic Man accurately presents the conventional explanation for Venus.
While surface gravity differences (Earth vs. Venus) have some effect, what matters most is PV = nRT
On Venus “n” is really large so the surface pressure is ~93 times higher than on Earth (1 bar).
http://www.drroyspencer.com/2011/12/why-atmospheric-pressure-cannot-explain-the-elevated-surface-temperature-of-the-earth/
GC: How does gravity or pressure block IR from escaping from the planet’s surface?
Could it be that the volcanos on Venus were by far more numerous than on earth, plus the CO2/H20 ratio off the charts higher coming out of the volcanos? I am trying to figure out why the CO2 content of the Venusian atmosphere is 220,000 times that of earth’s!
Let’s say if the atmosphere of Venus was mostly N2, but the same density and thickness (if that is possible), the surface temperature would still be close to the same would it not?
Rob asks: “Could it be that the volcanos on Venus were by far more numerous than on earth…?”
Good question, Rob. But “they” aren’t interested in science. They’re interested in pseudoscience.
There is a lot of evidence Venus still has molten lava on its surface. Molten lava could easily have a temperature over 500 °C.
The high energy solar would likely get through the atmosphere. UV photons have a Wien’s associated temperature of well over 1000 °C.
The high temperatures on Venus need to be explained by a thermodynamic heat source. CO2
is not a heat source.
JD, I have noticed that the human-caused global warming alarmists are much more interested in linking earth to Venus; and trying to show that we could have a runaway greenhouse effect that could turn us into Venus. Even Dr. Hawking entertained this hysterical scenario, which I consider to be shocking from a scientist of his stature to suggest.
I suppose scientists are not immune to authoritarianism. Some of them like to impose their political ideology onto others, and resort to scare tactics to get what they want on the policy front. As a simple weatherman for 38 years, I know that the idea of earth turning into Venus is complete garbage!
Rob: Scientists do not think Earth can have a runaway greenhouse effect. Neither do intelligent commentators.
–Rob Mitchell says:
December 5, 2018 at 2:13 AM
Could it be that the volcanos on Venus were by far more numerous than on earth, plus the CO2/H20 ratio off the charts higher coming out of the volcanos? I am trying to figure out why the CO2 content of the Venusian atmosphere is 220,000 times that of earths!–
Well there is a lot methane in our solar system, and Methane become CO2 in sunlight.
I [and others] tend to think Venus was a small gas giant.
Neptune has a lot methane, though it has far more hydrogen and helium in it’s atmosphere- but if assume the helium and hydrogen are removed when sun start main sequence and/or impactors, you are left small percent of this huge atmosphere.
Or Neptune is about 17 times more massive than Earth but the Neptune rocky core is thought to be about the mass of Earth. Or Neptune atmosphere and ocean/mantle are most the mass of planet, and most mass of planet is hydrogen and helium.
So one could have Venus forming before the sun became the sun, or small gas giant migrated from outer system at a later date.
–Lets say if the atmosphere of Venus was mostly N2, but the same density and thickness (if that is possible), the surface temperature would still be close to the same would it not?–
I think it depends the acid clouds, but Venus has about 3 times amount of N2 as compared to Earth’s atmosphere and could have acid clouds and have be mostly N2 rather than having CO2. And these clouds would be about same temperature.
With the CO2 one has a lower layer of Venus which has CO2 at critical pressure, wiki:
“The pressure found on Venus’s surface is high enough that the carbon dioxide is technically no longer a gas, but a supercritical fluid. This supercritical carbon dioxide forms a kind of sea that covers the entire surface of Venus. This sea of supercritical carbon dioxide transfers heat very efficiently, buffering the temperature changes between night and day (which last 56 terrestrial days).”
https://en.wikipedia.org/wiki/Atmosphere_of_Venus
That aspect of CO2 would some effects which you wouldn’t get by replacing CO2 with N2.
The four inner planets are rocky spheres. It seems to me they would be more or less made of the same stuff. Venus is closest to Earth in size. Yet, the atmosphere is off the charts denser. If methane turns into CO2 in sunlight, it would need a lot of O2 to go along with it since methane is CH4.
“Rob Mitchell says:
December 5, 2018 at 11:33 PM
The four inner planets are rocky spheres. It seems to me they would be more or less made of the same stuff. Venus is closest to Earth in size. Yet, the atmosphere is off the charts denser. If methane turns into CO2 in sunlight, it would need a lot of O2 to go along with it since methane is CH4.”
Sunlight oxidizes surfaces bodies in space, the Moon is dark grey/black due to the Sun oxidizing the top surface. And by the way, the sun also implants hydrogen and Helium into the Lunar regolith- something like 2 billion tonnes of hydrogen is implanted in top 1 meter of lunar surface. There are trillions of tonnes in top 1 meter of lunar surface- it’s a small concentration, though it has considered possible that one day, it could be mined.
Other than hydrogen and helium which most abundant, oxygen is higher trace amount in comparison to hydrogen and helium, and carbon is rare in comparison to oxygen. Since hydrogen is abundant and there is fair amount of oxygen, there is a lot water in the universe and in our solar system. Likewise because Hydrogen abundant, there is a lot methane or other hydrocarbons.
So, I think the question with CO2 is where the rarer carbon comes from. Though yes, most of mass of CO2 is the oxygen.
In terms of universe, wiki:
“Hydrogen is the most abundant element in the Universe; helium is second. However, after this, the rank of abundance does not continue to correspond to the atomic number; oxygen has abundance rank 3, but atomic number 8. All others are substantially less common.”
Though Carbon is fourth. Then Neon, Iron, and Nitrogen.
But neither of these four is produced/emitted by the Sun- as far as I know.
The relationship is 1st order on atm. pressure and not discernible on atm. composition, yet the GHE lives on.
Ftop, I agree that the relationship with pressure (total mass of the atmosphere) is much more dramatic and clear than the effect of CO2 concentration. Doubling the total mass of the atmosphere using a non-GHG like N2 would have much more impact than doubling CO2 concentrations within our current atmosphere. And yes, I agree that this is the reason Venus is so much warmer.
On the other hand, to first order, the total mass & pressure of the atmosphere on earth are constant with no discernible trends. Meanwhile, CO2 and CH4 are changing dramatically.
This at least makes it plausible that CO2 concentration could matter. Ie a weak correlation with a variable that changes greatly could easily be comparable to a strongly correlation with a variable that hardly changes at all.
@Tim,
IMHO you are right.
I am making good progress on a mathematical model that may resolve this question.
Tim Folkerts says, December 3, 2018 at 6:19 PM:
There exists, however, NO observational evidence whatsoever to suggest it does; quite the contrary, in fact:
https://okulaer.wordpress.com/2018/11/11/how-the-ceres-ebaf-ed4-data-disconfirms-agw-in-3-different-ways/
https://okulaer.wordpress.com/2018/03/24/the-data-sun-not-man-is-what-caused-and-causes-global-warming/
Kristian, I glanced at your link. It looks interesting but I have a few vague concerns that I haven’t have time to think about fully. Hopefully some point soon I can comment at your blog.
By all means. I would welcome it.
Kristin,
Solar TSI has been slightly but steadily decreasing since the 1960s. How does that cause warming?
“What’s causing global warming?”
https://www.youtube.com/watch?v=sKDWW9WlPSc
David Appell says, December 4, 2018 at 8:43 PM:
It doesn’t. And I haven’t claimed that it does. Which you know full well …
BTW, my name is Kristian, not Kristin.
Kristian,
I’m sorry for misspelling your name.
Frankly, you write such long and incomprensible comments that I can’t tell WHAT you mean.
But you seem to usually claim that ASR — absorbed solar radiation — is what’s causing warming. But every time I’ve asked you for evidence of that — like a time series — you don’t delivery and aren’t heard of again for a few more weeks.
David, the info. you seek was introduced here by Kristian earlier this year referring to Loeb 2018 Table 7 & graphically in Fig. 14. ASR has meaningfully (within CI) increased in the CERES/Argo era through mid-2016. In conclusions they note:
“CERES TOA fluxes exhibit pronounced interannual variability driven primarily by ENSO. SW TOA flux variations in the Arctic are noteworthy and are tied to changes in sea ice coverage.”
David Appell says, December 5, 2018 at 9:37 PM:
Hehe. No, David. It is YOU who tends to disappear whenever I point out that it is YOU who needs to come up with evidence that ASR has trended DOWN over the last 50-60 years. Because it is YOU who’s claiming a negative solar contribution to Earth’s temperature evolution over that timespan.
Pointing to TSI is irrelevant, since TSI isn’t what heats the Earth. ASR is.
Tim Folkerts says:
Doubling the total mass of the atmosphere using a non-GHG like N2 would have much more impact than doubling CO2 concentrations within our current atmosphere.
Why?
Here (briefly) is my reasoning. Basically it has to do with the lapse rate.
The lapse rate will be pretty much constant independent of the concentration of IR gases. Adding more GHG’s will (slightly) raise the altitude from which IR escapes from the atmosphere (the radiateve Top of Atmosphere, or TOA) to space. Higher = cooler due to the lapse rate, so less IR will leave. That means more IR must leave from the surface.
Adding more atmosphere in general should have MORE impact on the TOA altitude, and hence more of a reduction on the temperature and on the IR escaping. Which would mean more radiation has to leave form the ground to reestablish balance.
There are lots of other variables and details, but that is my conclusion anyway.
Entropic,
“Water disappeared early, dissociated by more intense UV than Earth gets.”
Water did not disappear…it is indeed disassociated and gets cooked off, but there is still some left…
What happens when Venus finishes cooking off its water?
I suggest, that without h20 to cool it, it will quickly (compared to the loss oof water) cook off the rest of its atmosphere…
The induced magnetic field caused by H and O ions may also disappear , or be seriously reduced, allowing the solar wind to strip off the rest of the atmosphere quite rapidly…
The stratosphere is where molecular density eg pressure gets, as on Mars for instance, too thin to slow radiative transfer effectively. Below it, convection and WV phase change do the heavy lifting because of the Equipartition Principle. That is also why ‘water finds its level’ by whatever means uses least energy.
No need to believe me, a mere Applied Scientist, but you ignore Maxwell etc.at your peril.
Of course convection and phase change continue into the stratospher, from observations. Also, direct non-window IR flux starts to be effective above c.6km. There are several paths or an ‘IR catastrophe’ could occur (grin).
I say the gas laws rule because average 2angstrom molecules in gaseous c.12angstrom cubes have enough free space for gas specie differences to be subsumed by their similarities of Physics ie Natures.
I guess even Iron gas would be similar if not plasmatic, maybe…….
brett….”I say the gas laws rule….”
Agreed. Provided you are talking about gas laws that equate pressure, temperature, and volume.
According to the Ideal Gas Law, the mass percent of CO2 in our atmosphere could provide no more warming that a few hundredths of a degree per degree C.
Gordon again completely ignores the effects and energy transfer of radiation, as if an entire field of physics doesn’t exist.
He does this because he doesn’t understand it.
Robinson & Catling have modeled temperature vs. altitude on the seven bodies in our solar system that have “Significant” atmospheres in our solar system.
Their analysis does not work for Mars given that the surface pressure (0.006 bar) is much less than 0.1 bar.
Funny Dr. Spencer you bring up Space.com. After last week’s Mars probe landing, I spent the past weekend reading up on the Mars environment and topography . Space.com impressed me with its ridiculous speculations about Mars’s extremely distant past and minimizing the reality that having very very little atmosphere and temperature stability make it a horrible prospect for people to ever live there.
Fake science was my impression of the site and now you are chipping in as well.
Keep educating us Dr. Spencer. I just wish you had a wider audience.
Wish he had an audience that could learn.
http://www.drroyspencer.com/2014/04/skeptical-arguments-that-dont-hold-water/
It’s hard to explain 2LoT to someone that believes this is relevant:
“So, yes, a cooler body can make a warm body even warmer still, as evidenced by putting your clothes on.”
I notice you didn’t refute that, only mocked it.
It both refutes and mocks itself.
Again you didn’t refute it.
Because you can’t.
That’s your opinion.
Still no refutation. Only dumb games.
Yes DA, at some point one would think you would get tired of your games. But, that appears to be all you’ve got, since you’re opposed to learning some physics.
Mr JDHuffman vs. that quote from Dr. Roy Spencer, PhD. Who to trust?
Oh my…. how so easily refutable that silly Venus article is. However, I do reserve the option to reverse my opinion when we find the oil rigs, the massive defunct factories … and millions of fossilized autos….. since all that heating must somehow be caused by an unnatural source( even if they can’t offer evidence of any heating what so ever besides natural solar cycle correlations). … however, it all looks perfectly natural…. .. so nothing to see here but moronic attempts to link apples to oranges for an agenda…
So you think there are no explanations for Venus??
Hmmm no- please re-read for comprehension and a little less agenda. My conclusion is that what has happened on Venus has an entirely natural explanation. Now, I will not act like you and inject some biased subjectivity spew and make a silly claim to suggest that you must think that it is human caused. Silly things like that just make one look, well, silly.
Martin says:
My conclusion is that what has happened on Venus has an entirely natural explanation.
No one disagrees with you on that!
whats the temperature of Venus’ atmosphere at 1 atmosphere altitude?
The same as ours but allowing for Venus’ closeness to the sun. Nasa orbital data.
That’s right.
https://en.m.wikipedia.org/wiki/Atmosphere_of_Venus#/media/File%3AVenusatmosphere.svg
However, if the temperature is controlled by the pressure, why
does the temperature rise above 100km?
Common 0.1 bar tropopause in thick atmospheres set by pressure-dependent infrared transparency.
https://media.springernature.com/full/nature-static/assets/v1/image-assets/ngeo2020-f1.jpg
Another junk chart from ren.
It does not show any P/T relationship at 0.1 bar.
EM, because putting more weight above squeezes gas atoms closer so more Kinetic Energy (vibrations) are contained in the same volumes. This is the Lapse Rate for each planet or moon etc.. Reduce insolation and there is less KE until it collapses into ice, or vice versa. Works for Titan, with its dense methane/nitrogen air/liquid/snow/ice. Just as it does here and on Venus. Zero GHE, only Atmospheric Thermal Effect. This does not need insolation to a surface though that helps. A few km, one ‘Optical Depth’ and over time the gas warms and bouyant convection starts its distributive magic. Water or other phase change eg methane enhances the process and reduces evaporative losses to Space. On Venus, there is that energy exchange around 50km IIRC involving Sulphur gas/liquid compounds for instance. Brett
1) Earth
2) Venus
Conclusion: can have very different climate dynamics.
Difficulty to understand: 2 of 10.
How does pressure block the massive amounts of infrared radiation coming off the Venetian surface?
David Appell
That is a question I have asked in the past and the pressure idea does not explain this. It would be around 16000 W/m^2. They seem to ignore this situation.
DA, the question you should be asking is:
“Why is the energy balance for Venus as messed up as Earth’s?
And the correct answer is:
“Both are from the same pseudoscience.”
Ger*an: Energy imbalance is a fact, an observable.
Nothing you’ve ever claimed about “bad physics” has been convincing in any way. You can’t even get beyond that label to explain why, let alone show how your “better physics” explains anything at all.
DA, until you understand the basics, like a horse is not rotating on its own axis, then you’re not ready for anything more advanced.
Learn some physics.
Blah blah. You couldn’t offer insightful analysis if your life depended on it.
Stick to trolling — it’s all you will ever be.
DA, you call my comments “trolling”. That’s because you have no science to offer.
You just ask stupid questions. I correct your stupid pseudoscience.
Think of it as a “symbiotic relationship”.
David Appell
You are correct about g.e.r.a.n he will NEVER give any valid physics to support his nonsense. Nor will he ever answer any questions you address him. He dodges and evades and he is a complete shameless phony pretending that JDHuffman is not the same person posting as g.e.r.a.n.
DA and Norman don’t understand the relevant physics, so in frustration, they call me ger*an!
Their closed minds keep them frustrated, confused, and ignorant.
Nothing new.
JDHuffman
David Appell has studied real physics. You are the one who lacks any knowledge of the subject but still pretends to know it. Remember it was Roy Spencer that told you to learn some physics. Too bad you were not able to take his advice. I think your poor reading comprehension prevents you from learning anything. You repeat the same stupid stuff over and over regardless on the mountain of evidence against your ridiculous ideas.
ibid.
CO2 in the lower stratosphere causes a loss of water vapor in the upper troposphere and cooling. During the solar minimum period in the lower stratosphere, GCRs produce large amounts of 14C and 14CO2.
http://www.cpc.ncep.noaa.gov/products/stratosphere/strat_a_f/gif_files/gfs_t100_nh_f00.png
What’s deeply concerning is the use of dramatic and scary terms of description such as ‘horrifying modern appearance’ and ‘evil twin’ to cite just a couple, as if most really see these same descriptions which I doubt they do or ever have.
You could stand naked in the basket of a balloon at the 1 bar level on Venus without boiling, freezing or exploding.
You would be suffocatng from the CO2 and lack of oxygen and in agony from sulphuric acid burns.
That sounds sufficiently horrifying for my taste.
Why are we thinking about a Mars colony when a Venus colony would be more technically feasible? It seems that radiation shielded floating colonies could be assembled on Venus, with plastic film and aluminum wire bags, filled with breathable air.
https://www.quora.com/Why-are-we-thinking-about-a-Mars-colony-when-a-Venus-colony-would-be-more-technically-feasible-It-seems-that-radiation-shielded-floating-colonies-could-be-assembled-on-Venus-with-plastic-film-and-aluminum-wire-bags-filled-with-breathable-air
“ren says:
December 4, 2018 at 9:19 AM
Why are we thinking about a Mars colony when a Venus colony would be more technically feasible? It seems that radiation shielded floating colonies could be assembled on Venus, with plastic film and aluminum wire bags, filled with breathable air.”
The problem with Venus is same problem of Earth, it’s hard to leave Earth and Venus.
The Moon is easy to leave. And if you don’t have bring rocket fuel from Earth, the Moon is a lot easier to leave.
What we need to go to Mars and Venus is to have a market for rocket fuel in space and one can make rocket fuel on the Moon, and you export lunar rocket fuel to Venus and Mars [or anywhere in space].
So once you have a market of rocket fuel on the Moon, you have rocket fuel market in space. But you even start a market for rocket fuel in space, before you mine lunar water and make rocket fuel. Or having a market for rocket fuel in space, makes it easier to start rocket fuel market on the Moon.
Anyways once you start a rocket fuel market on the Moon, the Moon become a gateway to the rest of solar system- and making settlements on Mars, Mercury, and Venus become possible. But before settlements, one first need to explore these planets, though exploring them becomes a lot cheaper.
So need to explore the Moon to determine where the best places to “commerically” mine lunar water. Then focus on exploring Mars- to find best locations for future settlements [commerical settlements- not governmental welfare states]. We explore Mars next, mainly due to political reasons- congress has passed laws to do this and lots space cadet want Mars to be explored.
And Mars can get enough money for exploration of Mars, if you has commercial ventures funding lunar water mining.
Or mars exploration will require decades of exploration, the Moon lunar poles [a tiny area] should require less than 10 years.
gbalkie wrote:
And Mars can get enough money for exploration of Mars, if you has commercial ventures funding lunar water mining.
No one is going to fund a trip/stay to/on Mars just to get water. That’s a loop that goes nowhere.
I’m willing to accept that the first N trips to Mars (N~2 or so) will just be exploratory missions. Why would any private enterprise fund such trip & stay after that?
–David Appell says:
December 5, 2018 at 11:34 PM
gbalkie wrote:
And Mars can get enough money for exploration of Mars, if you has commercial ventures funding lunar water mining.
No one is going to fund a trip/stay to/on Mars just to get water. Thats a loop that goes nowhere.
Im willing to accept that the first N trips to Mars (N~2 or so) will just be exploratory missions. Why would any private enterprise fund such trip & stay after that?–
I don’t support idea of a few trips to Mars, like the Apollo program. Or flags and footprint missions. I liked Apollo but Apollo was a PR stunt. A very successful PR stunt, but exploration was not it’s purpose, it’s purpose was to beat the Soviets to the Moon.
The reason NASA exists is because we have satellite market, which makes the space important. The satellite market is more 300 billion global market and NASA is small part of this market. Satellites are also important in terms of national security, but national security is also a small part of this market [though bigger than NASA]. And this global satellite market has been and will grow at about 5% per year.
Mining lunar water would increase this market. There is a lot of interest in having governmental bases on the Moon. Lunar water mining would decrease the costs of such bases by about a 1/3 of the total costs.
One could say the world is interested in lunar exploration and the US is interested in Mars exploration- US dominates the current Mars exploration. And I would say reason one would want to send crew to Mars is to increase the scale of Mars exploration- though flags and footprint type thing would not necessarily do this.
With Apollo it caused a lot of robotic exploration of the Moon, but after crew left, there was very little robotic exploration of the Moon- and one might expect a similar result with Mars flags and footprint missions.
A significance of lunar water mining is it enables/allows Mars settlements. Or if you have rocket fuel made on the Moon, the Moon become a gateway to rest of solar system [which includes Mars].
Some Mars fans think that Mars settlement will enable the use of the Moon, I sort of agree, but I think beginning with Moon, is the cheaper and faster way of doing this.
Going to Mars is a total fairy tale. We still can’t even get to the Moon (yes, the Apollo moon landings were faked). No one has figured out how to sufficiently shield humans traveling in space from the immense levels of solar and cosmic radiation. Nor even how to safely get through the Van Allen radiation belt surrounding the Earth.
What do you mean? The Apollo astronauts got safely through the van Allen radiation belts, and back.
Well, he thinks they were faked.
But I have been in higher radiation fields than the Van Allen belts.
It’s time distance and JDHuffman when dealing with radiation.
It also fails to mention that one Venusian day is equal to 180 Earth days because of its slow rotation. That means one side of the planet is facing the sun from dusk to dawn on Venus for 180 of our days. It also rotates counter-clockwise, the only planet in the solar system to do so. Scientists speculate it was struck by an object billions of years ago, causing its backward rotation.
Its CO2 atmosphere is so thick and heavy the pressure is what is actually causing so much heat (given the sun’s radiation can’t penetrate the sulphuric-laced upper atmosphere and then miles of CO2). This isn’t the first time Space.com has pushed these doomsday scenarios. They learned it from Neil DeGrasse Tyson.
Actually one day on Venus is 243 Earth-days.
Since Venus’s orbital period is 225 days, that means every day would be your birthday, if you were born on Venus.
“It also rotates counter-clockwise, the only planet in the solar system to do so.”
Oh how strange!
https://scienceblogs.com/startswithabang/2010/10/07/counterclockwise-but-there-are
EM, Oops! I misread – Stratospheres and above heat from incoming solar/cosmic energy colliding with the thin air especially the Ozone reactions resulting from high energy collisions. But the actual sensible heat we could feel is very small because of there being so few molecules relative to the lower atmosphere. That strato/meso/thermosphere acts as our first shielding from high energy radiation, and takes a hammering….Brett
Venus Atmosphere Temperature and Pressure Profiles
https://web.archive.org/web/20080205025041/http://www.datasync.com/~rsf1/vel/1918vpt.htm
The temperature of the particles in the stratosphere is the result of radiation.
http://ds.data.jma.go.jp/tcc/tcc/products/clisys/STRAT/gif/zt_nh.gif
There’s still the problem of the tropopause.
The pressure hypothesis does not explain it.
As shown above, other phenomena occurring in the troposphere depend on the pressure. The pressure must be well above 0.1 bar.
Numbers please.
How do you calculate the temperature profile?
ren, how does pressure or gravity impede and/or stop IR radiation coming up from the surface?
It doesn’t, DA.
Learn some physics, then you won’t have to ask so many stupid questions.
JDHuffman
David Appell has studied real physics. You are the one who lacks any knowledge of the subject but still pretends to know it. Remember it was Roy Spencer that told you to learn some physics. Too bad you were not able to take his advice. I think your poor reading comprehension prevents you from learning anything. You repeat the same stupid stuff over and over regardless on the mountain of evidence against your ridiculous ideas.
“David Appell has studied real physics.”
I have recommended DA try to get a refund.
Another commenter suggested he should sue for damages.
One of the best things you get from an education is how to tell fools when you see one.
Only if it’s a real education, DA.
Are you still working on getting a refund from your “education”?
how does pressure or gravity impede and/or stop IR radiation coming up from the surface?
Another stupid question, from DA.
JD,
You should look up pressure broadening.
If you could figure out the context, which is iffy.
There is no reason to assume Venus started out like Earth.
The reason the myth began was due to telescope observation and Venus looking like Earth. And now we know it isn’t like Earth, other than having about same gravity as Earth.
It’s possible Earth started out like Venus and Earth was transformed into present Earth by creation of our Moon.
Earth got water at it’s surface by impactors and plate tectonic activity. Venus lacks plate tectonic activity and has no evidence of early impactor formation of the planet or the Venus surface appears to be young.
Why Venus surface appears young is unclear.
It not clear to me that proto-Earth started out like Earth and seems possible we know less much about proto-Earth as we know about Venus.
It is said impactor which is pile of rocks and which has large mass and vaporize in our atmosphere rather impacting the surface is very dangerous because all it’s energy goes into the atmosphere. It seems to me, with Venus with large atmosphere that most impactors would mostly heat the atmosphere rather than impact the surface.
Roughly it seems the problem is plate tectonic and the shaping influence of impactors- [mostly as recent phenomenon] is new to science and the lack of this knowledge shaped our myths about Venus.
The Sun’s energy output is increasing by about 1% every 110 Myrs.
2 Byrs ago it would have been about 20% less. That’s a lot of W/m2. Why couldn’t Venus have been more placid then?
DA, is there any pseudoscience you DON’T fall for?
Why is this pseudoscience?
It’s a false belief. No one knows what the Sun was doing millions of years ago.
You just believe whatever your “church” puts out.
Now, no more stupid questions.
JDHuffman
Your ignorance is astounding and your inability to comprehend what science is about fantastic.
You do realize that astronomers have billions of stars to observe? Maybe you don’t know this. Science is based upon the concept of a Universe structured by certain predictable laws. The goal is to find the laws (relationships) and describe them in the precise language of mathematics. Once a Law is discovered (like gravity) and you have the mathematical relationship you can start using this law to predict things and then observe them. With Stars you can see evolution of countless stars with all types of masses and different compositions. You can use the known mathematical relationships to determine how stars change with time.
Also there are other factors that can help you determine conditions of the Sun millions of years ago. We have records in the Earth that cover vast amounts of time. You can find life in dated rocks over eons. That means the Sun would have had to be about the same then as is now. If it was much hotter life could not exist, if much cooler life could not exist.
It is odd that you claim to have 12 hours of higher education physics but you are totally clueless of what science is about. Sad but True.
https://www.youtube.com/watch?v=bSG_CjcbfKI
Blah-blah-blah.
Another rambling typing fest from Norman. Filled with misrepresentations, false accusations, and insults.
Nothing new.
JDHuffman
No your pet answer to all things you can’t refute is not rational to the comment I made about your lack of scientific understanding or intent.
Not only are you a phony you are also too lazy to come up with any new ideas. Lazy and phony.
Wrong, Norman. It’s not laziness that keeps me from wasting time with you. It’s an awareness that you have no desire to learn.
Nothing new.
JDHuffman
Accepting your totally wrong brand of physics is NOT learning. It is stupidity. You have yet to verify even one of your most stupid claims with a shred of actual evidence. Actual evidence: Established science, actual experiment and not stupid cartoons you make up, some actual measured values or observations.
You provide none of this and then you make the total stupid claim that I have NO DESIRE TO LEARN. You are a dork. Leader of team dork with your two allies DREMT (previously J-Halpless) and Gordon Robertson.
I am glad you are starting to quit responding to my posts though that seems a blatant lie. I post to other people, check a response and there is your stupid comment about me.
You are the one true poster that can’t learn and does not want to. Most posters seem willing to learn some things. Team Dork is the team that learns nothing but makes declarations over and over.
Just another rambling rant, from frustrated Norman-the-typist.
Nothing new.,
Ger*an/JDH tries to cover up his ignorance with snark.
It doesn’t work nearly as well as he thinks it does.
JDHuffman says:
Its a false belief. No one knows what the Sun was doing millions of years ago.
What evidence have solar scientists used to establish the 1%/110Myrs relationship?
DA demonstrates his obsession with Ger*an, and throws out another stupid question.
Nothing new.
JD,
another thing you should look up is the main sequence.
Or Hertzsprung-Russell
i read a book about Venus years ago and the author said there was not as much potassium on Venus as earth. They had some way of telling because of the amount of Argon in the atmosphere (decay product of potassium). He didn’t go into it and left it as a mystery why they two planets would have a different amount.
My question is what if the other alkaline metals are there in lower % compared to earth. It would explain the carbon and the sulfuric acid in the atmosphere. This is what keeps are “bad” chemicals in solid form here on earth.
If this is so stupid why bother responding to it?
What will you cleverly criticize next? Jane Fonda’s climate change workout?
Look Roy, don’t even bother until you have something worthy of attention. Like that IPCC report you are still struggling with
Stupid statements are the ones that most need rational replies. Jesus ascending into heaven without any means of propulsion. And had a sky alien for a dad. Amazing how many still buy that.
If a few more Romans had been scathing just after that surfaced, we might have skipped the Dark Ages and the Inquisition. .
PSR Social anthropologist for 43 years, Oz.
And – Jane Fonda is pleasant to watch, doing whatever.
Scathing, like killing believers?
No thanks.
Peter follows a trend in phony intellectualism. Attacking the Bible is a lazy way for them to feel smart.
The good thing about science is that its true whether or not you believe in it.
Neil deGrasse Tyson
Exactly!
As Launcelot said “…but in the end truth will out”.
The truth of global warming outed a long, long time ago.
I’m glad you finally admit it DA.
Maybe you should write a book about how you were so deceived.
It might not sell very well unless you include your years of stupid questions and comments.
Comedy sells.
Ger*an: One of the benefits of having a good education is gaining the ability to easily detect unknowledgeable people like you.
Though in truth, anyone can see right through you, degree or not. Even funnier, you seem not to realize this.
DA, obviously your “education” doesn’t help with your obsession with Ger*an. I hope he sees the effect he still has on you.
And I’m not sure what area you believe you are “educated” in, but it’s sure not physics.
–2 Byrs ago it would have been about 20% less. Thats a lot of W/m2. Why couldnt Venus have been more placid then?–
Now, about 2600 watts, and 2 billion years ago about 2000 watts.
Different question, what would Earth be like if instead of about 1360 watts, it was 2000 watts?
I think all would agree, that Earth would be much warmer.
Currently, the ground can heat to about 70 C, and that is pretty warm to walk on with bare feet. And ants (and etc) in desert on hot sand, die from this temperature when sun near zenith.
One could say the ground more commonly warms to about 60 C which much less unbearable to walk on.
With 2000 watts, probably anytime the sun gets above about 45 degrees above horizon, anything not wearing shoes, would have trouble walking on ground warmed by the sunlight.
In terms of climate or global climate and/or global temperature, what matters is the surface temperature of the ocean.
The tropical ocean can have surface waters as high as 35 C, with 2000 watt sunlight it seems highest ocean surface temperature would be between 35 to 40 C. And bigger difference would be the amount water which evaporated and a higher average tropical ocean surface temperature. The present average tropical surface temperature is about 26 C. And it seems a 2000 watt sunlight would increase the average to +30 C.
Outside of the tropics the average ocean surface temperature is about 11 C. This would very dramatically change.
We are in icebox climate, we could be in icebox with sunlight which was 2000 watts.
Or Earth has spent a lot of time not being in an ice box climate. Or a long time having no polar ice cap, and warm ocean surface temperature (and entire ocean warmer than 10 C (currently it”s about 3.5 C)).
So with 2000 sun, Earth will have a higher average temperature than it ever has had- if Earth in the past was caused to be warmed by sunlight).
So with 2000 watt sunlight, an ocean average temperature +10 C, and outside of tropics surface temperature of more than 20 C. But of course not exceeding average surface temperatures of the tropics.
-600 W/m2 isn’t a significant factor on Venus 2 Byrs ago??
–We are in icebox climate, we could be in icebox with sunlight which was 2000 watts.–
I missed an important word:
We are in icebox climate, we could NOT be in icebox climate with sunlight which was 2000 watts.
{Also missed the word, climate}
We *were* in an icebox climate. But now the dominant climate forcing is manmade GHGs, not Milankovitch forcings.
DA, there is NO climate forcing from manmade GHGs.
You have swallowed too much of the AGW kool-aid.
JDHuffman says:
DA, there is NO climate forcing from manmade GHGs.
More stupid bullsh!t.
Which you can’t prove.
DA, adding CO2 to the atmosphere does NOT increase surface temperatures.
Racehorses do NOT rotate on their axes. Radiative fluxes do NOT simply add. And “cold” does NOT warm “hot”.
Learn some physics.
No, we are in icebox climate- a cold ocean and polar ice caps.
We are of course in a warmer period of this Ice Age.
During our Ice Age, most of the time was in glacial periods, and the Ice Age has warmer periods called interglacial periods.
The last interglacial period was warmer then our current average global temperature. During the Eemian, sea level peaked at least 5 meters higher than current sea levels and ocean average temperature was about 1 to 2 C warmer than our current average ocean temperature of about 3.5 C
Wiki says:
“Sea level at peak was probably 6 to 9 metres (20 to 30 feet) higher than today, with Greenland contributing 0.6 to 3.5 m (2.0 to 11.5 ft), thermal expansion and mountain glaciers contributing up to 1 m (3.3 ft), and an uncertain contribution from Antarctica. Global mean sea surface temperatures are thought to have been higher than in the Holocene, but not by enough to explain the rise in sea level through thermal expansion alone, and so melting of polar ice caps must also have occurred.”
https://en.wikipedia.org/wiki/Eemian
We have had about 8 inch rise in sea level with some imagining that this very significant.
Are you aware that 8 in is the global average?
It’s about twice that rate on the US East coast.
Are you an American? If so, you are already paying for SLR.
As American I have already spent more than trillion dollars on the global warmers religious silly desires.
But per captia, I believe that being a German has been far more expensive.
Also in terms of removal from NASA budget, the silly beliefs have cost more than a manned lunar exploration program total costs.
Climate of Venus is a problem? For who or what?
And, who pronounced it “evil.” The March Hare?
–David Appell
December 4, 2018
-600 W/m2 isnt a significant factor on Venus 2 Byrs ago??–
I think Venus is hot because intense sunlight heats acid droplets.
And there is an acid cycle with the acid clouds.
And 2600 minus 600 watts makes sunlight significantly less intense. So the way I view it, a reduction of 600 watts could be a significant factor. Or lower surface temperature by 200 K or more.
Or droplets might warm to about 10 C rather than about 70 C, and elevation of clouds lower significantly.
gbaikie says:
Or droplets might warm to about 10 C rather than about 70 C, and elevation of clouds lower significantly.
You’re not doing science, you’re just speculating.
You asked:
-600 W/m2 isnt a significant factor on Venus 2 Byrs ago??
And say it is significant factor and why I think it is a significant factor.
Why do you think it would be significant?
How about say, because no sunlight would reach the surface??
Is -600 as a part of 2600 significant?
It’s 24%.
What if the Earth received 24% less sunlight?
We have had about 8 inch rise in sea level with some imagining that this very significant.
So why is the oil industry asking for $12 billion?
$12 B for 60 miles of seawall
Oil industry wants government to build seawall to protect refineries from climate change effects, Oregonian 8/22/18
https://archpaper.com/2018/08/rebuild-texas-seawalls-oil-gas-infrastructure/
–Peter
December 3, 2018
Funny Dr. Spencer you bring up Space.com. After last weeks Mars probe landing, I spent the past weekend reading up on the Mars environment and topography . Space.com impressed me with its ridiculous speculations about Marss extremely distant past and minimizing the reality that having very very little atmosphere and temperature stability make it a horrible prospect for people to ever live there.–
One can get temperature stability by having houses.
One can also get temperature stability and pressure by being submerged in water.
One could have houses and greenhouses underwater.
What is needed is lots of water.
And the cold desert planet could have lots of liquid water under the surface. And Mars does have slot of frozen water at or near the surface.
Many imagine that nuclear energy would be used to provide power. But even though Mars has 60% less sunlight at Mars distance from sun compared earth distance from the sun, one can get a large amount of solar power due to Mars having a thin atmosphere.
On Earth one has peak solar hours which average about 6 hours a day. And you could located in region which has cloudy weather.
So on Earth one gets about 25% of time (at best) when you can harvest solar energy. Whereas on Mars you can harvest solar energy about 50% of the time and get a fairly uniform amount energy during this time.
On average Mars is better place to harvest solar energy as compared to Earth surface.
And Mars could better place, if one can prevent global dust storms. But unless dust storm is severe, Mars is generally better than Earth. And certainly a lot better than Germany
But unless dust storm is severe, Mars is generally better than Earth. And certainly a lot better than Germany
Ha!
Only if you like wearing a spacesuit all the time. Get real.
–David Appell says:
December 5, 2018 at 10:26 PM
But unless dust storm is severe, Mars is generally better than Earth. And certainly a lot better than Germany
Ha!
Only if you like wearing a spacesuit all the time. Get real.–
I was talking about solar energy: 600 watts times 12 hours is 7.2 kw hours. Germany gets at best, about 3 kW hours.
Or average earth is something 170 watts over 24 hours:
170 x 24 = is about 4 kw hour per day. And solar panels 20% efficent, times by .2
And thermal solar is about 60% efficient, so times it by .6
Though non magnified sunlight at Mars distance will only warm water to about 30 to 40 C and on earth you get about 80 C.
So for hot showers, on Mars magnify the sunlight.
In terms not using spacesuit/pressure suit, be 5 meters or deeper under water.
–E
December 4, 2018
Where did the oxygen go?
It oxidised the surface rocks.
Oxygen only persists in an atmosphere if it is constantly replenished by photosynthesis.–
Venus lacks plate tectonic activity.
All rocky surface in our solar system have high amount oxidized rock.
About 40% of lunar surface is oxygen, as is Earth, and Mars. As is Mercury, Venus, and most space rocks.
The mass of water is mostly oxygen. In 9 kg of water has 8 kg is oxygen with 1 kg of hydrogen.
–Bobdesbond
December 3, 2018
Mr Spencer, by how much do the thick sulphuric acid clouds on Venus mitigate the warming?–
https://m.esa.int/Our_Activities/Space_Science/Venus_Express/Greenhouse_effect_clouds_and_winds
–On the global scale, Venuss climate is strongly driven by the most powerful greenhouse effect found in the Solar System. The greenhouse agents sustaining it are water vapour, carbon dioxide and sulphuric acid aerosols.
About 80% of the incoming solar radiation is reflected back to space by the cloud layer, about 10% is absorbed by the atmosphere and only 10% manages to get through it and heat the surface. —
Not particularly agreeing, but Venus acid clouds are considered to be a “greenhouse gas”
BTW I Don’t think 10% gets thru to the surface.
Not particularly agreeing, but Venus acid clouds are considered to be a “greenhouse gas”
How so?
-How so?-
The acid aerosols are a warming effect.
Or I agree with the religious, that the acid aerosols are not a cooling effect.
Or for the question: “how much do the thick sulphuric acid clouds on Venus mitigate the warming?”
I would answer: they don’t mitigate the warming, rather they cause the warming.
In my opinion they are not just part of warming, but rather the acid aerosols [also called by the religious, a greenhouse gas] are the control knob of the warming of Venus.
The sunlight would warm Venus, but the acid clouds at high elevation which are blocking about 70% of the sunlight, makes the surface hot.
It seems to violate cold causing something to hot, just as cooler land surface at higher elevation can heat the air at the basin of dried up Mediterranean sea to air temperature of about 80 C during a glacial period.
Or easily beating the current highest air temperature ever recorded, wiki:
” According to the World Meteorological Organization (WMO), the highest registered air temperature on Earth was 56.7 C (134.1 F) in Furnace Creek Ranch, California, located in the Death Valley desert in the United States, on July 10, 1913.”
gbaikie says:
Or for the question: how much do the thick sulphuric acid clouds on Venus mitigate the warming?
I would answer: they dont mitigate the warming, rather they cause the warming.
How so?
I don’t care what you believe, I care about what you can prove. So let’s see your proof.
–David Appell says:
December 5, 2018 at 11:28 PM
gbaikie says:
Or for the question: how much do the thick sulphuric acid clouds on Venus mitigate the warming?
I would answer: they dont mitigate the warming, rather they cause the warming.
How so?
I don’t care what you believe, I care about what you can prove. So let’s see your proof.–
I am saying the thick sulphuric acid clouds cause warming, rather than cooling. And you saying, you care that I can prove it.
I can’t prove CO2 causes warming in regards to planet Earth, nor can disprove it.
My position is that CO2 may cause warming on Earth and I believe that doubling of CO2 may cause within a time period of century as much as .5 C to global average temperature.
In regards to Earth, I don’t believe that high levels CO2 will increase the amount warming of ground temperature. Or Earth had very high levels of CO2 in the past, and this has not caused the ground surface temperature to increase in temperature. Or highest ground surface temperature we have presently are about 70 C.
But accept the idea that actual greenhouse or solar ponds can reach a temperature the bottom of the pond or surface of ground to be about 80 C. And this is due to reducing convectional heat losses.
But in past there has times when the ground has reached 😯 C or higher- as when the Mediterranean sea dried up and allowing there to large elevation difference- sea level to 3 to 4 km below sea level.
And it seems only way to heat Earth ground over 80 C is to heat large surface area at higher elevation. So heat air at elevation of say 4 km difference. So air can warm to 50 C. And at lapse rate of 6.5 per km, 4 times 6.5 = 26 K. Or if had dry air and lapse of say 8 C per km. 4 times 8 = 32 K.
Now without the higher surface, the lower surface could warm to 50 C. And addition of higher elevation land adjacent [or surrounding the lower land which otherwise warm to 50 C] the addition of higher elevation would add it’s temperature: 50 plus 26 or 32 C which gives the lower land air temperature of 76 or 82 C.
Having such high air temperature would prevent convectional heat loss of ground and allow ground to be 80 C.
If had the air density and it was 10 K higher rather 4 km, it add 65 K or 80 K, which increase the air temperature at lower elevation to 115 C or 130 C air temperature, which could warm the ground to +100 C.
And with Venus it’s 50 Km higher elevation of the clouds being a surface which is heated which is warming the air.
If the surface is water, the surface temperature depend on evaporation, but pure sulphuric acid or near pure acid has much higher boiling temperature than water. So acid evaporates less at low temperatures that water evaporates at- or the acid can reach higher temperature in sunlight than water.
We know on Earth sunlight heats droplets of water of the clouds, and we know on Venus sunlight would heat droplets of acid of the clouds
gbaikie says:
The acid aerosols are a warming effect
How so?
On Earth, aerosols are a cooling effect.
The sunlight would warm Venus, but the acid clouds at high elevation which are blocking about 70% of the sunlight, makes the surface hot.
How so? What’s your proof of this claim?
Gbaikie
Here’s the Venus energy budget.
https://davidappell.blogspot.com/2016/05/energy-balance-diagrams-from-around.html
Insolation is 656W/M^2.
The surface absorbs 22W/M^2.
That is 3.34%.
For comparison, these are the figures for Earth.
https://en.m.wikipedia.org/wiki/Earth%27s_energy_budget#/media/File%3AThe-NASA-Earth's-Energy-Budget-Poster-Radiant-Energy-System-satellite-infrared-radiation-fluxes.jpg
Insolation is 340W/M^2.
The surface absorbs 163W/M^2.
That is 48%.
Are you sure you can swallow all of that, Entropic man?
Be careful, that’s enough pseudoscience to choke a horse!
I’ll just hit a couple of points. The Venus isolation is after adjusting for albedo. The Earth insolation is not adjusted for albedo, but divided by 4, instead. And, “surface absorbs” is a guess.
Any comparisons are therefore bogus, as in, “pure pseudoscience”.
That turns out not to be the case.
Both are adjusted for a spherical body (divided by 4), but not for albedo. The figures I gave are for the total energy received at the top of the atmosphere, divided by the surface area.
A square metre perpendicular to the Sun at Venus’distance receives 2624 W.
A similar square metre at Earth’s distance receives 1360W.
Non of it is guessed. There are anough measurements from
Venus orbiters and landers to construct the energy budget.
If you were well enough protected to stand on the surface of Venus at local noon you would see light levels akin to twilight on Earth.
You’re not getting better, you’re getting worse.
* You can NOT divide by 4. You are trying to treat flux as energy. Can’t do that.
* You must adjust for albedo. That’s reality. (Not that it makes a huge difference, but NASA claims 2601 Watts/m^2, not 2624.)
* What the surfaces absorb is indeed “guessed”. There is no way to measure that.
Hard to swallow, huh?
I’m not even going to try and debate flux and energy.Your mistake is too deeply entrenched.
Measuring radiation absorbed by the suface is easy.
Use your lander to measure the amount of energy coming down from the clouds and the amount reflected from the surface.The difference is the amount absorbed.
E-man, you believe a snapshot like that could be applied to the entire planet?
Somebody’s been yanking your chain.
And, the reason you can’t divide flux is because flux is “power/area”, and “power” is a rate. Energy and power must be treated differently, because “energy” is a quantity, but “power” is a rate. (If you’re still confused, compare kilometers-per-hour with kilometers.}
I don’t see the problem. Energy budgets use flux in W/M^2 (ie power/area) throughout.
I tried counting landers. The Russians were the Venus lander specialists and successfully put down eight of them.
-JDHuffman says:
December 5, 2018 at 1:20 PM
You’re not getting better, you’re getting worse.
* You can NOT divide by 4. You are trying to treat flux as energy. Can’t do that.
* You must adjust for albedo. That’s reality. (Not that it makes a huge difference, but NASA claims 2601 Watts/m^2, not 2624.)
* What the surfaces absorb is indeed “guessed”. There is no way to measure that.–
I would say the rocky surface of venus absorbs near zero sunlight.
Simply because I believe a frying pan heated to 150 C is not going to warmed by sunlight at Earth surface. If something isn’t having energy added to it, it’s not absorbing energy.
It similar to cold doesn’t warm hot. Cold in this case is sunlight as compared to object which hotter than can be heated by sunlight at that distance from the sun. Of course if you could magnify the sunlight then it could be absorbed- or it can add heat, can make it warmer.
But I not even sure that the sunlight reaching the Venus surface can be magnified, because it’s not direct sunlight. Or if can magnify sunlight which going thru thick clouds on Earth, and burn a piece of paper, then I would be wrong in that regards.
But it seems that about 3% of sunlight does reach the surface and certain times of day it is like twilight, and when sun is near zenith or more 45 degrees above horizon it’s like overcast day- which comes different values of heavily overcast or lightly overcast day. I would say it’s when you locate the sun in the clouds, it’s a bright patch in the clouds. And would be like Venus day at it’s brightest. But most the sunlit hemisphere would twilight or night. Or only a small fraction of sunlit hemisphere has sunlight 45 degrees above horizon. And when sun at around say 30 degrees above horizon is twilight, at 15 degrees above horizon- and it’s most of the area of a sunlit hemisphere- and it would be night/dark.
E-man: “I don’t see the problem. Energy budgets use flux in W/M^2 (ie power/area) throughout.”
Yes, and that is why the energy budgets are so flawed.
Consider a car moving at 100 kph. You could imagine the car having 4 equal sections. Is each section moving at 100/4 = 25 kph?
Of course not.
And as for the Russian Venus missions, the Russians have not swallowed the AGW nonsense.
I tried to see if I was wrong, but I only found someone that agrees with me:
“This is why, on a cloudy day, you can’t burn things with a magnifying lens, even when its fairly bright outside. The energy is still there, its just scattered.”
https://www.askamathematician.com/2013/01/qhow-do-lenses-that-concentrate-light-not-violate-the-second-law-of-thermodynamics-if-you-use-a-magnifying-glass-to-burn-ants-arent-you-making-a-point-hotter-than-the-ambient-temperature-without/
That’s ugly link, I will tiny it:
https://tinyurl.com/y7j575mq
gbaikie, the high energy UV photons likely reach the Venus surface. Such photons could easily correspond to a temperature of over twice the Venus surface.
JDHuffman says:
gbaikie, the high energy UV photons likely reach the Venus surface. Such photons could easily correspond to a temperature of over twice the Venus surface.
Where is the data/proof showing that?
Venus’s solar constant is 2614 W/m2 and its Bond albedo is 0.77. So only 601 W/m2 remains going into the planet’s atmosphere. That could, at most, cause a surface temperature of 320 K, when the actual temperature is 735 K.
JDHuffman says:
Yes, and that is why the energy budgets are so flawed.
Consider a car moving at 100 kph. You could imagine the car having 4 equal sections. Is each section moving at 100/4 = 25 kph?
Talk about bad physics…. How is sunlight distributed over a planet’s surface? How is velocity distributed over a moving car?
JDHuffman says:
* You can NOT divide by 4. You are trying to treat flux as energy. Cant do that.
Somebody doesn’t understand what flux is.
Wow DA, 3 desperate comments, all within 4 minutes.
If you had taken time to compose yourself, and thoughtfully asked some responsible questions, maybe there would be some hope for you.
But, obviously not.
Seek therapy. And then learn some physics.
Then critique what I wrote, instead of avoiding doing so, as you usually do.
DA, you want a critique? Okay.
“Garbage.”
Learn some physics.
{smile}
You are without content.
Laughably so.
JDHuffman
I have to thank you. You just presented proof positive you are not adept at math or the logic necessary to use it.
Your irrational logic is hilarious to say the least. You think that dividing a flux would be equivalent to a car speed divided by 4. How incredibly irrational.
The solar flux at Earth distance is around 1361 Watts/m^2. That means a flat surface with Earth’s radius (at Earth’s distance from the Sun) would receive the following amount of energy from the Sun.
Area of a circle is A=(pi)r^2
Area =(pi)(6,371,000 m)^2 = 127,516,117,977,447 m^2
multiply this by 1361 W/m^2 to get the total Watts reaching the Earth.
173,549,436,567,305,460 Watts. That is the energy the Earth receives from the Sun every second a near constant rate.
Now you can’t have more than this amount of energy reaching the Earth. It is limited by its radius.
But the actual surface area of the Earth is A=4(pi)r^2. A sphere has more area than a circle with the same radius.
If you take the amount of energy reaching the Earth and want to average it over the entire area you have to divide by 4. It is very simple math. You are not able to figure out how it works. You fail at math. You fail at physics. Honestly what was your career? Failed comedian? Clown for parties? You have no physics knowledge or math. There I proved it, are you now satisfied?
“JDHuffman says:
December 5, 2018 at 6:10 PM
gbaikie, the high energy UV photons likely reach the Venus surface. Such photons could easily correspond to a temperature of over twice the Venus surface.”
Well you can generate UV light. “Safe” UV light are used for tanning booths and growing plants.
Does it have to be the unsafe band of UV light to work?
Anyhow I don’t think it’s going to warm a frying pan which is already warmed to 150 C.
What does the data say for how much UV reaches the Venusian surface?
There is an unknown UV absorber in Venus upper atm. so I doubt anyone can give an answer to surface UV from measurements, probably you can find some guesses.
Norman, that was another long dissertation, for nothing.
It’s okay to divide energy. It is a quality. You can NOT divide power. You will get erroneous results.
As I stated above, and you failed to comprehend: “Energy and power must be treated differently…”
DA and Norman were unable to understand my explanation about power vx. energy. They could not grasp the basic physics.
That’s probably why they always claim I don’t offer any physics. When I do, it goes right over their heads.
Nothing new.
Ball4
Sulphuric acid absorbs UV, with a strong peak at 250 micrometres.
https://www.sciencedirect.com/science/article/pii/S0140673662919128
JDHuffman
It is not that your comment goes above my head…I just think kilometers/hour is a super poor analogy to radiant flux. You do not transfer kilometers from one source to another.
A far better analogy would be the rate of gallons/minute. You can transfer gallons.
If you have a flow of 100 gallons/minute to a tank and the tank holds 1000 gallons you can calculate that the tank will fill in 10 minutes.
You can take the same 100 gallon/minute flow (it does not change) and divide it between 4 tanks. Each tank receives 25 gallons/minute. The total remains 100 gallons/minute but you can set it up so valving and piping split it up. Now each 1000 gallon tank will take 40 minutes to fill. Yes you can divide or add flows together.
You can have one source flowing 100 gallons/minute into your first tank and have an additional 400 gallon/minute flow also into the tank. The flows directly add, subtract and divide. Not sure why you would choose kph to represent a flow of energy?
JDHuffman
Also “flux” has multiple uses with radiant energy.
Here is a radiant flux:
http://hyperphysics.phy-astr.gsu.edu/hbase/vision/radiant.html
It is just in watts (joules/second). It is the amount of energy flow from a source. It does not have an area associated with it.
Then there is radiative flux:
https://en.wikipedia.org/wiki/Radiative_flux
This flux is given in Watt/m^2 it is the amount of power that passes through a given area.
You can certainly add, divide, multiply and subtract watts into different areas.
To help you understand. If you have a 100 Watt radiant flux from a source. If this energy passes through an area 1 square meter, the radiative flux is 100 W/m^2. If the same source passes through a 100 m^2 area the radiative flux is now 1 W/m^2.
It is all relevant to rational physics. If you have a 1 m^2 surface near the source it will reach a much higher temperature than the 1 m^2 surface farther away.
I am hoping you can try to understand some physics but I do not hope much.
EM, that’s a long known near IR absorber which can’t account for all the observed cloud absorp_tion by broadband IR. The H2SO4 clouds reflect ~76% of the incident solar radiation and absorb almost 50% of the solar flux deposited on Venus. Because most of the solar energy is deposited within the upper cloud, the heating rates within the upper cloud are very sensitive to uncertainties in the vertical distribution of the unknown UV absorber within that upper cloud (60-70km).
Sorry Norman but your initial tirade is your undoing. You clearly did not understand the basic physics. You just made a fool out of yourself pretending you did. And your two lengthy comments, trying to redeem yourself, just make you appear even more insane.
Great entertainment!
My mistake. I confused micrometres and nanometres.
JDHUffman
Wrong again (usual for you).
My points were quite valid and directly in relation with logic and reason and current definitions of flux.
You are out of your league on this blog. Other blogs some may think you know what you are talking about. Here everyone knows you are a blithering moron that doesn’t have any knowledge of physics at all.
On other blogs you can bluff your way through. On this one your ignorance stands out like a sore thumb.
YOU: “Sorry Norman but your initial tirade is your undoing. You clearly did not understand the basic physics.”
Quite the opposite. You don’t know any physics, you can’t calculate and you hide behind your ignorance with a stupid taunt. Not going to help you learn anything. Stay stupid it won’t bother me.
Norman, that last typing session just put you at 1000 words, for your mistake.
That’s what I like.
JDHuffman
What mistake? You should be more clear with your posts.
I am thinking you were thinking about your own terrible physics. I did see you were able to attempt some math.
You make about as much sense as a drunk baboon.
Norman, which mistake are you referring to? You make so many.
Your big mistake, just on this subthread, was not understanding my 100-kph-car example. Since 100 kph is the RATE the car is traveling, you cannot divide it by 4, and then claim that 4 equal parts of the car are each going 25 kph. You can’t divide the speed of the car. And you can’t divide the solar flux arriving Earth.
Unless you love pseudoscience….
JDHuffman
You have to be drinking heavily to think your speed analogy compares to a flow of energy. They are not the same at all.
With speed you are covering a distance in a time frame. You are not transferring kilometers to some other object. You must be so drunk you are stumbling around!
With energy or flow of material you are transferring material from one place to another. I cannot fathom how really stupid of a human you are!
Do you think you transfer miles from one place to another with you speed? Freaking idiot!
I do not think I can communicate with a foolish person like you. There is a saying:
“Never argue with a fool, onlookers may not be able to tell the difference.”
― Mark Twain
OR https://biblehub.com/proverbs/26-4.htm
Yes foolish person, you can divide radiant flux (Joules/second) in an area to get a value of watts/m^2 (radiative flux)
I have to be done with you. You are too stupid to communicate rationally with.
Bye and don’t write back.
Norman, you asked me to clarify what mistake you made. So, I did.
But, you still can’t understand the simple example. Whenever I try to teach you some physics, it goes over your head. You just don’t have the background to even understand the basics.
If you would stop your insults, misrepresentations, and false accusations, and try to learn some physics, you could be on the road to a better life.
And, it’s easy to argue with an idiot, without making yourself look like an idiot. All you have to do is use facts and logic, and avoid using the juvenile insults and name-calling. Responsible onlookers will be able to easily tell the difference.
JDHuffman
Okay then, no insult zone. I will just explain it to you in math. Follow the math and you will then understand.
The radius of the Earth is 6.371×10^6 meters
This creates an area of 1.27z10^14 m^2 that solar flux will be available for the entire Earth.
The radiative energy of the Sun at this distance is measured to be 1361 Watts/m^2.
The total solar energy reaching the Earth is then this number times the circle the Earth occupies. There can’t be more solar energy than this reaching the Earth.
You get 1.735z10^17 Watts. This is the amount of constant energy that reaches the Earth system. Of that 30% is reflected and not absorbed so you are left with:
1.2145×10^17 Watts.
The area of a sphere is 4(pi)r^2
The area of the Earth is 5.10z10^14 m^2
Divide the total energy being absorbed by Earth by the radiating area (the area of the sphere) and each m^2 would radiate 238 Watts of energy. The effective temperature if you assume a 0.96 emissivity gives you an effective temperature for a sphere receiving this much energy of 257 K.
I think the value used by Climate science gives you a 255 K value if they assume the Earth is a blackbody radiator.
Find flaw with the math. You won’t be able to.
Norman, your calculations are basically correct. But your interpretation of physics is still messed up. Maybe all the numbers confused you. Let’s simplify things a bit.
S = Solar flux
Sr = Solar flux Remaining after albedo
A = Area of “circle” Earth
Then, here were your steps:
S –> Sr –> SrA –> SrA/4A –> Sr/4
You just divided the flux by 4. You can’t do that, and end up with anything meaningful.
“Flux” has units of “power/area”, and “power” is “energy/time”. “Power” is a “rate”. You can’t treat “rates” as quantities. Consider the example of the 100-kph car. For equal sections of the car are not moving at 100/4 = 25 kph.
Perverting and corrupting physics, to get the results you want, is called “pseudoscience”.
JDHuffman
No insult but your speed analogy is not valid for this case.
Maybe you are confused with all this flux etc. Let us just ignore this and simplify so that you get the basics.
If you have a flat circular plate in space with very little external energy (say 3K). If the plate radius is one meter you have a total square surface area of (3.142 times 2…two sides) of 6.28 m^2 to radiate input energy from. If you add 1000 watts to this flat circular plate you would have a radiative flux of 159.2 Watts/m^2.
If this object were close to a blackbody its steady state temperature would be around 230 K.
If you added the same 1000 watts of energy to a sphere with the same radius you would now have a radiating surface area of 12.57 m^2 you would have a radiative flux leaving the sphere of 79.55 Watts/m^2 for a steady state temperature of around 193 K.
I think you are stuck on the Solar input. The effective temperature is based upon the total radiating surface based upon how much energy the object receives.
The Solar input only tells you how energy the object will receive but will not tell you how much the loss there will be.
Norman, as usual, you run from reality. Your problem hinders your ability to learn.
My last comment showed how you were wrong. All of your calculations ended up just dividing by 4, which is pseudoscience.
Now, you are off on another set of calculations, trying to distract from the fact that you were wrong. Stick with the issue!
If you want to wander off into space, consider a flat black body plate. The plate is perfectly insulated on the back side, so energy can only arrive/leave from the front. The plate has an area of one square meter, and receives 1000 Watts/m^2.
At equilibrium, it’s temperture is 364 K.
Now, the same plate is divided into 4 identical plates, each with the same area. Is each plate now receiving only 250 Watts/m^2?
Of course not.
Wise up and smell reality.
JDHuffman
Keeping in the NO INSULT ZONE.
I did give you a basic example of what is going on.
YOU: “Now, the same plate is divided into 4 identical plates, each with the same area. Is each plate now receiving only 250 Watts/m^2?
Of course not.”
You are not using the area correctly in your example. You do not calculate how much a surface receives in Watts/m^2
You use this value to calculate the amount of energy the surface receives. Don’t calculate backwards.
A 1000 W/m^2 source will deliver 1000 watts of energy to a surface 1 m^2. It will also radiate energy from a 1 m^2 surface and will reach a steady state temperature where it has a temperature that emits 1000 watts.
If you divide this square into 4 equal pieces and have the same 1000 W/m^2 source, each will receive 250 watts energy and they will warm until they emit 250 watts of energy. If you have a emitting surface area of 0.25 m^2 that emits 250 watts, it will be at the same temperature as a 1 m^2 surface emitting 1000 watts.
Look at the equation for Power from an emitting surface.
P=(Stefan-Boltzmann constant)(emissivity)(Area)T^4
Notice the Area in the equation. A surface with the same temperature but 1/4 the area will emit one quarter of the Power (watts).
The calculation you believe is pseudoscience only appears that way to you because you think the solar flux is being divided by 4. It is not so. The Solar flux reaching a flat circle or the surface of a sphere is the same. The input energy is the same in both cases. The area of emission is increased from a circle to a sphere. They both receive the same energy but the sphere has a larger emission surface.
If you need more, consider the tungsten filament in a light bulb. 100 watts of energy will raise the temperature of the filament to thousands of degrees. The very same energy input with a larger emitting surface will be considerably cooler.
The amount of energy reaching the Earth is a constant, the emitting surface is what you need to calculate the steady state temperature for a given input energy.
A 100 m^2 surface with 100 watts of energy added to it will not get super hot like the tiny surface area of a tungsten filament. Both receive exactly the same energy but the emitting surface area will give you the steady state temperature.
That is what they calculate to get the Earth’s effective temperature of 255 K.
Norman, you don’t believe you are insulting, but you are. You’re insulting reality.
You won’t admit you were wrong, but it’s evident you are trying to cover for yourself. You tried a diversion, but got caught. Now, you are trying another escape route.
You used units of “Watts/m^2” with the 1000 value, but you switched to units of “Watts” with the 250 value. IOW, you did not divide by 4. You now realize that you can’t divide the flux.
The plate is receiving 1000 Watts/m^2. If you divide it into 4 equal smaller plates, each smaller plate is STILL receiving 1000 Watts/m^2. You just can’t get around that issue, no matter how much you bang on your keyboard.
And, it’s the same for a planet. That’s why dividing by 4 to get 240 Watts/m^2, and then using that to find the temperature, is pseudoscience.
(See if you can reach the 1000-word goal, this time. You’re still falling short.)
JDHuffman
Your thought process is still backwards. You are not following what I am saying.
Here, I will help you. YOU SAY: “The plate is receiving 1000 Watts/m^2. If you divide it into 4 equal smaller plates, each smaller plate is STILL receiving 1000 Watts/m^2. You just cant get around that issue, no matter how much you bang on your keyboard.”
Nothing ever receives watts/m^2. That is not a valid thought.
If you have a radiant flux of 1000 watts/m^2 it means a 1 m^2 surface will receive the full 1000 watts of energy. There is no such thing as receiving 1000 watts/m^2. It is not a logical use of the values. Objects can only receive joules. The rate gives you how many joules a second an object is receiving. The m^2 has value only in finding out how much joules are received by and object with a known size.
As long as you think an object receives Watts/m^2 there is no further point debating with you. It is incorrect and not remotely logical.
JDHuffman
YOU SAY: “You wont admit you were wrong, but its evident you are trying to cover for yourself. You tried a diversion, but got caught. Now, you are trying another escape route.”
I am not wrong. It would be very foolish of me to admit I am wrong when I am not. Not at all trying to cover for anything or diverting anything or that you caught me in anything.
Try to think things through. Stop your line of thought and apply some logic.
YOU SAY: “And, its the same for a planet. Thats why dividing by 4 to get 240 Watts/m^2, and then using that to find the temperature, is pseudoscience.”
This is a very wrong thought. You are not using good logic.
The actual effective temperature does not divide the solar flux at all. I did explain this to you a couple of times. Try to think about it please.
The solar flux is the same. The amount of energy flowing to a circle the radius of the Earth and the sphere are the same. That is not a change. The energy received is not different or divided by four. The effective emitting surface of a sphere is 4 times that of a one-sided circle with the same radius.
Here again: Area of a circle = (pi)r^2
Area of a sphere = 4(pi)r^2. A sphere with the same radius has a surface area 4 times larger than one side of a circle. The amount of solar energy that goes through this area is only in watts.
It is basic math. When you take the solar flux at Earth’s distance it comes out as 1361 Watts/m^2. To find how much energy is passing through the area Earth occupies you take one side of a circle with the radius of Earth. Now you have m^2 value. To find how much energy the Earth receives a second you multiply 1361 Watts/m^2 times the area in m^2. The area cancels and you get a value of energy received by Earth. I already did all the calculations in a post above. Try to follow the math.
Norman, you have dropped the juvenile insults, but now you are starting back with the misrepresentations and false accusations. And, you’re still avoiding reality.
The pseudoscience definitely DOES specify dividing by 4. Where do you believe the 240 Watts/m^2 comes from?
960/4 = 240
The 240 is then used in the S/B equation to arrive at the 255 K.
You just can’t understand, or either you want to deny reality.
Let me know when you want to learn some actual physics.
(And your last two comments are only about 600 words total. You’re still not up to the 1000. Just shut down your brain and start banging on your keyboard. You’ll get there.)
And the “goldilocks zone” is left completely out of the discussion.
Feel free.
Sure. If the orbit of Venus was in the goldilocks zone it would not be the hell hole it is now.
A quick SB calculation suggests that if Venus had an Earthlike atmosphere it would have an average surface temperature around 9C warmer than Earth.
That is well within the goldilocks zone.
If your quick SB calculation were the same as is used in pseudoscience, then Venus with Earth’s atmosphere, would have a 44 °C hotter average surface.
My BOE calculation was that Venus receives about twice as much radiant flux as Earth it would thefore have a temperature of 4th√2 288 = 297C.
I’m not sure where you got 44C warmer, 332K.
Could you show your working?
Redid the calculation and found I had taken 5th√2.
New figure fo Venus 312K, 24K warmer than Earth. Still cant see how you calculate 44K warmer.
2601 * 0.7 = 1821 Watts/m^2
1821/4 = 455 Watts/m^2
Now use S/B equation, yields 299 K
Now add bogus 33K = 332 K
(Remember, this is from pseudoscience!)
JDHuffman
Of coure!
We are comparing TOA temperatures.
???
Not sure what that means. We were both calculating “average surface” temperatures.
Sorry, TOA stands for Top of Atmosphere.This is the altitude at which a planet radiates IR to space. The temperature at this altitude determines the amount of outward radiation. Alternatively, if you know the amount of radiation an SB calculation can give you the temperature.
This is also the surface temperature of a planet without an atmosphere.
“This is also the surface temperature of a planet without an atmosphere.”
Not neceassarily as that surface would be pounded to regolith and the albedo & surface emissivity would be significantly affected. The better atm. physics statement is “this is also the surface temperature of a planet with same surface pressure but ideally transparent atmosphere with no optical depth which allows for no change in either albedo and surface emissivity.
The issue I have with the ‘Venus is like earth’ false comparison is that Venus is ‘like earth’ pretty much only in size.
Venus sidereal day (~243 earth days) is longer than Venus’ orbital period (224.7 earth days). Technically, if you work out the difference between the sidedreal day and it’s orbital period, Venus’ ‘solar day’ is ~117 earth days.
Venus has a very low axial tilt compared to earth or Mars.
Venus has a retrograde rotation
Venus has no magnetic field, no Van Allen Belts to protect it.
Venus, as far as we know, never had water on it’s surface and has no water in it’s atmosphere.
Venus’ atmosphere is 90x denser than earth’s.
Attempting to compare the climate of the two planets and project the future of earth’s climate using Venus as an example is folly.
Bill…excellent points, Bill, especially the one about the lengthy Venusian solar day. It’s much closer to the Sun and with solar radiation entering every day 24/7, to use some Earth parlance, I think the ingredients are all there for a super hot surface. .
I have hypothesized with our much faster period of rotation, and solar input being only a fraction of the day, that when the Earth rotates away from the Sun, it should cool naturally due to the effect of atmospheric contraction.
Gay-Lussac, from the Ideal Gas Equation: V1/T1 = V2/T2. I think it’s possible we may have developed a balance between solar input causing the atmosphere to expand slightly and a natural cooling due to the subsequent contraction of the atmosphere when solar input is absent.
If I’m right, there would be hell to pay on Venus with a day extending to 117 Earth days. I wonder if the dark side of Venus cools, and by how much, when it is bereft of solar energy.
CO2’s greenhouse effect is at nearly 100% saturation at 1% here on earth. Another “doubling” will get us almost there. So talking about the effects of CO2 higher than 1% is only propaganda.
And if we all DO fork over $10 to $50 Trillion (and give up just a few more of our freedoms) according to the AGW plan, CO2 projected warming will decrease by less than 0.1C ACCORDING TO THEIR OWN INFLATED NUMBERS (it will, of course be less than half of that according to 25 years of actual temperature data).
One might reasonably suspect that the TRUE motivation of AGW proponents is other than reducing global warming. Which, of course it is…else we would be building about 5,000 to 10,000 nuclear power plants which is the only way (4th grade math on the back of an envelope calculations) to reach their stated emissions goals.
Hear, hear!
I was waiting for someone with better credentials to say this, but I haven’t seen it.
Most of the CO2 that we see is due to the temperature recovery from the Little Ice Age (LIA) with a lag of 300 years . Coincidentally, a much smaller amount is being added by humans. It’s an accident that warmists, the IPCC and their much-amplified propaganda machine have taken advantage of.
PhilF…”I was waiting for someone with better credentials to say this, but I havent seen it.
Most of the CO2 that we see is due to the temperature recovery from the Little Ice Age (LIA) with a lag of 300 years .”
I would not worry about credentials. You don’t need a degree in anything to pursue the truth.
Syun Akasofu, a prominent geophysicist who specialized in the solar wind, has a paper out on the effect of lowered temperatures during the LIA. He did not address CO2 per se but I have raised that issue.
If global temps were 1 to 2C below normal, it stands to reason the cooler oceans would have absorbed a good deal of CO2 from the atmosphere. As they re-warmed, CO2 should have been out-gassed.
Gordon. I didn’t mean it exactly like that. Why didn’t Salby, Harde or Berry(edberry.com) claim it. It’s just as good as what the warmists have (trace gas amplified by feedback using Bern model for retaining CO2, all ad hoc and going against paleontology findings). At least some of the resources they’ve wasted should have gone into finding out why/how CO2 lags temperature. But they aren’t interested in science.
philF…”I didnt mean it exactly like that”.
I was agreeing with you.
cam…”The high temperature on the surface of Venus was explained by Sagan & Pollock in 1967″.
Based on Sagan’s later circus act, in which he manipulated the media in a way that makes Bill Nye seem a neophyte, it seems fair to claim that Sagan was not a serious scientist.
I recall him driving me up the wall with his oft-repeated statement, “when the Big Bang occurred…”.
We have his counterpart here in Canada, in David Suzuki. He was once a good scientist who specialized in genetics related to fruit flies but along the way he adopted the religion of environmentalism and has since become myopic and intolerant of skeptical views.
I think it’s safe to claim that Sagan was wrong with his CO2-based runaway, greenhouse effect. Unfortunately, the idea was given fresh life by Hansen, circa 1988, who was a promoter of Sagan’s greenhouse theory.
The Pioneer data that revealed a surface temperature of 450C put the boots to that theory. As astronomer Andrew Ingersoll claimed, such a surface temperature contradicts the 2nd law if the cause is claimed to be a greenhouse effect.
The Pioneer mission to Venus found that there is more Deuterium in the atmosphere of Venus than on Earth.
This is evidence that there was once water on Venus.
There is about 20 ppm of water in the Venus atmosphere.
Or in terms of ppm, about 1/20th of the amount of CO2 in our atmosphere.
Of course Venus has about 90 times more atmosphere than Earth and so Venus has far more water in atmosphere as Earth has CO2 in it’s atmosphere.
Or there a few trillion tonnes of CO2 in our atmosphere and several trillion tonnes of water in the Venus atmosphere.
“Venus has a negligible amount of water vapour in its atmosphere, about 0.003% of its total mass”
And total mass of Venus atmosphere: ” ~4.8 x 10^20 kg
1.44 x 10^16 kg, 1.44 x 10^13 tonnes or 14.4 trillion tonnes
So, more than several trillion tonnes.
Say had space rock which had 25% of mass being water.
“C-Type asteroids (Carbonaceous chondrites) comprise about 75% of all known asteroids…. and have a high water content (often 10%, but as high as 22%)”
http://www.galleries.com/rocks/asteroids.htm
So C type and change it to 20% water chemical bonded [hydrates] but if exploding in the atmosphere, the chemically bonded water is released. A 1000 meter diameter sphere has 5.24×10^8 cubic meters
or about 1 billion tonnes and Earth doesn’t get 1 km rock impacting very often, and Venus would less often than Earth:
“Asteroids with a 1 km (0.62 mi) diameter strike Earth every 500,000 years on average.” – wiki
So as guess once every million years for Venus. And 20% means 5 impactors 1 km diameter give 1 billion tonnes of water
5 million years 1 billion tonnes water from 1 km diameter space rocks. Or 5 billion year, 1 trillion tonnes. So need to consider large impactors, say 10 km diameter: volume: 5.24×10^11
So about 1 trillion tonnes and 5 such rocks give 1 trillion tonnes of water. And say every 200 million year or 1 billion years for
1 trillion tonnes of water. Still not big enough. So has to be space rocks bigger than 10 km diameter which could add significant amount of water.
One 50 km diameter rock or larger could double the amount water in Venus atmosphere. 50 km: volume 6.54×10^13.
[And let’s see, what is largest NEO: 1036 Ganymed:
With a diameter of approximately 35 kilometers (22 miles), Ganymed is the largest of all near-Earth objects. The S-type asteroid has a rotation period of 10.3 hours. In October 2024, it will approach Earth at 56,000,000 km
[I don’t think the S type has much water- it has iron and magnesium silicates. S type the most common after C type.]]
Anyways space rocks bigger than 10 km impacting anything are quite rare and not easy give any kind of guess in terms of chance of hitting any planet- they kind like comets. And what largest comet, btw: apparently, Hale Bopp Comet but estimated range is 40 to 60 km in diameter [and some think it also had moon/satellite].
Anyway, suppose to have another comet you can see with naked eye, occurring soon:
“Dec. 4, 2018: On Dec. 16th, Comet 46P/Wirtanen will approach Earth less than 11.5 million km away–making it one of the 10 closest-approaching comets of the Space Age. It’s a small comet, with a nucleus barely 1 km wide, but such proximity makes even a small things appear large. The comet’s gaseous atmosphere is now as wide as a full Moon.”
“The 100-fold enrichment of the D/H ratio on Venus compared to Earth is thus a fundamental constraint on models for its atmospheric evolution.”
http://science.sciencemag.org/content/251/4993/547?sid=4172774e-a231-40c2-a57b-f33be14f020a
I was reading the a comet had twice Deuterium as Earth and that comet have more Deuterium than Earth, limits the amount of water, Earth could gotten from comets.
bobd…”The Pioneer mission to Venus found that there is more Deuterium in the atmosphere of Venus than on Earth.
This is evidence that there was once water on Venus.”
Deuterium as a straight isotope is hydrogen. You need oxygen to form deuterium oxide, which is heavy water.
I have never heard of a straight isotope, and strangely that is my line of work, the company I work for bought a company call Advanced Isotopes. I have probably handled more isotopes than the average bear.
That doesn’t address the question as to why there is an abundance of deuterium in the atmosphere of Venus.
Maybe because water with deuterium replacing hydrogen is heavier and less likely to be high enough in the atmosphere to encounter UV radiation, thus less likely to be lost to space, thus enriched in the atmosphere.
Because the enhanced greenhouse effect due to the surface temperature of Venus exceeding roughly 40 to 50 C causing more water vapor in the atmosphere of Venus.
There is a knee in a curve somewhere, which we should not encounter, if we can avoid it.
“That doesnt address the question as to why there is an abundance of deuterium in the atmosphere of Venus.”
You get more deuterium concentration only from sorting process involving hydrogen. If deuterium is in water, water boils at slightly higher temperature, so that could be a sorting process.
If had atmosphere of H2, and atmosphere out gassed hydrogen, the lighter hydrogen molecules would tends to leave first, thereby concentrating deuterium. Another sorting process.
It also related to preference of chemical bonding- if difference between hydrogen with neutron and hydrogen without neutron, that could be another sorting process. Ie:
–Abstract. We report measurements of the deuterium content
of molecular hydrogen (H2) obtained from a suite of air samples that were collected during a stratospheric balloon flight
between 12 and 33 km at 40◦ N in October 2002. Strong
deuterium enrichments of up to 400 versus Vienna Standard Mean Ocean Water (VSMOW) are observed, while the
H2 mixing ratio remains virtually constant. Thus, as hydrogen is processed through the H2 reservoir in the stratosphere,
deuterium is accumulated in H2
and:
Using box model calculations we investigated the effects of H2 sources and sinks on the stratospheric enrichments. Results show that considerable isotope enrichments in the production of H2 from
CH4 must take place, i.e., deuterium is transferred preferentially to H2 during the CH4 oxidation sequence.–
https://www.atmos-chem-phys.net/3/2015/2003/acp-3-2015-2003.pdf
So, it could H2, or some molecule with hydrogen as part of it: water, CH4, other hydrocarbarbon, acids, etc.
So if Venus was gas giant and had the hydrogen and helium boil off, you probably have higher levels of deuterium.
So if Venus was like Earth and had the water evaporate, and then the UV would disassociate the water into oxygen and hydrogen and the lighter hydrogen escapes to space.
And before it was all gone the warmer water dissolves the carbonate rocks and releases the CO2 from the water and so on and so forth. Like warm soda can not hold as much CO2 as cold soda.
All the carbonate rocks become CO2 in the atmosphere with the gravity of Venus able to hold that gas in place.
Otherwise one has to explain how a condensing planetoid of the size of Venus could capture such a thick atmosphere and how it could possibly be all CO2.
If someone could provide at least a plausible explanation for the 92 bars, I would like to hear it.
Otherwise, the dissolution of the carbonate rocks due to the enhanced greenhouse effect is the only explanation for the high pressure that I have heard.
bob, that’s quite an imaginative fairy tale.
And, as long as you include an “enhanced greenhouse effect”, it will remain a fairy tale.
bobd…”Otherwise one has to explain how a condensing planetoid of the size of Venus could capture such a thick atmosphere and how it could possibly be all CO2″.
There’s another option, ‘we don’t know’.
Scientists looking for a payday via grants, etc., may need to come up with bs to cover their butts but the rest of us can simply say we don’t know.
I prefer not knowing to stupid explanations for science that make no sense.
Well there was one scientist or was it Sherlock Holmes who said once you rule out all the possible explanations then you have to take a look at what you have ruled out as fairy tales.
bobd…”Maybe because water with deuterium replacing hydrogen is heavier and less likely to be high enough in the atmosphere to encounter UV radiation, thus less likely to be lost to space, thus enriched in the atmosphere”.
This is getting into serious atomic physics. How does deuterium get a neutron in its nucleus is the question when the hydrogen itself has none?
One would think hydrogen without the neutron would have been the source of the deuterium, even if the deuterium came from deuterium oxide, as heavy water.
I am thinking solar wind, which is a plasma of bare protons and electrons. The protons are hydrogen nucleii without the electron and vice versa. Maybe those protons and electrons interact somehow in the atmosphere of Venus and recombine to form deuterium with a neutron, which has a neutron, a proton, and an electron.
That’s not as easy on Earth since our magnetic field diverts much of the solar wind.
Deuterium is an isotope of Hydrogen
see this
https://en.wikipedia.org/wiki/Isotopes_of_hydrogen
Salvatore ! Salvatore !
Tell us again how everything is about to cool !
The global SSTs and the nino3.4 index are all heading upwards!
What is happening ? Why won’t the solar cycles effect kick in?
Tim Folkert,
You wrote (December 4, 2018 at 12:43 PM):
“* Convection-related processes lead to a fixed, stable lapse rate.”
1. Do you have a link to a paper that demonstrates that this is the case in margins of sufficient accuracy?
2. What is the value of this lapse rate?
3. How is this value related to the laws of physics?
phi…”2. What is the value of this lapse rate?
3. How is this value related to the laws of physics?”
I know one thing, it cannot explain why the air pressure at the top of Mt. Everest (~30,000′) is 1/3 the pressure at sea level, with the air temperature in proportion.
The Ideal Gas Law and gravity can explain it.
Gordon,
“The Ideal Gas Law and gravity can explain it.”
No, it does not work. Most of the volume of the tropspher is in subsidence and therefore should have a dry adiabatic lapse rate (10 K/km). The average value is close to 6.5 k/km. Compression causes a warming of 10 K/km, what can cool the column of 3.5 K/km?
psi….”Most of the volume of the tropspher is in subsidence and therefore should have a dry adiabatic lapse rate (10 K/km).”
I think the theory related to adiabatic processes is lacking. I don’t think columns of air rise without exchanging heat with air outside the column. It seems absurd.
I see a steady state condition covered by the Ideal Gas Law with dynamic processes superimposed on that steady state.
The lapse rate theory does not explain an air pressure at 30,000 feet that is 1/3 the air pressure at the surface. The IGL and gravity does. The drop in pressure is accompanied by an equivalent drop in temperature.
If that temperature is not linearly proportional to the pressure then you can factor in the dynamics aspects to explain it. The lapse rate theory is putting the cart before the horse. It ignores the physical cause of reduced pressure with altitude.
Once you establish a relationship between pressure and temperature at a relatively constant volume you can consider the dynamic processes riding on top of it.
I think gravity produces an effect on the atmosphere akin to a gas in a cylinder being compressed by a heavy piston. As the atmosphere experiences solar input for a portion of the day, the atmosphere expands and the piston rises. At night, without solar input, the atmosphere cools and the piston falls.
It’s all covered by the IGL. I think we have seriously complicated the process and lost tract of the basics.
BobD,
“Well, he thinks they were faked.
But I have been in higher radiation fields than the Van Allen belts.
Its time distance and JDHuffman when dealing with radiation.”
IF the radiation in VA belt is harmless, why have all manned space flights subsequent to Apollo stayed well below it? Moreover, why is NASA recently on record as being in the process of trying to develop technology that will allow them to safely travel through the VA Belts? If they already did it with virtually no protection with Apollo? Multiple times, perfectly. It makes absolutely no sense.
Even once through the VA Belts, the Apollo spacecraft had virtually no protection from Cosmic and Solar radiation, but none of the Astronauts who allegedly wen the moon ever suffered the slightest ill effects. But yet, now — even 50 years later — no one dares even go into the VA belts, let alone through them? Why would they have anything to worry about?
Come on, when are people going to wake up and apply some basic critical thinking skills to this. The illogic of it and so many other things is overwhelming. I wasn’t a believer in the Moon hoax, BTW, until I examined it with a very critical eye (only fairly recently).
“IF the radiation in VA belt is harmless, why have all manned space flights subsequent to Apollo stayed well below it?”
It’s not harmless, but at 10 km/sec, you spending less than 1 hour in the Van Allen belts, which as wild guess could equal to radiation of couple months living on ISS or say 5 years as airline pilot. If spent a day in it, you could get sick and have long term health effects. Compared surface of Europa:
“One is the high level of radiation from Jupiter’s radiation belt, which is about 10 times as strong as Earth’s Van Allen radiation belts. Europa receives 5.4 Sv (540 rem) of radiation per day, which is approximately 1,800 times the average annual (yearly) dose of a human on earth at sea level. Humans exposed to this level of radiation for one day would have greater than 50% mortality rate within 30 days”
And they count radiation as accumulative effect, the travel time of going to Mars has something like 3% increase in cancer in your lifetime, and the radiation of going thru Van Allen would be added to the trip time radiation level of going to Mars. And/or adds to total “allowable” level that a astronaut can receive in the career- and astronauts are allowed/permitted to have higher career radiation level than worker dealing with radiation or airline pilots and crew.
And reducing radiation does from Van Allen is easier than reducing radiation dose travelling to Mars- which is mostly due to galactic high energy particles.
RW, you are correct to question NASA. They have NO credibility, anymore.
But the Moon landings were done under a different NASA. There was a lot of talent collected at NASA back then, and the emphasis was on engineering accomplishment, not political agenda. There is overwelming evidence that the Moon landings occurred, as reported. The Apollo missions flew through the weaker portions of the Van Allen belts. As gbaikie mentioned, the time in the radiation was short. The spacecraft and spacesuits would handle the rest.
This new NASA has the primary mission of getting more funding. After that, their goal is all politcal agenda. Way down the list is anything to do with real science, and then they contract it out.
Yes, yes — I’m well aware they could get through it fairly quickly. Possibly as little as 30 minutes at the speeds they were traveling, but this still doesn’t address the spectacular illogic of why no manned craft since has ever attempted to go through it again or why NASA is on record recently saying they are currently trying to develop technology that will allow them to safely travel through it. In other words, the technology doesn’t yet exist.
It’s purported to be harmless, yet for nearly 50 years no manned space craft has been willing to go into it or through it, and this is the case still to this day. Come on. It’s harmless, but no one will go through it. Again, come on.
According to NASA, they have effectively lost the technology to safely travel through the Van Allen belts. You don’t lose technology. This would be the first time ever in human history that technology was lost. Technology is abandoned for better technology, but not lost. It’s spectacular nonsense, yet no one connects the dots.
Moreover still, once they are through the Van Allen Belts, i.e. Apollo allegedly, they would have virtually no shielding, i.e virtually no protection from solar and cosmic radiation. The hull of the ship was like 1/8th of an inch of Aluminum, and their suits had minimal levels of protection.
RW…”IF the radiation in VA belt is harmless, why have all manned space flights subsequent to Apollo stayed well below it?”
The VA belts interfere with communications signals.
Good question, why would one want to go in the Van Allen belts, what scientific questions could be answered by going there? And I never said it was harmless, only that I have been in higher radiation levels than the Van Allen belts. Obviously there is harm, so you measure it and don’t exceed limits.
Max radiation levels in the Van Allen belt is about 200 mrem per hour, which means the annual allowed dose for a rad worker in the US would be exceeded in about a day.
Almost 3 months for a lethal dose, but there’s an assumption there. Chronic doses are less severe than acute doses, so it’s likely longer than that.
Only one of the Apollo mission exceeded 1 Rad for the whole trip.
Takes at least 25 Rad for it to be detectable medically, radiation sickness not until 50.
So you can travel through the Van Allen belts without too much worry, but don’t put a space station there.
The reason they are worried about the van allen belts and space travel in general is the possibility of a Carrinton type event which would involve radiation levels much higher than normal.
Let me put this even simpler. Apollo allegedly went through the VA belts (and far beyond) with virtually no protection, yet everyone, i.e. every single astronaut who allegedly ever went through, never suffered the slightest ill effect from radiation exposure. Yet, no manned craft for nearly 50 years since has ever gone through it (or has been willing to go through it, even today). Hmmmmmmmm. Why would they have anything to worry about? If Apollo did it 50 years earlier with 1960s technology. And perfectly, i.e. 100% harmlessly, multiple times. Again, come on.
RW…”Let me put this even simpler. Apollo allegedly went through the VA belts (and far beyond) with virtually no protection…”
I presume the shell was made of metal, which will protect gainst all radiation except at the upper extremes of the EM spectrum. I presume also that their spacesuits would be worn while in flight which likely had more metal built in to protect against radiation.
No just no, metal does not protect against all radiation.
Different metals provide different levels of protection, Tungsten and Uranium being among the best, though gold, iridium, osmium, rhenium or platinum would probably be just as good, though much more expensive.
Light materials like aluminum would be much less efficient as shielding by a factor of about 10.
Then there are neutrons, got to be quite tricky to shield from neutrons, especially the faster ones, you need the combination of a moderator and an absorber plus shielding from the activation products involved with blocking neutrons. Layers of poly and borated poly are used for this.
Thanks bobdroege for your two comments about radiation.
Unlike the usual boasters, you seem to know about what you write about.
You write above about neutrons. I never read anything about them, with one exception: those generated by the fusion of deuterium and tritium
D + T -> He + n
where n has an energy of 14 MeV.
Do you mean such neutrons?
Not those, they would be found in the Sun, I think, off the top of my head I am not sure.
I was thinking of the neutrons produced when cosmic ray interact with matter.
bobd…”No just no, metal does not protect against all radiation.”
Sorry, bob, we use metal in electronics and communication to completely block EM radiation. Can you show me any visible light frequency that can penetrate metal? Microwaves cannot penetrate it otherwise we’d all be irradiated by microwave oven microwave EM.
The Van Allen belts are caused by an interaction between electrons and protons from the solar wind interacting with the Earth’s magnetosphere. That kind of radiation will be EM and metal will block it.
They use lead to block x-rays, and in the older TV sets before they changed the high voltage flyback transformer, they used simply metal shields to block x-rays from electrons cycling in the transformer under a 40 kilovolt EMF.
What about high energy gamma radiation?
We are not talking about shielding from visible light when transiting the van allen belts.
bobd…”What about high energy gamma radiation?”
From what I’ve heard there is no way to stop it. Gamma rays are zipping through the atmosphere all the time and likely right through us.
Gordon, you say
From what I’ve heard there is no way to stop it. Gamma rays are zipping through the atmosphere all the time and likely right through us.
The tenth thickness for gamma rays, ie the amount of shielding necessary to reduce the flux by a factor of 10, is 2 inches for lead, 12 inches for steel, and 24 inches for water. These are approximate numbers for the gamma ray window. The window is the energy of gamma rays that are the most penetrating, typical around 6 MeV.
So you haven’t heard much, and I used wiki for the ease of the search and it is the same as other sources from working with radiation for 40 years. So in this case it can be trusted.
bobd…”Then there are neutrons, got to be quite tricky to shield from neutrons, especially the faster ones, you need the combination of a moderator and an absorber plus shielding from the activation products involved with blocking neutrons. Layers of poly and borated poly are used for this”.
There are no such neutrons anywhere near the VA belts. The VA radiation is due to protons and neutrons in the solar wind and there are no neutrons in the solar wind. The Sun is made up mostly of hydrogen, which has no neutrons. Even if neutrons were ejected from Helium, they are neutral particles and would produce no radiation when interacting with our magnetosphere.
You are talking about neutrons accelerated in cyclotrons. I have no idea where such high energy neutrons would be found anywhere near the VA belts.
At least NASA was concerned about them
During a complete Apollo mission, astronauts are exposed to widely varying
fractions of radiations from the Van Allen belts, cosmic rays, neutrons, and other
subatomic particles created in high-energy collisions of primary particles with spacecraft
materials. In addition, the individual responsibilities of the crewmen differ, and,
therefore, radiation exposure may differ.
from this report
https://www.hq.nasa.gov/alsj/tnD7080RadProtect.pdf
bobd…”In terms of hazard to crewmen in the heavy, well-shielded command module, even the largest solar-particle event on record (November 12, 1960) would not have caused any impairment of crewmember functions or ability of the crewmen to complete their mission safely”.
Bob… thanks for article. The article can be summed up nicely by this quote from it:
“In terms of hazard to crewmen in the heavy, well-shielded command module, even the largest solar-particle event on record (November 12, 1960) would not have caused any impairment of crewmember functions or ability of the crewmen to complete their mission safely”.
The article is dated 1973 and I have heard nothing of astronauts dying or being harmed by any radiation or particles in space.
If you read further, the article gives their estimate of the astronauts dose from the solar event of 1960, it would have been short of radiation sickness but detectable from blood work.
Of course we haven’t gone that far out since.
Gents, the concern in Van Allen belts is primarily due to charged particles, i.e., protons and electrons.
JD…”Gents, the concern in Van Allen belts is primarily due to charged particles, i.e., protons and electrons”.
I have known about the Van Allen belts for decades. They are a concern with certain bandwidths in communications since the VA radiation deteriorates electronic communications signals.
If I had to fly through them I’d be far more concerned about direct bombardment by the electrons and protons in the solar wind. However, those are mainly diverted by a relatively weak magnetic field surrounding the Earth. The metal of a space ship would block them completely, as it would any radiation produced by the interaction of solar plasma and our magnetosphere.
Not so fast
The proton belts contain protons with kinetic energies ranging from about 100 keV (which can penetrate 0.6 m of lead) to over 400 MeV (which can penetrate 143 mm of lead).
from
https://en.wikipedia.org/wiki/Van_Allen_radiation_belt
thanks for playing
bobd…”Not so fast…”
Are we talking about radiation, or protons?
First of all, you should know that any idiot can submit a wiki article. The article claims protons and electrons are ‘trapped’ by our magnetosphere. How the hay does a magnetic field CAPTURE electrons and protons let alone trap them?
That’s akin to the AGW bs that something in the atmosphere traps heat. Since heat has to be associated with atoms, what is this mysterious substance capturing atoms? And how did those atoms carrying the heat get there via radiation.
We know that real glass in a real greenhouse can trap atoms but there is nothing in the atmosphere can do that. Therefore I presume that both glass and metal can block protons, unless of course, you accelerate them artificially.
When that is done in experiments, it’s usually a thin leaf of metal that is used, like gold leaf. Metal of a decent thickness will stop any protons that are not artificially accelerated.
I am alarmed at the amount of pseudo-scientific bs being spread on the Net by the likes of the wikis.
And how does a proton penetrate metal at the velocities involved with solar radiation? If solar protons could penetrate metal, every astronaut would be bombarded by the solar wind and all electronics in spacecraft would die.
Gordon,
do you remember your physics?
Electric current, magnetic field and relative motion, does that ring a bell?
Earths magnetic field causes the path of the incident protons of the solar wind to be bent into orbit around earth.
I work with electronics in radiation fields and yes they fry, but it takes a while, depending on the strength of the radiation field.
The fields we put electronics in are hundreds to thousands of times stronger than what a spacecraft would encounter during a solar flare similar to the Nov 20, 1960 event.
Remember Rutherford’s experiment with gold foil?
When Geiger and Marsden shot alpha particles at their metal foils, they noticed only a tiny fraction of the alpha particles were deflected by more than 90. Most flew straight through the foil.
That’s with alpha particles, less penetrating than protons, and the ones in solar flare are accelerated to energies up to several thousand MeV, those are quite penetrating.
bobd…”do you remember your physics?
Electric current, magnetic field and relative motion, does that ring a bell?
Earths magnetic field causes the path of the incident protons of the solar wind to be bent into orbit around earth”.
Remember my physics well, in fact I have been applying it in the electrical field, the electronics field, and the computer hardware field.
I read Syun Akasofu’s book on the solar wind, in which he specialized. He did not describe the magnetic field effects on electrons and protons as you have.
He described the interaction as one of induced EMFs which push currents (electrons) through the atmosphere, the oceans, and the surface. Not a word about protons going into orbit around the planet.
What do you think we’re running up there, a cyclotron, or a high frequency cavity resonator? What the heck would cause protons to go into orbit around the planet? When they are deflected, they move of on trajectories around the planet alright, but they keep on going into space not around the Earth in orbit.
I’ll answer this question
What the heck would cause protons to go into orbit around the planet?
The Earths magnetic field.
I operate cyclotrons, they keep charged particles in circular paths with a magnetic field.
If the Apollo project was a fake: how then can the Lunar Laser Ranging Project access the reflectors placed on the Moon by the Apollo 11, 14 and 15 crews)?
Did a few topsecret probes bring them to the Moon like did the Russian Lunakhod 2?
http://adsbit.harvard.edu/cgi-bin/nph-iarticle_query?1999A%26A…343..624C&defaultprint=YES&filetype=.pdf
But chhhht! This paper belongs, at least for a few very, very educated commenters, to pseudoscience, as it presents the calculation of lunar orbital and rotational parameters, and thus dares to mention the phenomenon of the rotation of the Moon around its axis!
Oh dear, I anticipate some funny reactions. The area of pseudoscience grows every day by a lot.
Be careful: the blog sofware used by Roy Spencer seems to have a very personla meaning of what a link is and shall not be!
Thus don’t simply click on the link. Copy and paste it entirely into the browser’s address field.
binny…”dont simply click on the link. Copy and paste it entirely into the browsers address field”.
I did, it still doesn’t work. There are dots in the URL which are not normally allowed in URLs.
Bindidon, you need to learn how to provide a link, or else use the “tiny” site.
With some extra effort, I was able to access the site, but it was a “password” only site. But I was able to read the abstract. They were talking about the orbital parameters. The used the word “rotation”, but even from the abstract you could tell they were referring to the orbit. If you still believe the paper mentioned “rotation on its own axis”, then please provide a readable version of the paper.
http://adsabs.harvard.edu/full/1993POBeo..44..119T
Oh, that’s annoying. You’ll have to copy and paste the whole line into your browser.
Thanks DREMT.
A great quote from the paper: “But many astronmers have accepted as a physical fact that such rotation takes place. It does not, but only appears so; it is an illusion, a most surprising one, too.”
And, I never thought about the Moon’s lack of a magnetic field. Just one more nail in the coffin for the Spinners.
Tesla was a genius.
It’s a shame, but I haven’t seen any spinners since Jerry died.
Well, you’ve come to the right place….
“Tesla was a genius”
Yes, indeed, here is a great quote from the man:
“Let the future tell the truth, and evaluate each one according to his work and accomplishments. The present is theirs; the future, for which I have really worked, is mine.”
Here is a link that works:
https://tinyurl.com/y87gs2jb
binny…”This paper belongs, at least for a few very, very educated commenters, to pseudoscience, as it presents the calculation of lunar orbital and rotational parameters”
You can’t even post a URL correctly, how the heck could you understand what’s written in the paper.
Robertson
I don’t like to insult anybody here, but I often had to counterinsult you each time you named me an idiot.
But this time at least you managed to keep polite, even if you did not understand even half a bit of the problem I had.
The problem of course isn’t in the link itself, but is located in the web site’s scanner, designed by a software firm working probably in Absurdistan.
*
I politely remind you by the way, genius Robertson, that though I tried many times to show you how to use
https://tinyurl.com/
in order to bypass this site’s problems e.g. with ‘d c’, you still need to explain us how to correctly edit a link containg such characters you separated by hand instead of using tinyURL as I proposed you to do.
Here is the last recent example of your lack of ability to grasp even simplest things:
http://www.drroyspencer.com/2018/11/the-sorry-state-of-climate-science-peer-review-and-kudos-to-nic-lewis/#comment-331404
There you see, probably for the hundredth time on Roy Spencer’s web site:
“Please note.the following URL needs to be copy/pasted into a browser with the *** in ncd***c removed before activation.
https://www.ncd***c.noaa.gov/monitoring-references/faq/anomalies.php”
At that time you still did not realise how simple it is to tinyurl the link instead, as I just did:
https://tinyurl.com/n2twzcm
Did you get it now? I hope so.
Sorry, I don’t think the Apollo missions add up. If they had technology in the 1960s that worked perfectly for multiple missions through the Van Allen belts (and far beyond), you don’t lose or abandon such technology unless better technology is first developed in its place. Logically, you would build and expand around the already working technology to make it even better and/or more advanced (within its limits) as the years since Apollo went by. You also wouldn’t ‘destroy’ fully working technology, or at least the plans and blueprints used to make it, unless you had first developed proven replacement technology, yet this is what NASA tells us is what happened and what they did. That is, they lost and/or destroyed the technology of Apollo. Here we are now nearly 50 years later and no manned craft has repeated even first obstacle just beyond low Earth orbit (i.e traverse the Van Allen belts), and they are apparently currently in the process of trying to develop technology (i.e. Orion) that will allow us to do what we did 50 years ago multiples times, perfectly each time. It doesn’t make sense. Look, I too believed in the Apollo missions like most everyone, and superficially at least — found the arguments against the possibility it was faked to make sense. That is, until I looked at it all with a much more critical eye. Now, I’m increasingly more and more skeptical they really happened the more I examine everything and process all the evidence and logic.
If Apollo really happened, i.e. we really landed man on the moon multiple times and safely returned each and every man perfectly (even Apollo 13 returned safely), it would be by far the greatest technologic and exploratory achievement man has ever done (by a country mile). And those who did it, i.e. NASA, destroyed or lost the technology to do it? And no one has even come close to repeating it in 50 years. It would be the first time ever in human history that technology was lost and/or went backwards. Or that a landmark scientific achievement has not been repeated.
The logic (or rather spectacular illogic) of this — I just can’t seem to get past. But in order to believe Apollo happened, this is what one has to believe.
“If Apollo really happened, i.e. we really landed man on the moon multiple times and safely returned each and every man perfectly (even Apollo 13 returned safely), it would be by far the greatest technologic and exploratory achievement man has ever done (by a country mile). And those who did it, i.e. NASA, destroyed or lost the technology to do it? And no one has even come close to repeating it in 50 years. It would be the first time ever in human history that technology was lost and/or went backwards. Or that a landmark scientific achievement has not been repeated.
The logic (or rather spectacular illogic) of this I just cant seem to get past. But in order to believe Apollo happened, this is what one has to believe.”
You are merely looking at Socialism/Marxism.
One ask the question, why did one richest countries become the rathole of Venezuela- how did they lose their technology?
NASA didn’t lose technology- they never had it. NASA is a government agency, they are suppose to govern/manage, yet NASA imagines it a rocket company. But it doesn’t end there, they also imagined they know more about mining- so best mining and rocket company in the world- they are delusional. And there much more that they imagine they are good at.
Everyone knows why the US beat the Soviets to the Moon. Everyone knows why US won World war II. And it was not because of governmental genius- it was the powerful private sector. A nation of smart and hard working people, free people.
One of biggest mistakes NASA had made, is imagining that NASA could lower the cost to get into space. All socialist imagine they are capable of doing this- it’s reason people think government should run the healthcare industry. Socialist have never lowered cost of anything. No one wants to lower cost, it’s like saying to you want to work your butt off, and get less money.
Lowering cost can only done by competition.
One successful aspect of Soviet Union in terms of space, is they purposefully had competition- they had different “agencies” striving to do the job of getting into space. One thing Russians [and Chinese] does is, copy the US, though they always seem to miss some important aspects.
The problem with US is those that govern imagine they know something and don’t need to learn anything.
Anyways the cost of cost into space has and is getting cheaper, and none this is due to NASA, if anything NASA has increased launch costs [and they have been trying to do this- so they have somewhat successful in this regard].
Now, the socialist of Europe can be said to have lowered launch cost [at least in comparison to NASA efforts of raising them]. France wanted to be independent of the US, so Europeans made Arianespace:
“Arianespace SA is a multinational company founded in 1980 as the world’s first commercial launch service provider. It undertakes the operation and marketing of the Ariane programme. The company offers a number of different launch vehicles: the heavy-lift Ariane 5 for dual launches to geostationary transfer orbit, the Soyuz-2 as a medium-lift alternative, and the solid-fueled Vega for lighter payloads.”
https://en.wikipedia.org/wiki/Arianespace
But US miltary did EELV:
https://www.afspc.af.mil/About-Us/Fact-Sheets/Article/249026/evolved-expendable-launch-vehicle/
SpaceX also got it’s start with US miltary- though SpaceX did head hunted a lot ex NASA personnel.
In todays $ the cost of the Apollo program was $200 Billion.
http://www.extremetech.com/extreme/186600-apollo-11-moon-landing-45-years-looking-back-at-mankinds-giant-leap
Having accomplished the incredible goal, and doing what Moon science they could do, there was no compelling reason to continue going back, at such great expense and human capital.
Meanwhile, the cheaper un-manned exploration of the planets, Viking, Voyager, etc. were equally impressive feats and produced even more impressive science.
–Nate says:
December 11, 2018 at 1:34 PM
In todays $ the cost of the Apollo program was $200 Billion.
http://www.extremetech.com/extreme/186600-apollo-11-moon-landing-45-years-looking-back-at-mankinds-giant-leap
Having accomplished the incredible goal, and doing what Moon science they could do, there was no compelling reason to continue going back, at such great expense and human capital.–
I would suggest exploring lunar polar region and doing this for about 40 billion dollars. And be finished within 10 years.
Then explore Mars at about 100 billion per 10 years and require couple decades.
The lunar poles have water at the surface and it could be mineable.
And NASA lunar exploration should determine where there could be mineable lunar water. A likely aspect of commercially mining lunar water will involve using a depot in lunar low orbit.
And part of lunar program should be to develop an operational depot in low Earth orbit. This hasn’t done before, resolving various problems related to this, would be helpful for commercial lunar water mining and be useful for Mars exploration.
The compelling reason is to start new markets in space.
When Apollo when to Moon, it essentially brought tanker truck rocket fuel to lunar surface, in order to leave the lunar surface [return crew to Earth]. If you have rocket fuel made on the Moon, you don’t need to bring a tanker truck of rocket fuel to lunar surface in order to return to Earth. You also export lunar rocket to low lunar orbit, which mean don’t need to bring rocket fuel from Earth in order to land on the Moon. It also means you reuse the space craft which lands and leaves the lunar surface- so you don’t need to throw away a 200 million dollar vehicle after one use.
You also export more than rocket fuel from Lunar surface at much lower cost.
Having lunar rocket fuel and lunar water, lowers the cost lunar bases and operational cost of such bases- lots of countries want to have lunar bases. Lots of people want and would pay to be lunar tourists. At the moment lunar dirt is worth more than gold- about 20 times more. You lower the cost of lunar dirt to price silver.
Lunar dirt has scientific value, it has unique non copyable signature so could be use as anti forgery device. People want it, because they wanted it [a collectable], and schools want some for the kiddies to play with. And if bring back 100 tons of dirt, it still would be quite rare.
One can export lunar water, for crew use going to Mars- they each need about 10 tons, and it be much cheaper than lifting water from Earth.
So NASA explores Mars with purpose of determining if and where towns could be on Mars. So people wanting to living on Mars is another market. And if you have lunar water mining and has to have lunar activity. The lunar water mining and other lunar activity allows mars settlement to become more viable as compared not having such lunar activity. And mars settlement add another market to lunar activity.
From the Moon one export stuff which used in GEO satellite market.
If want something big and not have it folded up to fit in a rocket, from Moon one lift anything of any size. It lacks atmosphere and doesn’t the gee loads of acceleration required to lift anything from Earth. Or gee load could less than something stationary on Earth surface- or less than 1/2 a gee.
So if had some huge radio disk on Earth surface, and it was 1/2 as strong and same size it could be lifted off the lunar surface.
Or quite simply, GEO sats are big, but they are limited in size- and from Moon they could be unlimited.
norman…”You have to be drinking heavily to think your speed analogy compares to a flow of energy. They are not the same at all.
With speed you are covering a distance in a time frame. You are not transferring kilometers to some other object. You must be so drunk you are stumbling around!
With energy or flow of material you are transferring material from one place to another. I cannot fathom how really stupid of a human you are!”
*********
Speed, or more accurately, velocity, in this context, is the effect of a force on a mass as observed by humans. When a mass is accelerated by a force from rest, there is a change of velocity while the force is applied, the degree depending upon the strength of the force and the amount of mass.
When the force is removed, the mass maintains a velocity due to its momentum and that velocity is a property of the mass, not time or distance. Therefore velocity is a property of the real phenomena, force and mass, not distance and time, which are human measurement techniques.
With velocity, you are transferring energy from one place to another. Kinetic energy = KE = 1/2mv^2. The v in that equation has nothing to do with time and distance, it is related to the property of a mass in motion. If you want to quantify it, you do so using the human derived systems of time and distance.
With regard to energy flow, if you mean radiation, as in EM, there is no transfer of a specific energy (heat) from one place to another. There is a conversion of energy at either end with the EM transferring nothing but itself, the energy related to its electric and magnetic field.
The only way you can transfer energy as described is in the same energy medium. For example, you can transfer heat within a fluid as heat to heat via the atoms/molecules of the fluid. You can transfer mechanical energy using a driveshaft or gears.
EM transfers nothing, it is its own energy medium. If you doubt that, consider that the heat producing EM is completely lost during the transfer to EM. The emitting body cools. If that EM reaches a cooler body, it can be absorbed and converted to heat, causing that body to warm.
That is not an actual transfer of heat, as you would expect when heat is transferred from the hotter end of a conductor to the cooler end. In that case, heat is actually transferred as heat (thermal energy).
With regard to flux, the name comes from Mewton’s fluxion, which is essentially a derivative in calculus. It describes an instantaneous and infinitesimal rate of change.
If you have an EM flux over an area, one line of flux could be described as the smallest changing part of the EM field and to get the total flux over an area you’d have to integrate all the infinitesimal changes of EM over an area.
When I studied magnetism as part of my electronics training, we talked about flux fields, and lines of flux. Now they have obscured that reality behind names like Webers, Maxwells, and Teslas.
correction:
sorry Isaac, the Mewton’s in “the name comes from Mewtons fluxion” should be Newton’s”.
Gordon Robertson
Why are you so obsessed with thinking distance and time do not exist. Yes you may have read some interesting philosophy ideas challenging these basic concepts. That is what philosophy is designed to do, make you think about things, look at things differently.
In science time and space both exist as quantities and in units that can be used to calculate and predict things.
In your case with a mass. If you add so much energy to a mass you will get a rate where this mass moves a given distance in a certain time frame. Again, distance and time are NOT human inventions. The units we use are our own. Distance and time both exist outside of our units we use. Any uniform cyclic motion can be used to create units of time to measure changing spacial positions.
You may think you are correct with you view. I consider it wrong.
No one on this blog has been able to convince you. But that still does not make you correct.
YOUR STATEMENT NEEDS CLARIFICATION: “When the force is removed, the mass maintains a velocity due to its momentum and that velocity is a property of the mass, not time or distance. Therefore velocity is a property of the real phenomena, force and mass, not distance and time, which are human measurement techniques.”
What does that even mean “property of mass”. How do you describe it, what units do you use to describe velocity. How do you compare one velocity with another?
In real science, a cyclic unit is used to determine the rate of change of position.
Distance is a measure of space between different positions of some object. Why do you feel it is a man made? What man made up distance and how exactly did he do this? Before some person created distance did any object move?
I hope you pay attention to French development , While you are bickering here about how many watts reach Venus surface The french are putting up the real fight against $8 per gallon of fuel carbon tax rip off
I say they should tar and feather Macron and run him out of town instead burning their own city down.
Eben
“… instead burning their own city down.”
You have no idea of what happens there.
The “gilets jaunes” (the yellow vest people) mostly are people really agressed by fuel cost increase. This is the consequence of decades during which gasoline fuel was sold far under real price, for trivial political and economical reasons.
But their legitimate claims are fully exploited by right-wing extremists, and these people are burning the cities down.
*
The same drama will happen when the French society suddenly understands what has been told since decades as well: namely that the costs for electricity produced by the 58 French nuclear plants were artificially kept down (the real dismantling costs of these plants were dissimulated all the time).
And then again, the cities will burn. But certainly not due to the people suffering from electricity cost increase.
It is amazing how people can become teachy when they see one’s small msitake.
I can duplicate the link I provide in my browser as long as I want, the source is visible at each time, as I could do that yesterday evening.
When I see a JDHuffman trying to teach me about the use of tinyURL, I get a big, big laugh: Robertson still hadn’t doscovered the feature as he wrote his most recent comment containing an NOAA link with the inevitable ncd_c sequence!
Mais les loups ne se mangent pas entre eux, n’est-ce pas?
Vous êtes tout simplement ridicules et arrogants en plus.
*
I hope it works now:
Determination of the lunar orbital and rotational parameters and of the ecliptic reference system orientation
from LLR measurements and IERS data
J. Chapront, M. Chapront-Touzé, and G. Francou
Observatoire de Paris/DANOF CNRS/URA 1125, 61 avenue de lObservatoire, F-75014 Paris, France
Received 27 April 1998 / Accepted 7 December 1998
https://tinyurl.com/y9u8cnva
And try to learn a bit from the contents, guys ‘n dolls, instead of robertsoning all the time!
Now come on Bindidon, try to learn a bit from the contents, guys n’ dolls, instead of Bindidoning all the time!
https://tinyurl.com/y87gs2jb
Make sure you look at p122, 1st paragraph and page 125, 2nd paragraph, at least.
As usual, a comment written by a ‘skeptic’ comparing what to compare makes no sense.
How can the words of an inventor, no matter how ingenious, outweigh the decades of experience of people highly specialised in their field?
And above all, to present an article written by two Serbs concerning an ardently revered Serb is very naive, and shows that the commenter seems to know very few about Serbia and its fanatic, extremely nationalist population.
Unfortunately I had to do with some of them.
*
I read in the text an ‘estimation’ of Lagrange’s incredibly deep work (Théorie de la libration de la Lune, 1780):
“A simple form of the equations can easily result in a confusion.”
Oh dear…
https://gallica.bnf.fr/ark:/12148/bpt6k2292245/
An utterly substance-free response.
Dr Spencer, I’ve discovered an objective, market-based metric to support your claims about extreme weather.
Property Insurance Rates Prove Climate Change and Extreme Weather is a Hoax
https://co2islife.wordpress.com/2018/12/07/property-insurance-rates-prove-climate-change-and-extreme-weather-is-a-hoax/
Norman,
“When it condenses the energy that it removed in evaporation is returned higher up in the atmosphere (warming it). With the radiant energy, water vapor with temperature radiates up and down”
As that condensed water cools via radiation, the now warmer air also cools via collision with that water…
The radiation escaping to space COOLS the atmosphere… The downward radiation is recycled energy which changes the total internal energy by exactly 0 joules…
Water is a much more effective emitter than co2 and thus a much better coolant… Losing water causes the planet to cool more ineffciently thus temps rise… This explains Venus.
If I replace water in the rad of my car engine with a liquid that conducts heat less efficiently the temp in my engine will go up… This is essentially what is happening on Venus, as the water is being cooked off ( about 99% so far judging by heavy water levels) co2 becomes the dominant coolant, as it is much less efficient an emitter, temps have risen…
–bobdroege
December 8, 2018
The reason they are worried about the van allen belts and space travel in general is the possibility of a Carrinton type event which would involve radiation levels much higher than normal.–
I think they have realized the political problem with sending crew to Mars. Sending crew seems to them cause crew to receive unlawfully high exposure to radiation.
Or radiation would not immediately kill crew but could or might kill them long term.
It similar to NASA requiring the crew to smoke 2 packs of cigarettes per day- modern politics will not permit this.
Plus we could have a lot of solar min in future which basically doubles the exposure to radiation.
Plus they been exposing large numbers of crew to radiation (ISS ) the numbers favor some bad medical results which could be thought to be due to this exposure.
Anyhow, NASA would not have this problem with manned lunar exploration, but dearly care about manned Mars exploration.
In some ways “Radiation Safety” is similar to “Climate Science”. It is another another field where tribalism and junk science rules. Dissent at your peril!
Regulations in the USA are based on the LNT (Linear No Threshold) model which leads to nonsensical policies such as ALARA (As Low As Reasonably Achievable).
Reality is more complex. Massive doses of ionizing radiation are deadly yet small doses may improve your health because your body is built to handle much more ionizing radiation than most people are exposed to. There are numerous studies demonstrating that the LNT model is false. Here are just two of them.
Dr. Sakamoto of Fukushima, Japan routinely gives his patients doses of radiation which are about one third of the LD50 dose. An LD50 dose has a 50% probability of killing the exposed person within 4 weeks. Yet Dr. Sakamoto still has his license to practice medicine and patients are still seeking him out. According to the LNT theory such huge doses of X-rays would significantly reduce life expectancy but the data suggests the opposite:
https://atomicinsights.com/wp-content/uploads/Sakamoto-2012_ANSconf-June23.pdf
The number of people treated by Dr. Sakamoto is small. Is there some way to scientifically determine the effect of non-lethal doses of ionizing radiation on a much larger group? While it is hard to imagine such a study being planned there have been several unplanned studies following nuclear accidents and the detonation of nuclear weapons.
Seventeen hundred apartments in Taiwan were built with re-bar contaminated with Cobalt 60, a strong gamma ray source. Here is what a research study concluded:
“Approximately 10,000 people occupied these buildings and received an average radiation dose of 0.4 Sv, unknowingly, during a 9–20 year period. They did not suffer a higher incidence of cancer mortality, as the LNT theory would predict. On the contrary, the incidence of cancer deaths in this population was greatly reduced—to about 3 per cent of the incidence of spontaneous cancer death in the general Taiwan public. In addition, the incidence of congenital malformations was also reduced—to about 7 per cent of the incidence in the general public.
The average dose (0.4 Sv) is significant given that the LD50 does for gamma radiation is 5 Sv. You can find the rest of the report here:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2477708/
–gallopingcamel says:
December 9, 2018 at 1:33 PM
In some ways Radiation Safety is similar to Climate Science. It is another another field where tribalism and junk science rules. Dissent at your peril!–
Our world is filled to brim with pseudoscience.
Some think mad scientists are a problem, but it’s mad pseudo scientists which are ruin of the world. But science is also not the answer to the world, but that science works is better doing things that don’t work,
So when the insane are made more insane- things are not working.
Or neurotic get other neurosis, it’s not working.
And economics, wow! If it worked socialism might have a slim chance of working [though human nature would probably prevent it].
–Regulations in the USA are based on the LNT (Linear No Threshold) model which leads to nonsensical policies such as ALARA (As Low As Reasonably Achievable).
Reality is more complex. Massive doses of ionizing radiation are deadly yet small doses may improve your health because your body is built to handle much more ionizing radiation than most people are exposed to. There are numerous studies demonstrating that the LNT model is false. Here are just two of them.–
Yes, perhaps, but the pseudo science is working against NASA desires, and they are hapless against the political reality.
Also their obsession with finding alien life, works against them, politically. They think it is a cash cow, and it’s a stinky septic tank [which btw, you might find alien life in, if you looked].
The pseudoscience about Venus gives us another example of the flawed physics from NASA, the “temple” of the false religion.
Being closer to the Sun, Venus gets almost twice the solar irradiance. But, the atmosphere is so thick it is believed less solar energy gets to the surface, than on Earth. Yet the surface is over 400 °C (700 °F) hotter!
So with an atmosphere composed almost entirely of CO2 (which pseudoscience claims “warms the planet”), and being much closer to the Sun, and with areas of molten lava on its surface, the “high priests” at NASA inform us that Venus has a colder effective (black body) temperature than Earth.
Venus — 226.6 K
Earth — 254 K
(Source NASA — “Venus Fact Sheet”)
Obviously too much CO2 is the cause of all the cooling on Venus….
JDHuffman vs. NASA again. Tough choice.
Svante, were you unable to understand my comment, or do you just automatically reject reality?
I was with you all the way up to the word “pseudoscience”. You have given us enough of that already.
Svante, you can’t get enough of pseudoscience. That would be like a drug addict saying “don’t give me anymore drugs”.
Again, I was with you all the way up to the word “pseudoscience”. You have given us enough of that already.
See….
@JDH:
“……the high priests at NASA inform us that Venus has a colder effective (black body) temperature than Earth.
Venus 226.6 K
Earth 254 K
(Source NASA Venus Fact Sheet)”
I thought those TOA (Top Of Atmosphere) temperatures were measured from spacecraft and therefore not subject to significant error. So I did a search to find out how many independent measurements have been made. For Earth we have UAH and RSS. For Venus, NASA is not the only outfit that has launched Venus missions. The Russians have something to say on the subject.
IMHO the TOA temperature should not be strongly dependent on radiation issues (TSI, Albedo, IR emissivity). What matters most is the precise composition of the clouds. What are the partial pressures of each gas and vapor present in the upper atmosphere?
Thus far I have found the TOA for Venus claimed to be anything from 220 to 232 Kelvin. The NASA estimate is almost dead center in the range. This may be in part be caused by semantics. Are we talking about “Cloud Tops” or the “Top Of the Atmosphere”? Given what Dr. Roy does for a living I hope he can tell us which of the estimates are credible.
Galloping Gazelle, those black-body temperatures are not measured.
It’s incoming power times albedo, translated to a black-body with the same output power.
WRONG Svante.
You forgot the bogus “divide by 4” that comes from the pseudoscience that you can not get enough of.
That’s right, the are of the sphere is four times the area of it’s shade.
GC asks: Are we talking about “Cloud Tops” or the “Top Of the Atmosphere”?
We are talking “effective temperature”. It is a term from pseudoscience in which the after-albedo solar flux is divided by 4., and then used in the S/B equation The concept realizes that the surface area of a globe is 4 times the surface area of a disk with the same radius. But, the concept ignores the fact that you can NOT divide flux.
See discussion here:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-332740
cam…”the high priests at NASA inform us that Venus has a colder effective (black body) temperature than Earth”.
I hate terms like effective and average since they can be seriously abused to make a point.
The rotational period of Venus is much slower than that of the Earth. Off the top of my head, I think a day on Venus is something like 173 Earth days. That suggests the night side is pretty darned cold.
So, here we have an alleged blackbody where one side has a surface temperature around 450C and the night side who knows what?
There are suggestions that the two sides do not differ much thermally. More consensus bs from the paradigm crew.
We are discussing something that Dr. Roy gets paid to measure every day. Nobody is going to pay me one red cent to explain what the effective black body temerature of Earth or Venus is.
While I respect all your opinions it seems likely that Dr. Roy could explain this in a convincing way so that most of us would accept his view.
gallopingcamel, search for the word venus here to learn what Dr Spencer thinks:
http://www.drroyspencer.com/2014/04/american-thinker-publishes-a-stinker/
“The rotational period of Venus is much slower than that of the Earth. Off the top of my head, I think a day on Venus is something like 173 Earth days. That suggests the night side is pretty darned cold.”
The “effective” rotation of Venus is much faster- 4 to 5 days- due a vast amount air [say about 1/10th of entire atmosphere] rotating.
A chance to check these figures from first principles.
For Venus
Solar irradiance 2601W/M^2
Divide by 4 to get irradiance per M^2 of TOA.
Albedo 0.77
The planet is absorbing 2601(1-0.77)/4=149W
The planet will also emit 149W of longwave radiation Fromm the Top of the Atmosphere.
Sigma, the SB constant is 5.6710^-8
The SB equation is E=sigmaT^4 or T=3rd√(E/sigma)
For Venus T=3rd√(149/5.6710^-8)=226.7K
For Earth solar irradiance is 1361W/M^2.
Divide by 4 as above.
Albedo is 0.3
The planet is absorbing 1361(1-0.3)/4=238W/M^2
The planet will also emit 238W/M^2 from TOA.
For Earth T=3rd√238/0.85.6710^-8=254K
The SB equation accurarely predicts Top of Atmosphere black body temperatures, or surface temperature without atmospheric effects.
Whatever theory is put forward to explain atmospheric
effects must explain why Earth’s surface is 33K warmer than the black body temperature and Venus is 496K warmer.
The idea that the GHE (Greenhouse Effect) for Earth is 33 K is false. An airless Earth would have an average temperature of 255 K if the surface temperature was uniform. That is an unrealistic assumption.
Our (airless) Moon has an average temperature of 197 K and some people contend that raising the rate of rotation by a factor of 29.5 would have no effect so an airless Earth could also have an average temperature of 197 Kelvin.
Along with Scott Denning (shameless name dropping) this camel can show that the average temperature of an airless body rises as the rotation rate increases. My calculations suggest that the average temperature of an airless Earth would be 209 Kelvin which corresponds to a GHE = 79 Kelvin.
“Along with Scott Denning (shameless name dropping) this camel can show that the average temperature of an airless body rises as the rotation rate increases. My calculations suggest that the average temperature of an airless Earth would be 209 Kelvin which corresponds to a GHE = 79 Kelvin.”
I think it’s accepted that rotation rate effect the average temperature. It’s also accepted the moon does not absorb much energy due to it’s highly insulative surface. Or if bare rock, the Moon would absorb more energy. And different type of bare rock absorb more.
And transparent surface can absorb more sunlight than any kind of bare rock. And Earth ocean water absorbs a lot more sunlight than any kind of bare rock.
An ocean covered by atmosphere which diffuses sunlight will also absorb diffused or indirect sunlight. Or our ocean absorbs 1120 watts rather than 1050 watts of sunlight.
Atmospheres can cause planet to absorb more sunlight as compared to vacuum and bare rock. Atmosphere of a planet also radiate less energy as compared to a planet with vacuum and bare rock.
I would say Earth without considering greenhouse gases and having clouds would have average temperature of about 5 C [278 K].
And Earth without clouds reflects a considerable amount of sunlight. Added in the clouds and increasing the amount of sunlight reflective may decrease or increase average global and temperature. And radiant effect of all greenhouse gases might add about 10 to 15 K to the surface air temperature.
And it seems to me, the land areas have a net cooling effect in terms of global average temperature. And arrangement of land and other aspect of land, can result in less cooling effect caused by land areas.
Oh, also volcanic activity under ocean does increase the ocean temperature, and different amount of volcanic activity increase the ocean temperature by varying amounts.
Changes in the temperature of entire ocean require a long amount of time [generally] and long amount of time is about 1000 years- or many centuries.
And average ocean temperature determines Earth average temperature and determines global climate. And is it currently about 3.5 C and it’s this cold due to the much long periods of glacial periods which is a period in which the Earth land areas cause a lot global cooling.
Wrong E-man.
You can NOT divide by 4. Doing so just indicates you do not understand radiative physics.
You can divide by 4 if you don’t care about getting the right answer.
And it appears they don’t care about getting the right answer. They only seem to care about going along with the pseudoscience.
If they were able to think for themselves, they would see that something was wrong with the figures. How could a much hotter planet be much colder than Earth?
Venus 226.6 K (-46.5 °C, -51.8 °F)
Earth 254 K (-19 °C, -2.5 °F)
Their false religion keeps them blind to facts and logic.
“How could a much hotter planet be much colder than Earth?”
It is called the GHE effect.
Dr Roy Spencer says:
http://www.drroyspencer.com/2014/04/american-thinker-publishes-a-stinker/
Of course, the “magic gas”!
It can heat! It can insulate! It can defy all laws of physics!
Order now, while the supply lasts.
Order two volumes, and get a magic “green plate”, at no extra charge.
Don’t blame the messenger, that was Dr. Roy Spencer, PhD.
Why are you acting so guilty then, Mark.
Did you get caught again?
It’s probably worth repeating again, since Svante is unable to understand.
NASA claims:
Venus 226.6 K (-46.5 °C, -51.8 °F)
Earth 254 K (-19 °C, -2.5 °F)
The institutionalized pseudoscience claims that because Venus has more CO2, it is colder!
JDHuffman says:
NASA claims:
Venus 226.6 K (-46.5 C, -51.8 F)
Earth 254 K (-19 C, -2.5 F)
These are brightness temperatures, not actual temperatures. As you know, or should know.
JDHuffman says:
Of course, the magic gas!
It can heat! It can insulate!
The atmosphere is an insulator. CO2 is around 90 to 750 times as opaque to some wavelengths of light as N2 or O2.
– Mike Flynn, June 18, 2017 at 3:34 AM
http://www.drroyspencer.com/2017/06/a-global-warming-red-team-warning-do-not-strive-for-consensus-with-the-blue-team/#comment-251624
DA takes my quote out of context, and misrepresents it.
Without a knowledge of the relevant physics, that’s all trolls can do.
Nothing new.
I quoted you exactly.
You don’t know the difference between brightness temperatures and actual temperatures.
gbaikie says:
Oh, also volcanic activity under ocean does increase the ocean temperature
By how much (to the nearest order of magnitude).
Show your work.
JDHuffman says:
Of course, the magic gas!
It can heat! It can insulate! It can defy all laws of physics!
Why do you sleep under a blanket at night?
Why do you sleep under a bridge at night?
“The pseudoscience about Venus gives us another example of the flawed physics from NASA, the temple of the false religion.”
A tax payer supported temple of false religion.
If the temple needed to get public donations, NASA would probably be a more practical religion.
I think it would a better world if religions were more focus on space [or also called, the heavens]. They probably get more donations and do more for the poor.
“Being closer to the Sun, Venus gets almost twice the solar irradiance. But, the atmosphere is so thick it is believed less solar energy gets to the surface, than on Earth. Yet the surface is over 400 C (700 F) hotter!”
The rocky surface is cover by thick global clouds. But even if the skies were clear, very little sunlight would reach the surface.
Earth surface even if Earth were cloudless likewise gets a lot less sunlight as compared to the Moon. And considering a lot people aren’t living in deserts, and so they have more rain and more clouds, the thick atmosphere of Earth plus the clouds, makes harvesting solar as a non-viable way to generate electrical power.
So cloudless Venus is worst than cloudy Earth in terms of harvesting solar energy at it’s rocky surface. And global coverage of very thick clouds of Venus makes Venus far worse than Earth to harvest solar energy. So Earth is not the worst rocky surface to harvest solar energy from, it’s only the second worst rocky planet to harvest solar energy from.
Of course most people who lived on Venus, would live near the cloud level, and so sky cities would beat out getting solar energy from the Martian surface. In that case, Earth is the worst, followed by Mars, and Venus would be the third worst planet to get solar energy from.
NASA publishes their science in peer reviewed journals.
Where do you publish yours?
Probably on a blog, like you do. Now, PST.
Old news: AUGUST 22, 1968
“PASADENA, Calif. — The surface air pressure on the planet Venus may be 75 or 100 times that on Earth–or four to five times greater than the Venus pressure reported recently by Soviet scientists–Jet Propulsion Laboratory researchers have revealed.
Dr. Arvydas J. Kliore and Dan L. Cain concluded in an article for the Journal of Atmospheric Sciences that the Russian Venera IV spacecraft either landed on a 15-mile-high Venusian peak undetected by ground radar or stopped transmitting before it reached the planet’s solid surface.”
https://www.jpl.nasa.gov/news/news.php?feature=6100
I was thinking the old news might be mostly correct. It matters a lot where on the surface that you talking about when say what Venus air pressure is.
Earth makes it simple, we have sea level pressure and you can say what Earth pressure is at sea level. And if Earth lacked ocean where would you say “sea level” was. You would average the elevation of the land, so if nothing changed Earth “sea level” would be lower by km or so. If Earth had 1/2 as much ocean as it has, Earth would still have huge ocean, and of course sea level would be lower- by over a km.
I don’t see anyone attempting to say what Venus air pressure is “at sea level, though they do this with Mars and say this is what Mars air pressure is: ” 600 pascals (0.087 psi; 6.0 mbar)” at “sea level”.
Anyways I was thinking of a question of what would surface temperature of Venus be, if it had 1/2 as much atmosphere.
Or it’s estimated the mass of Venus is about 4.8 x 10^20 kg, so what if Venus had about 2.4 x 10^20 kg of atmospheric mass.
It seems obvious to me at at the non stated “sea level” it would be much cooler. But if had different reference point of air temperature at 1 atm pressure, I don’t that the air temperature at 1 Atm would change much.
Though it seems there things which could alter this assumption, it seems it could have large upon the global wind of Venus. And another large effect is Venus wouldn’t CO2 at critical pressure.
So it seems to me either or both these factors would have largest effect upon the air temperature at 1 atm on Venus.
But generally I don’t think these would be large effects. Something like a possible difference of as much as 10 K. Or rough similar to effect of the presence of CO2 upon temperature of Earth- fairly small and would hard to measure.
Now say you had long pole sticking though atmosphere of Venus and it was marked at point of 1 atm pressure. Or had it marked in terms of it’s range- highest and lowest point over period of 5 years. Then removed 1/2 of Venus atmosphere, and it would lower in pole in which is would at the level of 1 atm of pressure. How much lower?
And it seems if halving atmosphere had significant effect upon global wind, one would have a greater range of marked pole.
The effect of reducing pressure in the Venusian atmosphere has been modeled by Robinson & Catling with impressive accuracy from the surface to the top of the stratosphere. The model also works for the other six bodies in our solar system that have significant atmospheres.
Here are R&C (model) estimates of global temperature for Venus at various pressures. They are in close agreement with the Jenkins et al. radar measurements:
@ 92 bars, T~730 K
@ 46 bars, T~642 K
@ 23 bars, T~565 K
@ 11.5 bars, T~498 K
@ 5.75 bars, T~438 K
@ 2.88 bars, T~383 K
@ 1.44 bars, T~339 K
@ 0.72 bars, T~309 K…..Eureka! The Goldilocks zone.
@ 0.36 bars, T~263 K
“Here are R&C (model) estimates of global temperature for Venus at various pressures. They are in close agreement with the Jenkins et al. radar measurements:
@ 92 bars, T~730 K
@ 46 bars, T~642 K ”
So 90 K difference. Hmm.
I suppose it has no parameters regarding Venus critical pressure which is suppose to cause more uniformity in temperature, nor global wind, which also causes more uniformity of surface air temperature.
Or roughly speaking the air at surface should get more turbulent- currently it’s supposedly tomb like, or sluggish.
730 K = 457 C and 642 K = 369 C
And NASA fact sheet say 92 bars with
Average temperature: 737 K (464 C)
Wiki, Venus atm:
0 height 462 C 92.10 atm
10 km 385 C 47.39 atm
15 km 348 C 33.04 atm
So no rocky surface at 15 km and tiny amount of rocky surface at
10 km, and some [or tiny amount] rocky surface below 0.
And wiki:
“Ninety percent of the atmosphere of Venus is within 28 km of the surface; by comparison, 90% of the atmosphere of Earth is within 10 km of the surface.”
Half atmosphere on earth is below 5.5 Km- oh wiki says 5.6 km, I thought it might be 5.3 km. Of course Earth has lot elevation at sea level, and Venus would be different though got lots of land area which fairly flat- or as said about 80%.
Anyhow if remove 1/2 of Venus atmosphere, 90% of atmosphere would be less 15 km rather than less than 28 km. Though not really allowing for cooler atmosphere nor less pressure- perhaps they cancel each other. But I will go with less than 15 km unless someone can correct me. And critical pressure at 72.9 atm so that gone with 1/2 atmosphere.
Got to go, late.
For people interested in wether or not the Moon rotates around its axis, here is a link to a historical info. It is a book digitised by Google:
Histoire des Mathématiques
par J.F. Montucla, de l'Institut national de France
Tome Quatrième
Achevé et publié par Jérôme Lalande
An X (mai 1802)
Cinquième Partie
Livre Sixième
The source is here:
https://books.google.de/books?id=mh8zvmWDKqUC&pg=PA273#v=onepage&q&f=false
As this book is not behind paywall, Google provides us with the OCR source. I had thus the opportunity to translate the text in Chapter X by adapting it to today’s French and then using Google’s translator. It was a quick shot:
https://drive.google.com/file/d/1-_TZMxnDO8G7SYFbcpQL-p65YZrkVsey/view
The main contributors imho were
– Tobias Mayer
– Jean d’Alembert
– Joseph-Louis (de) Lagrange
Tobias Mayer was the first person able to exactly determine the longitude of a place on Earth. He was the author of remarkable data concerning the Moon.
From Jean d’Alembert:
https://gallica.bnf.fr/ark:/12148/bpt6k623951/f316.image
Especially Lagrange was the author of impressive work:
– https://books.google.de/books?id=7NQ4VbUofzgC&pg=PA5&lpg=PA5#v=onepage&q&f=false
– https://gallica.bnf.fr/ark:/12148/bpt6k2292245/
What is amazing at the end of all this is that
– while the ‘skeptics’ pretend that the Moon doesn’t rotate around its axis because it shows us the same face all the time
– all these scientists mentioned here, having observed the Moon during the XVIIIth century, say the contrary.
For those who ask why the duration of Moon’s rotation actually is exactly the same as that of its revolution, that’s no coincidence a all! It was quite different a billion years ago, and will be in a billion years as well.
Don’t guess: rather search for information…
Bindidon, is there ANY pseudoscience you won’t swallow?
binny…”For people interested in wether or not the Moon rotates around its axis, here is a link to a historical info. It is a book digitised by Google:”
Hate to tell you this, binny, but this book talks about libration, not rotation about the Moon’s axis. The author makes a big deal of trying to prove a property of an orbit.
As we turn on the Earth, we can view the Moon from various angles. In the extremes of those angles we are able to peak around the corners of the Moon and over it’s top a bit.
Those prone to illusions think the Moon is turning back and forth whereas is is our observation platforms moving back and forth.
A well known illusion is the apparent retrograde motion of Mercury. From certain points in our orbit, due to relative motion between us and Mercury in its orbit, Mercury appears to move backwards for a bit then resume moving forward.
That illusion confounded observers for the longest time. Some tried to explain it as Mercury performing some kind of loop during its orbit. That sounds a lot to me like the horse we have discussed turning in loops as it translates around a track.
Does this horse always face in the same direction?
If not, it rotates. (That’s the definition of rotation.)
DA, you can’t even understand the issue. It’s “rotate on its own axis”, as a toy top spins in place.
Learn some physics.
Does a rotating horse always face the same stars?
Does a David Appell stop trolling?
Gbaikie
There were eight successful Venus landers.
https://en.m.wikipedia.org/wiki/List_of_missions_to_Venus
They generally agreed on 93 bar atmospheric pressure and 730K surface temperatures.
If Venera 4 reported maximum pressure of 75 bar, then it may well have stopped reporting early.
–Entropic man says:
December 9, 2018 at 3:55 PM
Gbaikie
There were eight successful Venus landers.
https://en.m.wikipedia.org/wiki/List_of_missions_to_Venus
They generally agreed on 93 bar atmospheric pressure and 730K surface temperatures.–
Yeah.
Wiki, Venera 7:
“The probe transmitted information to Earth for 53 minutes, which included 20 minutes from the surface. It was found that the temperature at the surface of Venus was 475 °C (887 °F) ° ± 20 °C Using the temperature and models of the atmosphere a pressure of 9 Megapascal ± 1.5 MPa was calculated.”
And 475 C = 748 K. And about 90 atm. And at location of:
05°S 351°E
And I don’t know elevation of surface at 05°S 351°E and probably need more detailed coordinates even had accurate map of surface to use. Or since 1 degree = 111 km. One say 05°S 351°E +/- about 100 km.
“If Venera 4 reported maximum pressure of 75 bar, then it may well have stopped reporting early.”
Or could have stopped because it hit a mountain.
But when you say, “93 bar atmospheric pressure and 730K surface temperatures” at what elevation?
I sure of elevation difference but even much less than Earth, the elevation difference of ocean floor and mountain peaks is quite large.
If google: elevation differences of Venus:
Topography. Sharp elevation differences characterize Earth’s surface: the continental areas are much higher than the ocean basins. Venus also has a broad range of elevations, but unlike Earth most of its surface lies within a few kilometers of the mean elevation.”
Which about what I thought, but let’s see if more info.
“Mar 4, 2007 – Overall, Venus looks fairly flat. Elevation differences vary little, only 2 to 3 km, except for a few highland regions. The continents rise to only some 10 km, compared to a 25-km difference on Mars and a 20-km one on the Earth. Only some 10 percent of the surface extends above 10 km.”
And:
“Venus’ surface, in contrast, has little variation in terms of elevation, with the majority covered by smooth, volcanic plains. In fact, it is estimated that if a terraforming event began to allow for water to accumulate on the surface, roughly 80% of the planet would be below sea level. The majority of the above ground landmass would be in the form of two that formed from the planet’s two main highland regions – Ishtar Terra, located in the northern hemisphere, and Aphrodite Terra, just south of the equator.”
https://www.universetoday.com/22551/venus-compared-to-earth/
Chances are Venera 7 didn’t land in Aphrodite Terra, but see how big it is and where. Google: Aphrodite Terra.
Yup, looks opposite side of the planet:
https://en.wikipedia.org/wiki/Aphrodite_Terra
Aphrodite Terra:
“It is about the same size as Africa, and much rougher than Ishtar Terra.”
Oh, here:
https://en.wikipedia.org/wiki/Ishtar_Terra#/media/File:Map_of_Venus.png
from here:
https://en.wikipedia.org/wiki/Ishtar_Terra
Might be in a depression.
The veriation in elevation on Earth from the deepest subduction trench to the top of Everest is about 12km.
With similar size and gravity Venus probably has a similar elevation range. If you take the 92bar as the lowest elevation, then at 15km altitude you would get 33bar. At 5km it is 66bar If Venera 4 got to 77bar it would have been within 5km of the reference surface.
Unfortunately if you look at the mission report here
https://en.m.wikipedia.org/wiki/Venera_4
the data does not seem to be as reliable as originally thought.
“The veriation in elevation on Earth from the deepest subduction trench to the top of Everest is about 12km.
The deepest trench known is about 11 km. below sea level Everest is about over 8 km above sea level.
So the greatest variation is close to 20 km.
Everest height 8848M
Marianas Trench -10,994M
Difference 19842M
You got it.
Finally….
we could go over again, the Equatorial bulge. wiki:
“As a result of Earth’s equatorial bulge, the highest point on Earth, measured from the center and outwards, is the peak of Mount Chimborazo in Ecuador rather than Mount Everest. But since the ocean also bulges, like Earth and its atmosphere, Chimborazo is not as high above sea level as Everest is.”
But let’s skip it.
JDHuffman
Not radiative physics, but geometry. You divide by 4 to convert from the solar irradiance, the amount of radiation hitting a 1 Metre square plate at the planet’s distance fromthe Sun to the amount of radiation hitting each square metre of the planet’s atmosphere.
That is the quantity you need to begin caculating the blackbody temperature
The total amount of energy intercepted by a planet is the solar irradiance multiplied by the area of a disc of the same diameter as the planet.
That is πr2, where r is the radius of the planet in metres.
The surface are of a planet in square metres is 4πr2.
To convert that total intercepted energy into the flux per square metre of planetary surface you use the equation
Flux per square metre of surface = solar irradiance πr^2/4πr^2
That simplifies to solar irradiance 1/4.
Hence the divide by 4.
Still wrong, E-man. All of your geometry is correct, but you can’t treat “flux” as “energy”. You’re still confusing a “quantity” with a “rate”. You’re still trying to divide the 100-kph car into 4 equal parts and claim each part is traveling at 100/4 = 25 kph.
That makes you wrong.
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-332740
Flux is no energy.
But a distribution of flux over an area can be, in the long-term where flux is constant, averaged into a flux per unit area.
Incoherent rambling, as often seen from physics-deprived trolls.
What is the definition of “flux?”
Trollvid Appell, please stop David-ing.
JDHuffman
The unit of quantity of energy is the Joule.
Watts are the rate at which Joules are delivered.1 Watt delivers 1 Joule for each second of delivery.
I illuminate a square metre of ground with a light. This has a power of 1 Watt. The surface receives a flux of 1W/M^2. The quantity of energy delivered is 1 Joule for each second of illumination.
In one minute the surface would receive 60 Joules
If I divide the flux by 2 the power becomes 0.5 W/M^2 and it takes two minutes to deliver 60 Joules.
In your view this is an impossible calculation. Please explain why.
E-man, Watts are Joules/second. Power is energy/time. Power is a “rate”. Power is a “rate”, energy is a quantity.
When you divide flux by 4, you are treating it as energy. But, then you can not use that energy in the S/B equation. You are confused, because you want to believe in pseudoscience.
Back to the simple example of the car traveling at 100 kph. The car has total kinetic energy of E. Now consider the car is composed of 4 equal masses. Each individual mass now has energy of E/4, but the same rate, 100 kph. You can divide the energy up, but the rate stays the same.
The “effective temperature” is one of the cornerstones of the AGW pseudoscience. Once people understand “effective temperature” is nonsense, then the 33K becomes nonsense. Then, the entire hoax is exposed.
That’s why people refuse to see the simple truth.
Ger*an says:
When you divide flux by 4, you are treating it as energy.
False. It’s simply noting that the daily flux of sunlight on Earth is distributed, on average, over its full surface area.
DA doesn’t even understand his pseudoscience!
THe Sun’s daily flux isn’t distributed over the Earth’s entire surface area??
The David Appell wont stop trolling??
Entropic man
Best of luck trying to explain science to g.e.r.a.n.
I have tried but it was not successful. He is stuck in some type of strange logic loop with his car and speed example. It has nothing at all to do with radiant flux but he thinks it does and continues to use it.
I pointed out to this poster that you can divide gallons a minute up quite easily into different tanks. You can have one tank receive 1000 gallons per minute. If it is a 10,000 gallon tank it will fill in 10 minutes. You can take this exact same water flow and break it up into 4 flows that go into different tanks, you now have effective flows to each tank of 250 gallons/minute. If each tank has a 10,000 gallon capacity it will now take 40 minutes to fill a tank.
He thinks the solar flux can’t be divided because it is a rate. He must no live on Earth or go outside. You have considerably less energy reaching the surface when the Earth tilts away from the Sun in Winter. The solar flux does not change but the area the Sun must illuminate grows larger. g.e.r.a.n is lost and no one can lead him to the light. One poster said the best thing to do is ignore him completely. When he posts DO NOT RESPOND. DNFTT. It is hard to do, sometimes you think there is a hope of reaching him but soon you find it is an illusion.
If people come to this blog, read his posts and agree with him, what is that to me. Most people can’t understand science and never will. He only can con people that have no knowledge of science. Anyone who has any actual education know he is bogus.
This link provides all the evidence to show that g.e.r.a.n is a phony fraudulent poster, more interested in stirring things up. A troll.
http://www.physicalgeography.net/fundamentals/6i.html
Same solar flux, larger area, less Watts/m^2 reaching the surface. Simple and correct. It won’t change g.e.r.a.n.
Norman always loses to facts and logic, and must resort to Insults, misrepresentatins, and false accusations.
That’s all he has.
Nothing new.
JDHuffman
If you are not g.e.r.a.n why do you feel insulted. Why are you such a dishonest phony? What does it do for you?
The actual reality and not your twisted version is that my facts and logic are correct. Yours are bogus BS that have no meaning. But being a phony everything you post is false.
Now consider that your car example and speed is not valid at all to pertain to radiant energy.
A more realistic example would be to take the Power (watts or horsepower units) that is required to keep a vehicle at 100 kph. The speed is more akin to temperature (which is average kinetic energy of the molecules of the body) and no you can’t divide the temperature and come up with different temperatures. If you divide a 100 C body into 4 each part is still 100 C. But you can take the power needed to keep a car at 100 kph and divide that into 4 cars. The same energy but the speed will not only be 25 kph for each vehicle.
If you take 1000 watts of power and the size of the receiving surface is 1 m^2 you will receive 1000 watts. If you have this same 1000 watts reach a 4 m^2 surface each m^2 will only receive 250 watts.
You are so totally ignorant. Think of the Sun itself. At the solar surface the flux is millions of watts/m^2. This same flux is 1361 W/m^2 at Earth distance. The same amount of energy covers much more area and the flux drops. The total energy is the same as it is on the solar surface, the flux is divided among all the many square meters making up the sphere at Earth distance. You are a compete and total dork! You don’t even understand simple ideas and confuse it all. You are a liar and fraud when you claim to have studied actual physics. You don’t know a thing at all.
Some say Gordon Robertson is the dumbest poster. I think you might be the one. So stupid.
Norman, you get rattled because you can’t understand the basic physics. That frustrates you, and you don’t have the maturity to deal with it. So you lash out with insults, misrepresentations, and false accusations.
The solution, of course, is to learn some physics. And, grow up.
Or, just keep banging on your keyboard, hopelessly hoping something will change.
Entirely you choice.
entropic…”I illuminate a square metre of ground with a light. This has a power of 1 Watt. The surface receives a flux of 1W/M^2. The quantity of energy delivered is 1 Joule for each second of illumination”.
How far away is the illuminated surface? Flux density decreases as the square of the distance from the source.
A light rated at 1 watt is rated for the electrical power it consumes. That does not translate to EM flux. Depending on the type of lamp, a certain amount of the 1 watt is lost as heat.
It doesn’t matter how far away the surface is — a illumination of 1 W/m2 is 1 W/m2.
It’s like you never took kiddie physics in high school.
David, it’s like you never took kiddie “not-trolling” in high school.
Here’s a link to the full paper by Tesla on the Moon’s rotation. It is a part of a general article in which he exposes myths.
I have never seen the article before and felt amused that Tesla used the same demonstration I offered here on Roy’s blog recently, using coins. He used washers.
What he claims is absolute proof that the Moon does not rotate on its axis.
https://teslauniverse.com/nikola-tesla/articles/famous-scientific-illusions
Again, I defy anyone to mark two coins side by side marked at the same point, then move one coin around the other while keeping the mark against the perimeter of the other coin. It is impossible to rotate that coin about its axis while keeping the same face against the perimeter of the other coin.
And, in response to criticisms, he wrote two further articles on the subject:
https://teslauniverse.com/nikola-tesla/articles/moons-rotation
https://teslauniverse.com/nikola-tesla/articles/moons-rotation-0
dremt…”In this article Dr. Tesla proves conclusively by theory and experiment that all the kinetic energy of a rotating mass is purely translational and that the moon contains absolutely no rotational energy, in other words, does not rotate on its axis. EDITOR”.
Interesting. Link two opens with the caption:
“In this article Dr. Tesla proves conclusively by theory and experiment that all the kinetic energy of a rotating mass is purely translational and that the moon contains absolutely no rotational energy, in other words, does not rotate on its axis. EDITOR”.
The Moon, in its orbit meets all the definitions of curvilinear translation.
There’s an interesting quote towards the end of the article, too, on the same subject:
“A motion of this character, as I have shown, precludes the possibility of axial rotation. The easiest way to free ourselves of this illusion is to conceive the satellite subdivided into minute and entirely independent parts, as dust particles, which have different orbital, but rigorously the same angular, velocities. One must at once recognize that the kinetic energy of such an agglomeration is solely translational, there being absolutely no tendency to axial rotation.”
Do you agree that a body that doesn’t rotate always points in the same direction?
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
You could have looked in the first paper.
“The moon does rotate, not on its own, but about an axis passing thru the center of the earth, the true and only one.”
What “first paper?”
Whoever says it was wrong — the Moon rotates about both its polar axis and the Earth-Moon center of mass. (And the Earth also rotates about *its* polar axis and the Earth-Moon center of mass.)
Just look here:
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
David displays his inability to follow the sub-threads he responds on, or to understand the word “rotation”
Dr Roys Emergency Moderation Team says:
I showed you four astronomers in agreement.
There isn’t a SINGLE astronomer named at any of your links.
What are their names? 1. 2. 3. 4.?
That’s because you are replying in the wrong place, David. The link with the names was elsewhere. Here it is:
https://tinyurl.com/y87gs2jb
The two authors of the paper, and Savic and Kasanin (see reference to them on page 125, 2nd paragraph).
1) Tomic, A.
2) Jovanovic, B.S.
3) Savic
4) Kasanin
Google “Savic-Kasanin theory” for further information on those two, if you like.
Dr Roys Emergency Moderation Team says:
https://tinyurl.com/y87gs2jb
The two authors of the paper, and Savic and Kasanin
That’s only 2 people, not 4.
Is either of them an astronomer? If so, how do you know?
PS: The very first sentence in the article contains a typo!!
How about citing some *living* astronomers, alive and working today.
Got any 3 of those?
DREMT: I googled “Savic-Kasanin theory.” First three links are about Neptune.
Where do they discuss the Moon? I don’t see any….
You are pathetic, David. The article itself is published in an astronomy journal, in 1993. So yes, the authors count as astronomers, and are presumably still alive. The article has typos and grammatical errors because English is not the native language of the authors. There is a quote from a Savic-Kasanin paper, re the Earth’s moon, in the article, page 125, 2nd paragraph. The paper is from the 60s. They might still be alive. But why shift the goalposts, anyway?
You have your 4 astronomers. You only asked for 3.
1. I dont believe the Earth is 4 billion years old. Its at most, 50 million years old.
Lord Kelvin proved it in the 1890s.
2. I also think the Universe is static, not expanding.
A fact demonstrated by Einstein in 1920.
3. There is definitely a luminiferous ether.
Maxwell demonstrated in the 1860s.
You see what I did there? I have a belief. So I go back and find one authority figure who at some point in the past agreed with me.
Then I neglect all developments on that subject, by countless others since then. Cuz they just don’t matter.
Well that’s all very interesting, Nate, and I’m sure it’s true.
But the rest of us think independently…
Great point Nate.
Lord Kelvin was a truly accomplished scientist yet he could not imagine how limited his understanding of physics was. He may have said this:
“There is nothing new to be discovered in physics now. All that remains is more and more precise measurement.”
Even if you believe that the above was actually said by Michelson it does illustrate the overconfidence of physicists in the late 19th and early 20th centuries.
While physics has made amazing strides in the last 100 years our we are still in the shallows with a mighty ocean before us.
If one could return 100 years from now it seems likely that discoveries will abound that we can hardly imagine today.
‘But the rest of us think independently’
Good, then no point in bringing up what a dude said about your belief 100 y ago.
Lagrange? That was longer ago than that. Oh, but you don’t criticize your own. Bindidon can do as he pleases.
Nate, there is absolutely no problem with me bringing up Tesla, or Bindidon Lagrange. It’s been of interest to people, and people can read the arguments without having to listen to us bicker. I do not have a belief.
Lagrange is a towering figure in mathematics and orbital mechanics. Precisely the topic being discussed. His work is still useful and undisputed today.
OK Nate.
BTW, Your buddies JD and possibly Gordon don’t believe this is correct:
‘ the satellite subdivided into minute and entirely independent parts, as dust particles, which have different orbital, but rigorously the same angular, velocities.’
They think all parts of the moon have the same velocity.
‘One must at once recognize that the kinetic energy of such an agglomeration is solely translational, there being absolutely no tendency to axial rotation.’
Well, each piece has translational kinetic energy. However when summed, the total KE is GREATER THAN 1/2MV^2, where M is total mass and V is velocity of the center of mass.
That means that there is MORE energy than just the translational kinetic energy of a Mass with velocity V.
Hmmm, what is that extra energy due to? Maybe rotation about the cm?
That could work.
Read the third paper. That could work.
I am responding to what you posted DREMT. If there is more to the argument that you left out, that is relevant to my post, feel free to explain it or quote it.
And feel free to read the third paper.
And having read it, I see that all his mathematical arguments are correct. He uses the parallel axis thm correctly. And in all his examples prove mathematically there IS indeed axial rotation and kinetic energy due to rotation.
In all of examples, such two masses in orbit linked by a string or similarly the bullets fired and linked by a string, he correctly finds that the two objects have DIFFERENT velocities. And when they leave orbit (or guns) they end up ROTATING around their center of mass.
So in the end his argument as to why the Moon doesnt rotate is completely mysterious, and doesnt agree with any of his examples.
It is truly baffling.
OK Nate.
‘So in the end his argument as to why the Moon doesnt rotate is completely mysterious.’
Yep.
Maybe you can illuminate us about what his argument in the end actually is, and why it is convincing to you?
Nate, I acknowledge that you are “baffled” by the third paper. That’s OK. Others won’t be. I acknowledge that you are desperate to discuss it with someone, and are repeatedly attempting to bait me into another long, pointless discussion. That’s OK too. Try to find someone you are less obsessed with, and try to bait and goad them into a discussion instead. That would be super-OK.
‘attempting to bait me into another long, pointless discussion. ‘
I’m sorry?
Who posted the Tesla articles and asked people to look at them? That be you. You did so because you thought they confirm your views.
You say it was not appeal to authority. But in the end you cannot explain nor defend his flawed arguments.
This is a forum for discussion. If you are unwilling or unable to discuss this topic, perhaps you should stop posting on it.
That’s some top-notch baiting, Nate, as expected. But like I said, try it on someone you are less obsessed with. You need to realize there are other people on this website to talk to besides me. If I want a two-week long pointless exchange with a professional sophist, I will let you know.
Sorry to ruin your weird fantasies, DREMT/Halp/Tesla, I have been talking with many others besides you.
This is your usual MO. Post something by someone else you admire, Tesla, Postma, etc. Urge people over and over to just look at it. But you yourself dont actually understand it.
It is puzzling why you believe arguments that you don’t actually understand.
Nate continues to bait and goad.
I suggested you read the third paper. I didn’t say I was going to spoon feed it to you.
You are the one saying you find it “baffling”, so that’s your problem. If I thought you had a genuine bone in your body, I might have helped. But you are what you are.
Stop making stuff up about me and go and bother someone else, you bizarre, desperately obsessive little creep.
Ok, DREMT. You post an article, tell people to read it. I read it and tell you what I think is wrong with it.
Anybody else here will have a response to that, or refute it, or agree.
You get all flumoxed and take it
take it personally, and look for ulterior motives.
Sad.
You don’t understand the paper, and that is supposedly my problem. You will respond until the end of time. Sad.
(Ended up explaining the papers a little further down, anyway, beginning here):
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333731
Gordon Robertson
Not sure about your coin example or if it compares to the Moon.
You have been suckered by JDHuffman. You fell for his nonsense.
Tesla was a brilliant inventor, that does not mean everything he claims is true. He is mistaken about the Moon and wrong.
Here is total evidence for you. Accept it or not, that is up to you. It is proof positive Tesla and JDHuffman are wrong. The debate is over. Science prevails, evidence wins. Unsupported declarations and JDHuffman’s pseudoscience are shown to be false.
https://www.forbes.com/sites/quora/2017/10/03/how-does-the-iss-travel-around-the-earth/#78c92a6d141f
The ISS has to be rotated on its axis to keep the same side always facing Earth.
Norman found another link that he doesn’t understand.
Nothing new.
norman…”Not sure about your coin example or if it compares to the Moon.
You have been suckered by JDHuffman. You fell for his nonsense.
Tesla was a brilliant inventor, that does not mean everything he claims is true. He is mistaken about the Moon and wrong”.
*****
Nope…I have used reasoning that I acquired at university studying engineering. We did hundreds of problems related to angular velocity alone.
Just happens JD and I agree.
Tesla also agrees with JD and me and they say that great minds think alike. Also, you claimed Tesla was brilliant.
Did I tell you I’m good looking too?
If you could roll one coin around the other, that would be representative of the Moon rotating on its axis. You can’t. You need to slide it around to keep one face of the sliding coin against the perimeter of the stationary coin.
That sliding motion is translation, not rotation.
This makes the rotation of the moon obvious, unless you are blind:
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
norman…”Imagine it this way your friend is standing in front of you. If you walk in a circle around your friend, you are revolving. If you do not change your location, but turn around to not face your friend, you are rotating. If you walk in a circle around your friend and constantly turn so that you are continually facing him, you are both revolving and rotating”.
The guy in the article is plainly wrong. You don’t need to constantly turn as you walk around the friend, all you have to do is walk sideways. Or walk straight around an orbit with your left shoulder, for example, always pointing in.
If you did what he suggests you’d be rotating quarter of a turn each instant in order to keep yourself facing the friend.
Is that what the Moon does, keep oscillating back and forth by 90 degrees as it turns?
The Moon is revolving with a constant momentum (and velocity) and it does not rotate at all as it revolves. It doesn’t have to. The Earth’s gravitational field keep nudging it around into the orbit.
That nudging is not a rotation about the Moon’s axis since the same side always points to the Earth.
Think of it as a rope attached between the Moon and the centre of the Earth. The Moon wants to go straight but it can’t because the rope won’t let it. At the same time, it can’t rotate or it would wind itself around the rope.
ps. substitute a bucket of water for the Moon on the rope. The bucket always has its handle pointing in along the rope. It can’t rotate and if it did, the water would spill out.
The water is held in by the bottom of the bucket and the sides. The water wants to go straight and the bucket pulls it into the orbit. If the bucket rotated on its COG, there would be nothing to hold the water in once the bucket rotated 180 degrees.
Gordon, has it occurred to you that the Moon is not a bucket of water with an attached rope?
Explain why the Moon isn’t rotating:
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
Gordon Robertson says:
The Moon is revolving with a constant momentum (and velocity) and it does not rotate at all as it revolves.
How does the Moon “revolve” if it does not rotate? How does it not rotate if it revolves.
Gordon, you are as stupid as a rock.
DA…”How does the Moon “revolve” if it does not rotate? How does it not rotate if it revolves.
Gordon, you are as stupid as a rock.”
**********
Revolution is orbiting you idiot.
Gordon Robertson says:
The Moon is revolving with a constant momentum (and velocity)
Good lord, did you ever even to high school?
Velocity is not a constant for a revolving object.
Hence neither is momentum.
It’s like you understand ZERO physics. None at all.
Gordon, you are dumb.
‘We did hundreds of problems related to angular velocity alone.’
Then how is it that you get translation, rotation, and angular momentum so horribly wrong?
binny…”Robertson still hadnt doscovered the feature as he wrote his most recent comment containing an NOAA link with the inevitable ncd_c sequence!”
Did you not suggest people copy/paste your link into a browser? That’s all I’m doing, but mine work.
Gordon, we all know about the units of physics. Stop pretending that your cutting-and-pasting is some kind of revelation. Like ren does.
DA…”Gordon, we all know about the units of physics.”
We????? I think you majored in Home Economics. Take that back, you’re not even smart enough to get a degree in that, although you are good at cooking up bs.
All on topic wrt general discussion of heat, work, and electromagnetic energy.
Just laying this out there, if anyone disagrees, please disagree using math and physics.
Just going over the somewhat confusing distinctions between horsepower, watts, and joules as related to heat and work.
I have seen a joule defined as a watt/sec but that makes no sense. It’s the other way around, a watt is a joule per second.
A joule is a measure of work (mechanical energy) and is measured in newton-metres (N-m).
Found this old page in the archives that show interesting definitions and relationships.
https://web.archive.org/web/20080219020838/https://www.unc.edu/~rowlett//units/siderive.html
examples:
newton (N) – force – kg·(m/s2) -> kg·m·s-2
pascal (Pa) – pressure – N/m2 -> kg·m-1·s-2
joule (J) – energy or work – N·m -> kg·m2·s-2
watt (W) – power – J/s -> kg·m2·s-3
coulomb (C) – electric charge ->A·s ….A = amps
volt (V) -electric potential – W/A -> kg·m2·s-3·A-1
Both HP and Watts are rates of doing work (power), or the rate at which energy is expended.
A newton is a force and force = ma. Since a Newton is a force it equals ma where the mass is in kilograms and a = acceleration.
Therefore, F (newtons) = Kg x acceleration = Kg-m/s^2
However, work = force x distance = N-m
a joule = F.d = N-m = (ma)-m = (Kg.a)-m = (kg-m/s^2)-m
= (Kg-m^2)/s^2
(all in the form ‘force x distance’)
A watt = joule/sec = [(Kg-m^2)/s^2]/sec = (Kg-m^2)/s^3.
You cannot reverse that and claim a joule is a watt/sec because that would have an s^4 in the denominator.
sorry this table came out with wrong formatting. Use table at link.
newton (N) force kg·(m/s2) -> kg·m·s-2
pascal (Pa) pressure N/m2 -> kg·m-1·s-2
joule (J) energy or work N·m -> kg·m2·s-2
watt (W) power J/s -> kg·m2·s-3
coulomb (C) electric charge ->A·s .A = amps
volt (V) -electric potential W/A -> kg·m2·s-3·A-1
Binny will be calling me an idiot, and rightly so,
Part 2
First, the work approach.
The horsepower is based on the British foot-pound system, and was defined by James Watt as 33,000 ft-lb /minute. Divided by 60 that is 550 ft-lb/sec.
The actual measure here is energy/sec or energy as work/sec. Therefore a ft-lb is a unit of work and can be converted to kilogram-metres, Kg-m.
Remember, a joule is work expressed in N-m, which can be expressed in Kg-m. That is a Kg can be expressed as a mass or a force.
As a force, it is equated to the force produced when a Kg mass is accelerated at a gravitational acceleration of 9.8 m/s^2.
Same with the pound, it can be expressed as a pound-mass or a pound-force, the latter requiring an acceleration factor of 32 ft/s^2.
The watt is defined using kilograms (as a force) and metres rather than feet and pounds. The standard electrical watt is defined in terms of horsepower as 1 HP = 746 watts.
Just to confuse the issue there is also a metric HP and a another definition for the watt as well.
I tried to work this out on my own by simply converting ft-lb to N-m but I was out by a factor of 10. I forgot that the pound as a mass is defined on the force of gravity and that gravitational acceleration at 32 ft/sec^2 (Imperial system) or 9.8m/s^2 (CGS system) has to be factored in when converting HP to watts.
Work = force x distance = in ft-lb or Kg-m
F = ma, therefore
Work = mass x acceleration x distance
A pound or a kilogram is a mass and to convert either to force, one has to use the standard definitions of the pound or kilogram, which are related to the gravitational acceleration of 32 ft/sec^2 or 9.8 m/s^2 respectively.
A Kg is based on the gram, which was originally defined as the AMOUNT of water (mass) in 1cc of water at the freezing point of water. The WEIGHT of that gram depends on the gravitational attraction exerted on it, therefore the weight of a pound or a kilogram of MASS depends on a gravitational field.
*****
Back to the HP.
1 HP = work/sec = (f x d)/sec = (mg x d)/sec
1 HP = 550 ft-lb/sec = (mg x d)/sec
converting Imperial to metric…
1 ft = 0.3048 m
1 lb = 0.45359237 kg
g = 9.8 m/sec^2
1 HP = (550 ft-lb/s)(0.45359237 kg/lb)(9.80665 m/s^2)(0.3048 m/ft)
= 745.69987 (kg-m^2)/s^3
~746 watts.
1 Watt = (1kg-m^2)/s^3
and the watt seems to me a metrification of the HP.
therefore 1 HP = 746 watts approx.
That’s the relationship between HP and watts, which are both measures of work rate or the rate of energy dissipation.
Of course, watts are also used in electrical dissipation. In that case, the watt is defined as the current through a resistance produced by 1 volt across the resistance per second.
The work done is equivalent to mechanical work in that a force acts over a distance on an electrical charge. The force is the voltage E and the distance is the distance over which the charge moves.
The total energy used, or consumed, is calculated in watt-hours, or kilowatt-hours.
Note that the w-hr or Kw-hr are not rates. It is not watts per hour, it is watt-hours.
About heat.
Heat is defined by the calorie, which is the amount of heat required to raise 1 gram of water by 1 degree C at such and such a temperature.
Neither HP, watts, nor joules are the native measurement for heat, they are the mechanical equivalent of heat.
The scientist Joule worked out the equivalence between work in joules and heat in calories.
4.184 joules of work are EQUIVALENT to 1 gram calorie of heat.
They gloss over that important fact in the wiki, giving the impression that heat is measured directly in joules. It’s not, a joule is only a mechanical equivalent of heat.
*****
About electromagnetic energy.
EM is measured in electron volts. That’s because it is generated by electrons with a value related to the energy difference between electron orbitals in atoms which are measured in eV. Those energy levels are also related to the frequency of the electron, which is related to its temperature.
The eV gained popularity in work related to electric charges where the energy E = qV. That is E is the energy gained by a charge when it moves through an electric field with voltage V, hence the electron-volt.
1eV = 1.6 x 10^-19 joules.
Again, this is an equivalence.
My point is that people should be very careful before they begin relating EM to heat using measures like joules. May look good on paper, but unless you understand what these terms mean, you could get yourself into deep doo doo.
The impression can be gained that EM and heat are one and the same because they can be expressed using the work equivalent of heat. They are very different forms of energy and one cannot presume that one can be added to the other, or worse still, one can be expressed in lieu of the other.
If you hook up a team of horses to a wagon and the horses pull the wagon over a distance, generating 10 HP, you can equate that 10 HP to heat. But you would not want to call the team and the wagon heat, would you?
Of course, Rumford did hook up a team of horses to a large drill apparatus so the blunted drill was turning inside metal. He had the metal inside a barrel of water, and the horses turning the drill generated enough heat to boil the water. He was the first to note a relationship between work and heat.
The scientist Joule later worked it out exactly.
Rumford’s experiment converted mechanical energy to thermal energy. You would not want to confuse mechanical energy with the thermal energy that boiled the water.
In the same way, you would not want to confuse EM with heat. They are entirely different energies and have nothing in common. You simply cannot treat both as generic energy and add one to the other.
–The scientist Joule worked out the equivalence between work in joules and heat in calories.
4.184 joules of work are EQUIVALENT to 1 gram calorie of heat.–
Joules based on air rather than water.
gbaikie…”Joules based on air rather than water”.
It was Joule who did the experiment in which the equivalence between work and heat were determined. He set up a small paddle in water and drove it with something. As the paddle turned in the water, the water heated. He calculated how much work it took to raise the temperature of the water.
Have you got a source for the air thing? Doesn’t make sense to me when they tend to name units after the scientist who developed them. The named watts after James Watt, who worked out the HP thing with 33,000 ft-lb/min = 1 HP.
Air is 1 joule per K per gram.
It could be a coincident or perhaps, K or gram came after joule.
gbaikie says:
4.184 joules of work are EQUIVALENT to 1 gram calorie of heat.
No.
1 cal = 4.184 J
*calories*, not “gram-calories.” That latter is not a unit of heat.
Gordon Robertson says:
Have you got a source for the air thing? Doesnt make sense to me when they tend to name units after the scientist who developed them. The named watts after James Watt, who worked out the HP thing with 33,000 ft-lb/min = 1 HP.
You solder circuits for a living but think in terms of hp and not Watts?
That is so strange that I highly doubt it.
DA is all messed up about “gram calories”.
(He shouldn’t attempt physics by himself. He could get hurt.)
DA…”gbaikie says:
4.184 joules of work are EQUIVALENT to 1 gram calorie of heat”.
GB didn’t say that, I said it. That equivalence comes from the scientist Joule, who devised an experiment to prove it.
correction: “I forgot that the pound as a mass is defined on the force of gravity…”
Should read: “I forgot that the pound as a WEIGHT is defined on the force of gravity…”
In other words, a pound-mass is the amount of matter in a volume, whereas a pound-force is the force gravity exerts on that mass. Or the force that mass exerts on the Earth’s surface or whatever may come between them.
If you stick a weighing scale between the mass and the surface, it will read one pound of force. The mass is exerting a force of one pound vertically onto the scale.
If you take the combo into space, where there is no gravitational attraction, the scale will measure nothing. The mass will still consist of 1 pound of matter but it will have no weight.
GR: there is no such thing as “pound-mass.” There is mass, and there is a pound. If you don’t know the difference go back to high school.
DA…”GR: there is no such thing as pound-mass. There is mass, and there is a pound. If you dont know the difference go back to high school”.
You have already proved you’re an idiot, no need to embellish it.
There is a pound, there is mass, and there is weight. Weight is a force due to gravity. A pound-force is a force due to gravitational acceleration of a mass (weight). A pound-mass, is the amount of matter in a mass of one pound.
A pound of mass in space would have no weight but would still be a pound-mass.
Somebody’s a slug
Horsepower is an archaic unit that no one uses anymore, except maybe musclehead car owners.
Power is now expressed in watts.
DA, obviously you have no experience in the real world. Consequently, you may not know “HP” stands for “horsepower”. It is how electric motors are rated.
Face reality, and learn some physics.
https://www.baldor.com/catalog#category=110
hp is an archaic unit. The rational world, except for a few backward Americans, uses watts.
DA believes “denial” is a river in Egypt….
Cliche much?
DA…”Horsepower is an archaic unit that no one uses anymore, except maybe musclehead car owners.
Power is now expressed in watts”.
That right? In the electrical industry, motors are rated in HP. I have never heard of a 1 HP motor referred to as a 746 watt motor. In fact, I don’t recall hearing the word watt used to measure anything but electrical power.
JD…”Face reality, and learn some physics.
https://www.baldor.com/catalog#category=110”
Baldor is now Allan-Bradley, is it? Interesting.
The “Standard” when I was in high school was the CGS (Centimeter, Gram, Second) system. It really stunted my growth when the MKS system took over while I was was at university.
Some CGS units I never liked (e.g. dynes and ergs). Give me Joules and Watts!
However when it comes to mechanical units I am still having trouble with Newtons and Pascals.
Are any of you old enough to sympathize with me?
Robertson
I hate to insult persons, but you are the origin of my reaction.
You are really the dumbest commenter on this site: you are not even able to carefully read a comment let alone the associated documents.
No, Robertson: neither Galilei not Cassini nor d’Alembert nor Mayer let alone the famous mathematician Lagrange did confound libration and rotation of the Moon around its axis.
What they wrote is that this rotation of the Moon around its axis is a consequence of what they observed, analysed and calculated concerning Moon’s libration.
How can a simple person line you doubt about the accuracy of Lagrange’s equations?
That is the problem with pseudoskeptics like you and JDHuffman: you simply pretend
– this mathematician is wrong;
– prove me wrong
instead of YOU (!) proving him wrong.
This is inimaginable.
I just looked at
https://teslauniverse.com/nikola-tesla/articles/famous-scientific-illusions
and got a good laugh.
I have lots to do in other contexts, but your primitive, stubborn and arrogant denialism disturbs me so much that I will translate some parts of Lagrange’s work as published in
https://gallica.bnf.fr/ark:/12148/bpt6k2292245/
and present the translation on this site.
I’m sure you and your pseudoskeptic friend JDHuffman both will continue to deny the fact, but that is a secondary aspect.
Bindidon, if you left out all of the insults, and links that you don’t understand, you wouldn’t have anything in your comment.
But, you likely knew that. That’s why, like Norman, you must include the insults and links.
Nothing new.
I understand what is behind the links, JDHuffman.
And I guess that, unlike Robertson, you perfectly know that Moon’s libration is not at all the kernel of Lagrange’s work.
Bindidon, if you understood the links, and if Lagrange was not about “rotating on its own axis”, then why did you link to it?
binny…”What they wrote is that this rotation of the Moon around its axis is a consequence of what they observed, analysed and calculated concerning Moons libration.
How can a simple person line you doubt about the accuracy of Lagranges equations?”
**********
Simple person??? Me, JD, and Tesla all agree on this problem, that the Moon does not rotate on its axis. Surely a simply person could not be in agreement with Tesla?
I have not claimed laGrange is wrong I am claiming the application of his equations to rotation about an axis is wrong.
LaGrange is known for several basic principles related to calculus. One of them is related to F(b) – F(a) always being positive along the range of x=a to x=b with b>a. That’s the basis of establishing a derivative at a point on a function.
He also developed series, in which a polynomial is expressed as a series of parts related to their powers. He introduced a means of computing the maxima and minima of a function, along the lines of Newton-Rhapson.
The only related math he produced related to this Moon problem, was his work on surface area. That fits libration but has nothing to do with angular velocity. AFAIAC,the article deals solely with libration even though they do natter on about the Moon’s rotation.
I am picking them up as thinking the Moon rotates back and forth producing libration, which is nonsense.
Once again, you are appealing to authority.I don’t see anything in the equations of LaGrange that apply to this problem. They may apply to libration but not to the angular momentum about it’s own axis that would have to be proved for the Moon to rotate about said axis.
In fact, in the article, I saw no application of LaGrangian math, just a reference to it.
Why would libration have anything to do with rotation about an axis?
binny…”I have lots to do in other contexts, but your primitive, stubborn and arrogant denialism disturbs me so much that I will translate some parts of Lagranges work as published in
https://gallica.bnf.fr/ark:/12148/bpt6k2292245/
and present the translation on this site”.
Still waiting for the translation but I would save my breath if I were you. The title of the book says it all, The Theory of the Libration of the Moon”. Even my high school French made that clear.
The book is about libration, NOT rotation of the Moon. Maybe that’s why you have not gotten back to us, you have read it further and realized that.
I did look up LaGrange points, which the book is likely about. LaGrange is describing points in the orbit of a four body system where the effects of gravity of the other three affects the orbit or trajectory of a 4th object.
If you look at the Sun, Earth, and Moon orbits, there are several points, according to LaGrange where the gravitational fields of either body affect another body differently.
For example, when the Moon is on the far side of the Earth from the Sun, the Sun’s gravitational effect on the Moon ‘should’ be different than when the Moon is on the Sun side of the Earth.
That has nothing to do with the Moon allegedly rotating on its axis.
Gordon Robertson says:
For example, when the Moon is on the far side of the Earth from the Sun, the Suns gravitational effect on the Moon should be different than when the Moon is on the Sun side of the Earth.
Not by much.
Why don’t you calculate the difference for us, Gordon?
DA, learn some physics.
Then you wouldn’t have to ask such stupid questions.
Why don’t you calculate the difference for us, Ger*an?
Simple calculation, DA.
If I do it for you, will you promise not make a comment on this blob, again?
Not at all. Why do you need credit just for doing a high school-level calculation?
Another stupid question brom DA.
That’s all he has.
brom indeed.
I know why you avoid questions….
Wow, DA finds a typo!
No knowledge of physics, just typing!
Same as someone else….
Everyone here knows why you avoid questions.
I avoid your stupid questions, as most everyone here does.
I ask very good questions. You can’t answer them just like other deniers can’t.
Wrong DA, you make up your own “reality”.
You are opposed to truth, facts and logic.
DA…”Why dont you calculate the difference for us, Gordon?”
Don’t have to, LaGrange already did it well.
I wonder if he knew ZZTop?
Don’t forget to have “a good laugh” at these two papers Tesla wrote in response to criticisms of the first paper. Although, as with the first, it might be more beneficial if you read them rather than laughed at them. You wouldn’t want to come across as “an arrogant boaster”.
https://teslauniverse.com/nikola-tesla/articles/moons-rotation
https://teslauniverse.com/nikola-tesla/articles/moons-rotation-0
I have read these nice elucubrations.
Did you understand what Lagrange wrote?
Bindidon, what did Lagrange write that is relevant to “rotating on its own axis”?
JD…”Bindidon, what did Lagrange write that is relevant to rotating on its own axis?”
I think he said, but don’t quote me, that he agrees with Tesla.
Then he got it right.
☺
If the Moon weren’t rotating, it wouldn’t always face the Earth:
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
DA, learn orbital motion.
Or, remain a physics-deprived troll.
Your choice.
Does a rotating body always point in the same direction?
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333177
You avoid questions because you don’t have the courage of your convictions.
DA is running out of stupid questions, so he resorts to false accusations.
Typical.
Why are you afraid to answer questions? You do anything to wiggle out of them.
Another stupid question.
Keep dancing. It’s all you’re good at.
David, you don’t even seem to understand your own argument, let alone anyone else’s. Further upthread, you say to me:
“Whoever says it was wrong — the Moon rotates about both its polar axis and the Earth-Moon center of mass. (And the Earth also rotates about *its* polar axis and the Earth-Moon center of mass.“
Those who insist our moon has two motions (the “Spinners”) actually think that the moon TRANSLATES about the Earth-Moon center of mass, and ROTATES on its own axis. That is your argument. There, now you should understand it. You’re welcome.
Obviously, if you believed that our moon both rotated about the Earth-Moon center of mass AND rotated about its polar axis, then you would believe that we can see all sides of the moon from Earth.
The simplest way to describe the moon’s motion is as ONE rotation about the Earth-Moon center of mass. If you define that off-center rotation as translational motion, as Tesla does (with full explanation as to why in the third paper), then axial rotation of the body is separate and independent of that motion.
Dr Roys Emergency Moderation Team says:
The simplest way to describe the moons motion is as ONE rotation about the Earth-Moon center of mass.
Describe the difference between an orbiting moon that IS rotating, and one this is NOT rotating.
Your tidal-locking moon GIF on the right vs. the one on the left, respectively. The one on the right is rotating on its own axis CW once per CCW orbit.
Our tidally-locked moon, on the left, is orbiting but not rotating on its own axis.
Coming soon: though retired I have much more to do every day than simply explaining to anglo-saxon pseudoskeptics how to find it out.
Luckily, there is this wonderful tool called ‘Google Translator’. All but perfect, but.
binny…”Luckily, there is this wonderful tool called Google Translator. All but perfect, but”.
First of all, you need to be able to copy the text to paste it into Google. Your link leads to an article in jpeg format where the text cannot be copied directly. I’d have to copy the jpeg then use a translator on the Net to convert it to text.
Did not need to, I figured it out indirectly. It’s likely about LaGrange points not rotation. I base that on the title which is about libration and hid LaGrange points are about libration.
Robertson
Once more you prove how thoroughly incompetent you are.
The link here
Historical origin: https://gallica.bnf.fr/ark:/12148/bpt6k2292245/f4.highres
was only to show at the document’s begin!
If you were a bit more intelligent, you would simply extract
https://gallica.bnf.fr/ark:/12148/bpt6k2292245/
out of the front page, instead of trumpeting your redundant blah blah.
The following pages are exactly what I translated using Google’s tool.
How is it possible to be so dumb?
binny…”Did you understand what Lagrange wrote?”
I do now, but like JD said, it has nothing to do with the Moon rotating on its axis. LaGrange is talking about the effect of three orbiting bodies on a 4th body’s orbit or trajectory based on their relative positions.
Now some wag at NASA, a theoretical physicist, is calculating a freeway through the planets using those LaGrange points. I say, put your foot down on the accelerator and forget them.
GR says:
“Now some wag at NASA, a theoretical physicist, is calculating a freeway through the planets using those LaGrange points. I say, put your foot down on the accelerator and forget them.”
You wouldn’t if you had to pay the cost of the fuel.
Salvatore du Prete
Sorry, your Grand Minimum may not happen.
https://earthsky.org/space/solar-cycle-24-25-sunspot-predictions
more modelers spewing pseudo-science. They are likely funded by the eco-alarmists.
“[Wei-Hock “Willie” Soon] has accepted more than $1.2 million in money from the fossil-fuel industry over the last decade while failing to disclose that conflict of interest in most of his scientific papers. At least 11 papers he has published since 2008 omitted such a disclosure, and in at least eight of those cases, he appears to have violated ethical guidelines of the journals that published his work.”
NY Times, Deeper Ties to Corporate Cash for Doubtful Climate Researcher, 2/22/15
https://www.nytimes.com/2015/02/22/us/ties-to-corporate-cash-for-climate-change-researcher-Wei-Hock-Soon.html
Translation. David Appell and his Alarmist friends can’t debate Willy Soon’s brilliant presentations so they resort to “Ad Hominem”, accusing him of being a shill in the pay of the fossil fuel industry.
The Alarmists on the other hand can’t be criticized for accepting billions from the leftists who distribute our tax dollars their lick spittle researchers.
It is a disaster when politics and religion ride together. It is also a disaster when politics and “Science” ride together. We live in a time where junk science informs public policy on a scale that would delight Lysenko.
Every time I post something the climate shysters return with ad hominems , there is always something wrong with the person who made the post , something wrong with the website it was posted on , something wrong with people commenting on it, etc ,so I will do what they do now .
In India where this study come from about half the population defecate out in the open – that’s the level of their development and capabilities as they can’t even built an outhouse over the hole in the ground,
I would not pay any attention to “climate science” coming from that country.
Eben, how does human defecation compare to non-human animal defecation in India? Wildlife defecation? How does open defecation affect climate? Climate change?
DA, that’s 4 stupid questions on one comment.
You’re getting worse as you get older.
Theorem: Calling a question “stupid” is a lot easier than answering it.
Theorem: Making up theorems is a lot easier than learning the valid ones.
Yet again, you won’t answer a legitimate question because you know what the answer implies. It’s a old pattern with you.
DA, learn some physics, and face reality.
Then you won’t have to resort to stupid questions and false accusations.
Or, stay a pathetic troll.
Always your choice.
Still again. I’m glad I don’t have to dance around questions and can answer any that come.
DA, all you have is your “dance”.
You can’t deal with reality.
In northern hemisphere, about 40% is in tropical zone, 52% in temperate zone, and 8% is in the arctic zone. And obviously, the same goes for southern hemisphere.
What is the average temperature of the northern hemisphere?
Does northern hemisphere have higher or lower average temperature as compared to southern hemisphere.
As guess I would say the northern tropics average temperature is about 26 C, temperate zone is about 11 C and arctic is about -15 C.
And southern hemisphere about the same but southern polar region is about -30 C.
If it was the same except cooler southern polar region, then Southern hemisphere would be slightly cooler.
But I don’t think it’s exactly the same, and I would tend to guess that southern hemisphere is warmer than northern hemisphere.
So google it:
“If global warming were a race, the Northern Hemisphere would be winning. It is warming faster than the Southern Hemisphere, with some of the most rapid warming rates on Earth located in the Arctic, where sea and land ice is shrinking and thinning. Not only is the North winning now, but projections show that, largely due to the influence of manmade greenhouse gas emissions, it is likely to widen its lead in the coming decades.”
http://www.climatecentral.org/news/in-global-warming-northern-hemisphere-is-outpacing-the-south-15850
Which is interesting as I was also interested in question of what regions are going to warm the most- I was thinking tropics vs temperate and/or arctic zones.
But is there agreement that northern hemisphere will warm faster than southern?
Or if global temperature increase by 1 C, will most of warming be in northern Hemisphere?
And then, which part of northern hemisphere- such as the arctic region?
So are ever going to get point having polar sea ice being ice free during summer, and/or are the thousands years old frozen tree stumps going get some trees starting to grow around them?
But at the moment is southern hemisphere have average of about 14 C and northern hemisphere about 16 C.
Or is there less difference or more difference?
Though I continued to google:
“In Earth’s present-day climate, the annually-averaged surface air temperature in the Northern Hemisphere (NH) is ? 1.5C higher than in the Southern Hemisphere (SH). This interhemispheric temperature difference has been known for a long time, and scientists have pondered over its origin for centuries. Frequently suggested causes include differences in seasonal insolation, the larger area of tropical land in the NH, albedo differences between the Earth’s polar regions, and northward heat transport by the ocean circulation.”
https://www.researchgate.net/publication/258778329_Why_is_the_Northern_Hemisphere_warmer_than_the_Southern_Hemisphere
Hmm, I didn’t know it was known for a long time, though obviously south polar region has been know to colder for long time. As is has been known that Antarctica is global cooling effect.
Next item:
Why is the Northern Hemisphere Warmer than
the Southern Hemisphere?
ABSTRACT
The question of why, in the annual mean, the Northern Hemisphere is warmer than the Southern Hemisphere is addressed, revisiting an 1870 paper by James Croll. We first show that the ocean is warmer than the land in general which, acting alone, would make
the Southern Hemisphere warmer because of its greater fraction of ocean. Croll recognized this and thought it was caused by greater humidity and greenhouse trapping over the ocean than over the land. However, for any given temperature, greenhouse trapping is actually greater over the land than the ocean…”
what else?
“Over the last 1000 years, temperature differences between the Northern and Southern Hemispheres were larger than previously thought. Using new data from the Southern Hemisphere, researchers have shown that climate model simulations overestimate the links between the climate variations across the Earth with implications for regional predictions.”
https://www.sciencedaily.com/releases/2014/03/140330151320.htm
And:
Why is the southern hemisphere warmer than the northern?
https://socratic.org/questions/why-is-the-southern-hemisphere-warmer-than-the-northern
Why is the southern hemisphere warmer than the northern?
Present your data.
“David Appell says:
December 11, 2018 at 7:18 PM
Why is the southern hemisphere warmer than the northern?
Present your data.”
Apparently, it isn’t.
But I thought the region around 30 degree south would be warmer, and this would enough off set the colder southern polar region- due to this general region being larger than polar region.
But it seems northern hemisphere get ocean warmth from Southern hemisphere, and apparently it is significant enough to make some difference.
gbaikie says:
The question of why, in the annual mean, the Northern Hemisphere is warmer than the Southern Hemisphere is addressed, revisiting an 1870 paper by James Croll. We first show that the ocean is warmer than the land in general
What volume of land are you using for comparison?
Otherwise you’re comparing a volume to an area, which doesn’t make any sense in terms of temperature.
Moon’s rotation around its axis
Here is a translation of Lagrange’s work published 1780 in Berlin
https://tinyurl.com/y7e67j42
Of course: only the informal part of Lagrange’s paper has been translated. The equations are here:
https://gallica.bnf.fr/ark:/12148/bpt6k2292245/f16.item
The formal stuff begins below ‘SECTION PREMIERE’.
*
The translation was performed using Google Translator, and I hadn’t very much motivation to correct the text a posteriori.
As indicated in the pdf file: though the title is on libration, the contents deal about what everybody in 1750 considered be the main origin of libration, namely the equality in time of Moon’s rotation around its axis with Moon’s revolution around Earth.
**
For the lovers of Belgrade’s school (‘It is only the gravitation’):
http://www.doiserbia.nb.rs/img/doi/1450-5584/2013/1450-55841301135T.pdf
Bindidon, you cannot assume that the Moon is rotating on its own axis, in sync with its orbit, then observe that it only keeps one side facing Earth, and then conclude your assumption was correct!
At least you get to learn about “circular reasoning”.
https://en.wikipedia.org/wiki/Circular_reasoning
No assumptions. The Moon is obviously rotating, because it is not pointing in a constant direction.
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
DA, if you had ever studied physics, you would have learned about gravity, and how it affects orbital motions.
Maybe someday….
Is an object that always points in the same direction rotating?
You don’t even know how to ask a meaningful question. All you can do is ask stupid questions.
Let’s see if you can phrase that question correctly.
Why do you always decline to answer questions?
I only refuse to answer stupid questions, DA.
I always answer responsible questions.
I think you’re afraid to answer my questions, because you know what they imply.
I think you are unable to phrase your question so that it makes sense.
That’s why you can only ask stupid questions. You don’t know enough physics to ask a responnsible question.
You’re afraid to answer questions no matter how I phrase them.
Are your pants on fire?
You’re wasting time. Let me know when you have the cojones to answer questions like a man.
Wrong DA, you are the one wasting time. You are the one faking it. All you have are your tricks. You use stupid questions, erroneous and irrelevant links, and false accusations, to avoid reality.
Your question is not worded properly. It’s just another stupid question. You don’t know enough physics to phrase it correctly. And, you won’t try to learn.
Here, you demonstrate you don’t even understand simple units:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333157
How many examples of your incompetence do we need?
JDHuffman
David Appell is correct about you. You do not answer questions at all. You avoid them. If you attempted to answer them others would see how fake you are. Avoiding answering questions gives you two supporters the idea that you have something of value to contribute.
Here is a question I have asked you that you provide no answer for. You have a cartoon drawing you post about the blue/green plate radiant energy exchange. I have asked you to link to any valid science that supports this nonsense you made up. You will not do it. It is a valid and good question. You avoid it because your supporters would see how phony you actually are. Most know you are a total fraud and don’t have the slightest clue of any physics. You fake it for a couple of crackpots who post and then go around avoiding all questions and faking your way through a string of comments.
Will you ever answer any questions asked of you. I don’t think so, you are not capable of doing it. Pretending is more fun for you.
Norman, you only insult, falsely accuse, and misrepresent because you don’t understand the relevant physics.
No wonder you worship DA….
norman…”You have a cartoon drawing you post about the blue/green plate radiant energy exchange”.
Put your money where your mouth is and prove there is an energy exchange.
There is no thermal energy exchange for the simple reason that heat cannot be transferred from a colder body to a warmer body without some kind of external compensation.
There is no EM exchange because a hotter body cannot absorb EM from a cooler source.
This is all backed up by the 2nd law yet you, swannie, and Eli Rabbett are arguing that the 2nd law is null and void, that heat can be transferred from a cooler body to a warmer body.
Prove it!!!! Put out a paper on it, if you’re right, you’ll get a Nobel prize.
You are seriously confusing blackbody theory which does not apply. The basis of BB theory, the Stefan-Boltzmann equation, supplies no provisions for a two way heat transfer or a two way EM exchange. S-B applies only to a one way transfer of both, from hot to cold.
Any equation you see to the contrary, related to S-B, is pseudo-science, unless applied at thermal equilibrium where there is no heat transfer.
DA…”I think youre afraid to answer my questions, because you know what they imply”.
We know how you operate. You pose a dumb question and when someone tries to answer it you throw out an even dumber question. Or you ask for peer reviewed papers.
You are unable and unwilling to engage in an intelligent conversation based on physics. You respond solely with ad homs and pseudo-science about radiation from the Earth raising the temperature of the Sun. Or repeatedly asking dumb questions about how heat gets to Earth from the Sun.
In your charade as a science writer, you interview only alarmist scientists, completely ignoring skeptics with integrity, like Roy and John at UAH. Of course, they’d likely decline an interview with you based on your antics on this blog.
Recently, you have taken shots at Willie Soon, based purely on alarmist dogma. Willie is right in what he claims and no amount of alarmist denial will change that.
JDHuffman
You do actually confirm David Appell is totally correct about you. Once again you don’t answer a good and valid question.
Instead of answering my question which is for you to provide valid supporting physics for you cartoon radiant heat transfer this is how you respond:
“Norman, you only insult, falsely accuse, and misrepresent because you dont understand the relevant physics.”
A very weak and meaningless post that absolutely does NOT even attempt to answer the question posed. I have interacted with you a lot and you will post several times and not once provide valid information.
This time is no exception.
DA…”Is an object that always points in the same direction rotating?”
Another dumb question. If you had even the slightest understanding of rotation you’d know it is defined by the angular velocity/momentum of a body about an axis, not the direction in which it points.
Even at that, there are different angular velocities applied based on whether a mass is a particle or a rigid body made up of particles. One is an angular velocity related to a particle, which would be it’s tangential velocity to a curve. The other is the angular velocity of a radial line through a rigid body.
Several people posting here have gotten that confused, claiming that particles further away from the axis of rotation are moving at higher velocities. That’s true on the Earth as well but they all turn 360 degrees in 24 hours.
How can anyone engage in a discussion with you when your understanding of basic physics is so flawed?
“Norman, you only insult, falsely accuse, and misrepresent because you don’t understand the relevant physics.”
Norman,
Here’s another one that can’t give a straight answer.
He came up with the correct plate temperatures, then went into circular evasive maneuvers.
http://www.drroyspencer.com/2018/11/the-sorry-state-of-climate-science-peer-review-and-kudos-to-nic-lewis/#comment-331046
Incorrect, liar.
http://www.drroyspencer.com/2018/11/the-sorry-state-of-climate-science-peer-review-and-kudos-to-nic-lewis/#comment-333304
Gordon Robertson says:
“the Stefan-Boltzmann equation, supplies no provisions for a two way heat transfer or a two way EM exchange.”
With P for power [W], e for epsilon, s for sigma:
Pt = Ph-Pc = esATh^4 – esATc^4 = esA(Th^4 – Tc^4)
Ph goes from hot to cold. It is not heat.
Pc goes from cold to hot. It is not heat.
Pt goes from hot to cold. It is heat.
DREMT, no responsible person would fall for the nonsense Svante spews. He/she is just another clown addicted to pseudoscience.
The prisons are filled with addicts….
He/she/it is certainly an exceptionally relentless troll.
So what are the correct blue/green plate temperatures DREMT?
Svante, so glad you asked.
https://postimg.cc/image/jcotys8e3/
DREMT?
Svante, please stop trolling.
No answer, afraid of contradicting granddad?
Svante, I already told you my exact position on the plates, BEFORE we even got into that discussion. What the hell is wrong with you? What does it say at the bottom of this comment I have linked to? Look at the date of the comment. For crying out loud you are a CLOWN.
http://www.drroyspencer.com/2018/11/the-sorry-state-of-climate-science-peer-review-and-kudos-to-nic-lewis/#comment-330706
The “thought exp.” is Eli’s plates. Just so you don’t get “confused” somehow.
What plate temperatures do you get in the thought experiment then?
😂
No straight answer from DREMT.
Incorrect. You have had your answer,
No formulas, no numbers => no physics.
OK Svante.
DA…”The Moon is obviously rotating, because it is not pointing in a constant direction.”
Draw an imaginary line through the dark spot to the other side of the left gif and watch that line as the Moon orbits. It does not turn about the Moon’s centre.
You see it turning because you have programmed yourself to see it. Snap out of your dream world and watch it closely.
Gordo, suppose you are living on the surface of the Earth at some mid-latitude location. Ever day, the light from the Sun shines during the day and it’s dark at night, indicating that the Earth rotates once every 24 hours relative to the Sun.
Now, since the Moon is tidally locked to the Earth, one can easily see that if one were situated on the Moon on the side facing the Earth, the sunlight would illuminate that location with a similar cycle as that on Earth, only that the period is almost a month between dates of maximum illumination.
The obvious conclusion is that the Moon rotates relative to the Sun, just as does the Earth, that is, the Moon rotates about it’s axis.
Wrong, Swanson.
The Moon orbits Earth. That results in the rotation relative to the Sun.
Huffingman, “Rotation” must be defined and measured relative to some fixed reference. Ultimately, the basic reference is what’s called an “inertial reference frame”, that is to say, a coordinate system which is NOT ROTATING. Such a reference frame can be created by using three gyroscopes mounted orthogonally, which is to say, one each mounted with axies of rotation pointing along one of the three axies of a 3D coordinate system.
The dynamics of inertial reference frames is central to the navigation of satellites and ballistic missiles, which are by now very well understood problems in physics. An inertial measuring system placed on the surface of the Moon would show that the Moon rotates in that inertial reference frame.
All your posts about orbits are pure crap with no basis in physics. You really need to LEARN SOME PHYSICS as I did when I took the graduate level dynamics course while working to build satellites almost 50 years ago. You are nothing but a useless troll, wasting everyone’s time.
Swannie River, please stop trolling.
Swanson, I’m glad to see you finally expose yourself as an unprofessional incompetent. Your unscientific “experiment” was funny, but only fooled the fools.
Now, you’re just another useless clown, entertaining us with your arrogant pomposity.
Nothing we haven’t seen before.
Huffingman, You offer no physics to refute what I wrote, just more bluster and ad hominem. Your ignorance is profound, as you continue to cling to your lunatic physics, insisting that all the other scientists who have studied astronomy and astrophysics are incorrect. Sad but expected from one who continually proves that he/she is a brainless troll.
Emotional Swansong, please stop trolling.
binny…”Here is a translation of Lagranges work published 1780 in Berlin….”
With all due respect to LaGrange, I think he is seriously mistaken. Another example of someone getting lost in the math and failing to see the reality.
Same thing happened with the orbit of Mercury, in which Mercury seemed to move backwards at certain times. It’s an illusion due to relative motion. Libration is similar, at certain orbital points the Moon appears to be rotating slightly to expose more of its surface. It’s not rotating at all, it’s the human mind being sucked in by relative motion.
Tesla proved the Moon does not rotate about its axis using washers. I proved it myself independently using coins. Like Tesla, I arrived at the conclusion that the apparent rotation is a complete illusion.
If the same face of the Moon always faces the Earth, that Moon face cannot turn through the 360 degrees required to complete a rotation.
As Tesla pointed out, the motion of the Moon is translation. If you took it’s orbital path and straightened it into a line, you’d get exactly the same motion. In order for the Moon to rotate while translating across the line, it would have to roll.
You can test that directly using two coins and marking them both at one point. The only way to get one coin around the other coin while keep its mark on the other coin is to slide it around. The moment you roll it, replicating rotation, the mark has to leave the other coin’s perimeter.
You have to remember the reality. The Moon is trying to move in a straight line at considerable speed. Gravitation force holds onto it like a rope attached to the same side and force the Moon into an orbit.
As I have pointed out several times, in order for the Moon to rotate it would have to wrap itself around the rope.
Gofdo, You need to read Tesla’s second note again. He points out that when the outer mass is released, it exhibits two different velocities, with the inner point moving slower than the outer point. The result is the center of mass moves along a straight line with the velocity being the average of that for those two points, but the body is also rotating about it’s center of gravity. You have, yet again, confused translation with rotation.
“The rotation is, however, not due to an exclusive virtue of angular motion, but to the fact that the tangential velocities of the masses or parts of the body thrown off are different.”
Yes Tesla has a very odd story here.
If he looks at the parts of a rotating object, he can show that each piece is TRANSLATING.
Therefore, he proves that motion of ANY rotating object is just a sum of translating pieces! Duh.
Hence the moon behaves like any other rotating object.
‘ the tangential velocities of the masses or parts of the body thrown off are different.’
Thats pretty much the definition of a rotating object.
So Tesla is saying “The rotation is, however, not due to an exclusive virtue of angular motion, but to the fact that” it is a rotating object!
“The moon does rotate, not on its own, but about an axis passing thru the center of the earth, the true and only one.”
Gordon,
“in order for the Moon to rotate it would have to wrap itself around the rope.”
Gordon, if the rope is tied on its other end to a fixed object (like a star). It will end up wrapping around the moon.
Thus relative to the inertial frame of the stars, the moon is clearly rotating.
There is no way around it (pun intended).
“As will be seen, we arrive at precisely the same results whether the movement is rectilinear or in a circle. In both cases the total kinetic energy can be divided into two parts, respectively of the same numerical values, but there is an essential difference. In angular motion the axial rotation is nothing more than an abstract conception; in rectilinear movement it is a positive event.”
Tesla discusses two balls orbiting at different radii from a center point held by strings,
‘Let us now inquire what will happen if the two masses are rigidly joined, the connection being assumed imponderable.’
This is like the near and far side of the moon.
‘The moment the strings are broken and they are thrown off they will begin to rotate’
The moon, with its gravity cut, will behave similarly.
Hence, his statement that ‘the axial rotation is nothing more than an abstract conception’ makes no sense.
“In estimating the kinetic energy of a rotating mass in any of the ways indicated we arrive, thru suitable conceptions and methods of approximation, at expressions which may be made quantitatively precise to any desired degree, but do not truly define the actual condition of the body.”
On the use of the parallel axis theorem specifically:
“Still more remote from palpable truth is the equation of motion obtained in the manner indicated in Fig. 4, in which the first term represents the kinetic energy of translation of the body as a whole and the second that of its axial rotation. The former would demand a movement of the mass in a definite path and direction, all particles having the same velocity, the latter its simultaneous motion in another path and direction, the particles having different velocities. This abstract idea of angular motion is chiefly responsible for the illusion of the moon’s axial rotation, which I shall endeavor to dispel by additional evidences.”
Then you have the discussion of Fig. 5, the contraption with the spokes and rotating balls with adjustable friction, from complete freedom of movement up to “unable to rotate on their own axes”…
As Galileo showed, describing a cannon ball’s motion as a combo of falling and horizontal motion is USEFUL. It had predictive value.
Nonetheless, Tesla doesnt like describing the Moon’s motion as a combo of rotation plus translation. He thinks the rotation is abstract or conceptual, not real in his opinion.
But he agrees with spinners when he says that the motion and kinetic energy of the Moon CAN BE described MATHEMATICALLY as a combination of a translation part, PLUS a rotation about-its-own-axis part.
He also shows that objects orbiting just like the moon, continue to rotate after leaving orbit, for example:
‘If a metal ball, attached to a string, is whirled around and the latter breaks, an axial rotation of the missile results which is definitely related in magnitude and direction to the motion preceding. By way of illustration — if the ball is whirled on the string clockwise ten times per second, then when it flies off, it will rotate on its axis ten times per second, likewise in the direction of a clock.’
So the rotational part seems quite real. Clearly, describing things this way has has been quite useful for landing things on other planets.
PS, Tesla clearly disagrees with you guys on the hammer throw.
“To begin with, observe that when a mass, say the armature of an electric motor, rotating with the angular velocity ω, is reversed, its speed is -ω and the difference ω (-ω) = 2 ω. Now, in fixing the ball to the spoke, the change of angular velocity is only ω; therefore, an additional velocity ω would have to be imparted to it in order to cause a clockwise rotation of the ball on its axis in the true significance of the word. The kinetic energy would then be equal to the sum of the energies of the translatory and axial motions, not merely in the abstract mathematical meaning, but as a physical fact. I am well aware that, according to the prevailing opinion, when the ball is free on the pivots it does not turn on its axis at all and only rotates with the angular velocity of the frame when rigidly attached to the same, but the truth will appear upon a closer examination of this kind of movement.
Let the system be rotated as first assumed and illustrated, the balls being perfectly free on the pivots, and imagine the latter to be gradually tightened to cause friction slowly reducing and finally preventing the slip. At the outset all particles of each ball have been moving with the speed of its center of gravity, but as the bearing resistance asserts itself more and more the translatory velocity of the particles nearer to the axis O will be diminishing, while that of the diametrically opposite ones will be increasing, until the maxima of these changes are attained when the balls are firmly held. In this operation we have thus deprived those parts of the masses which are nearer to the center of motion, of some kinetic energy of translation while adding to the energy of those which are farther and, obviously, the gain was greater than the loss so that the effective velocity of each ball as a whole was increased. Only so have we augmented the kinetic energy of the system, not by causing axial rotation of the balls. The energy E of each of these is solely that of translatory movement with an effective velocity Ve as above defined such that E = M Ve2. The axial rotations of the ball in either direction are but apparent; they have no reality whatever and call for no mechanical effort.“
On an unrelated matter, Tesla didn’t think the moon would rotate on its axis if the gravitational “string” between the Earth and the moon were cut. He thought this because (and this is from the end of the second paper, not the third this time):
“But from the character of motion of the satellite it may be concluded with certitude that it is devoid of momentum about its axis. If it be bisected by a plane tangential to the orbit, the masses of the two halves are inversely as the distances of their centers of gravity from the earth’s center and, therefore, if the latter were to disappear suddenly, no axial rotation, as in the case of a weight thrown off, would ensue.”
It’s because, he says, the mass of the half of the moon nearer to Earth is greater than the further half.
‘Only so have we augmented the kinetic energy of the system’
TRUE.
‘not by causing axial rotation of the balls.’ ‘The energy E of each of these is solely that of translatory movement’
Unproven assertions.
If he had bothered to consider ANGULAR MOMENTUM of this system, he would have found that tightening the balls produced an increase in angular momentum of the system.
Unlike kinetic energy, angular momentum is undeniably ANGULAR motion. It cannot be mistaken for LINEAR momentum.
And because the angular momentum has increased, a TORQUE was required. Again Torque is undeniably an angular quantity.
“If he had bothered to consider ANGULAR MOMENTUM of this system, he would have found that tightening the balls produced an increase in angular momentum of the system.”
He did.
‘ the masses of the two halves are inversely as the distances of their centers of gravity from the earths center’
This makes little sense.
However he chooses to slice up the Moon, the same slicing can be done to the steel ball.
The steel ball, like the Moon, is a solid sphere, is orbiting while keeping the same face to the center, and held in orbit by a centripetal force.
Therefore their behavior when the centripetal force is removed should be the same.
If the steel ball rotates about its own axis as it leaves its orbit, then the Moon should do the same.
This own-axis rotation as it leaves orbit is DIRECT EVIDENCE that it had own-axis rotation while in orbit, since cutting the centripetal force produces no torque.
“However he chooses to slice up the Moon”
Precisely in half, by a plane tangential to the orbit. His point was he thought the nearer half to the Earth was of greater mass than the farther, unlike with the steel ball.
‘His point was he thought the nearer half to the Earth was of greater mass than the farther, unlike with the steel ball.’
Dubious assertion, but if true the difference is rel. small. And how would this small difference so radically change its motion after gravity is cut?
Not obvious. Shows you he is capable of making a flimsy argument.
In any case, you guys have been arguing that ANY mass (horse, car, train, steel balls, etc) with the same kind of motion as the Moon, will not be rotating about its own axis while in orbit, or after leaving orbit.
Here Tesla is arguing that steel balls and moons behave differently. This is contradiction.
He did not discuss ‘that tightening the balls produced an increase in angular momentum of the system.’
“He did.”
Where??
Nate, ignoring your usual attempts at putting words in my or others mouths, this:
“Here Tesla is arguing that steel balls and moons behave differently. This is contradiction.”
Is a bit silly. He is not arguing that steel balls and moons behave differently. He is arguing that an object with more mass on the side closest to the axis of rotation (the Earth, or the person swinging the ball) will behave differently, when released, to one where the mass is distributed evenly.
We start to get into that familiar territory where I remember why I was so adamant further upthread NOT to discuss this with you…
Where? Here…in the very next line or so after what I had quoted (if I don’t quote everything from the paper, it’s ignored):
“Incidentally it should be pointed out that in true axial rotation of a rigid and homogenous mass all symmetrically situated particles contribute equally to the momentum which is not the case here. That there exists not even the slightest tendency to such motion can, however, be readily established.”
And that’s where he moved on to discussing Fig. 6.
But more to the point…how can you quote, “Only so have we augmented the kinetic energy of the system”, which is a line he says after explicitly making clear that the augmentation is solely translational, and say it’s “TRUE” and then quote “not by causing axial rotation of the balls” and say that’s an “unproven assertion”? There is either translational motion, or axial rotation. If you agree the augmentation is solely translational then you also agree it’s not by causing axial rotation!
As I already said his statement about the Moon’s mass distribution is:
‘Dubious assertion, but if true the difference is rel. small. And how would this small difference so radically change its motion after gravity is cut?’
He does not explain why this should make all the difference between the Moon and a metal sphere. Can you?
This appears to be completely made up.
‘We start to get into that familiar territory where I remember why I was so adamant further upthread NOT to discuss this with you’
Good discussion so far. I see no need for you to be offended. But if you choose to be, that’s up to you.
‘In any case, you guys have been arguing that ANY mass (horse, car, train, steel balls, etc) with the same kind of motion as the Moon, will not be rotating about its own axis while in orbit, or after leaving orbit.’
This is my impression of what you guys have been arguing. Correct me, if its wrong.
‘Incidentally it should be pointed out that in true axial rotation of a rigid and homogenous mass all symmetrically situated particles contribute equally to the momentum which is not the case here. That there exists not even the slightest tendency to such motion can, however, be readily established’
He is not analyzing angular momentum here, nor discussuing the need for torque.
‘But more to the pointhow can you quote, “Only so have we augmented the kinetic energy of the system”, which is a line he says after explicitly making clear that the augmentation is solely translational, and say its “TRUE” and then quote “not by causing axial rotation of the balls” and say thats an “unproven assertion”? There is either translational motion, or axial rotation. If you agree the augmentation is solely translational then you also agree its not by causing axial rotation!’
No, I agree only that there is augmentation of the kinetic energy of the system when the balls are tightened.
The augmentation comes from adding rotational energy to each ball, that wasnt there when they were loose.
Again, he is saying parts of the balls are translating. As I said before, this is true for the parts of ANY rotating object, so irrelevant.
I also think he has not stated that there is ALSO augmentation of the ANGULAR MOMENTUM of the system. He doesnt want to say this, though it is absolutely true, because, unlike kinetic energy, the extra angular momentum comes from the rotation of the balls.
“This appears to be completely made up.“
OK Nate. I acknowledge receipt of your opinion.
“Correct me, if its wrong.”
I don’t have to do anything.
“The augmentation comes from adding rotational energy to each ball, that wasnt there when they were loose.
Again, he is saying parts of the balls are translating…”
I acknowledge receipt of your opinion on what he is saying. However, what he is actually saying is what he’s written.
“He doesnt want to say this…”
I acknowledge receipt of your opinion that Tesla is dishonest, and your apparent opinion that he doesn’t discuss angular momentum anywhere in the paper.
“Good discussion so far. I see no need for you to be offended”
I acknowledge receipt of your opinion that this has been a discussion, or a good discussion, and that you think I said I was offended.
I will now express my opinion that any discussion with you is pointless, and wish to stop.
Ok, so you randomly decide to become an asshole again. What a shocker.
This one is all on you.
The reason for the discussion of the two weights, joined together, and released from rotation about an off-center axis (Fig. 7), or fired from the two rifle barrels (Fig. 8) seems to have been missed by some. On discussing Fig. 6, the ball rotated about an off-center axis, Tesla shows that:
“there is no torque or rotary effort about center C [the center of mass of the ball] and that the kinetic energy of the supposed axial rotation of the ball is mathematically equal to zero.”
and
“These are the distances from center O [the off-center axis of rotation] at which the masses of the half spheres may be concentrated and then the algebraic sum of their energies — which are wholly translatory those of axial rotation being nil — will be exactly equal to the total kinetic energy of the ball as a unit”.
In the discussion of Fig. 7, angular motion of the two masses about an off-center axis, Tesla explains:
“Let us now inquire what will happen if the two masses are rigidly joined, the connection being assumed imponderable. Here we come to the real bug in the question under discussion. Evidently, so long as the whirling motion continues, and both the masses have precisely the same angular velocity, this connecting link will be of no effect whatever, not the slightest turning effort about the common center of gravity of the masses or tendency of equalization of energy between them will exist. The moment the strings are broken and they are thrown off they will begin to rotate but, as pointed out before [in the second paper], this motion neither adds to or detracts from the energy stored.”
Whereas in the discussion of Fig. 8, rectilinear movement from the two rifle barrels, Tesla demonstrates:
“Now, this whirling movement is real and requires energy which, obviously, must be derived from that originally imparted and, consequently, must reduce the velocity of the balls in the direction of flight by an amount which can be easily calculated.”
This is why Tesla comes to the conclusion:
“As will be seen, we arrive at precisely the same results whether the movement is rectilinear or in a circle. In both cases the total kinetic energy can be divided into two parts, respectively of the same numerical values, but there is an essential difference. In angular motion the axial rotation is nothing more than an abstract conception; in rectilinear movement it is a positive event.”
When the strings are broken, in Fig. 7 (angular motion), the rotation doesn’t detract from the energy stored (hence the “abstract conception”). When fired from the rifle barrels, in Fig. 8 (rectilinear movement) the rotation does (hence the “positive event”).
That’s why he has added Fig. 7 and 8. It’s just further evidence for his overall conclusion that there is no true physical axial rotation in an object rotating about an off-center axis, such as the moon.
‘the kinetic energy of the supposed axial rotation of the ball is mathematically equal to zero.’
He does lots of calculation, but in the end I can find no proof of this. It is just made-up AFAIK.
‘Now, this whirling movement is real and requires energy which, obviously, must be derived from that originally imparted and, consequently, must reduce the velocity of the balls in the direction of flight by an amount which can be easily calculated.’
This is misleading. The velocity of the center of mass of the balls cannot be reduced in the direction of motion, by conservation of momentum. The faster one is reduced and slower one sped up. Same happens in balls with strings case.
‘As will be seen, we arrive at precisely the same results whether the movement is rectilinear or in a circle. In both cases the total kinetic energy can be divided into two parts, respectively of the same numerical values’
Yep.
‘but there is an essential difference. In angular motion the axial rotation is nothing more than an abstract conception; in rectilinear movement it is a positive event.’
Nope does not follow.
He seems to readily accept as REAL a combination of linear motion of center of mass, plus rotation about that center (situation after gun firing, or string cutting)
But he does not accept as real a combination of orbiting of cm plus rotation about cm.
But this is just his opinion, no math or physics backs it up.
“Nope does not follow.“
It does follow if you accept that in Fig. 7 (the whirling masses) the rotation doesn’t detract from the energy stored, and when fired from the rifle barrels, in Fig. 8, the rotation does.
“This is misleading. The velocity of the center of mass of the balls cannot be reduced in the direction of motion, by conservation of momentum. The faster one is reduced and slower one sped up. Same happens in balls with strings case”
So how come Tesla’s math, which further upthread you were happy to state was all correct, results in this conclusion:
“so that V3 – V4 is the loss of velocity in the direction of flight owing to the rotation of the two mass points.“
?
His math appears correct but what he says about it is not always..
‘V3’ is a made up ‘effective’ velocity that never existed.
V4 was the velocity of cm before and after the link is established.
If i throw a ball at 50 mph and it is attached by a string to another ball on the ground, when the other ball gets yanked by the string, both will be travelling at 25 mph.
Before and after the yank, the cm velocity is 25 mph.
Nate, you either agree with his math, which includes his use of the effective velocity of the common center of gravity of the two balls at the moment of discharge from the rifles, or you don’t.
No, I agree with parts, dont agree with other parts.
There is a center of mass velocity V4. Same before and after. V3 is an abstract velocity that he is comparing to.
OK Nate.
–David Appell says:
December 11, 2018 at 7:45 PM
gbaikie says:
The question of why, in the annual mean, the Northern Hemisphere is warmer than the Southern Hemisphere is addressed, revisiting an 1870 paper by James Croll. We first show that the ocean is warmer than the land in general
What volume of land are you using for comparison?
Otherwise you’re comparing a volume to an area, which doesn’t make any sense in terms of temperature.–
I didn’t say it. I was quoting it. From here:
http://ocp.ldeo.columbia.edu/res/div/ocp/pub/seager/Kang_Seager_subm.pdf
Apparently Northern Hemisphere is warmer. And they provide their various explanations as to why.
What I said was I asked question which hemisphere is warmer. And I said I would guess southern was warmer, despite the colder polar region. And then I looked to see if could get an answer.
And as above, apparently the northern Hemisphere is warmer.
And also this, unbeknownst to me, as been argued about for some time:
“…‘‘In few departments of Natural Philosophy have Philosophers differed more widely in opinion, than in the comparison of the temperatures of the two hemispheres.’’(Harvey 1834, p. 29). Indeed, the scientific debate on the surface air temperature difference between Earth’s hemispheres and its causes has a long and rich history that can be traced back to the beginning of the sixteenth century. …”
https://www.researchgate.net/publication/258778329_Why_is_the_Northern_Hemisphere_warmer_than_the_Southern_Hemisphere
Also at this last ref they give amounts and explanations of why.
The amounts vary a bit depending data set used, Ie:
CRU: 14.0 average global
14.6 northern Hemisphere
13.3 southern hemisphere
So difference of 1.3 C and also give number for adjustments for elevation differences.
One data set is as high as 1.5 C difference.
What wanted to say is question about the amount warming the religious feel will be forthcoming in various regions of tropical, temperate and polar regions.
But before could ask, I realized I din’t know what difference if any between the two hemispheres.
Currently I am not confident that know the answer, but it appears there some belief in idea, that northern hemisphere is measurably warmer.
Apparently Northern Hemisphere is warmer. And they provide their various explanations as to why.
Again, this statement makes no sense unless you specify what volume you’re discussing.
A 2-D manifold, like a surface, doesn’t even HAVE a temperature, because a 2D surface contains no molecules, so it can’t have an average molecular kinetic energy, viz it has no temperature.
This is why ocean heat content is BY FAR the best measure of global warming, as Roger Pielke Sr has been saying for over a decade now….
(See his Physics Today article)
RP Sr’s 2008 Physics Today article:
https://pielkeclimatesci.files.wordpress.com/2009/10/r-334.pdf
Read the first couple of paragraphs in the section “Climate System Heat Changes”
And here are the data on ocean heat content:
http://tinyurl.com/jbf2xco
I read:
https://pielkeclimatesci.files.wordpress.com/2009/10/r-334.pdf
And very little I disagree with.
But I don’t agree that governments should control CO2 emissions.
If I thought governments should control CO2 emission, I think governments work towards getting more nuclear power.
I think it’s a good thing that India might be going in the direction of using more nuclear power. But I am unable to predict how successful India will be. Same goes for China.
I think what US should do, is get serious about space exploration
More pseudoscience:
“The 2007 IPCC report estimated that global average total net anthropogenic radiative forcing in 2005 was 1.6 (+0.8, −1.0) Wm−2 with 0.12 (+0.18, −0.06) from solar irradiance. This estimate corresponds to a heat accumulation in the climate system of 2.8 (+1.6, −1.7) 1022 joules per year.”
The imaginary “forcing” is corresponding to an imaginary energy increase of 2.8 X 10^22 Joules/yr!
The IPCC creates 28,000,000,000,000,000,000,000 Joules out of thin air!
Pure pseudoscience!
JDHuffman
Not pseudoscience at all! You still can’t understand what the claim is. Rather than attempt (even a little) to learn what is being stated you react with no thought process.
The forcing is a concept between different states. In both cases you have the same input solar flux. In the case with GHG you have energy emitted by the heated GHG. The surface IR heats the gases and some of this energy is then returned to the surface, in a case with no GHG zero energy is returned. This energy is then absorbed (and will be absorbed despite your false claims it won’t be) and will increase the internal energy of the surface. It now has the same solar input plus the added returned energy from the heated GHG. It is really simple physics.
Norman, not only do you deny reality, you can’t even read!
“The IPCC creates 28,000,000,000,000,000,000,000 Joules out of thin air!”
See if you can count the zeroes, at least….
JDHuffman
I can read fine. Remember that is your problem not mine. There is NO creation of joules out of thin air. The original energy came from the Sun. If you have no GHG that energy is not directed back to the Earth.
Oh ignorant poster. Do a simple experiment with a thermos bottle. Have a heater in a thermos that continuously adds energy to the liquid in the thermos and have a normal non insulating container and with same amount of fluid and identical heating. Have a thermometer in each and let me know what you get. In the thermos a silvered inner lining reflects IR back into the fluid. It is not creating new energy but it is redirecting energy back into the fluid. With a continuous input (which the Earth system has that you seem to ignore or something) of energy this redirection process will increase the temperature and the energy of the fluid considerably over the non-thermos container.
Why are you so ignorant on basic heat transfer. I have linked you to dozens of real and valid physics. Your only comment is that I don’t understand what is stated. I certainly do understand the material and am completely correct with my understanding. All your physics comes from blogs from crackpots that make up things with zero proof or evidence. Real physics does not work that way.
JD…”estimated that global average total net anthropogenic radiative forcing in 2005 was 1.6 (+0.8, −1.0) Wm−2 with 0.12 (+0.18, −0.06) from solar irradiance”.
The key word is ‘estimated’, which would be better represented as ‘guesstimated”.
The error margins are so generous that the fictitious forcing could be anywhere from 0.6 W/m^2 to 2.4 W/m^2 and 0.3 to 0.6 W/m^2 for the other.
It’s sad when people have to bend numbers to make them fit a ridiculous theory.
Gordon, those were NOT my words.
If you insist on misquoting me, get in line.
Norman, until you can realize that a racehorse is not rotating on its own axis, you will never be able to learn any physics.
But, at least you can type.
JDHuffman
I learn physics just fine. Just not your pseudoscientific unsupported declarations. Such as the Moon does not rotate on its axis. Pseudoscience with zero support.
Alert for anyone who does not know your character.
This is how JDHuffman thinks: Valid experimental or observational physics is pseudoscience to this poster.
When JDHuffman puts physics into his posts he means unsupported, non-experimental, non-observed declarations. Basically he defines pseudoscience (things he or others make up) as “physics” and all experimental verified physics is pseudoscience to him.
Once you understand this about JDHuffman you will be able to follow his complete nonsense. And I do mean complete nonsense. He can’t solve the easiest of physics problems and he barely knows math. He can’t follow calculations and refers to posting of such as rambling.
He has an idiot follower that posts as Dr Roys Emergency Moderation Team says: Which many believe may be him pretending to be someone else.
Norman, you just about got all your usual nonsense included. But you left out your phony “experiments”.
Guess you get to bang on that keyboard some more.
One more observation about JDHuffman (g.e.r.a.n)
When you point out his flawed reasoning he resorts to mindless meaningless posts. He can’t even attempt to defend his indefensible nonsense but he pretends he can.
norman…”In the thermos a silvered inner lining reflects IR back into the fluid. It is not creating new energy but it is redirecting energy back into the fluid…”
Norman…your version of physics is seriously strange. Are you now going to tell us that the reflected IR warms the fluid in the thermos to a higher temperature than it is heated by your internal heater?
I would like to know how effective the silver interior lining could be with a hot liquid pressing against the lining. If the thermos was half full, you might have a point, but a full thermos cannot radiate IR against that lining. All heat transfer in that case would be through the inner wall by conduction and by convection to the bottom and top of the flask where heat can be conducted directly.
Heat from the liquid escapes through the inner wall of the thermos, then IR can radiate from the outer wall of the inner shell. There is a vacuum in between flask walls that resists conduction through the vacuum, however, that vacuum is unlikely to be perfect. Also, there is physical contact between the thermos flask at the top and the bottom through which heat can be conducted.
The flask cools through conductive leakage over time and radiation likely plays an insignificant role.
You should get this nonsense about back-radiation out of your mind. It’s not happening.
JD…”Gordon, those were NOT my words.
If you insist on misquoting me, get in line”.
Did not imply they were your words, I realized that. I was agreeing with you and adding an observation about the quote.
Unlax, Doc, these pseudo-scientists would like nothing better than to see you and I on bad terms. ☺
Gordon Robertson
Here:
https://www.explainthatstuff.com/vacuumflasks.html
This will answer your question: “I would like to know how effective the silver interior lining could be with a hot liquid pressing against the lining”
The inner lining is heated by the liquid an radiates toward the outer silver lined surface which reflects back nearly all the IR.
Gordon Robertson
YOU: “Normanyour version of physics is seriously strange. Are you now going to tell us that the reflected IR warms the fluid in the thermos to a higher temperature than it is heated by your internal heater?”
I would not say it in that fashion. The correct wording is that the combination of the heat source and the reflected IR will warm the fluid to a higher temperature than the internal heater would do without the reflected IR. That is correct and you could do the experiment yourself, you will see the fluid in the thermos gets hotter than the non thermos fluid. Try it, try it to prove me wrong. You won’t be able to.
“He has an idiot follower…”
Norman, please stop trolling.
Gordon Robertson says:
The flask cools through conductive leakage over time and radiation likely plays an insignificant role.
Gordon, how does radiation play a role in cooling if it carries no heat, as you often like to claim?
Gordon Robertson says:
JDestimated that global average total net anthropogenic radiative forcing in 2005 was 1.6 (+0.8, −1.0) Wm−2 with 0.12 (+0.18, −0.06) from solar irradiance.
“The key word is estimated, which would be better represented as guesstimated.:
Gordon, here are the data:
http://www.esrl.noaa.gov/gmd/aggi/aggi.html
There error bars are quite small because the calculations are not complicated.
“This is why ocean heat content is BY FAR the best measure of global warming, as Roger Pielke Sr has been saying for over a decade now.”
I agree with Roger Pielke Sr, but I would say it is global temperature. And it determines global climate.
It is why we are in an icebox climate.
If you think it is global temperature (of what?), then you are DISAGREEING with RP Sr.
“If you think it is global temperature (of what?), then you are DISAGREEING with RP Sr.”
No, I am agreeing with him.
I say it, my way, that air surface global temperature is a clue, but the ocean temperature is actually, global temperature.
RP Sr says IPCC is overly dependent on air surface temperatures and there is more than just global CO2 levels.
gbalkie, you changed your answer from above. Kudos for seeing the light.
David, I think you are being a bit too pendantic here. In these discussions, a “global temperature” would be some sort of average of the temperature of the surface air layer — say 1-2 meters above the ground. The place where people live. The temperature recorded inside a Stevenson screen.
Certainly the oceans are critically important in terms of total thermal energy absorbed/retained, but that is a different issue.
Finally, I can aim an IR thermometer at a surface and get a temperature, so “surface temperature” is a perfectly valid concept.
tim…”In these discussions, a global temperature would be some sort of average of the temperature of the surface air layer say 1-2 meters above the ground”.
Makes sense. However, one version of the AGW theory claims back-radiation from CO2 can raise the actual solid surface area, causing more WV to be released.
I want to know how heat from CO2, at 0.04%, in a cooler region than the surface, can transfer heat to the surface so as to raise its temperature higher than the temperature the surface is heated by solar energy.
That represents a recycling of heat so as to raise the temperature of the heat source. Does that no strike you as being perpetual motion?
No, you DON’T want to know how CO2 can raise temperatures. It has been explained repeatedly to you here. There are descriptions all over the internet. There are dozens of textbooks that address this. There are classes you could take.
If you wanted to learn, you would have already made the effort. You would already know that the *SUN* provides a transfer of energy *TO* the surface. GHG’s merely restrict the energy flow *FROM* the surface.
tim…”No, you DONT want to know how CO2 can raise temperatures. It has been explained repeatedly to you here. There are descriptions all over the internet. There are dozens of textbooks that address this. There are classes you could take”.
You’re talking like a well-schooled little alarmist.
Prove it!!! Never mind what the text books say, and your fellow alarmists, prove it.
You’d better get in line, however. Twits have been trying to prove it since Arrhenius and even today the IPCC, with all its reviewers have offered no proof, just probabilities.
Tim…it’s a lot more fun when you let go of the anal approach to science and just enjoy it. If you cannot prove something just say, “I don’t know”. Or, at least, offer a theory that has a semblance of fact.
Right now you are claiming it has been explained to me. Horsebleep!!! All I have heard is a load of pseudo-science based on Tyndall’s lab experiment that proved CO2 can absorb IR.
Your opinion, and those you have alleged to have told me, are based on consensus, not fact.
Show me proof of the warming caused by CO2 that cannot be explained by natural means and a re-warming from the Little Ice Age.
Tim, CO2 can NOT raise temperatures.
It is NOT a heat source.
You’re still trying to boil water with ice cubes….
“[CO2] is NOT a heat source.”
Very true. I agree 100%
“You’re still trying to boil water with ice cubes….”
Very false. That does not accurately describe my position.
Rather than argue semantics, lets see if we can agree on some very basic calculations first. Suppose we have a sphere with a surface area of 1 m^2. It is coated with a material that makes it a black body, ε = 1. Inside this sphere is a 800 W electric heater that uniformly heats the surface.
If this sphere is suspended in a large, evacuated chamber where the walls are cooled to 77 K by liquid nitrogen, what will the surface temperature of the sphere be (once everything has stabilized)? [345 K = 72 C]
If this sphere is suspended in a large, evacuated chamber where the walls are cooled to 0 C by ice, what will the surface temperature of the sphere be? [374 K = 101 C]
If this sphere is suspended in a large, evacuated chamber where the walls are room temperature (20 C), what will the surface temperature of the sphere be? [383 K = 110 C]
Do you agree??? If not, show your calculations.
(That “ε” should be an epsilon. )
Tim, I agree that you can do the math. But, bogus equations with imaginary devices is NOT reality.
How the “black body” is used is very revealing. In your examples here, the black body is absorbing all IR, regardless of any emission temperature. That is the way the bogus equation corrupts science. But, the green plate somehow magically transforms from a black body into some type of insulator, to corrupt science in that situation.
Whenever you’re corrupting science, it’s pseudoscience.
HD, until you can make actual predictions, your words mean nothing.
I can’t even tell from your reply if you think my calculations are correct in my simple setting.
Are my numbers correct for the situation given? If not, what numbers would you calculate with your ‘correct equations’ to replace any ‘bogus equations’ I used?
I only checked your first calculation, and got the same results as you. I assumed your other calculations were also correct, unless you entered a wrong number somewhere. Your math appears correct. That’s why I stated: “Tim, I agree that you can do the math.”
The math is correct, but the physics is wrong.
And, I’ve got plenty of predictions. Here are just two:
1) You can NOT boil water using only ice cubes.
2) Put a pot of boiling water in an isolated room. Allow the temperatures to equalize. The energy in the room is enough to re-boil the water. But, it will not self-organize to do so.
JD can skip the black-body, use a real grey body material of your choice.
Good idea, Svante.
Using real objects just destroys pseudoscience that much faster.
JD, here’s the thing. Your two predictions are directly in agreement with me and standard science — INCLUDING the science of the ‘greenhouse effect’.
Ice @ 0C cannot — by itself — warm anything above 0C.
CO2 @ -10C cannot — by itself — warm anything above -10C.
A room @ +20C cannot — by itself — warm anything above 20C.
But in conjunction with other sources, these cool objects can have an impact.
For example, did you note that adding ice in the second example raised the temperature of the sphere above 100C. Adding the ice (and removing the liquid nitrogen) *caused* (in conjunction with the 800 W heater) the water inside the sphere to boil!
Tim, here’s the thing. Your efforts are just another bust.
You said I couldn’t predict anything, and then you agreed with my predictions. (I got many more, too.)
And even though your math is correct, your physics is wrong. Yet, you continue to push your pseudoscience: “For example, did you note that adding ice in the second example raised the temperature of the sphere above 100C.”
No Tim, that’s what your “math” showed. But you had to violate the laws of physics.
That’s called pseudoscience.
Tim clearly knows his stuff.
And, Ger*an, you didn’t prove him wrong, you just wrote some words.
Tim got some things right, and some things wrong.
You, however, got NOTHING right!
Another comment with no content.
You’re very good at those.
Ger*an says:
1) You can NOT boil water using only ice cubes.
Who said you could?
DA, with enough irrelevant comments and stupid questions, you will become irrelevant and stupid.
If not already….
The Southern Hemisphere is much closer to the Sun at perihelion. The seaonal tilt allows for stronger solar there.
But the Earth’s velocity is fastest at perihelion, meaning it spends less time there and near that point.
DA, the Southern Hemisphere has much more ocean surface than the north. Water absorbs higher energy much better than land. Photons travel at the speed of light.
Just a little more reality for you to deny.
You just changed your argument.
Nope.
It’s still the same reality.
Yes, you certainly did change your argument.
No confidence in the first one.
Always confident with reality, DA.
You should try it.
Ifs and buts….
I wouldnt call a 2.6% annual increase exponential. Maybe you do though.
Well, if each year is 2.6% greater than the last year, that is the definition of exponential growth!
x = x0(1+0.026)^n
tim…”Well, if each year is 2.6% greater than the last year, that is the definition of exponential growth!”
Are you talking about CO2 concentrations or the alleged heating effect of CO2 in our atmosphere?
I have not seen an exponential growth in the global average during the entire UAH record. There is no proof that any warming that did occur is related to CO2.
Pretty sure I was simply talking about exponential growth. The post I responding to was literally describing the exact mathematical definition exponential growth and saying it was not exponential growth.
Gordon Robertson says:
There is no proof that any warming that did occur is related to CO2.
False.
“Increases in greenhouse forcing inferred from the outgoing longwave radiation spectra of the Earth in 1970 and 1997,” J.E. Harries et al, Nature 410, 355-357 (15 March 2001).
http://www.nature.com/nature/journal/v410/n6826/abs/410355a0.html
Radiative forcing measured at Earths surface corroborate the increasing greenhouse effect, R. Philipona et al, Geo Res Letters, v31 L03202 (2004).
http://onlinelibrary.wiley.com/doi/10.1029/2003GL018765/abstract
“Observational determination of surface radiative forcing by CO2 from 2000 to 2010,” D. R. Feldman et al, Nature 519, 339343 (19 March 2015).
http://www.nature.com/nature/journal/v519/n7543/full/nature14240.html
Just as Gordon indicated, “There is no proof that any warming that did occur is related to CO2.”
Why don’t the above papers prove that?
Can you explain the proof in more than two sentences? Take as much time as you need.
Tim is right (again) — growth at a constant percentage is exponential growth.
On another tack
I tried several methods to work out how far you have to go back before the slowing rate of rotation produced a ridiculous previous spin speed/ day length. I must be doing something stupid because it’s hard to get to a six figure digit, in years, even five digits look ridiculous with some assumptions. Then the question arises how much heat does that loss of angular momentum generate?
johno…”Then the question arises how much heat does that loss of angular momentum generate?”
The heat is not coming from the angular momentum, it’s coming from the Sun. If the Sun disappeared all of a sudden, the Earth’s temperature would drop to near 0K, and no amount of angular momentum would warm it.
I presume you are referring to the relation between solar input and angular momentum. If you have a longer day then more heat should be generated by the Sun. Is that what you are saying?
Don’t forget you’d have a longer night for heat to be dissipated.
I think you’ve got a much tougher problem than you may realize, given the tilt of the axis and the eccentricity of the Earth’s orbit.
Some of the loss of angular momentum[and Venus’ spin is slowing] will be expressed as heat, http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html the worst case I came up with is Venus slowed from a 24hr. day to its present rotation in @ 2,500 yrs possibly two orders of magnitude more, but before we can establish how it became so hot we need answers to bigger questions. What’s causing the slowdown? being one.
If the sun disappeared ‘suddenly’ and the earth kept spinning as is does then i’d agree with you, but if the Earth ‘suddenly’ stopped spinning it would warm up very quickly.
What assumptions are reasonable when it comes to the rate of slowing? Is it like an applied brake where the rate of slowing accelerates to exponential or should we assume that the rate of slowing eases as it approaches ‘tidal equilibrium’ or perhaps it just slows at a constant rate?
The rate of rotation of Venus is very low and it is retrograde. You can’t explain this in terms of tidal energy/momentum transfer.
The most likely explanation is a collision with a massive body.
“The rate of rotation of Venus is very low and it is retrograde.” It’s also slowing it’s the rate of slowing and from what/when that i’m trying to get to grips with. If the Earth suddenly stopped spinning for instance i’d expect all the platinum group metals[at least those not in the core] to instantly boil. If Venus has slowed from an Earth like rate of spin to it’s present state something similar, over a longer period, should be expexcted.
” You cant explain this in terms of tidal energy/momentum transfer.” I’m looking at kinetic energy>heat energy
johno…”If the Earth suddenly stopped spinning for instance id expect all the platinum group metals[at least those not in the core] to instantly boil”.
Why? What mechanism would cause the boiling?
Gordon, what happens to all the Earth’s rotational kinetic energy if it suddenly stops spinning?
It’s stupid question time again.
It’s only a stupid question to those too stupid to answer it.
So answer it yourself.
johno…”Some of the loss of angular momentum[and Venus spin is slowing] will be expressed as heat, http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html the…”
Work can be converted to heat but where is the Earth doing work on anything? The reason it keeps spinning, and the Moon, is that it encounters no resistance in space. I mean, where would heat be transferred from/to?
Kinetic energy is a generic terms that covers all energy in motion. It’s true, heat is the kinetic energy of atoms, but just because there is kinetic energy involved does not mean there has to be a heat loss.
If a bolder falls off a cliff, there is a gradual conversion from potential energy to kinetic energy. However, there won’t be any significant heat created until the bolder strikes the ground.
At much higher velocities (and altitudes), like with a space craft returning to Earth’s atmosphere, there is significant friction between the spacecraft and atoms in the atmosphere. That produces a load of heat. I don’t think there is much heat generated by the Earth rotating on its axis.
“Work can be converted to heat but where is the Earth doing work on anything? The reason it keeps spinning, and the Moon, is that it encounters no resistance in space. I mean, where would heat be transferred from/to?”
Not Earth, Venus, which slows despite encountering no resistance, and warms too. So how much has it slowed? since when? and how would it dissipate the resultant heat. There is no transfer, what slows warms.
I’m guessing @ 200,000 years, longer and the numbers get really silly, but that represents a lot of slowing and a lot of kinetic energy converted to heat. That said i may be doing something stupid.
I have avoided commenting on the fact that the Moon is in a 1:1 phase locked orbit relative to Earth knowing that whatever I say there will be all kinds of flak and much of it will be “Ad Hominem”.
No matter what the amazing Tesla said on the subject (thank you Dr Roys Emergency Moderation Team), the Moon is rotating at the precise rate needed to present the same face to us Earthians at all times.
That is a remarkable situation so how did it come about? Imagine two bodies orbiting each other with random rates of rotation. As they rotate they will transfer energy and momentum to each other via tidal forces. That momentum transfer will stop when one of them is in a 1:1 phase lock orbit relative to the other.
Given the large disparity in the relative masses of the Earth and the Moon (80:1) we should not be surprised that the Moon phase locked to the Earth rather than the opposite.
Then there is the 3:2 phase lock orbit of Mercury to the Sun. Is this situation stable or will it degrade to a 1:1 phase lock?
Galloping camel
Note that there is still momentum transfer from the Earth to the Moon.
This is now accelerating the Moon in its orbit by transferring angular momentum from Earth’s rotation.
Thus the Moon is moving further from Earth into a longer period orbit, while Earth’s rotation is slowing.
The Moon will be in a longer period orbit in the future.
Therefore it will be not rotating more slowly than it is now.
😉
E-man, you are correct in your pseudoscience, but wrong in your physics.
Angular momentum is not transferred by the force of gravity.
NASA is convinced the Moon is moving away from Earth. But, their belief is based on the measurements from reflectors left on the Moon. Such precise measurements are very questionable, due to the great distances and many variables. Even if it turns out the Moon is actually moving away, it is best explained in more logical ways than “transferring energy”.
It is a consequence of general relativity, known as frame-dragging or the Lense-Thirring effect.
https://en.m.wikipedia.org/wiki/Frame-dragging
In summary, an object moving through the gravitational field of a large rotating object will experience an acceleration.
There is also the tidal bulge effect.
The Moon induces a tidal bulge in the oceans. The Earth’s rotation carries this bulge ahead of the line between the centre of the Moon and the centre of the Earth.
This slightly offsets the vector of the Earth’s gravitational pull on the Moon, slightly accelerating the Moon in its orbit.
Like I indicated E-man, you have the pseudoscience right, but your physics is wrong.
JDHuffman says:
Angular momentum is not transferred by the force of gravity.
Of course it is.
For one thing, neither the Earth or the Moon has a constant density — their densities vary in all three spatial coordinates, even if slightly. These differences exert torques on each body and hence a change in angular momentum (torque=dL/dt).
DA, learn some physics.
Everytime you can’t provide any science you have the same answer.
DA, everytime I provide any science, it goes right over your head.
Learn some physics.
Parroting “learn some physics” time after time shows you have no better answers.
It’s funny you think people don’t realize that.
Learn to stop asking the same already answered questions over and over again.
Just because you type words doesn’t make it an answer.
Does a nonrotating astronomical body always point in the same direction, that is to say, toward the same stars?
Ent man: tidal locking has nothing to do with general relativity or the Lense-Thirring effect (frame dragging). It’s a classical phenomenon.
“Just because you type words doesn’t make it an answer“
No, the fact that my answers directly answer your questions is what makes them answers.
cam…”No matter what the amazing Tesla said on the subject (thank you Dr Roys Emergency Moderation Team), the Moon is rotating at the precise rate needed to present the same face to us Earthians at all times”.
Have you tried the experiment offered by Tesla, and most humbly, by myself?
Place two coins (washers) side by side and mark them both at the same spot. Try to roll the right coin around the perimeter of the left coin without that mark leaving the perimeter of the other coin.
Do you agree that the mark should turn through 360 degrees wrt to the (RH coin) COG/axis over a full revolution? Try rolling the RH coin just once through 360 degrees.
If you don’t agree, take the RH coin and place it against a ruler with the mark against the ruler. Now roll it along the ruler. Does the mark turn through 360 degrees? Does the mark leave the ruler? It has to or it cannot roll.
Now bend the ruler into a slight arc AWAY from the coin (I’m kidding, just visualize it). Repeat the process. Keep bending the ruler till it forms a full circle and repeat.
You can’t get the RH coin around the perimeter by rotating it while keeping the mark against the perimeter of the LH coin. In fact you cannot even move it by rolling it while that mark remains against the LH coin’s perimeter. So how the heck does the Moon do it? ☺
If the mark stays on the LH coin’s surface, the RH coin cannot rotate. It can slide and that sliding motion is curvilinear translation…by definition.
It’s an illusion, as Tesla claimed, that the Moon is rotating.
Your mind is playing with you, Cam, or maybe, like me, you have taken a few too many shots to the head while playing sports. ☺ ☺
Gordon, look at the moon here on the left:
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
Keep your eyes on its polar axis, and follow the dark patch. The dark patch is rotating about it polar axis, viz, the Moon is rotating.
“Does The Moon Rotate?”
Fraser Cain, publisher, Universe Today:
https://www.youtube.com/watch?v=VGnIuqYKnTE
Does The Moon Rotate?”
The Moon orbits. It does not rotate on its own axis.
If the Moon doesn’t rotate, why doesn’t it always face in the same direction?
Stupid question.
It’s a good question. Clearly you have no answer.
Question answered:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333170
and
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333249
What question was answered?
Just answer the questions! You’re only trying to confuse things.
This question has been answered:
“If the Moon doesn’t rotate, why doesn’t it always face in the same direction?“
In the linked comments. As you know. So stop trying to confuse things.
GC, you are siding with Institutionalized Pseudoscience, by believing the Moon is rotating on its own axis. That is the safe stance, if the flak bothers you. You likely won’t get as many ad hominem attacks from Skeptics. Typically, it is the fanatics that resort to insults, false accusations, and misrepresentations.
For the Moon to be rotating on its own axis, definitions must be changed from the original forms. If you stick with the original definitions of “orbiting” and “rotating on its own axis”, then the Moon is only “orbiting”. It is not rotating on its own axis.
https://postimg.cc/06WdBkGB
Unfortunately, if you side with the “Spinners”, you must then also believe a racehorse is rotating on its own axis, as it runs the oval track, which they all claim. When you realize you cannot swallow aspects of the false religion, you may find the flak is not so bad after all….
As this animation shows, the Moon (real moon on the left) is rotating about its axis, whereas the moon on the right is not, because its dark patch always points in the same direction.
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
It’s easy to imagine slowing the Earth’s polar rotation rate down from 1/day to 1/year. The Earth would then still clearly be rotating, but the same face would always present to the Sun. That’s orbiting + rotation.
DA, you’re changing the definition of orbiting to match your pseudoscience.
That’s easier than learning some physics, I guess….
No — I didn’t say a thing about orbiting. The orbit in the above stays the same: once per year.
Okay then, you don’t know the definition of orbiting.
Huffingman, Actually, it is YOU who doesn’t know the definition of “rotating” in classical physics (aka: Dynamics).
Swanson, the discussion involves “rotating on its own axis” and “orbiting”.
Now, do you wish to try again?
OK – what is your definition of orbiting?
Give an example of a body that orbits but doesn’t rotate.
The moon.
If the Moon isn’t rotating, why doesn’t its dark patch (left animation) always point in the same direction?
https://en.wikipedia.org/wiki/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
Question answered:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333170
and
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333249
I’m not interesting in hunting through past links to try to infer what you meant. Because these questions are so simple it’s easier to just answer them than to hunt for what you claim are your previous answers.
If the Moon isnt rotating, why doesnt its dark patch (left animation) always point in the same direction?
https://en.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
You don’t have to “hunt”, David, you just have to click in the links. Or just scroll down up or down to the many times now where we have already discussed the contents of those links. Your “repetitive questions” technique has been made to look very silly indeed.
Re: tidal locking.
With the way the “Spinners” see things, a moon moving as per the moon on the right, in this GIF:
https://en.m.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
would have its rate of axial rotation INCREASED by tidal forces, until it is moving like the moon on the left.
The way the “Non-Spinners” see the moon on the right, is that it is rotating once on its axis, CW, per CCW orbit. So we would see, in the transition from right GIF to left GIF due to tidal forces, that the rate of axial rotation is DECREASED.
“Spinners” have to believe tidal forces can increase, in some cases, and decrease in others, the rate of axial rotation in an orbiting body, until you have your exact 1:1 phase lock. “Non-Spinners” only ever have to believe tidal forces gradually decrease the rate of axial rotation in an orbiting body, to nil (in every possible case).
People in general have to understand the torques involved in tidal locking to predict the motions. If you had bothered to read and understand the wikipedia article, you would have learned that same mechanism that slows a fast moon will speed up a slow moon.
What do YOU think happens to the moon on the right due to the tidal forces???
Why are you asking me a question that I answered in my comment?
“The way the “Non-Spinners” see the moon on the right, is that it is rotating once on its axis, CW, per CCW orbit. So we would see, in the transition from right GIF to left GIF due to tidal forces, that the rate of axial rotation is DECREASED.“
“If you had bothered to read and understand the wikipedia article,”
I did Tim. That’s why I said:
““Spinners” have to believe tidal forces can increase, in some cases, and decrease in others, the rate of axial rotation in an orbiting body, until you have your exact 1:1 phase lock.”
Precisely because I was aware that you guys have to think:
“the same mechanism that slows a fast moon will speed up a slow moon”
And…that’s funny. Makes so much more sense the “Non-Spinner” way.
Dr Roys Emergency Moderation Team says:
The way the Non-Spinners see the moon on the right, is that it is rotating once on its axis,
If the moon on the right is rotating, why does it always face in the same direction?
Question answered:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333170
and
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333249
Dr Roys Emergency Moderation Team says:
Obviously, if you believed that our moon both rotated about the Earth-Moon center of mass AND rotated about its polar axis, then you would believe that we can see all sides of the moon from Earth.
No, that’s not obvious.
And it’s not true for a 1:1 tidal locking. See this animation (on the left):
https://en.wikipedia.org/wiki/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
David, the “Spinner” side of the argument (your “side”) is that the moon translates about the Earth-Moon cm and rotates on its own axis. Your argument is not that the moon rotates about the Earth-Moon cm and rotated on its own axis. That’s why you are getting confused.
Dr Roys Emergency Moderation Team says:
David, the Spinner side of the argument (your side) is that the moon translates about the Earth-Moon cm and rotates on its own axis. Your argument is not that the moon rotates about the Earth-Moon cm and rotated on its own axis.
Your two sentences describe the same motion — a Moon that orbits around the Earth-Moon center-of-mass and a Moon that also rotates about its polar axis.
BTW, the same is also true for the Earth — it both orbits about the c.o.m. and rotates about its polar axis.
Wrong, David, the two sentences do NOT describe the same motion.
Dr Roys Emergency Moderation Team says:
With the way the Spinners see things, a moon moving as per the moon on the right, in this GIF:
https://en.m.wikipedia.org/wiki/Tidal_locking#/media/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
would have its rate of axial rotation INCREASED by tidal forces, until it is moving like the moon on the left.
1) What Tim said.
2) The moon “on the right” isn’t rotating about its axis. You can tell because its dark patch always faces in the same direction.
DA, learn some physics.
For the umpteenth time, the discussion is pretty much just semantics (at least for a perfectly circular orbit). The moon or the horse on a merry-go-round can be described perfectly well as
EITHER
1) rotating about the center, with no additional rotation about its own axis.
2) a simple circular translation about the center plus an additional rotation about its own axis.
Both perfectly well describe the situation — for example, both say one point on the moon is always pointing ahead along the path, with one side always facing toward the earth. So there is no real reason to say either is “wrong” here.
[The interesting bit comes when you try describing an elliptical orbit. In that case, no point on the moon is always pointing toward forward along the path, and no face is always directly toward the earth. I have never seen any coherent explanation for the orientation of the moon from the “it only rotates about the center, it doesn’t rotation on its axis” crowd.]
Except 2) is not how gravity forms an orbit.
But hey, in pseudoscience the laws of physics are meant to be broken….
Ger*an says:
DA, learn some physics.
Written like someone who has no science, but insists on taking up space anyway.
ME: 2) a simple circular translation about the center plus an additional rotation about its own axis.
JD: Except 2) is not how gravity forms an orbit.
Why? I am not ready to accept this on your authority just because you ‘want’ this to be the answer.
Suppose I push an object (for example, a uniform sphere 10 m in diameter) at ~ 17000 mph so it is the right speed to orbit. Suppose when I let go of it, it is not rotating relative to the fixed stars. I just created a new, small moon.
What torque will start it rotating relative to the stars? Will the torque act ‘all at once’ so the sphere will immediately start rotating relative to the stars once every ~ 1.5 hours? Or maybe it will ‘spin-up’ to this rate over the course of a few mintues? a few weeks?
Explain the source of the torque, and show that it is the right magnitude.
Tim, gravity does not create a torque. That’s just your pseudoscience. Get away from the idea of “torque”.
An object is falling toward Earth. The force due to gravity is instantaneously perpendicular to the velocity. The resultant vector creates the orbital path.
The racehorse is a simple model of orbiting.
Does the Moon always face in the same direction?
Question answered:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333170
and
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333249
I’m not interested in trying to dissect your answers in links you give, which as far as I can tell never answer the question.
Does the Moon always face in the same direction? It’s a simple question, with a yes or no answer.
The answer is “no”, because it is “orbiting”. And yes, those linked comments do answer and explain your endless repetitive questions.
Then again, it’s not quite as simple as that. Relative to the “fixed stars”, the moon is changing direction, because it is “orbiting” (rotating about the Earth-moon barycenter, with zero axial rotation).
Relative to the Earth, the moon is always moving west to east, about the Earth’s polar axis. So in another way, the moon only travels in one direction, east. So the answer to your question is also “yes”.
‘Get away from the idea of ‘torque’.
An object is falling toward Earth. The force due to gravity is instantaneously perpendicular to the velocity.’
Torque is pseudoscience for JD, best avoided.
He thinks a force applied thru the center of an object can still make it turn.
He must have an awful lot of trouble using wrenches properly.
” ‘Non-Spinners’ only ever have to believe tidal forces gradually decrease the rate of axial rotation in an orbiting body, to nil (in every possible case). ”
Yes!
But if understand well: in a planet/satellite constellation, the decrease of the planet’s axial rotation speed results in an increase of the satellite’s revolution speed and hence of its distance to the planet.
Correct?
Bindidon, I am a “Non-Spinner”. You are a “Spinner”. So why are you saying, “Yes!”. Have you now changed “sides”?
“Have you now changed ‘sides’ ?”
No, of course! You wrote something intelligible:
“…tidal forces gradually decrease the rate of axial rotation in an orbiting body…”
Maybe I misunderstood you?
Dommage!
Read my translation of Lagrange’s work.
But please: entirely, not like Robertson, who reads 5 lines and decretes “Lagrange is wrong”, or “It is not about Moon’s rotation” or similar nonsense.
https://tinyurl.com/y7e67j42
I am no physicist, but decades ago I had some courses in Mechanics theory, and therefore can understand Lagrange’s equations up to a certain extent.
Wonderful work. Take it or leave it!
I do not want to convince you, let alone would I feel the need to do.
Bindidon, “Spinners” have to believe that tidal forces act to increase, in some cases, and decrease, in others, the rate of axial rotation of an orbiting body, until it is “rotating” at the exact same rate as it is “orbiting”.
“Non-Spinners” only have to believe that tidal forces act to decrease the rate of axial rotation of an orbiting body, until it is reduced to nil, in ALL cases.
If you don’t understand why, keep reading my post further above, and keep thinking about it, until you do understand.
DREMT: if the Moon isn’t rotating, why doesn’t it always face in the same direction?
[repeats same answer as the last time you asked that question]
It’s a good, fair question. Where did you previously answer it?
Question answered:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333170
and
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333249
dremt…”The way the “Non-Spinners” see the moon on the right, is that it is rotating once on its axis…”
Neither moon is rotating about its axis.
Draw an imaginary line through the dark spot to the other side of the image. Follow that imaginary line and it is obvious that the line is not turning about the axis of the moon object.
The point of the axis is on the line and is itself describing a circle about the other body. That is curvilinear translation. Exactly the same as if the left object was sliding along a straight surface.
In the left gif, that line points perpendicular at all times to a tangent line on the surface of the body it is orbiting. In the right gif, the line is always horizontal.
The line in the left gif does go through an angular displacement of 360 degrees per orbit but that is due to the effect of curvilinear translation, not rotation about an axis.
It’s the same motion of the moon object one would expect if it was attached rigidly to a rigid body rotating on the other body’s axis. Exactly the same as wooden horses turning on a merry-go-round.
Wrong, Gordon, the moon on the right is rotating on its own axis, CW, once per CCW orbit. The moon on the left is not rotating on its own axis (that one represents our own, tidally-locked moon, and you have no need to persuade me that it is not rotating on its own axis).
Here, GC is right — the Moon does rotate, as every astronomer on the planet will tell you.
DA, you don’t know the Moon ever rotated on its own axis. It’s just your belief. You have no facts. You have no evidence, let alone proof. You have to pervert the facts today to claim it rotates.
You wallow in your pseudoscience.
Nothing new.
If the Moon doesn’t rotate, why doesn’t it always face in the same direction?
Show me three astronomers who agree with you, and I’ll begin to take your claim seriously.
https://tinyurl.com/y87gs2jb
OK, theres the authors of this 1993 paper on the Tesla papers, plus see page 125, 2nd paragraph, Savic and Kasanin.
So now you have to take it seriously. 😊
Again, if the Moon doesnt rotate, why doesnt it always face in the same direction?
I showed you four astronomers in agreement. You now have to begin to take the claim seriously.
Question answered:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333170
and
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333249
I don’t see the names of any astronomers at your links.
What are their names?
1.
2.
3.
4.
Dr Roys Emergency Moderation Team says:
You could have looked in the first paper.
The moon does rotate, not on its own, but about an axis passing thru the center of the earth, the true and only one.
If you stand on the surface of the Moon, will you always face in the same direction? Will you always see the same stars in the same places without deviation?
Re: astronomers names, you posted elsewhere and I responded:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333646
The stars appear to move from the lunar surface because the moon is “orbiting”, i.e rotating about the Earth-Moon barycenter.
DA, if you had a basic understanding of the relevant physics, and could think for yourself, you wouldn’t need a list of names to be your security blanket.
DREMT, yes, the Moon (and the Earth) are rotating around the Earth-Moon barycenter. We call this “orbiting.”
At the same time, does the Earth also rotate w.r.t its polar axis?
The Earth does rotate on its own axis, yes.
And no, you (the “Spinners”) do NOT call rotating about the Earth-moon barycenter “orbiting”. We (the “Non-Spinners”) do. Once again, with the way the “Spinners” see things, you think “orbiting” is a TRANSLATION about the Earth-moon barycenter, and that the moon then also rotates on its own axis. That is the only way it works for you.
If the moon was rotating about the Earth-moon barycenter AND rotating on its own axis, we would see all sides of the moon from Earth.
Thanks gallopingcamel for this nice exposé.
Millions of years ago, the Earth day’s duration seems to have been quite a bit less than today’s. Evidence for that was shown e.g. when looking at Nautilus fossile stuff.
I read somewhere that tidal records laid down in ancient estuaries show cycles in alternating deposits of sand and silt: according to these observations, the day was 21 hours 620 million years ago.
The Italian-German scientist Ludmila Carone, working at the Max-Planck-Institute for Astronomy in Heidelberg/Germany, has shown years ago, in a little graphic she computed the source data for, that it would take about 50 billion years for Earth and Moon to reach the state of Pluto and Charon, a planet/satellite system that has reached final tidal locking, as the two have lost any rotation energy and both face eachanother for eternity.
I recently asked her if she could not have some time to spend on a reedition of her graph, this time including the 4 billion years before now. Unfortunately, she simply has too much work (she is tracking exoplanets actually).
Millions of years ago…Earth day’s duration seems to have been quite a bit less than today’s.
Evidence for that was…at Nautilus fossile stuff.
I read somewhere that…the day was 21 hours 620 million years ago.
Ludmila Carone… has shown…in a little graphic she computed the source data for, that it would take about 50 billion years for Earth and Moon to reach the state of Pluto and Charon
Bindidon, do you prefer red or white wine with all that pseudoscience?
Why do you allow yourself to discredit the work of so many people by calling it pseudoscience, without ever giving a proof of your dishonest claim?
Who are you, JDHuffman? Show us your work!
Bindidon
If you remember how JDHuffman interprets words it will help you deal with him.
He calls established experimental based physics pseudoscience.
So when he uses the term “pseudoscience” rest assured it is good valid science and true.
When he uses the word “physics” he is talking about his own or other unsupported declarations. When someone makes a declaration he will call this “physics”. When he says “learn physics” he just means go to blogs like Postma’s or PSI and read the made up unsupported material on these blogs and you will have all the “physics” you can handle.
Also remember you are trying to reason with someone who has no knowledge of real science.
Bindidon, things like “Earth’s day was 21 hours 620 million years ago” is NOT science. It is a belief. People may have done some actual investigation. They may have been sincere. They may have been talented. But, their work proves NOTHING.
I’m not discrediting their work. I’m just pointing out how easily people accept such as “proof”.
This is how we end up with such non-scientific hoaxes as AGW. As indicated just above, poor Norman does not know the difference between physics and pseudoscience!
JDHuffman says:
Bindidon, things like Earths day was 21 hours 620 million years ago is NOT science. It is a belief. People may have done some actual investigation. They may have been sincere. They may have been talented. But, their work proves NOTHING.
What evidence do they cite, and why is it wrong?
Just another stupid question.
JDHuffman
Sorry you are the one who thinks actual experimental based physics is pseudoscience. And likewise you are the one who thinks your absurd and made up unsupported declarations are science. Case of point your illogical cartoon you post to about temperature of plates.
Case of point:
https://postlmg.cc/HrxkJyBB
Your cartoon is crap pseudoscience that is proven false by actual experimental evidence (which of course you are not able to accept)
https://app.box.com/s/5wxidf87li5bo588q2xhcfxhtfy52oba
You reject experimental physics but accept your made up CLOWN ideas and call them physics. NO SUPPORT nothing, just your declarations.
No sir, you are the one who does not understand at all what pseudoscience is. You use the word all the time but do not have the slightest idea of its meaning.
JDHuffman
Here is some real science that you call pseduoscience:
5.5 Radiative exchange
“In heat transfer by radiation, energy is not only transported from hot to cold bodies; the colder body also emits radiation that strikes the warmer body and can be absorbed there. An exchange of energy takes place, in contrast to the transfer that occurs in heat conduction and convection. This radiative exchange depends on the mutual position and orientation of the radiating surfaces, their temperatures and there radiative properties. In the following sections it is assumed that the radiating surfaces are separated by a medium that has no effect on the radiative exchange, that neither absorbs, emits nor scatters radiation. This condition is exactly satisfied by a vacuum, although most gases also have little effect on radiative exchange”
From Textbook on Heat Transfer page 588
https://drive.google.com/file/d/1Vl02ky5h40xDygiCIvDHFrj5c9hYvcNN/view
Norman, you’ve used that link before. And, I’ve helped you to understand it before. But, you refuse to learn. You still believe you can twist physics so that “cold” warms “hot”.
And, you refuse to admit the racehorse is NOT rotating on its own axis.
You deny reality.
norman…”In heat transfer by radiation, energy is not only transported from hot to cold bodies; the colder body also emits radiation that strikes the warmer body and can be absorbed there”.
-contradicts the 2nd law
-contradicts the Stefan-Boltzmann equation
-contradicts quantum theory
-contradicts basic intelligence.
Gordon Robertson
YOU: “norman…”In heat transfer by radiation, energy is not only transported from hot to cold bodies; the colder body also emits radiation that strikes the warmer body and can be absorbed there”.
-contradicts the 2nd law
-contradicts the Stefan-Boltzmann equation
-contradicts quantum theory
-contradicts basic intelligence.”
Wrong on all counts!
You don’t understand the 2nd Law at all (I am certain you never studied any higher level physics, you get yours from blogs only).
I have linked you to Clausius himself saying the same thing. You are a science denier.
Stefan-Boltzmann equation? Wrong it says the very thing (if you ever took the time to read things)
q = ε σ (Th4 – Tc4) Ah
The Th4 is the energy radiated by the hot object, the Tc4 is the energy it absorbs by the cold object.
You don’t understand quantum theory at all or that most surface molecules and atoms are in ground states at room temperature.
It does not contradict basic intelligence at all, it contradicts a crackpots version of science. Everyone else can understand it.
JDHuffman
YOU: “Norman, youve used that link before. And, Ive helped you to understand it before. But, you refuse to learn. You still believe you can twist physics so that cold warms hot.
And, you refuse to admit the racehorse is NOT rotating on its own axis.
You deny reality.”
Wrong! The only thing you are capable of doing is making unsupported and incorrect declarations. I have seen nothing else from you. You don’t understand real physics and make up garbage. That is all I have ever seen you do!
Norman, that is not the S/B equation. The S/B equation comes from the Stefan-Boltzmann Law, which only deals wiht one “T”, the temperature of the emitting surface. The equation you mentioned is a mis-application of the S/B Law.
I would recommend you learn some physics, but we know about your learning disability.
At least you were able to learn how to type.
JDHuffman
Just another one of your incorrect unsupported declarations. Is that all you can do?
YOU: “Norman, that is not the S/B equation. The S/B equation comes from the Stefan-Boltzmann Law, which only deals wiht one “T”, the temperature of the emitting surface. The equation you mentioned is a mis-application of the S/B Law.”
Yes it is the S/B equation used when you have more than one radiating surface around. You don’t know enough about actual physics to have a valid opinion. It is the totally valid and correct use of the Stefan-Boltzmann Law to find heat flow when you have another radiating surface. Learn some physics not your made up pseudoscience.
YOU: “I would recommend you learn some physics, but we know about your learning disability.” (Correct interpretation of your use of the word “physics”…learn some unsupported declarations JDHuffman makes)
The S/B Law applies only to the radiative emission from a surface. It says NOTHING about absorp.tion.
Norman demonstrates, again, his inability to grasp simple concepts.
JDHuffman
The Stefan-Boltzmann equation gives the energy radiated by the two bodies. If they were blackbodies all the IR from the cold body would be absorbed by the hotter one. In real world gray bodies the emissivity takes care of the difference. If a hot body can radiate a frequency of IR it will also absorb the IR. That is why in the equation you only need the emissivity of the hot surface. The frequency it can emit is the same as the frequency it will absorb.
Your ignorance of actual science does not make your incorrect statement correct:
YOU: “The equation you mentioned is a mis-application of the S/B Law.”
It is NOT at all a mis-application of the S/B Law. It is very useful and valid application of the Law as it will tell you the correct heat flow of a hot or cold surface. It is used in all engineering applications that deal with Radiant Heat Transfer and it is most successful. It works as stated and is used as stated.
Wrong again.
See Norman, this is just another example of you refusing to learn.
As I stated: “The S/B Law applies only to the radiative emission from a surface. It says NOTHING about absorp.tion.”
But, that does not fit your belief system. And to you, your belief system trumps reality. Your belief system keeps you from learning.
Nothing new.
JDHuffman
Lord are you an obstinate poster.
YOU SAID: “Norman, that is not the S/B equation. The S/B equation comes from the Stefan-Boltzmann Law, which only deals wiht one “T”, the temperature of the emitting surface. The equation you mentioned is a mis-application of the S/B Law.”
I specified directly which point I disagree with in your statement! How dense do you have to be?
YOU: ” The equation you mentioned is a mis-application of the S/B Law.”
With this equation: q = ε σ (Th4 – Tc4) Ah
It IS NOT a mis-application of the S/B Law. It is a useful and productive application of the Law!!
Only using the amount of energy emitted by a surface will not help you in engineering any heat transfer devices! You have to know the amount of Heat transfer which is the amount of energy emitted by a body minus the energy it is receiving from the surroundings.
Not only have you flunked physics but you also flunk engineering.
Norman, you’re always pathetically predictable and predictably pathetic.
Nothing new.
JDHuffman
Will you make a meaningful comment to address what I stated?
I talk about how useful the application of the Stefan-Boltzmann equation is to figuring out heat transfer.
You can’t defend your position so you resort to an attempt at cleverness which kind of fails.
YOUR response to my very good and excellent post: “Norman, you’re always pathetically predictable and predictably pathetic.
Nothing new.”
So what does that explain or answer? You have not even come close to explaining why you think that the equation for heat transfer between two surfaces is a mis-application of the Stefan-Boltzmann Law.
Talk about predictable and “Nothing new”. I challenge your dork statement (which demonstrates your lack of knowledge of physics). Rather than defend it or attempt to support your declaration you go into lawyer spin and pretend you are winning the debate. How predictable is that? You do it with nearly every poster that presses you to explain your dork physics.
Team Dork loses again. Team Dork cannot defend their stupid declarations. Team Dork, trying to save face, posts a dork comment with zero meaning but attempts a clever word play. Team Dork is dorky.
Norman, I don’t take time for you anymore because you haven’t shown any ability to learn. I’ve explained before that your attempt to twist the S/B Law into the bogus heat transfer equation is invalid, due to its derivation. You cannot understand that and it frustrates you.
All you can do is just bang out another juvenile rant:
“Team Dork loses again. Team Dork cannot defend their stupid declarations. Team Dork, trying to save face, posts a dork comment with zero meaning but attempts a clever word play. Team Dork is dorky.”
Nothing new.
JDHuffman
The reason you are a dork is because you make really stupid comments like this one.
YOU: “Norman, I don’t take time for you anymore because you haven’t shown any ability to learn. I’ve explained before that your attempt to twist the S/B Law into the bogus heat transfer equation is invalid, due to its derivation. You cannot understand that and it frustrates you.”
It is NOT my twist of the S/B Law nor is the heat transfer equation bogus. It is an established equation used in engineering of heat transfer problems that involve radiant energy. Please don’t be as dumb as you are with these types of posts.
You are the lunatic that shows zero ability to learn anything.
Norman, the only thing you got right was my quote.
Nothing new.
Gordon Robertson says:
normanIn heat transfer by radiation, energy is not only transported from hot to cold bodies; the colder body also emits radiation that strikes the warmer body and can be absorbed there.
-contradicts the 2nd law
How?
What happens to radiation from a cold body that approaches a warmer body? Prove your claim with experimental data.
JDHuffman says:
The S/B Law applies only to the radiative emission from a surface. It says NOTHING about absorp.tion.
Sorry bub, but it does. By definition, a blackbody ab.sorb.s all radiation incident upon it.
Sorry dud, but it doesn’t. You must be confusing “absorb” with “emit”.
Obviously you need to learn some physics.
What is the definition of a blackbody?
Obviously you need to learn some physics.
What does the definition of a blackbody say about its ab.sorp.tion?
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333608
That link didn’t answer the question.
What does the definition of a blackbody say about its ab.sorp.tion?
(This is baby physics.)
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333608
I’m always curious how long you will keep asking the same stupid question. It’s obvious you can’t learn, even when someone tries to help you.
Try a new therapist.
You avoid questions when you realize you’re trapped in a corner.
I’ll keep asking until you’re man enough to admit when you’re wrong.
norman…”The Th4 is the energy radiated by the hot object, the Tc4 is the energy it absorbs by the cold object”.
If that was the case the Tc4 would have its own area, emissivity, and the S-B constant.
That equation comes from Boltzmann himself who took up Stefan’s equation, r =(fi)T^4 to prove it using statistical mechanics. Stefan’s equation is based on the work of Tyndall in which he observed different colours being emitted from a platinum filament as ever-increasing electrical current was run through it.
A platinum filament radiating to air is obviously a case of radiation from a much hotter device transferring heat to a much cooler medium. I dare say, that radiation from a filament with a temperature of 1500C – 3000C would heat the nitrogen and oxygen in the air.
In your version, Boltzmann added the second T term to create a temperature gradient between the hotter body and the cooler body. The second T term has nothing to do with radiation from that body, it’s about a temperature gradient that supports the 2nd law, from hot to cold.
Wrong, DA. I just avoid answering your stupid quesions because I know they are just your tricks. You have now admitted you don’t really want answers. You just use your stupid questions as attempted “gotchas”.
You have nothing else.
So some say land area have higher increase in average temperature as compared to ocean area from higher CO2 levels.
Which makes some sense to me because land area cool quicker than ocean area.
And most people think CO2 reduces heat loss [rather than somehow increase the surface temperature].
One also look at it this way, one cause cause land area to increase in average temperature due to UHI effects. And UHI effect are about reducing heat loss.
And UHI effect aren’t going to increase ocean air temperatures.
Though islands will increase air temperature- land temperature can higher during day, and ocean keeps air warmer on island during night- and could call that an effect like the UHI effect.
Plus warmer day land air temperature extends in ocean area near the island.
Plus shallower waters can warmed more as compared to deeper waters, and have warmer surface waters around island.
But anyhow if land warms more, and ocean less by CO2 effect, and Earth is mostly ocean and having larger area be ocean, it means a reduced effect upon global air temperature caused higher levels of CO2.
gbaikie says:
So some say land area have higher increase in average temperature as compared to ocean area from higher CO2 levels.
Which makes some sense to me because land area cool quicker than ocean area.
It’s true — warming over land is about 50% higher than the global number — but it’s because the land absorbs all the heat incident upon it, whereas the heat ab.sorb.ed by the ocean surface can more pass to depth.
gbaikie
“One also look at it this way, one cause land area to increase in average temperature due to UHI effects.”
Sorry: this such an old stuff. If you would spend time in evaluating absolute temperature time series, you would quickly understand that this problem has been solved since longer time.
UHI effects exist only if you average absolute data arising from UHI suspected stations with the rest.
If you construct your anomalies station by station, and then average the anomalies cell by cell, latitude band by latitude band and then month by month, the problem no longer exists.
Because even if a station shows spurious increases here and there, the departures from the computed mean will look as those for stations lacking UHI effects.
Btw: why do ‘skeptics’ speak all the time about UHI, but never mention its opposites, e.g. RCC aka ‘rural cooling corner’s?
Maybe simply because their main rule is: ‘Warming is bad; cooling is good’ ?
“Btw: why do ‘skeptics’ speak all the time about UHI, but never mention its opposites, e.g. RCC aka ‘rural cooling corner’s?
Maybe simply because their main rule is: ‘Warming is bad; cooling is good’ ?”
Well if you worried about higher temperature, UNI causes much higher temperature air temperatures than “global warming”.
I am not worried about higher air temperature or global warming, but some believers are very frighten about a possibility warmer air temperatures.
The German people have spent more than 1 trillion dollar due to this fear.
And average air temperature of Germany is less than 10 C.
Any Germans leave the country on vacation to find warmer places.
And should be noted that Indians are not afraid of global warming and the country of India has average temperature of over 24 C.
But if they were concerned about warmer air temperatures, I would imagine, the Indians are wise enough to do something about UHI effects, which causes higher air temperatures than global warming.
The main rule of anyone who is sane is that global warming is good and global cooling is bad.
“… to do something about UHI effects, which causes higher air temperatures than global warming.”
Sorry gbaikie, you are incredibly exxagerating here.
And you should know that, because you often have presented here information originating from the Berkeley Earth Group, without discrediting them, like do some pseudoskeptics on this site.
Why don’t you read their paper?
http://static.berkeleyearth.org/papers/UHI-GIGS-1-104.pdf
And if you mean the heating caused by the sum of all urban corners, of all industrial sites and of all energy plants on Earth with electricity production based on thermic processes (fossile burning, nuclear): these quantities are negligible in comparison witn Sun’s energy.
–Bindidon says:
December 14, 2018 at 9:23 AM
“… to do something about UHI effects, which causes higher air temperatures than global warming.”
Sorry gbaikie, you are incredibly exaggerating here.–
No. I am not
I am talking about people being affected, and I am not talking about polar bears or dolphins.
Most people live in urban area and most people get more warming from UHI effect as compared any increase in global temperatures.
“Take London as an example. Temperatures in the UK’s capital are, on average, one to three degrees celsius hotter than the surrounding countryside.”
https://www.wired.co.uk/article/uk-heatwave-london-urban-heat-islands
“These surface urban heat islands, particularly during the summer, have multiple impacts and contribute to atmospheric urban heat islands. Air temperatures in cities, particularly after sunset, can be as much as 22°F (12°C) warmer than the air in neighboring, less developed regions.”
https://tinyurl.com/y9rd2guq
LA area has highest urban heat island effect in California
“We call it not an urban heat island but an urban heat archipelago because it’s like a whole chain of urban heat islands that run into each other,” Solomon said”
https://tinyurl.com/yckukawv
–And you should know that, because you often have presented here information originating from the Berkeley Earth Group, without discrediting them, like do some pseudoskeptics on this site.
Why don’t you read their paper?
http://static.berkeleyearth.org/papers/UHI-GIGS-1-104.pdf —
Introduction:
“The Urban Heat Island (UHI) effect describes the observation that temperatures in a city are often higher than in its rural surroundings.
London was the first urban heat island to be documented but since
then many cities have been identified as urban heat islands.
A well-known example is Tokyo where the temperature has risen much
more rapidly in the city than in nearby rural areas: Fujibe estimates excess warming of almost 2°C/100yr compared to the rest of Japan the warming of Tokyo is dramatic when compared to a global average as seen in Figure 1. ”
I wouldn’t imagine London or Tokyo having much, I would guess Paris has more.
Tokyo:
“Japanese climate researchers at the Tokyo Metropolitan Institute for Environmental Sciences have found that in the last one hundred years, average temperatures in Tokyo have risen three degrees (Celsius), versus one degree for Japan as a whole. Of these three degrees, two are attributed to the urban heat island effect and one to global warming.”
https://tinyurl.com/y77qwe5x
Paris:
“The UHI has an ellipsoidal shape and stretches along the prevailing wind direction. The maximum UHI intensity of 6.1K occurs at 23:00UTC located 6km down-stream of the city centre and this largely remains during the whole night.”
https://tinyurl.com/yaalplu6
The Impact of Heat Islands on Mortality in Paris during the August 2003 Heat Wave
“Conclusions: Our results support the influence of night temperatures on the health impact of heat waves in urban areas. Urban heat exposure indicators based on satellite imagery have the potential to identify areas with higher risk of death, which could inform intervention decisions by key stakeholders.”
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3279432/
–And if you mean the heating caused by the sum of all urban corners, of all industrial sites and of all energy plants on Earth with electricity production based on thermic processes (fossile burning, nuclear): these quantities are negligible in comparison witn Suns energy.–
I don’t know what you mean by all urban corners…
But Urban heat island isn’t “electricity production based on thermic processes” and etc.
Urban areas are very small percentage of land area and land area is small portion of Earth surface.
And I am talking about people worried it getting to get too warm.
Animals and all of life, have lived in much warmer periods than today’s global average temperature. And no animals are dying from global warming.
gbaikie…”Most people live in urban area and most people get more warming from UHI effect as compared any increase in global temperatures”.
Of course they do. The radiation from a blacktopped road in summer can be felt, all global warming combined could not.
Ben dit donc!
UHI effects exist only if you average absolute data arising from UHI suspected stations with the rest.
If you construct your anomalies station by station, and then average the anomalies cell by cell, latitude band by latitude band and then month by month, the problem no longer exists.
Absolutely not. UHI is supposed to add a parasitic trend to stations observations. Nothing to do with the more or less urban character at a given moment. The anthropogenic disturbances of thermometers concern all scales. A rural station is not less sensitive to the phenomenon.
phi…”UHI is supposed to add a parasitic trend to stations observations”.
That’s especially true, as Anthony Watts found out, when the hot air exhaust from an air conditioning unit was blowing right onto a Stevenson screen containing a thermometer.
Those boxes are designed to protect the thermometer from direct solar radiation, and to a degree, from lateral winds. However rising hot air from a sidewalk, or winds above a certain level, can penetrate the box and affect the temperature.
If an air conditioner exhaust is blowing onto the box from nearby the air temperature measured by the thermometer is bound to be affected.
You write like scientists aren’t aware that concrete might increase a thermometers readings. (This is one reason why adjustment occur.)
How does someone “write like” they think something without actually writing that they think it?
Who here follows Nir Shaviv’s work? He believes much of the warming of the last 100 years is caused by various amplifications of solar irradiance, combined with Cosmic Ray Flux (CRF), which is affected by both our own Sun and our solar system’s orientation within the Milky Way. Among the solar forcings that Shaviv emphasizes is cloud formation (confirming Roy’s belief), as can be read about in one of the latest entries on his website, ScienceBits.
Shaviv is a good and interesting scientist, but I don’t see a lot of support for his claims among the broad climate scientific community. Mostly because, I think, the evidence that increased cosmic or solar flux influences climate via clouds just isn’t there.
tom…”Who here follows Nir Shavivs work?”
It’s been a while but I’m onside with Nir.
Why, Gordon?
DA…”Why, Gordon?”
He’s a smart researcher and a skeptic, as all smart researchers and scientist should be.
Lots and lots of scientists are smart researchers, and all are skeptics.
Why is Shaviv right and everyone else wrong?
You seem to have nothing to say about his science. Have you ever read even one of his papers, Gordon? Be honest….
Why is Shaviv wrong and everyone else right?
You seem to have nothing to say about his science. Have you ever read even one of his papers, David? Be honest…
JDHuffman
“E-man, you are correct in your pseudoscience, but wrong in your physics.”
Why are you speaking all the time about ‘pseudoscience’ without proving it is?
*
“Angular momentum is not transferred by the force of gravity.”
Do I understand you right?
Do you clearly doubt that tides raised in the oceans by the Sun and Moon on Earth dissipate significant energy, and transfer angular momentum from the spin of the Earth to the orbit of the Moon?
Such an opinion is new to me.
*
“NASA is convinced the Moon is moving away from Earth.”
NASA? Why NASA? Why do you reduce the world to the US?
*
“But, their belief is based on the measurements from reflectors left on the Moon. Such precise measurements are very questionable, due to the great distances and many variables.”
If these measurement are ‘highly questionable’: why don’t you show this, by either giving a source, or your own proof of it?
Since 40 years, people work at the Observatoires in Paris and Grasse on Lunar Laser Ranging, and what you name ‘questionable’ in fact is their daily work.
I’m a bit sad of translating French material into English all the time.
What about you using this time Google’s Translator to have an idea about what these people exactly do, and how they solve what you comfortably describe as ‘highly questionable’?
http://wwwrc.obs-azur.fr/cerga/laser/laslune/Webquestions.htm
“Even if it turns out the Moon is actually moving away, it is best explained in more logical ways than ‘transferring energy’.
Maybe, but then… why don’t you show us this ‘best explanation’, right here and now?
I would enjoy.
bindidon, you can’t understand the Moon’s motion if you believe a racehorse is “rotating on its own axis”, running the oval track. You can’t even begin to understand, with such a mindset.
What’s the French expression for “closed mind”?
Who told you I would? Maybe I misunderstood the question?
I am not at all interested in these stupid discussions about racehorses running on oval tracks, or similar stuff.
I am interested in science, e.g. that developed by Joseph-Louis Lagrange concerning an explanation for the reason why we can see Moon’s librations (an explanation based on decades of observations by several people).
If you think what this man wrote is ‘pseudoscience’, or is based on ‘circular reasoning’, then I invite you to give us a proof of it: here and now.
Why do you never consider your mind possibly being closed?
Well, when you commented as “La Pangolina”, you claimed a racehorse rotates on its own axis.
Are you now wanting to be removed from the list of Spinners?
“Well, when you commented as La Pangolina, you claimed a racehorse rotates on its own axis.”
That, JDHuffamn, is YOUR simple-minded belief, based on you being a gullible follower of Robertson’s egocentric nonsense.
Rose is Rose and JP is JP.
She never had read anything written by Cassini, Mayer, d’Alembert, Laplace, Lagrange, etc etc.
Btw, Rose is absent till mid of March next year, exactly as I was last year till March of this year… and my absence was what led the poor Robertson genius to think we are one and the same person.
Even Salvatore del Prete had understood I was on one of the Canary Islands!
Okay, I’ll leave La Pangolina on the Spinners list. Do you want to also stay on it?
If the racehorse isn’t rotating, why doesn’t it always face in the same direction?
DA, your stupid questions just keep getting stupider.
binny…”Rose is absent till mid of March next year, exactly as I was last year till March of this year… and my absence was what led the poor Robertson genius to think we are one and the same person.”
Don’t you find it a burden to carry two personalities?
Robertson
“Don’t you find it a burden to carry two personalities?”
Not more / not less than you finding it a burden to be appreciated by many people here as the most ignorant and most pretentious commenter, whose average comment size is inversely proportional to their relevance.
.
Again, if the racehorse isn’t rotating, why doesn’t it always face the same direction?
Because the racehorse is orbiting the track, DA.
Learn some physics.
Yes, obviously the racehorse is orbiting.
But why isn’t it also rotating, since it doesn’t always face in the same direction?
Question answered:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333170
and
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333249
Dr Roys Emergency Moderation Team says:
The moon does rotate, not on its own, but about an axis passing thru the center of the earth, the true and only one.
Does the Moon rotate about its polar axis?
No.
Dr Roys Emergency Moderation Team says:
Obviously, if you believed that our moon both rotated about the Earth-Moon center of mass AND rotated about its polar axis, then you would believe that we can see all sides of the moon from Earth.
Here’s exactly where you go wrong.
You claim isn’t true if (and only if) the Moon’s rotational period = its orbital period.
If you can’t envision that in your mind, take two tennis balls, mark one, and turn it as you let it orbit the first.
Nope. You just don’t understand rotation about an off-center axis. An object rotating about an axis far from its center of mass, always shows the same face to that axis as it does so.
Where is the off-center axis in the Earth-Moon system?
The Earth-moon barycenter.
DREMT:
The barycenter is, in classical physics, just a fancy word for the center of mass.
What’s an example of a nonrotating Moon that’s orbiting?
Dr Roys Emergency Moderation Team says:
An object rotating about an axis far from its center of mass, always shows the same face to that axis as it does so.
Really? Is this true for the Earth in the Earth-Sun system?
(No.)
That’s because the Earth is also rotating on its own axis.
DREMT: Wait a minute. You claimed:
“An object rotating about an axis far from its center of mass, always shows the same face to that axis as it does so.”
But you agree this isn’t true for the Earth, right?
So your claim fails.
It’s not a “claim”, David. It’s just rotation about an off-center axis. It’s a fact. The Earth doesn’t always show the same face to the axis it rotates about because it is ALSO rotating on its own axis. An object that is rotating about an off-center axis and NOT rotating on its own axis, always shows the same face to the axis it rotates about. This isn’t complicated.
DREMT: If the Moon isn’t rotating, why doesn’t it always face in the same direction?
There is the Theory of the Moebius, a twist in the fabric of space, where time becomes a loop…
…where time becomes a loop…
…where time becomes a loop…
…where time becomes a loop…
…where time becomes a loop…
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333609
I see, JDHuffman: you are not able to falsify Joseph-Louis Lagrange.
Please don’t try to escape using the superficial words of Nikola Tesla. These are far away from any valuable contradiction to Lagrange’s work.
Bindidon, I know you have difficulty with both English and physics, but do you realize you have created a “straw man” and a false accusation?
binny…”I am not at all interested in these stupid discussions about racehorses running on oval tracks, or similar stuff”.
You were just discussing angular momentum a couple of replies back. How could you not be interested in the angular momentum of a horse attached to a rotating platform, which is zero wrt to the wooden horse axis/COG?
The Moon isn’t attached to a rotating platform.
Neither is a racehorse going around the track.
David struggles to get to grips with the concept of an “analogy”
“The giant impact hypothesis is currently the favored scientific hypothesis for the formation of the Moon. Supporting evidence includes:
** Earth’s spin and the Moon’s orbit have similar orientations.
** Moon samples indicate that the Moon’s surface was once molten.
** The Moon has a relatively small iron core.
** The Moon has a lower density than Earth.
** There is evidence in other star systems of similar collisions, resulting in debris disks.
** Giant collisions are consistent with the leading theories of the formation of the Solar System.
** The stable-isotope ratios of lunar and terrestrial rock are identical, implying a common origin.”
https://en.wikipedia.org/wiki/Giant-impact_hypothesis
I tend to favor the idea.
And if true, it indicates that Earth and Venus are not similar or “twins”. Or those that believe the fairy tale that Earth and Venus started out the same, would be wrong. Or Venus didn’t start like Earth and something went wrong. If anything Earth started out like Venus and something went wrong with Earth- a terrible thing happened proto earth was hit by a very huge space rock.
Now it’s commonly assumed the planets are hit by big rocks during their formation, but Earth was hit by a particularly large rock or a small planet or perhaps using modern lingo, a large dwarf planet.
Though maybe this common assumption of Late Heavy Bombardment is wrong or over stated. Perhaps one can say it overly earth centric- rather than solar system event and it was mostly a Earth related event.
Late Heavy Bombardment
“During this interval, a disproportionately large number of asteroids are theorized to have collided with the early terrestrial planets in the inner Solar System, including Mercury, Venus, Earth, and Mars.”
https://en.wikipedia.org/wiki/Late_Heavy_Bombardment
Or there no evident of it, in regards to Venus. And off hand can’t think evidence of it on Mercury. Though both Moon and Mars have a large impact crater, Hellas basin and Aitken basin on the Moon.
Now big rock- 100 km diameter or larger will kill everyone on Earth, but one could say it does little in terms of “damaging” Earth or vast damage is largely effects the surface of Earth- the thin surface we are wholly dependent upon. So things like boiling ocean and the sky in complete ruins. Much Worse than the Hell on Earth.
Though a rock bigger than 500 km in diameter starts to do geological damage or geological transformation of Earth or effects Earth. So the biggest asteroid or also called the dwarf planet Ceres, has diameter of 945 km. And if hit Earth, would affect Earth. And giant impact hypothesis suggests larger rock than Ceres.
Mercury:
“Caloris Planitia is a plain within a large impact basin on Mercury, informally named Caloris, about 1,550 km (960 mi) in diameter. It is one of the largest impact basins in the Solar System. “Calor” is Latin for “heat” and the basin is so-named because the Sun is almost directly overhead every second time Mercury passes perihelion.”
…
“The impacting body is estimated to have been at least 100 km (62 miles) in diameter.
Bodies in the inner Solar System experienced a heavy bombardment of large rocky bodies in the first billion years or so of the Solar System. The impact that created Caloris must have occurred after most of the heavy bombardment had finished, because fewer impact craters are seen on its floor than exist on comparably-sized regions outside the crater.”
https://en.wikipedia.org/wiki/Caloris_Planitia
Test pilots take Virgin Galactic SpaceShipTwo to space for first time, land back on Earth
Supersonic space plane reaches 51 miles above Earth
“Mission official Enrico Palermo said the pilots reached reached Mach 2.9 and took SpaceShipTwo to an altitude of 51 miles before beginning its gliding descent. It landed minutes later.
The 50-mile mark is what earns astronauts their space wings, according to NASA. The International Space Station is about 200 miles above the planet.”
https://www.clickorlando.com/news/space-news/virgin-galactic-spaceship-first-trip-to-space-a-success
I would say US military counts 50 miles as space- one gets space wings.
Linked from:
http://www.transterrestrial.com/
Which says:
“At the Galloway Symposium last week, Jonathan McDowell made a good case that this, not the traditional Karman line of 100 km, is the right altitude. If one accepts that, it is the first flight of humans to space from American soil since the Shuttle retired over seven years ago. Heres hoping that Blue Origin does the same thing next year (except theyre designed to get to 100 km).”
Blue Origin had hoped to have a manned flight this year.
They must be running behind schedule.
Gordon Robertson says:
“the Stefan-Boltzmann equation, supplies no provisions for a two way heat transfer or a two way EM exchange.”
Can you get this:
With P for power [W], e for epsilon, s for sigma:
Pt = Ph-Pc = esATh^4 esATc^4 = esA(Th^4 Tc^4)
Ph goes from hot to cold. It is not heat.
Pc goes from cold to hot. It is not heat.
Pt goes from hot to cold. It is heat.
Pt = Ph – Pc = esATh^4 – esATc^4 = esA(Th^4 – Tc^4)
Two terms, two directions, the total is from hot to cold.
Svante, you are mis-applying the S/B Law. The S/B Law does not involve absorp.tion.
You have not been paying attention.
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333516
I didn’t say S/B law, cross it out if you like.
“The S/B Law does not involve absorp.tion.”
Aha.
1. Wikipedia in English
A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity, ε < 1.
2. Wikipedia in German
Ein Schwarzer Körper ist ein idealisierter Körper, der alle auf ihn treffende Strahlung vollständig absorbieren kann (Absorp.tionsgrad = 1).
Nach dem kirchhoffschen Strahlungsgesetz erreicht daher auch sein Emissionsgrad ε den Wert 1, und er sendet die bei der betreffenden Temperatur maximal mögliche thermische Leistung aus.
Did Wikipedia’s S/B page (which you referred to more than once) suddenly become ‘pseudoscience’ too?
Interesting!
Aha.
Did Bindidon look before jumping off that cliff?
Interesting!
By definition, a blackbody absorbs all radiation incident upon it.
By definition, irrelevant is “unrelated to the matter being considered”.
You wrote “The S/B Law does not involve absorp.tion.”
That’s wrong — 100% ab.sorp.tion is a precondition for applying the Law to the body in question.
No, it’s right.
As usual, you don’t understand the relevant physics.
(How many examples do you need?)
G*:What’s the definition of a blackbody?
dud, maybe you should ask your therapist next time you are getting counseling about your obsession with your G*.
Then, learn some physics.
I’m asking you, because I don’t think you know.
No dud, you’re asking me because you have nothing else. You don’t understand the relevant physics. You can’t think for yourself.
Your endless questions are just a cover for your incompetence.
I’m asking you, because I don’t think you know.
You have yet to prove that wrong.
dud, when you have convinced yourself that a racehorse is also rotating on its own axis, as it runs an oval track, then you don’t understand the relevant physics.
All you have are your empty debate tricks, such as your irrelevant questions.
Just keep providing more evidence of your incompetence. It’s here for all to see.
David Appell
It will help you understand g.e.r.a.n when you realize how he uses words.
When he says the word “physics” he means his unsupported declarations. When he says you have to learn physics it means you must accept his unsupported declarations.
Also you must realize he never studied a day of actual physics in his life. You can tell by reading his posts. He does not know even the simple ideas in physics and makes up tons of stuff. He is a product of Principia Scientific blog and still posts there. On this blog most the articles are this form of declarations with zero supporting evidence and often contradictory to real physics based upon experimental and observational evidence. All kinds of unsupported claims are made on this blog.
Again you are correct and g.e.r.a.n shows his complete lack of knowledge.
Here:
https://www.britannica.com/science/blackbody
Here:
https://www.learner.org/courses/physics/glossary/definition.html?invariant=blackbody
Here:
https://whatis.techtarget.com/definition/blackbody
Thanks Norman, but I know how Ger*an avoids questions. “Learn some physics” is a favorite excuse, and others. I’m not so much looking for a real answer as making it clear that his arguments are pathetic. They are all he has left.
JDHuffman says:
dud, when you have convinced yourself that a racehorse is also rotating on its own axis,
If the racehorse isn’t rotating, why doesn’t it always face in the same direction?
Just answer, finally.
David Appell
Yes you have proved a true point. He will not answer a direct science question posed to him.
g.e.r.a.n never answered questions I asked when he went by that name.
Your questions have been answered, David.
I’m glad to see DA finally admit he is not really looking for answers. That has been obvious for a long time, but now we have his exact admission.
And Norman jumps in to help DA. That’s always funny also. Like DA, Norman always uses a standard set of tricks. He likes to insult, falsely accuse, and misrepresent. Nothing new with these two.
Now that DA has admitted he does not really want an answer, I will answer his last question.
“If the racehorse isn’t rotating, why doesn’t it always face in the same direction?”
DA, the racehorse IS always facing the same direction. It is the instantaneously direction of the resultant vector. This is why the racehorse, or race car, is an adequate model of orbiting. A moon follows its orbital path due to the sum of the vectors acting on it.
To someone that does not understand the relevant physics, it appears as if the racehouse is “rotating on its own axis”, but that is not happening. For example, a person inside the track would always see the same side of the horse, just as we always see the same side of the Moon. But in some other frame of reference, it appears the Moon is rotating. But, frames of reference do not matter with “its own axis”. Certain people learned this when the example of an airplane prop was used. You can find some special frame of reference where the rotating prop appears to not be rotating, but you wouldn’t stick your arm in it. Reality always wins.
Try something new. Face reality. Learn some physics.
Huffingman, Yes, if you arbitrarily pick a coordinate system, you can produce a situation in which there appears to be no rotation, as you describe. For an obvious example, select Earth based coordinates with the X axis pointing from the center toward 0.0 longitude, the Y axis pointing toward 90.0 East longitude and thus the Z axis points toward the North Pole. Using those coordinates, the Earth does not appear to rotate because the coordinate axes are themselves rotating about the Z axis. But, it’s rather obvious that the Earth rotates, don’t you agree?
You obviously don’t comprehend the need to use a non-rotating coordinate system as the basis of your claims regarding the Moon’s rotation. Conservation of angular momentum requires that a spinning gyro within a 3 axis gimbal mount, which is pointing in a certain direction in space, will continue to point in that direction absent some external force to move the gyro. Add 2 more gyros mounted orthogonally within the same mount and you have a non rotating reference, which when placed on the surface of the Earth or the Moon, would prove that both are rotating about their axes.
Learn Some Physics and Please Stop Trolling.
Yes Swanson, the Earth rotates on its own axis, as demonstrated in various ways.
But, the Moon does NOT rotate on its own axis, as demonstrated in various ways.
PS Get over yourself.
Huffingman, Here’s a video which “proves” the Earth doesn’t rotate, that is, if one uses Huffingman’s flawed version of reality. The video shows the same perspective with which one views the Moon, that is, the camera location above the same side of the planet always captures the same view for the observer.
Of course, just like the Earth in the video, the fact that Moon’s image doesn’t change does not prove that the Moon lacks rotation. Please Stop Trolling and Learn Some Physics.
JDHuffman says:
But, the Moon does NOT rotate on its own axis, as demonstrated in various ways.
So demonstrate it.
JDHuffman says:
DA, the racehorse IS always facing the same direction.
Loco.
Let’s say the racehorse starts out, out of the gate, facing and running to the north.
What direction is it facing after the first turn?
PS: Let’s call the “first turn” the first quarter turn.
What direction is the horse facing after the first quarter turn?
Swanson believes a geostationary satellite proves the Earth is not rotating on its axis!
And DA still can’t understand the difference between orbiting and rotating on its own axis.
The clown competition is fierce tonight….
Again….
JDH wrote:
DA, the racehorse IS always facing the same direction.
Let’s say the racehorse starts out, out of the gate, facing and running to the north.
What direction is it facing after the first quarter turn? Still north?
binny…”A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity, ε < 1".
The only reason that body is absor.b.ing is because it is cooler than the body that radiated the EM it is absor.b.ing.
The S/B law does not involved absorp-tion. It is based on the EM radiated from a very hot body to its environment. Or, maybe to a body surrounding it, like a sphere or cylinder.
If the sphere or cylinder was significantly hotter, the equation would be applied from the hotter body to the cooler body.
Only at thermal equilibrium can there be a two-way transfer.
Gordon Robertson says:
The only reason that body is absor.b.ing is because it is cooler than the body that radiated the EM it is absor.b.ing.
Really??? How does the ab.sorb.ing body know where the radiation came from?
svante…”Can you get this:
With P for power [W], e for epsilon, s for sigma:
Pt = Ph – Pc = esATh^4 – esATc^4 = esA(Th^4 – Tc^4)
*****
Your equation is not quite right. There is no power transmitted since power = work/sec. You could use power in relation to the power dissipation of the body of temperature, Th, but not in relation to the environment represented by Tc.
The temperature gradient, Th – Tc is intended to represent the temperature gradient in a solid where heat flows from hot to cold.
In place of Pt, you could have something like R for radiation intensity. The equation is not about two bodies radiating at each other, it’s about one body with Th radiating to an environment at temperature Tc.
The only reason you can pull esA out of the equations as a common factor is that they apply to the same radiation source, which has an area A, an emissivity, e and the S_B constant, s is a constant anyway.
There is no reference in this equation to Tc^4 having an area or emissivity. This form of Stefan’s original equation,
R = sT^4, has Th^4 radiation EM and Tc^4 as the temperature of the environment.
If Tc^4 was a separate body radiating, it would require it’s own A and e, and the terms A and e could not be pulled out as common terms.
Equations on the net showing a separate body of a cooler temperature radiating to a hotter body are bogus. The authors are confused about the meaning of S-B.
When I look at the two temperature version of S-B, I think JD is right, there’s something amiss.
The T^4 term comes from a very hot body with an average temperature around 1000C. In order for the second T term to apply to a surrounding environment, the T value would have to be much lower than T^4.
Stefan got the T^4 relationship from data guestimated by a third party when the colours described by Tyndall on the colour of his glowing platinum filament wire changed colour from a reddish colour to a much hotter orangy-white colour.
The third party guestimated the actual temperatures from the equivalent colour temperatures of other heated bodies then Stefan used those temperature and related temperature to EM radiation.
If the temperature gradient in S-B is to be applied, it can only be applied to a heated filament of equivalent temperature.
In other words, if one end of the filament was at 1500C and the other end at 1200C, the temperature gradient could determine the radiation intensity.
I would guess that as the lower temperature approaches the higher temperature, or vice-versa, the limiting radiation intensity would be described by Stefan’s equation,
R = fi.T^4
I need to study this more but have other things on my mind right now. Like what kind of snack to have.
We need to keep in mind that Stefan and Boltzmann thought heat flowed through the air as heat rays. In those days, some kind of aether was suspected to transport the heat. We now know that is incorrect, that heat cannot flow through air other than at a very low conductive level via conduction, or as a transport of mass, as in convection.
Somehow I think the lads visualized heat flow through air when they devised their equation. EM should not be expressed in W/m^2.
EM as energy can do no work and it carries no heat. After conversion to another form of energy, like heat or electricity, work can be done by the heat or the electricity by converting each to mechanical energy.
I suppose one could argue that the movement of electrons in a conductor is work but that would be internal work, which cannot be harnessed. EM can cause electrons to move in a conductor, but that conductor needs to interact with a magnetic field or another current-carrying conductor to do external mechanical work.
There is no way to convert EM straight to mechanical energy as work. I don’t think watts should be associated with EM and the only way to associate watts with heat is by a conversion of heat to its mechanical equivalent.
The native measurement of heat is the calorie, not the watt.
Gordon says: “I dont think watts should be associated with EM…”
Gordon, you’re likely familiar with a spectrum analyzer. Recall that typical choice of units, for the display, is “milliwatts”. EM is easily measured in (associated with) units of Watts.
Gordon Robertson wrote:
I dont think watts should be associated with EM and the only way to associate watts with heat is by a conversion of heat to its mechanical equivalent.
It doesn’t matter what you think — you’re wrong.
Explain the well known solar constant at Earth: 1365 W/m2.
The W (watts) is simply the energy passage per unit time. Power.
JD…”Recall that typical choice of units, for the display, is milliwatts. EM is easily measured in (associated with) units of Watts”.
I know, but the watts being measured is related to the electrical current produced by the EM in a measuring device. In other words, the mW is a measure of the effect of EM on electrical charges in a conductor that encounters the EM, not a measure of EM itself.
If you want to measure EM, you put up an antenna, or a coil, and the EM induces an EMF in the antenna/coil. The EMF generates a current and the power is the product of the current and the EMF that produced it.
It’s telling you the EM can produce a certain current in a conductor when it encounters a conductor, but the EM itself can do no work as EM, it must first be converted to electrical energy, which can produce a current that can do work (and be measured in watts/sec). As electrical current, it is no longer EM.
I know this sound nit-picky but it is what is being over-looked by alarmists when they claim EM from a colder atmosphere can raise the temperature of a warmer surface that warmed it in the first place. Alarmists are also freely inter-changing EM and heat, as if they are one and the same.
That leads to confusion re the 2nd law.
DA…”Explain the well known solar constant at Earth: 1365 W/m2.
The W (watts) is simply the energy passage per unit time. Power”.
The watt is not a measure of generic energy, it is a measure of the mechanical energy related to work. The watt is derived from the horsepower, which is a measure of the mechanical energy of horses doing work.
The 1365 W/m^2 is a measure of the effect of solar EM on the instruments used to measure it. Although the watt is a measure of mechanical energy is can used as an equivalent of the work done on an electric charge by a voltage difference. EM reacting with a meter’s pickup produces and EMF in the coil and that EMF produces a current.
The product of that current through a load (the meter) and the voltage across the meter, which is produced by the interaction of the EM and charges in the circuit, is measured as the power created by the resultant electric current.
In other words, the current produced is doing work on the meter to deflect its needle in an analog meter. The EM is not doing the work, the current it produces is doing the work.
EM cannot do work if it does not interact with mass, therefore describing EM as having a power is ingenuous. It may be useful for visualization but it doesn’t mean anything when applied to solar EM at TOA.
1365 W/m^2 is the power it can potentially produce, but as EM in free space it has no power.
Gordon Robertson says:
The watt is not a measure of generic energy, it is a measure of the mechanical energy related to work.
Completely and idiotically wrong.
Power = dE/dt, regardless of the energy source.
Gordon, you so often show that you don’t even understand the most basic of physics. Pathetic. Luckily you have a very generic Canadian name so no one knows who you are.
DA…”Power = dE/dt, regardless of the energy source”.
Every time you respond, you reveal a deeper level of ignorance than the time before. I would not think it possible.
Power is a measure of work, not generic energy. There is no such thing as generic energy unless you are talking philosophically. As such, energy has no form and no units.
If you insist on using math, try thinking a bit before you use a generic energy term.
Power = work/unit time = energy/unit time
1 HP = 550 ft-lb/sec
1 HP = 746 watts with watts measured in Kg-m^2/sec^3
The word ‘watt’ is named in honour of James Watt, who developed the HP based on the work done by real horses per minute (33,000 ft-lb/minute)
Work is energy….energy is work.
Gordon,
There is also that Ph and Pc are only virtual quantities. These are the powers that would radiate these bodies if they were in an environment at 0 K. P=f(Th, Tc) is the only valid power, the only one to have a physical sense.
phi…here’s a good explanation of how Stefan determined the T^4 relationship between EM radiation and a certain temperature.
http://www.applet-magic.com/stefanlaw.htm
“Tyndall had found that the total radiant energy emitted by a platinum wire at a temperature of 1200C is 11.7 times that radiated at 525C. A temperature of 1200C is an absolute temperature of 1473K and 525C is equivalent to 798K”
“1473/798 = 1.849
and
(1.849)^4 = 11.61
which is very close to 11.71”
If E = fi.T^b, then based on Tyndall’s temperature to radiation relationships:
E1/E2 = (T1/T2)^b
Note: this relationship suggests the ratio of radiation intensities versus the ratio of temperature difference is not linear.
so…b = log(E1/E2)/log(T1/T2)
Since, according to Tyndall, E1/E2 = 11.7
and T1/T2 = 1473k/798K = 1.849
then, b = log(E1/E2)/log(T1/T2) = log 11.7/log1.849
=4.0016
b = 4.0016
There’s your 4th power.
However, that 4th power was derived from very hot temperatures at 1473K and 798K. I don’t see why the 4th power would still apply if the Tc temperature is at room temperature.
For one, we don’t know that the relationship between radiation at temperatures in the 1473K range would have a ratio of 1.849 with room temperature at 293.15K.
In fact, that ratio is 1473K/293.15K = 5
That suggests Stefan-Boltzmann might not apply across the temperature spectrum.
The equation….E = esA(T^4 – Tc^4) is meant to measure radiation heat loss between bodies at the respective temperatures, not the net radiation between two bodies of different temperatures.
SB is derived analytically via integration of the Planck distribution. It is not merely empirical.
Gordon Robertson says:
That suggests Stefan-Boltzmann might not apply across the temperature spectrum.
What experimental evidence supports your supposition?
(My guess: none.)
DA…”Gordon Robertson says:
That suggests Stefan-Boltzmann might not apply across the temperature spectrum.
What experimental evidence supports your supposition? ”
I just explained it in the post above your reply. Is it that far over your head?
I supplied a link to the derivation of the 4th power used by Stefan in E = fi.T^4. The 4th power relationship is based on Tyndall’s observation that a certain ratio of EM radiation existed between a narrow range of temperatures in a heated platinum wire.
Why don’t you supply the proof that it applies across the entire temperature range from 0K to whatever? A genius like you should find no problem with that.
Gordon, please stop the gish gallop.
Let’s do one step at a time please?
You know how to calculate with units, do you?
In metric units:
e: 1
s: W/(m^2*K^4)
A: m^2
T: K
e*[1] * s*[W/(m^2*K^4)] * A*[m^2] * (T*[K])^4 =
esAT^4 * [1*W*m^2*K^4 / (m^2*K^4)] =
esAT^4 * [W]
Watt is a unit for the property of power.
You can convert it to any other unit of power, for example horse power.
Power is energy per unit of time, or work/sec as you said.
Are you with on that?
svante…”Power is energy per unit of time, or work/sec as you said.
Are you with on that?”
Of course I am. However, the equation you provide has no time factor in it. It is a measure of radiation intensity, not power, therefore it should not be measured in watts.
The presumption that EM has power is based on the old notion that it is heat rays, and as such, presumed to have the same ability as heat to do work.
The units of watts is kg-m^2/s^3, clearly a measure of mechanical energy. You can write that as:
(Kg-m)/s . m/s^2. Remember, the watt is a measure of mechanical energy per unit time.
The first part, Kg-m/s describes the work done or energy expended. Work = fd where f = Kg and d = m. However, that use of Kg as a force means Kg-mass has been converted to a Kilogram-force, which requires the Newtonian definition of force as f = ma = mg.
In this case, Kg means Newtons of force.
A kilogram-force, in newtons, can be defined as the force created by a Kg-mass by the acceleration of gravity, which is expressed in m/s^2.
Therefore watts are described in the form:
work/sec = kg-m/sec = f.d/sec = mg.d/sec
= (kg.m/sec^2)/sec . m = kg-m^2/sec^3
Now, show me how anything in that definition of power applies to R = esA(Th^4 – Tc^4)
The S-B constant is rated in watts/m^2/T^4 but I am questioning their measurement of EM in watts and how Stefan-Boltzmann came to incorporate the watt in their constant. I would agree that power can be used to measure the loss of heat at the surface due to heat dissipation, but I don’t think it’s proper to claim that lost heat is now flowing through space as EM.
I would agree there is a relationship between the intensity of the EM produced and the heat loss, however EM and heat are entirely different forms of energy with EM measured in electron volts and defined by its frequency. Heat has no frequency.
The radiated EM is not doing work until it encounters mass. Even then, it’s not the converted EM doing the work, it’s the energy to which the EM is converted doing the work, or causing the work.
Work = force x distance and EM is not a force. EM can do work if it encounters a charged particle and causes the charge to move. Until then, it is nothing more than potential energy, which has no power.
Gordon Robertson wrote:
EM can do work if it encounters a charged particle and causes the charge to move. Until then, it is nothing more than potential energy, which has no power.
EM carries energy.
If it impinges on a surface, it transfers an amount of energy dE/dt per unit time.
That’s a power.
Per unit area that’s dE/A*dt. It’s not “potential energy” — it’s actual energy. It can be measured.
DA…”EM carries energy.
If it impinges on a surface, it transfers an amount of energy dE/dt per unit time”.
EM ‘IS’ energy, it does not ‘carry’ energy. It is electromechanical energy consisting of an electric field with a perpendicular magnetic field and is defined by its frequency.
When it contacts a surface, if the conditions are right, it is CONVERTED to another form of energy by electrons in the the atoms of the material. That energy is either heat or electrical energy, or both.
Gordon Robertson says:
“However, the equation you provide has no time factor in it. It is a measure of radiation intensity, not power, therefore it should not be measured in watts.”
You do not need to think about EM, think only about the result: Tk.
The same physical property can be measured by many units.
Units are irrelevant as long as they measure the same physical property.
1) Power is the property energy divided by the property of time.
2) Time can be measured in the unit s.
3) Energy can be measured by the unit J.
4) Power can be measured by the unit W.
5) W is defined as J/s, or kg*m^2/s^3 like you say.
6) Tk has the unit Watt so the property is power.
Are you with me?
Roy: Drought is back in California. And even worse in Oregon.
https://twitter.com/DroughtCenter/status/1073224192634576897
This is not unusual.
Though some ‘commenter’s might try to identify the following with alarmism, it has nothing to do with it.
2018 was an unusually warm year in Western Europe, an especially in Northwest Germany. We really had a centennial summer.
Corn only grew there up to about 1.2 meters high (over 2 meters otherwise). And corn harvest was mid August instead of October.
To talk about almost empty dams and rivers in Germany this year is imho superfluous.
But that is far from a reason to talk about climate catastrophes, just as a cold November in Northwest America does not mean a global slowdown:
https://s3-us-west-1.amazonaws.com/www.moyhu.org/2018/12/map.png
–To talk about almost empty dams and rivers in Germany this year is imho superfluous.–
Emptying dam water is a choice. Or any dam can be drained fairly quickly. Dams are about retaining extra water and using the over abundant of water as a power resource and a water resource.
If want talk about nature rather than choices by humans, one should a look lake levels of lakes which are not dammed or not controlled by management choices.
You also wonder if dam resources are enhanced by adequate dredging- another management choice.
gbaikie…”Emptying dam water is a choice”.
Around the Vancouver, Canada area it comes down to water management. The local water authority makes bad decisions at times by allowing dam levels to get too low during the rainy season.
The data says CO2 has little if any effect on climate. Temperature is now about what it was in 2002. CO2 has increased since 2002 by 40% of the increase 1800 to 2002. By similarity, none of the other ghg (except water vapor) have any significant effect on climate either. https://pbs.twimg.com/media/DuaFJw7VYAEz-mU.jpg
Dan Pangburn says:
The data says CO2 has little if any effect on climate. Temperature is now about what it was in 2002.
Not accounting for natural variations — and cherry pickin — here is dishonest and the same as lying.
THe UAH LT v6.0 annual average for the decade of the 2000s was 0.14 C.
For this decade-to-date it’s 0.23 C.
For N.O.A.A. surface temperatures: 0.61 C vs 0.76 C.
But the real tell is ocean heat content, increasing almost monotonically:
http://tinyurl.com/jbf2xco
DA,, Because of low effective thermal capacitance, TLT responds much faster (about 30X faster) than the surface, which is 71% ocean.
15*10^22 J added to the top 700 m of ocean would raise its temperature only about 0.138 K. Considering the roiling of the surface, http://www.youtube.com/watch?v=1ir1w3OrR4U it is unclear how well that graph represents SST.
The more significant tell is that CO2 has increased since 2002 by 40% of the increase 1800 to 2002. This is a large change in CO2 compared to T change, even by the T measures that you consider important. Another 40% increase in CO2 would increase T by only about 0.14 K even using the values for T that you consider important.
The linear trend of UAH LT v6.0 from mid-2002 to present (Oct 2018) is +0.19 C/decade. That’s a warming of +0.32 C (= +0.57 F).
(And easily statistically significant.)
Based on what statistical model? If you don’t know the model, or if the model even applies, then you don’t know what it signifies.
DA,, The assumption that a linear trend might be meaningful is bogus when obviously there was an el Nino.
Exactly. These trend fetishists annoy me no end.
I believe that Touma ans Wisdom 2001 are spot on with their analysis about Venus.
http://iopscience.iop.org/article/10.1086/321146/pdf
In short; planetary pertubations on orbits and cycles has led to entry of the chaotic zone (google Laskar chaotic zone) in which the planet fluctuates wildly. This has severe consequences for the alignment of the core axis with the mantle axis eventually leading to a spinning stop whilst converting rotational energy to heat. And there is plenty of energy to explain all of Venus enigmatic geologic hmmm Venologic melting/heat features.
andr…”planetary pertubations on orbits and cycles has led to entry of the chaotic zone (google Laskar chaotic zone) in which the planet fluctuates wildly”.
I think such a theory might also better explain the epicentric behavior of earthquakes.
The current theory, based on the plate tectonics hypothesis, explains earthquakes as due to movement of large plates moving underneath each other. I would think that if a plate of several hundred miles, or maybe several thousand miles, slid under another, there would be countless earthquakes all along the faces between the plates.
As it stands, earthquakes can be pin-pointed to within miles distance and feet of depth. That is based on the triangulation of seismic waves emitted by the quakes. I think the turbulent motion described inside the Earth in this article could better explain the localized action of earthquakes.
The quakes could better be explained by fluids in certain regions being forced in localized faults, causing local action.
I would think that if a plate of several hundred miles, or maybe several thousand miles, slid under another, there would be countless earthquakes all along the faces between the plates.
It doesn’t matter what you think. What do you have evidence for?
DA…”It doesnt matter what you think. What do you have evidence for?”
Where’s the evidence to support the plate tectonics theory? Some bleary-eyed geologist saw patterns between the horn of Africa and South America and concluded the continents must have drifted apart. That’s why plate tectonics theory used to be called the continental drift theory.
Scientist with more sense have noted that the sediments in the oceans off such areas don’t match, for a start. And, no one has ever witnessed a phantom plate move under another plate, or buckle to cause mountain ranges.
“Wheres the evidence to support the plate tectonics theory?”
The plate boundaries can be drawn.
Ocean floor spreading
Most of earth surface is less than 1 billion years old.
The moon doesn’t have plate tectonic movement
There are no boundary which are drawn
Most/all of surface is more 1 billion years old.
No attempt made to make a complete list, rather let’s start there.
–By Stephanie Osborn
The Osborn post is a lengthy explanation of Dr. Zharkova’s model, model updates and predictions, with some additional example of how the ‘barycentric wobble’ influences the earth’s temperature. For readers who found Dr. Zharkova’s GWPF Presentation confusing, this article will help with the understanding of her model’s significance, and the output is worth considering. Osborn’s bio is HERE.–
https://nextgrandminimum.com/2018/12/12/the-latest-on-the-double-dynamo-solar-model-and-dr-zharkovas-predictions-of-a-grand-minimum/
Linked from:
https://tallbloke.wordpress.com/
I didn’t get anything new from this, but maybe some will like the explanation better.
Oh, I guess this clarifies it better:
“The gist of the matter is that all three main cycles are entering minimum phase, beginning with the end of this current solar cycle (Cycle 24). Cycle 25 will be even lower than 24, with 26 being very nearly flat-lined. Cycle 27 will begin to show a few signs of life, then there will be a gradual rise to full activity over several more solar cycles, even as the last three cycles have slowly decreased in levels. “
It clarifies it better, in sense of what she meant by knowing if theory correct in 5 to 10 years.
But not sure it will prove the theory, but it could disprove the theory. Though the chance of disproving it, is an important aspect.
What could prove it better is a continuing to improve the model over the next 5 to 10 years.
Does Stephanie Osborn have any track record in correctly predicting future solar cycles?
“Does Stephanie Osborn have any track record in correctly predicting future solar cycles?”
Does STEPHANIE OSBORN, Interstellar Woman of Mystery, have track record in correctly predicting future solar cycles?
I don’t think so. Her bio includes:
“Now retired from space work, Stephanie has trained her sights on writing. She has authored, co-authored, or contributed to over 35 books, including the celebrated science-fiction mystery, Burnout: The mystery of Space Shuttle STS-281. She is the co-author of the Cresperian Saga book series, and has written the critically acclaimed Displaced Detective Series, described as “Sherlock Holmes meets The X-Files,” ….”
http://www.stephanie-osborn.com/Bio.html
I doubt anyone in the world has had successful track record of predicting solar cycles. But that might not include a “Interstellar Woman of Mystery”
–David Appell says:
December 15, 2018 at 10:11 PM
Or Zharkovas? Talk is cheap. —
So far there is “1,039 Responses to Can Space.com Teach Us Anything Useful about Climate? and as such is evidence that does prove your point.
I don’t think Space.com can teach us anything useful about climate. Nor would I even say Space.com can teach us anything useful about space, though Space.com can have some stuff which is interesting regarding space, and might even be said to be above average in comparison with other space news sites.
For instance:
How to See Ghostly Green Comet 46P/Wirtanen Fly by Earth This Weekend
By Elizabeth Howell, Space.com Contributor | December 15, 2018 08:00am ET
“You can expect a ghostly green blob to grow brighter in the sky near Orion in the coming days, as Comet 46P/Wirtanen makes it closest approach to the Earth in 20 years this weekend.
Shining just bright enough to be glimpsed with the naked eye, this fuzzy visitor of ice and rock will be easily visible through late December. NASA has even sponsored an observing campaign (led by the University of Maryland) to track the comet with professional and amateur astronomical groups.”
https://www.space.com/42751-see-comet-46p-wirtanen-earth-flyby-december-2018.html
Could be interesting, and perhaps useful for some people for various reasons.
Or Zharkovas? Talk is cheap….
gbaikie says:
For readers who found Dr. Zharkova’s GWPF Presentation confusing….
Does she ever present to real science groups, not fake ons like GWPF? Or publish anywhere?
David, I assume you know how publishing works. And don’t think she is going on some sort of speaker circuit.
The general idea of grand min is hardly new, and eventually NASA and etc are going to venture a guess about cycle 25.
swannie…”Huffingman, Yes, if you arbitrarily pick a coordinate system, you can produce a situation in which there appears to be no rotation, as you describe. For an obvious example, select Earth based coordinates with the X axis pointing from the center toward 0.0 longitude, the Y axis pointing toward 90.0 East longitude and thus the Z axis points toward the North Pole. Using those coordinates, the Earth does not appear to rotate because the coordinate axes are themselves rotating about the Z axis. But, its rather obvious that the Earth rotates, dont you agree?”
*********
It’s the opposite from what you claim, the Moon is not turning about it’s axis but from other frames of reference it can APPEAR to be rotating. It’s an illusion. It does not appear to be rotating about it’s own axis, however, it appears that a radial line on the Moon is turning. That is exactly what one would expect with curvilinear motion.
Your analogy of the Earth with the coordinate system inside the Earth, and turning with it, is wrong. You would analyze the Earth’s rotation from a fixed axis which is not turning.
You can see that better using polar coordinates. Set up a fixed X-y coordinate system with a radial vector positioned with its tail at 0,0 and pointing along the x-axis in the +ve direction. As the radial line, r, rotates from the x-axis with angle theta between it and the x-axis, the instantaneous angular velocity of the line is d(theta)/dt.
That line represents one of the lines you defined, say from the Earth’s centre to the zero meridian (longitude). The line is turning on a fixed x-y coordinate system. In the case of the Earth, the angular velocity of such a line is fixed at 360 degrees every 24 hours.
It does not matter where you are on the planet, the angular velocity is the same. No matter where you are it will take 24 hours to complete a 360 degree rotation.
If you now apply that to a radial line on the Moon, from its centre to the point where it meets the Earth, that line can never turn about its axis. If you use the same fixed x-y axis with the earth at it’s centre, you can now describe a moving axis for the Moon that is centred on the Moon’s centre.
That Moon x-y axis will move around the Moon’s orbit with the radial line described above always pointed at the Earth. That Moon axis is not rotating it is translating with curvilinear motion. There is no rotation of the radial line from the Moon’s centre about the 0,0 centre of the Moon coordinate system.
There can’t be, that radial line is a major axis on the Moon coordinate system and it is always pointed at the Earth. The Moon coordinate system is translating in orbit with no angular velocity about it’s origin.
Exactly Gordon.
Swanson attempts another smokescreen with the gimbals. But, as usual, the angular momentum created by gimbals just prove him wrong.
The Earth rotates on its own axis at such great angular momentum that the axis maintains the same orientation, relative to the stars. The Moon demonstrates no such rotation.
But Swanson will trick the uneducated typists….
If you stood on the surface of the Moon, would you always see the same stars?
Stupid question time, again.
Everyone knows that you avoid questions you can’t answer by calling them “stupid.”
What’s comical is that you think people can’t see through you. Fool.
Yes, JDHuffman is the king of stupid since Flynn left:
No. Words Avg Unique% Stupid Name
206 5344 25 21.4 6 David Appell
191 6469 33 20.5 33 JDHuffman
98 14782 150 15.2 4 Gordon Robertson
81 2310 28 25.4 0 Dr Roys Emergency
62 12450 200 17.1 0 gbaikie
45 6870 152 16.4 16 Norman
34 892 26 39.8 0 Svante
31 1196 38 35.4 0 Entropic man
28 2965 105 34.2 1 Bindidon
24 1356 56 37.6 0 bobdroege
21 1973 93 31.2 0 Tim Folkerts
20 1875 93 37.1 0 gallopingcamel
20 1946 97 29.1 0 Ball4
19 1005 52 45.2 0 Nate
19 3009 158 26.5 0 Geoff Wood
12 838 69 40.1 0 phi
11 270 24 61.9 0 Jack Dale
10 1183 118 37.1 0 E. Swanson
Thanks, Mark.
If you get time, how about a quick check on the word “dork”?
Wrong, DA.
Your questions are just stupid. You don’t understand the relevant physics, but you want desperately to defend your false religion. So you come up with distracting questions hoping that when someone answers, you can come up with even more distracting stupid questions.
You don’t want an answer. You just want to pervert and corrupt truth.
JDHuffman
YOU: “If you get time, how about a quick check on the word “dork”?”
Interchangeable with poster that goes by JDHuffman. If you want to understand the word “dork” read the many nonsense posts of JDHuffman (previously g.e.r.a.n).
You could not get closer to what a dork is than reading the posts.
Poor frustrated Norman Grinvalds. He has nothing to offer except his obsession with g.e.r.a.n and his failed pseudoscience.
At least he can type.
Case proven
Norman, clowns make fools of themselves.
I just point out their antics.
And, of course, that does “prove the case”.
Gordon Robertson says:
Its the opposite from what you claim, the Moon is not turning about its axis but from other frames of reference it can APPEAR to be rotating.
Wrong.
Look at the Moon on the left in this animation:
https://en.wikipedia.org/wiki/File:Tidal_locking_of_the_Moon_with_the_Earth.gif
Does its dark patch always face in the same direction?
If you stood on the Moon’s surface, would you always see the same stars?
Gordo, Huffingman and other confused Lunatics.
See my reply with video above.
Also, in 24 hours (1 Day), the Earth rotates 360 + 360/365.25 degrees. That extra bit of rotation is necessary to account for the motion of the Earth around the Sun during that 24 hour period. Learn Some Physics.
False.
That false logic would lead you to believe the racehorse is making two laps for every lap.
But, your comedy is definitely improving.
Have you considered sticking your head in a rotating airplane prop? That would be funny!
JDHuffman says:
That false logic would lead you to believe the racehorse is making two laps for every lap.
Uh… No.
Ignore leap years and assume the Earth orbits circularly around the Sun in exactly 365 days.
Today the Earth rotation by 1 cycle/24 hour. Slow it down to 1 cycle/365 days.
Does the same side of the Earth always face the Sun?
Is the Earth still rotating? (Yes)
Uh …No.
If the Earth were to slow down to one rotation per orbit, it would not always face the Sun. It would just take a year to complete a “day”.
Learn some physics.
JDHuffman
JDHuffman: “Learn some physics.” Translation: “Learn my unsupported ideas that go against all established physics…learn my pseudoscience…I am the only person on Earth that understands science, the rest are a bunch of clowns pretending to study math and science for several years and all the experiments they do that contradict my divine edicts are bogus. Blindly follow me and you will “Learn some physics”.”
Poor Norman.
He must be so frustrated. He can never come up with a responsible comment. All he can do is insult, misrepresent, or falsely accuse.
JDHuffman
YOU: “Poor Norman.
He must be so frustrated. He can never come up with a responsible comment. All he can do is insult, misrepresent, or falsely accuse.”
Very responsible and totally correct comment from me. All you offer is made up physics and your own unsupported declarations. You reject links to actual science and you call experiments bogus.
I am not frustrated at all. I am making these comments to let new posters understand how shallow you are and how much of a complete phony you are. It will help them if anyone believes you have a shred of valid points. You don’t. Your made up nonsense may work on the idiot DREMT or Gordon Robertson. It seems to work on goofy Postma’s blog and Principia Scientific. On this blog you have actual scientists that know how phony you are.
Norman, the reality is you don’t have the technical background to understand the issues. You have your beliefs, but you fail every time you try to support them. Your uneducated interpretations of wiki articles always get shot down. You can’t think for yourself. You actually believe a racehorse rotates on its own axis running an oval track. You believe the atmosphere can defy 2LoT. You believe a black body is also a heat shield. You believe ice can radiatively warm a warmer object. You don’t even understand basic algebra.
You must be so frustrated.
JDHuffman
YOU: “Norman, the reality is you don’t have the technical background to understand the issues. You have your beliefs, but you fail every time you try to support them. Your uneducated interpretations of wiki articles always get shot down. You can’t think for yourself. You actually believe a racehorse rotates on its own axis running an oval track. You believe the atmosphere can defy 2LoT. You believe a black body is also a heat shield. You believe ice can radiatively warm a warmer object. You don’t even understand basic algebra.”
All a totally stupid comment and lies within.
First direct lie: “You have your beliefs, but you fail every time you try to support them.”
Idiot! I put links to physics textbooks in many of my links. That is a dishonest statement from you!
Yes a horse rotates on its axis as it runs around a track. If it did not turn it would run straight ahead. Your ignorance is great and you logic is nonexistent that you can’t see this.
A black body will absorb all the radiant energy it surrounds. It will start to emit on its own when it heats up. It stops all the IR from the object is surrounds. It shields this energy. If you keep the black-body cold it will work as a radiant shield.
YOU: “You believe the atmosphere can defy 2LoT. You believe a black body is also a heat shield. You believe ice can radiatively warm a warmer object. ”
No I do not think the atmosphere defies the 2LoT. I know it defies your ignorant understanding of it, not that actual law.
Ice will warm a HEATED warm object more than colder surroundings. It is established fact, that you are too stupid to comprehend real physics is your own stupidity no one else’s.
You are just a complete Dork that knows nothing. A phony fraudster that comes to troll.
ibid.
swannie…”Also, in 24 hours (1 Day), the Earth rotates 360 + 360/365.25 degrees. That extra bit of rotation is necessary to account for the motion of the Earth around the Sun during that 24 hour period. Learn Some Physics”.
Have you been hanging out with your buddy, DA?
There are two measures of the Earth’s rotation, one wrt the Sun and one wrt the stars. Look up solar time versus sidereal time.
The solar day = 24 hours (noon till noon)
The sidereal day = 23 hours 56 minutes 4.091 seconds
Gordo wrote,
Yes, that’s exactly my point. Over 24 hours of the solar day, the Earth’s rotation is:
360 * 24/(23 hours 56 minutes 4.091 seconds) = 360.9852 degrees
We have a winner! Gordo has just admitted that an orbiting body adds one rotation for each orbit. Which, for the Moon, says that the Moon rotates once an orbit.
Great work, Gordo. Now, if only Huffingman had half a brain, the discussion would be over. Learn some astrophysics and stop trolling!
Wrong Swanson. Gordon did not “admit” any such thing. You are misrepresenting him.
But, what you have demonstrated is more of your confusion about orbital motion. A “solar day” is referenced to the Sun. A “sidereal day” is referenced to the stars. Applying the same referencing to the Moon, an “Earth day” would be referenced to the Earth. And since the Moon is NOT rotating on its own axis, an “Earth day” equals a sidereal day.
Back on Earth, if the “solar day” were equal to the “sidereal day”, then the Earth would not be rotating on its own axis.
You’re invited to continue making a fool of yourself….
Huffing man continues with his delusional physics.
The “Sidereal Period” for the Moon is not the same as the Lunar Month. The Sidereal Period is ~27.3 Earth days, while the the Moon’s Synodic Month is ~29.5 Earth days. As a result, there are about 365.25/29.5 = 12.38 Full Moons a year but about 365.25/27.3 = 13.38 rotations of the Moon measured against the stars. The Moon, being tidally locked to the Earth, rotates once an orbit, according to widely understood astronomical measurements.
Learn some Astrophysics and stop trolling, idiot…
Wrong again, Swanson. You have trouble with facts and logic.
Your link is comparing the Moon to the Sun. The Moon defiinitely has two motions, relative to the Sun. It is orbiting around Earth, and the Earth/Moon system is orbiting around the Sun.
So you are trying to compare two different things, continuing to demonstrate your confusion.
You still are avoiding the simple fact that, back on Earth, if the “solar day” were equal to the “sidereal day”, then the Earth would not be rotating on its own axis. As I indicated in my comment, if you treated the Moon with the same reference system, it would show the Moon is not rotating on its axis.
You’re in WAY over your head so now you must resort to juvenile name-calling.
Huffingman wrote:
Of course, given that the Earth is orbiting the Sun, the two periods aren’t going to be equal because the Earth/Sun vector rotates in inertial (i.e., astronomical) space, unless the Earth’s rotation is reversed and slowed to once per year. That’s what DA’s animation showed. That’s not what we see when we look at the Moon.
Furthermore, HuffingGuy, the Moon’s Sidereal and Solar periods are different, as I indicated. The inverse of your comment tells us that tells us that the Moon rotates in the star based coordinate system, which is the only system within which one can properly describe astronomical rotation.
Learn some Astrophysics
Swanson, not only do you have trouble with facts and logic, you are a slow learner.
Earth’s sidereal day and solar day are different because there are TWO different motions.
The Moon’s sidereal period and solar period are different because there are TWO different motions.
The Moon’s “Earth period” (one orbit around Earth) is the same as its sidereal period because there is only ONE motion. Which amounts to just one more verification that the Moon is NOT rotating on its own axis.
Huffingman wrote:
Apparently, many astronomers disagree with you, (1) and (2) .
The Moon rotates once every ~27.3 Earth days relative to the stars. Learn some Astrophysics.
Swanson, I can’t speak for the “many astronomers” you claim to speak for, but I doubt they would disagree with my comment.
Your problem is that you don’t understand the relevant physics. And consequently, you can’t understand my comment.
But you’re funny, so please continue making a fool of yourself.
Huffingman, If you had taken the time to read the two references I gave, you would see that both state that the Moon rotates once every ~27.3 Earth days and provide a description of the reasoning behind these statements. It is your responsibility to refute these standard astronomical claims, yet all you do is throw out a comment that I somehow “don’t understand the relevant physics”.
As pointed out by DA and also by the references, it’s obvious that the Moon rotates because it can be seen simply by observing the moon against the stars. For example, take the vector between the Earth and the Moon and extend it thru the moon toward the background of stars. That vector is approximately fixed in the Moon, since the Moon is tidally locked with the Earth, with some wobbles. Then, at some point in the moon’s phase, say, the new Moon, chart the position of the stars behind the moon on a star map using that vector. Repeat this procedure for a year of New Moons and you will see that the reference vector rotates relative to the background reference of stars.
That proves beyond a shadow of doubt that the Moon rotates once each orbit, relative to the stars. Your denial of theses obvious facts does not dispute those facts, it only makes you appear as an ignorant moron.
Swanson, your links are from institutionalized pseudoscience (IP). This debate is pointing out that institutionalized pseudoscience is WRONG. Trying to support IP, by linking to IP, is “circular reasoning”. You can only use established physics, facts and logic, as I do.
The Moon does NOT rotate on its own axis. To claim such, the definitions of “orbiting” and “rotating on its own axis”, must be changed, as you and others have attempted. The apparent “rotation” relative to the stars is due to the orbiting motion. That motion is NOT “rotating on its own axis”.
The Moon only has one motion, orbiting. People that believe it is both orbiting and rotating on its own axis must also believe a racehorse does the same. The fact that people believe the racehorse is making both motions should tell you something about the influence of IP, and the willingness of people to let others think for them. In familiar terms, it’s called “being sheep”.
But, as always, thanks for the humor.
Huffingman wrote:
Yet again, you repeat your empty assertions without any reference to physics. It’s obvious to everyone except you (and Gordo?) that the Moon rotates relative to the Sun and it only takes little more effort to understand that the Moon also rotates WRT the stars.
No, the Moon’s rotation is proven and is easily measured against the stars. Learn some Astrophysics, like, which proper non-rotating reference frame have you selected against which you use to measure the rotation. Next time you ride a horse around a track, take a compass along and record the heading at points along the circuit. The needle will swing around one turn as the horse rotates in the other direction each lap.
Swanson, you still don’t even understand the issues.
“…the Moon rotates relative to the Sun and it only takes little more effort to understand that the Moon also rotates WRT the stars.”
“…the Moon’s rotation is proven and is easily measured against the stars.”
Again, that motion is due to “orbiting”. You are seeing one motion, and imagining another. The motion is the motion of a racehorse, which is NOT also rotating on its own axis. The motion is also similar to a tennis ball tied to a string and swung around your body. It is NOT rotating on its own axis, as it orbits around your body.
“Learn some Astrophysics, like, which proper non-rotating reference frame have you selected against which you use to measure the rotation.”
“Rotating on its own axis” defines the frame of reference. A rotating airplane prop is “rotating on its own axis”. If you adjusted a second prop to the same RPM as the first, the first prop would not be rotating, relative to the second prop. So would you stick your head in the first prop? If not, why not? It’s not rotating, based on an arbitrary frame of reference!
“The needle will swing around one turn as the horse rotates in the other direction each lap.”
No Swanson, the needle will always face the same direction, if it is free to remain aligned with the magnetic field.
You’re not just funny, you’re pitifully funny.
HuffingBoy, ROFLMAO
You’ve just totally failed basic Boy Scout navigation. Your race horse description is completely goofy. The body of the compass may be thought of as being attached to the horse. The needle of the compass points toward the magnetic pole, thus, as the horse runs, the body of the compass will rotate with the horse, the resulting apparent change of the needle from the rider’s perspective will be a rotation. The horse will rotate once per lap.
But, you don’t need a horse and a race track. Get a compass (or GPS device) and place it in fixed position within a car. Drive around a parking lot, making a circuit and returning to your starting point. As you move the compass heading will change, eventually returning to your beginning heading. If you can’t do that, take a hand held compass and walk around the block, observing the needle’s apparent rotation in your hand.
Swamson, you can argue with yourself:
“The needle will swing around one turn as the horse rotates…”
“The needle of the compass points toward the magnetic pole…”
HuffingBoy, I was attempting to clarify the earlier statement which you jumped on, since you have no other reply.
The needle will APPEAR to ROTATE to the observer on horseback, in the car or while walking. That’s because the body of the compass is rotating along with the horse, the car or your hand. The needle points in a constant direction and thus provides a reference against which one can measure rotation.
You still don’t get it (or, can’t admit you’ve been wrong for a long time). Your horse rotates once per lap just as the Moon rotates once per orbit.
You will need a lot more “attempting to clarify” to make up for your gross incompetence.
So keep banging on your keyboard. It’s fun watching a clown, tangled in his own web.
Maybe another keyboard banger will come help you out. The more, the merrier….
HuffingBoy, As you apparently are now unable to provide a reasonable, physics based critique of my analysis, instead offering nothing but empty personal attack, I presume that you’ve given up trying to promote your bogus, moronic claims about non-rotating race horses and the Moon. As any astronomer will agree, the Moon does rotate as it orbits.
Swamson, that’s a hoot!
You haven’t offered ANY physics. You believe the Moon rotates because it looks like it rotates! You got confused about both the compass and the sidereal day. You don’t understand the difference between orbiting and “rotating on its own axis”.
All you can offer is links to pseudoscience, and mentioning “astronomers”.
Ever heard of a “braindead groupie”?
HuffingBoy, A compass is the most basic geophysical instrument which has been used to measure direction on the surface of the Earth for more than a thousand years. A simple compass shows that your claims based on your race horse model are obviously wrong. Live with it.
Yes Swamson, a simple compass is easy to understand.
But, obviously not for braindead groupies….
HuffingBoy puts out more troll poop. He presented his claims regarding the Moon’s rotation and added his race horse analogy, but lost to physics, so now all he has left is empty bluster. Game Over, troll.
Swamson, you don’t even understand the issues. You are so lost you don’t understand there are TWO motions being discussed. The Moon is only making ONE motion. To claim there are TWO motions, yet we only see one side of the Moon, you MUST believe there are TWO motions. And that means you clowns must believe orbiting is represented by Figure 2.
https://postimg.cc/06WdBkGB
Of course, that is pure pseudoscience. The Moon only has ONE motion, as represented by Figure 1. It is orbiting, but not rotating on its own axis.
But please, don’t let facts and logic keep you from making a fool of yourself. You’re doing such a great job….
When JD or Gordon can address elliptical orbits, then there will be something to discuss.
So JD or Gordon, consider a small moon following a highly elliptical orbit that returns to the same orientation every perigee and keeps the same face generally toward the earth. Find a crater on the moon directly facing the earth at perigee (ie 90 from the direction the moon is heading at that point. One orbit later, the same crater will again be directly toward the earth.
As the moon orbits, which way will that crater face according to your understanding?
a) always directly toward the earth (ie toward one focus of the ellipse).
b) always directly toward the center of the ellipse.
c) always perpendicular to the direction of travel (like a racehorse on an elliptical track),
d) something else.
Note that for a circular orbit, a, b, and c are the same and are all true. But a, b, and c are all DIFFERENT for an elliptical orbit and no more than one could be true. If your theory can handle elliptical orbits, it is useless.
tim…”Note that for a circular orbit, a, b, and c are the same and are all true. But a, b, and c are all DIFFERENT for an elliptical orbit and no more than one could be true. If your theory can handle elliptical orbits, it is useless”.
I think this is a red-herring argument and an obfuscation. We are talking about our tidally-locked Moon whose face is always pointed to the Earth.
I am sure there are many different situations with moons and satellites throughout the universe but my head is hurting as it is with other distractions.
How about a moon tidally-locked to two different planets as in the Lagrangian points? It’s conceivable that somewhere there is a Moon that is tugged one way by one planet then another way at another time by another planet. It could be rocking back and forth.
–How about a moon tidally-locked to two different planets as in the Lagrangian points? —
Our twin system of the Earth and the Moon has two sets of 5 Lagrangian points.
And could twin system of two planets plus a Moon. Or could two planets in orbit around sun, like Earth and Venus and that they would share a Moon.
Earth system already has what people have decided to call quasi moons.
Wiki:
“Claimed moons of Earth
…
Although the Moon is Earth’s only natural satellite, there are a number of near-Earth objects (NEOs) with orbits that are in resonance with Earth. These have been called, inaccurately but provocatively, “second”, “third” or “other” moons of Earth.
2016 HO3, an asteroid discovered on 27 April 2016, is possibly the most stable quasi-satellite of Earth. As it orbits the Sun, 2016 HO3 appears to circle around Earth as well.”
https://en.wikipedia.org/wiki/Claimed_moons_of_Earth
I think they are simply, moons of Earth.
And I think Ceres and Pluto are simply planets rather than dwarf planets. But if trying to follow the definitions [which fairly vague and crazy] that some people want to use, it seems in both cases your moon in orbit around two planets: with twin planets or two planet like Earth and Venus, would be called a quasi moon. Or that moon would be more of quasi moon then what people are attempting to call, Earth’s quasi moons.
Now, one aspect of our quasi moons is they all appear to be temporary- though quite permanent in terms of a human lifespan or time periods of decades to centuries. So, quite temporary in terms of millions of years [to a billion years] and all NEOs appear to be temporary in this sense. As are main belt asteroids and the Trojans of Jupiter- we have millions of temporary space rocks.
So one aspect of two planet sharing a Moon is question do they have to be permanent and what time period are we calling permanent.
And “manmade” moon which sharing two planets could be called a cycler. Wiki:
A Mars cycler (or EarthMars cycler) is a special kind of spacecraft trajectory that encounters Earth and Mars on a regular basis. The term Mars cycler may also refer to a spacecraft on a Mars cycler trajectory. The Aldrin cycler is an example of a Mars cycler.
https://en.wikipedia.org/wiki/Mars_cycler
So one could move Ceres so it’s a cycler, between say Earth and Venus, or easier, Venus and Mars.
To keep it simple, you could have two planets orbiting each other at same distance to each other [and distance from Sun] as our Earth and Moon, but have the Moon be much larger, say the size of Mars [or about 1/9th of the Earth’s mass].
And add moon which about 1/2 mass of the Moon [1/160th the mass of Earth] at L-3 [or opposite side of orbit as Mars mass planet which replaced the Moon].
And both Mars sized planet and smaller moon could be tidally locked with Earth mass planet.
You also add small moons in L-4 and/or 5.
Though more likely and/or stable is having Earth size planet, instead be about twice of Earth’s mass.
Tim, your red-herring is a dead fish.
Kepler pointed out that most orbits are elliptical, with his laws of orbital motion. Newton followed on years later, improving our understanding by combining Galileo, Kepler, and his newly developed calculus, especially as how it applies to gravity. Orbital motion is well understood, by those that have studied the relevant physics.
As explained several times before, the Moon, or any moon, would be moving in the direction of the orbit, i.e., the direction as produced by the resultant vector. That means the same side basically always faces the inside of the orbit, just as a racehorse has one side facing the inside of the track. That means an onlooker would be able to detect a “wobble”, if the orbit or track were highly elliptical. The higher the eccentricity, the more the “wobble” would be apparent.
“That means an onlooker would be able to detect a wobble, if the orbit or track were highly elliptical. The higher the eccentricity, the more the wobble would be apparent.”
There is a wobble.
It is called libration.
https://en.m.wikipedia.org/wiki/Libration
Good job, E-man.
You finally got something right!
You are aware, that not all moons are tidally locked. Some do in fact rotate on their axis more than once per orbital period.
You are aware, that irrelevant comments mean the commentor isn’t able to offer any relevant comments.
Get your facts straight and I won’t offer any irrelevant comments.
Deal?
Remember, you said any moon … always faces the inside of the orbit
that’s wrong, go back and finish your physics studies
Gordon says: “We are talking about our tidally-locked Moon whose face is always pointed to the Earth.”
JD says: “the same side basically always faces the inside of the orbit just as a racehorse has one side facing the inside of the track”
So you to disagree fundamentally with each other, since you give two different answers (and both of you think your answer is so obvious it doesn’t warrant discussion)! Time to Gordon & JD to figure why they disagree.
No Tim, Gordon and I agree that the Moon does not rotate on its own axis, along with others. You are desperately trying to find some quotes, from different people, to which you can apply your pedantic measurement.
I guess that’s because you have no meaningful arguments.
And since you love pedantry, “So you to disagree…” should be “So you two disagree…”
JD, you directly disagree with the direction a moon will face in an elliptical orbit!
So you don’t even agree on what it means to “not rotate on its axis”.
Tim, you don’t even knew the difference between “to” and “two”.
When ya got nothing, pick on typos. 🙂
https://upload.wikimedia.org/wikipedia/commons/9/94/Elliptic_orbit.gif
The side facing the planet at perigee (the left side of the orbit) will be facing
1) the center of the orbit = straight ‘up’ [JD]
2) the planet = diagonally ‘up and to the left’ [Gordon]
once the moon has traveled 1/4 of the distance around the orbit.
Apparently “to agree” can now mean to give distinctly different answers.
Tim Folkerts says:
Like 3% of the climate scientists.
Tim, it appears both our versions are correct, based on your poorly worded situation. It’s only your interpretation that is wrong.
I’ll try to unconfuse your interpretation of my statement, and let Gordon unconfuse you about his.
Again, the racehorse is a simple way to explain to people that do not understand orbital motions.
The racehorse’s head always faces down the track. Even in the turns, the horse’s head instantaneously faces tangent to the track. That means the same side of the horse always faces the inside of the track. If the track is an elongated ellipse, with major axis much larger than minor axis, it does not change the relative position of the moving horse.
Consequently my statement stands:
“As explained several times before, the Moon, or any moon, would be moving in the direction of the orbit, i.e., the direction as produced by the resultant vector. That means the same side basically always faces the inside of the orbit, just as a racehorse has one side facing the inside of the track. That means an onlooker would be able to detect a “wobble”, if the orbit or track were highly elliptical. The higher the eccentricity, the more the “wobble” would be apparent.
The race horse running around a track is indeed a simple analogy for orbital motion. But it is TOO simple in some key ways.
A horse on a merry-go-round is indeed a simple analogy for orbital motion. But it is TOO simple in some key ways.
Here is (the start of) a stronger analogy. (We start with a circular orbit, but will look at elliptical orbits soon.)
1) To a very good approximation, the moon is an isotropic sphere.
2) Gravity can exert no torque on an isotropic sphere. So there is approximately no torque on the moon.
3) A good analogy (for a simple circular orbit) would be a horse on a merry-go-round where the platform is turning at a constant rate.
4) The pole is mounted on a frictionless pivot — required to model the no-torque situation for the moon.
5) This means that however fast the horse is spinning initially, it will keep spinning that fast as the merry-go-round rotates. If the platform rotates around ‘n’ times per minute and the nose points due north ‘m’ times every minute, those numbers are constant (within the no-torque approximation).
Any disagreement so far?
In 4), I assume the “pole” is a vertical pole supporting the horse at its center of mass. And in 5), I assume “spinning” is referring to “rotating on its own axis”.
Otherwise, I have no problems, so far. Don’t try any tricks. I’ll be watching.
4) Yes .. vertical through the CoM
5) I am trying to avoid the contentious term “about its own axis”. I meant what I said — the orientation of the nose relative the ‘fixed ground’ changes at a constant angular speed if it is on the vertical, frictionless pole through the CoM.
(omega) = m rev/min = m/60 rev/sec = 6m deg/sec = pi*m/30 rad/sec
This is the angular speed in the inertial reference frame of the ground (ignoring the slow rotation of the earth, etc). [The horse will have an different angular speed in the non-inertial reference frame of the merry-go-round platform, but we are not addressing that at the moment].
6) NEW CLAIM: (omega) will only change in an inertial reference frame if some external torque is applied. This is “Newton’s First Law of Rotation”.
tim…”D, you directly disagree with the direction a moon will face in an elliptical orbit!”
Hopefully you are nor referring to your hypothetical Moon in a highly elliptical orbit. If so, you don’t understand local angular momentum.
By local angular momentum, I mean the momentum related to a radial line from the moon’s centre to the surface pointed at the Earth. The angular momentum of the line is the rate at which the entire line would turn through 360 degrees, not individual particles on the line.
If a moon has no angular momentum to begin with, it won’t suddenly develop angular momentum due to the eccentricity of its orbit. In an eccentric orbit, perihelion is the point closest to the central mass and aphelion is the point furthest from the mass.
At perihelion, the orbiting body has been accelerated by attraction from the central mass. The orbiting body wants to move in a straight line past the mass but gravitational force from the mass nudges it around into an orbit.
As the body is pulled toward the mass heading for perihelion, it accelerates, reaching maximum velocity and momentum at perihelion. Then it is pulled around the mass, heading in the other direction. As it leaves the mass, the body decelerates as the central mass pulls back on it, but due to it’s average momentum, it cannot fly off into a parabolic or hyperbolic orbit.
That’s what keep a body in orbit around a central mass. If its momentum is too high, it will sling shot off on a parabolic or hyperbolic orbit. If the momentum is too low, the body will spiral into the central mass.
Let us pray that neither the Moon nor the Earth are impacted by a collision large enough to reduce their momenta to that degree.
If you have a body that is tidally-locked, the same face will always face inward toward the central mass. There are no forces acting on it in an angular direction to cause it to turn about its axis.
The non-rotating body wants to move in a straight line with rectilinear translation. Gravitational force from the central mass changes that translation from rectilinear to curvilinear, therefore the same face is always pointed in toward the central mass.
In that case, the body is exhibiting curvilinear translation, which is a translation with no angular momentum (rotation) in this case.
tim…”Any disagreement so far?”
Yes. You are trying to obfuscate a simple problem.
This is not a philosophical debate, it is an exercise in basic engineering physics. Never mind inertial frames, or any other kind of reference frames, anyone observing from Earth always sees the same side of the Moon.
If you have the horse on the merry-go-round, it is fixed in position and cannot rotate about an axis, pole, or whatever you envision. It orbits the centre of the MGR with the same side always pointing in, exactly the same as the Moon.
If you had several people positioned around the central part of the MGR, all would report seeing the same side of the wooden horse. None would report that the horse is turning on the pole upon which it is mounted.
Why do you have so much trouble with that?
Tim…”…the orientation of the nose relative the fixed ground changes at a constant angular speed….”
Tim…you are confusing local angular velocity about a COM with the change in angle of a line tangent to a radial line from the central axis of the merry-go-round to the horse.
A line through the horse, from tail to nose is what you’re talking about. That line is tangential to a normal line (radial line) from the centre of the MGR to the horse.
Of course it’s angle will change, just as the angle of any tangent line at any point on a circle or curve will change.
It’s the same with the Moon. If you draw a radial line from the Earth’s centre through the Moon, through the face pointing at Earth and through the Moon’s centre, a line tangent to that radial line through the Moon’s centre would change angle constantly as the Moon orbits the Earth.
That is curvilinear translation, by definition. The curvilinear translation would actually reference the portion of the radial line from Earth’s centre that fell within the Moon circumference. That line would always turn with all points on it at a constant angular velocity. Furthermore, all points on the line would turn in parallel, concentric circles about the Earth’s centre.
As you know, you can describe that line with the tangent line minus 90 degrees. The tangent line will change angle at the same rate as the line through the Moon inline with the radial line to the Earth’s centre.
That motion is not rotation about the axis, through which both lines intersect on the Moon’s centre. As proof, you can see that the wooden horse does not turn about the pole upon which it is mounted, even though a line through the horse from tail to nose does change angle.
Gordon,
First I am not sure at what points you are talking about your “local angular momentum” vs just angular momentum in general. it seems you are jumping back and forth, but I can’t tell.
We could borrow nomenclature from atomic physics/chemistry used for electron orbits and talk about “orbital angular momentum” and “spin angular momentum”. Orbital angular momentum would be L = |rxp| = mvr. This quantity is conserved since gravity is a central force. This is the starting point to derive Kepler’s 2nd Law (as r = distance from the planet increase, v = the speed forward in the orbit, decreases).
Spin angular momentum would then be S = 2/5 mr^2 (omega), the angular momentum of a rotating sphere (which I think is what you are calling the ‘local momentum’).
The total angular momentum is then J = L + S.
******************
Second, I think you have a fundamental misunderstanding of “tidal locking”. You state “If you have a body that is tidally-locked, the same face will always face inward toward the central mass.”
This is empirically incorrect, since our tidally locked moon does NOT show exactly the same face to the earth all the time. Tidally locking is a weak, slow process. It creates only a very small torque to cause the friction to slow the rotation of a moon. It gradually causes the same face to *generally* face the planet, but it does NOT ‘lock’ the orientation of moon the way a pole on a merry-go-round locks a horse to the platform. A better analogy would be a very GOOD bearing for the pole that only gradually slows the rotation of the horse relative to the platform.
And, as explained, tidal locking makes much more sense from the “Non-Spinner” perspective:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333359
Gordon shrieks:
“If you draw a radial line from the Earth’s centre through the Moon, through the face pointing at Earth and through the Moon’s centre, a line tangent to that radial line through the Moon’s centre would change angle constantly as the Moon orbits the Earth. That is curvilinear translation, by definition.”
NO NO NO. 100% WRONG!
You keep making up phony definitions of curvilinear translation.
The real definition:
“Translation, rectilinear and curvilinear: Motion in which every line in the body remains parallel to its original position. [http://www.brown.edu/Departments/Engineering/Courses/En4/notes_old/RigidKinematics/rigkin.htm]
Look at page 3 of the following link:
https://www.asu.edu/courses/kin335tt/Lectures/Kinematics/Kinematics%20I%20Spring%202005.pdf
The curvilinear translation figure shows the arrow remaining pointed in the same direction as it translates along the curve. It DOES NOT keep tangent to the curve.
Your tangent line keeps changing its direction as the radial line changes it angle, which does not meet the real definition of translation.
Gordon, please quit making a complete idiot of yourself. You continually display your SUPREME ignorance in the field of kinematics.
JD shrieks,
“No Tim, Gordon and I agree that the Moon does not rotate on its own axis, along with others.”
What others? You mean your “mini-me” appendage DREMT?
Well. certainly not Joseph Postma, who states:
“The moon rotates on its axis, which orbits the Earth.”
Total bummer, man! LMAO.
“What others?”
JD, Gordon, DREMT, gbaikie, AndyG55, Ftop, Nikola Tesla, A. Tomic, B. S. Jovanovic, Savic & Kasanin…
tim…”Gordon says: We are talking about our tidally-locked Moon whose face is always pointed to the Earth.
JD says: the same side basically always faces the inside of the orbit just as a racehorse has one side facing the inside of the track
So you to disagree fundamentally….”
The two statements sound to me like they are exactly in agreement.
JD’s statement that the same side of the horse always faces the inside of the track is the same as my point that the same side of the Moon always faces the Earth.
I don’t get your point that we are in disagreement. Maybe you are viewing the statements from one of your sci-fi dimensions.
Gordon, can you really not understand the difference? I explained it once above, but here goes again.
Look at this animated gif: https://commons.wikimedia.org/wiki/File:Elliptic_orbit.gif
At perigee (the leftmost point of the orbit), one side of the moon will be facing directly toward the planet. Call that “Side A”.
Later, the moon will be 1/4 of the total distance around the orbit (ie at the “bottom” of the ellipse). Which way will Side A be facing now?
* According to the JD’s “horse on a track” answer, SIDE A would be facing straight toward the center of the ellipse (ie “straight upward”).
* According to Gordon’s “Face the planet” answer, SIDE A would
facing the planet (ie “up and to the left”).
Clearly no more than one of these could be correct. Side A cannot simultaneously point “up and to the left” and “straight up”.
PS. My answer based on “no torque/constant angular speed” would be less than 90 degrees CCW from the starting orientation (ie “up and to the right”). This would make BOTH of you wrong.
Tim, you are spinning, again.
Just look at the simple images. If you believe orbiting motion is represented by Figure 2, you need to ask for a full refund from any institution that claims to have taught you physics.
https://postimg.cc/06WdBkGB
Believing Figure 2 is “orbiting” means you haven’t a clue about the relevant physics.
tim…”Gordon, can you really not understand the difference?”
I have already made it clear to you that I have no interest in obfuscations, I am focused only on the reality of our Earth-Moon system. The Moon has a nearly circular orbit around the Earth whereas your ellipse has an orbit more like Halley’s comet. You are exaggerating out of all proportion in an obvious attempt to win a philosophical point.
We know about libration, which allows an observer on Earth to see a tiny bit more of the Moon’s surface at certain points of the Moon’s orbit. From that perspective, I get your point, but it has nothing to do with the points made by JD and myself that the Moon always has the same face pointed to the Earth, or that a horse running a track always has the same face pointed to the track centre.
If you observed the horse with binoculars from a vantage point at the centre of the track, and the track was extremely elliptical, you would see more of the horse at different positions on the track. However, you would never see most of the other side of the horse.
You might see more of its butt as it ran away from you, and more of the face as it turned toward you, but you would never see the side pointing away from the track.
I don’t know why you are arguing this point.
JD…”Believing Figure 2 is orbiting means you havent a clue about the relevant physics”.
I agree. Figure 2 is one of the most stupid rebuttals I have encountered.
This exercise about the Moon and it’s non-rotation has revealed to me that alarmists have a serious problem with basic physics and logic.
JDHuffman
In your image:
https://postimg.cc/06WdBkGB
You claim Figure 1 represents the original definition of “orbiting”
They are both legitimate orbits. Where do you pull out your nonsense from?
Here:
https://www.thefreedictionary.com/Orbit+(physics)
The definition of orbiting is only that some object moves around an another. It has nothing at ALL to do with how the object rotates as it moves along a curved path. You just make up everything don’t you?
You can’t even get the definition of “orbiting” correct and you come here and lecture people on what it means. Really sad.
You don’t know the definition of radiative flux, black body, orbiting. Why don’t you get a dictionary and read up on words. Making up your own definitions and then calling people wrong, because they do not use your made up version, but prefer the standard definition, is not rational by any means.
Orbiting — as defined by every scientist everywhere — is simply a point moving around a path (typically due to gravity). Both of your diagrams are orbiting — just with differing amounts of spin angular momentum. Find any source (all the way back to Newton if you dare) that serves as an “origin” for your “original” definition. In fact, you are presenting YOUR definition.
You can demand that every scientist everywhere redefine “orbit” in every textbook everywhere to include a specific bit of spin angular momentum, but it ain’t happening.
The fact you still can’t deal with elliptical orbits (either of you) shows that your understanding lacks true depth. How would you drawyour diagram for an elliptical orbit? The arrows always pointing toward the planet (like Gordon wants)? The arrows always pointing forward (like JD wants)?
Yes Gordon, this Moon issue is a huge issue for the promoters of pseudoscience. They must do whatever is necessary to defend it. You can see their fanaticism. They will do anything to protect their false religion. That’s why you see so many actually claim a racehorse rotates on its own axis as it runs the track!
They’ve lost all regard for reality.
And Tim begins his spin: “Both of your diagrams are orbiting — just with differing amounts of spin angular momentum.”
Nice try Tim, but I didn’t say anything about “rotating on its own axis”. The “spin” was all added by you. The simple images are orbiting ONLY. Figure 2 represents orbiting as represenseted by the “Spinners”.
“Figure 2 represents orbiting as represenseted by the “Spinners”.”
No — “orbiting” as represented by ALL standard textbooks and all trained scientists — is a point moving along a path. Points have no orientation. For both of your diagrams the “orbit” would be the motion of the center of mass or the arrow.
You can say you don’t like this definition. You can say you like your definition that includes an orientation of an arrow pointing straight ahead in the direction for travel. But you can’t say “your” picture is any sort of “original” definition of “orbit” because you can’t find a single textbook from any point in history that agrees with you.
*******************************
And again, semantics is not the focus of science — predictions that agree with nature for new situations are. Everyone can predict the correct orientation of a tidally locked moon along a circular orbit. Test your theory — predict the orientation of a moon along an elliptical orbit. Draw a new diagram for your predicted orientation of an arrow along the ellipse. [Hint: simply drawing an ellipse with the arrows pointing ‘forward’ is not how a real moon would orbit.]
Keep spinning Tim.
Realists are enjoying it immensely!
Especially your ignorance of how vectors form the orbital path.
Once again, the Spinners get caught in their own web. Figure 2 shows that the Moon has to do in order to ALSO be rotating on its own axis. It would require BOTH false moments to result in the Moon only showing one face to Earth.
Of course, they will try to spin their way out.
That’s when it gets especially funny.
https://postimg.cc/06WdBkGB
Tim, whether you choose to accept it or not, Figure 2 is the “Spinner” concept of “orbiting” or “revolution” or even (as we’ve seen in my discussions with David) “rotation about the Earth-moon barycenter”. You are all quite clear that the moon has two motions, you think it both “orbits” or “revolves” AND that it “rotates on its own axis”.
The moon’s motion is of course as per Figure 1. For the “Spinners” to claim the moon has two motions, which they do, MUST mean that they believe the part of the moon’s motion absent the “rotation on its own axis” is as in Figure 2. Then, you add the “rotation on its own axis” to get from Figure 2 to the motion as per Figure 1. FACT. That is the only way you can conceive of the moon’s motion as consisting of two motions.
Another FACT is that “rotation about the Earth-moon barycenter” (in other words rotation of an object about an off-center axis) would result in the motion in Figure 1. Not Figure 2. To get to the motion in Figure 2 would require the object to be rotating about the Earth-moon barycenter AND rotating on its own axis (in opposite directions, at the same rate).
DREMT buries Tim under 10 feet (3 meters) of “facts and logic”!
Nothing new.
And the funny thing is Tim wanted so much to “define” his way around the relevant physics. But, the simple images are too clear and understandable for the Spinners to confuse.
Again, just because we wouldn’t want any Spinners to avoid seeing:
https://postimg.cc/06WdBkGB
DREMT says “Tim, whether you choose to accept it or not, Figure 2 is the “Spinner” concept of “orbiting” or “revolution” … “
Yes. Exactly. That is EVERY scientist’s concept of “orbit”.
“… or even (as we’ve seen in my discussions with David) “rotation about the Earth-moon barycenter”>
No. This is a DIFFERENT concept than what you just called “orbit”. In fact, Figure *1* illustrates “rotation about the Earth-moon barycenter”.
What some are calling the “Spinner” view (Figure 2) should more accurately be called the “Orbitalist” view or the “Revolutionist” view. Orbitalists view Figure 2 as the “basic motion” — a simple translation of the center-of-mass. Figure 1 is this basic translation PLUS a rotation.
The other position would then be the “Rotationalist” position. In this paradigm, Figure 1 is the “basic motion” — a rotation of the universe around the barycenter, carrying everything (particularly the moon) with it. Figure 2 is then the basic motion plus a counter-revolution.
************************************
Two reasons to prefer the Orbitalist view.
1) Figure 2 is a lower energy motion (by an amount KE = 1/2 I(omega)^2). The lowest energy state is generally the starting point; the simplest motion; the ground state.
2) No Rotationalist has yet been able to correctly extrapolate to an elliptical orbit. When they tried, they came to different conclusions and cannot reconcile those difference.
Reason 1 is suggestive; Reason 2 is fatal. If the Rotationalist theory gives an incorrect prediction for moons in elliptical orbits, then it is wrong. That is how science works. You have to agree with nature!
“No. This is a DIFFERENT concept than what you just called “orbit”. In fact, Figure *1* illustrates “rotation about the Earth-moon barycenter”.”
Yes, Tim. Indeed. Not sure how you got the impression I disagreed.
“What some are calling the “Spinner” view (Figure 2) should more accurately be called the “Orbitalist” view or the “Revolutionist” view. Orbitalists view Figure 2 as the “basic motion” — a simple translation of the center-of-mass. Figure 1 is this basic translation PLUS a rotation.
The other position would then be the “Rotationalist” position. In this paradigm, Figure 1 is the “basic motion” — a rotation of the universe around the barycenter, carrying everything (particularly the moon) with it. Figure 2 is then the basic motion plus a counter-revolution.”
Tim you are mostly just agreeing with and restating everything I just explained, except by ridiculously stating that we think the entire universe is rotating around the barycenter, which is NOT what the “Non-Spinners” think. Maybe read my comments a bit more carefully, before you respond.
Let’s make it simpler still.
Figure 1 = how the “Non-Spinners” view “orbiting” (meaning orbital motion with no axial rotation). Figure 2 = how the “Spinners” view “orbiting” (same meaning).
“Axial rotation” is then separate and independent from those motions, for both camps.
Why pretend this is more complicated than it is, unless your intent is to obfuscate?
Tim’s forte is obfuscation. He is a “spin-artist”.
Above, he now posits two “reasons” why he prefers pseudoscience over physics.
1) Energy consideration.
2) The simple images are “circles”, not ellipses.
Tim, energy is not a consideration because Figure 2 is NOT orbiting. It is not how gravity affects an object in orbit. It is your pseudoscience. You can’t avoid reality no matter how hard you avoid logic.
And since you are trying to hide behind there not being any ellipses, this will scare you out of your hiding place:
https://postimg.cc/jDwZ97z1
DREMT asks: “Not sure how you got the impression I disagreed.”
I was reading your statement (emphasis added)
Figure 2 is the Spinner concept of
** “orbiting”
or
** “revolution”
or
** … “rotation about the Earth-moon barycenter”.”
To me, that said you thought Figure 2 represented all three, and that all three were identical thoughts.
DREMT
Let me address once again a critical issue — elliptical orbits.
You say: “Then, you add the rotation on its own axis to get from Figure 2 to the motion as per Figure 1. FACT. That is the only way you can conceive of the moons motion as consisting of two motions.”
Yes, that is exactly how “Orbitalists” conceive of the motion on the right side of JD’s figure.
Turning the tables, how do “rotationalists” conceive of an elliptical orbit? The FACT is that you can’t just ‘stretch’ the circle into an ellipse with the arrows pointing forward. That is NOT how real moons move. You also cannot have one side of the arrow always directly toward the planet — that is a DIFFERENT wrong answer. Both of these have the moon rotating relative to the fixed stars at varied rates — but there is no torque that could cause this change in rotation rate.
If we redraw the “arrows around the planet” diagram with an elliptical orbit, you descrition “Then, you add the rotation on its own axis to get from Figure 2 [now an ellipse] to the motion as per Figure 1. FACT. That is the only way you can conceive of the moons motion as consisting of two motions.” This gives the right answer!
SUMMARY
** For circular orbits: both Orbitalists and Rotationalists get the right answer. Rotationalist’s description *might* be slightly easier or more intuitive.
** For elliptical oribits, Orbitalists get the right answer. Rotationalists get the wrong answer.
The edge Rotationalists might have for circular orbits is destroyed when they get the wrong answer for elliptical orbits.
“To me, that said you thought Figure 2 represented all three, and that all three were identical thoughts.“
Tim tries to tell me what I wrote. Look, David told me in no uncertain terms that the moon is BOTH rotating about the Earth-moon barycenter, AND on its own axis. I told him he was wrong. You would agree he was wrong. You and I agree on what “rotating about the Earth-moon barycenter” looks like. Figure 1. So this is what I wrote:
“Tim, whether you choose to accept it or not, Figure 2 is the “Spinner” concept of “orbiting” or “revolution” or even (as we’ve seen in my discussions with David) “rotation about the Earth-moon barycenter”.”
Notice I said “even”, and mentioned David, indicating it was particularly stupid to put that under Figure 2.
Tim quotes me: “Then, you add the rotation on its own axis to get from Figure 2 to the motion as per Figure 1. FACT. That is the only way you can conceive of the moons motion as consisting of two motions.”
Then bizarrely says:
“Yes, that is exactly how “Orbitalists” conceive of the motion on the right side of JD’s figure.”
No Tim. That is how the “Spinners” conceive of the moon’s motion, the motion on the LEFT side of JD’s figure. Figure 1.
Until you can start accurately conveying my words, I feel no reason to discuss anything further with you.
Tim has grasped at so many straws, there is going to be a world-wide shortage of straws!
Barycenter, frames-of-reference, Orbitalist, Revolutionist, Rotationalist, ellipses, the list goes on and on.
Like all the other clowns, it’s all just spin, bluster, and reality-avoidance.
No physics, just pseudoscience.
Nothing new.
All the options Tim gives are just “that’s wrong”, “you can’t think that”, or “that’s just a different way of being wrong”. Then he just says “or there’s my way, which is right”. It’s a lot like “heads I win, tails you lose”. I’ve yet to see him show an alternative diagram to your “2a” as to what the “correct” orbital path would look like according to Tim.
norman…”The definition of orbiting is only that some object moves around an another”.
Not in the way we have used ‘orbit’ in this debate, for want of a better word. In this context, an orbit has to be under the influence of a gravitational force.
When you run on a track you don’t claim to be orbiting the track, although you could. We can use words any way we want. You claim to be running laps, or just running around a track.
You would not claim the Moon is lapping the Earth. You can claim it is revolving about the Earth, even though it doesn’t rotate about its axis as it revolves.
There seems to be WAY too much confusion about who says what. So let me just give my position; my understanding.
ORBITALISTS: (aka “spinners”)
An “orbit” is simply a path — typically a circle or an ellipse in astronomy. It has no orientation associated with it; just a series of locations. This tells you the period of REVOLUTION. (And specically, an orbit is not a “rotation” around the barycenter).
Any information about the orientation relative to the ‘fixed stars’ is a separate issue, defining the period of ROTATION. If the ‘fixed stars’ never rise or set, the object is not rotating. If the ‘fixed stars’ rise once every 23hr,56min (like earth), the object is rotating once every 23hr,56min. If the ‘fixed stars’ rise once every 27.3 days (like the Moon) the object is rotating once every 27.3 days.
In this view, the total KE can be found by simply adding the translational KE of the orbit (1/2 mv^2) to the rotational KE of the object (1/2 Iω^2).
In this view, the total angular momentum can be found by simply adding the orbital L (mvr) to the rotational L (Iω).
In this this view, the rotation rate will be basically constant (other than a small bit of “tidal friction” that can gradually change rotation rates).
In this view, the rotation relative to the fixed stars will be constant in an elliptical orbit.
In this view, the motion of the center of the arrows in JD’s diagram is the orbit. The orientations of the arrows is the rotation (one rotation per orbit on the left where the arrows always point forward; zero rotations per orbit on the right where the arrows always point up.)
“ROTATIONALISTS”
(I hate to speak for this groups, since I find their position to be not self-consistent. )
An “orbit” is a path with an orientation associated with it. That orientation is alternately described as “always toward the planet” or “always forward along the direction of travel”. [For circular orbits these are the same; for elliptical orbits they are different and hence cannot both be correct definitions]. Orbits are “rotations about the barycenter”. An object orbiting once per month is automatically also once per month relative to the ‘fixed stars’.
Any change in orientation in addition to the rotation about the center is a rotation about the object’s axis.
In this view, circular orbits are intuitive for many people.
In this view, for circular orbits you could use the parallel axis theorem to find the moment of inertial and hence the total KE & L.
In this view, I see no way to handle elliptical orbit. Both “always toward the planet” or “always forward along the direction of travel” fail to accurately describe the orientation of the object.
“There seems to be WAY too much confusion about who says what.”
That is because you have tried to confuse the issue as much as possible, which is a part of your job. I’ll just repeat my comment again, to unconfuse the issue.
“Tim, whether you choose to accept it or not, Figure 2 is the “Spinner” concept of “orbiting” or “revolution” or even (as we’ve seen in my discussions with David) “rotation about the Earth-moon barycenter”. You are all quite clear that the moon has two motions, you think it both “orbits” or “revolves” AND that it “rotates on its own axis”.
The moon’s motion is of course as per Figure 1. For the “Spinners” to claim the moon has two motions, which they do, MUST mean that they believe the part of the moon’s motion absent the “rotation on its own axis” is as in Figure 2. Then, you add the “rotation on its own axis” to get from Figure 2 to the motion as per Figure 1. FACT. That is the only way you can conceive of the moon’s motion as consisting of two motions.
Another FACT is that “rotation about the Earth-moon barycenter” (in other words rotation of an object about an off-center axis) would result in the motion in Figure 1. Not Figure 2. To get to the motion in Figure 2 would require the object to be rotating about the Earth-moon barycenter AND rotating on its own axis (in opposite directions, at the same rate).”
There you go, problem solved. You love to waffle on about elliptical orbits. Show, in a simple diagram similar to JD Huffman’s, how our moon moves throughout its orbit.
JDH wrote:
DA, the racehorse IS always facing the same direction.
Let’s say the racehorse starts out, out of the gate, facing and running to the north.
What direction is it facing after the first quarter turn? Still north?
What about on the back stretch? Still north?
Or, on the back stretch, maybe south???
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-333823
Wiki:
Runaway climate change or runaway global warming is hypothesized to follow a tipping point in the climate system, after accumulated climate change initiates a reinforcing positive feedback. This rapid acceleration in climate change may lead to potentially irreversible damage to the climate system, making timely mitigation efforts unfeasible. This is thought to cause the climate to rapidly change until it reaches a new stable condition. These phrases may be used with reference to concerns about rapid global warming. Some astronomers use the expression runaway greenhouse effect to describe a situation where the climate deviates catastrophically and permanently from the original state—as happened on Venus.
Although these terms are rarely used in the peer-reviewed climatological literature, that literature does use the similar phrase “runaway greenhouse effect”, which refers specifically to climate changes that cause a planetary body’s water to boil off.
https://en.wikipedia.org/wiki/Runaway_climate_change
Does anyone think climate change can cause Earth ocean’s [or even some mud puddles] to boil?
“Does anyone think climate change can cause Earth oceans [or even some mud puddles] to boil?”
In about 1 billion years.
gbaikie…”a tipping point in the climate system, after accumulated climate change initiates a reinforcing positive feedback”.
Problem is, GB, no one ever defines positive feedback. The kind referred to for a tipping point requires an amplifier.
This amplifier is ‘supposed’ to come from CO2 back-radiation, which is allegedly supposed to raise the temperature of the surface thus releasing more WV. Where does the extra heat come from? There is no heat amplifier.
All over the Net you see explanations in which feedback is mistaken for amplification, or gain. Feedback is not amplification, it is part of an amplified system. It can only work if the feedback is added to an input signal then amplified. That’s where the tipping point comes from, the gain is exponential and runs away to the tipping point.
The word feedback in positive feedback means to feed back part of an amplified signal. Where’s the amplifier?
They have many things they worry about- huge increase in Methane in Atmosphere. And much larger increased level of CO2.
My question “allows” any fevered possibility as long as it fits it into “global warming issue”, rather say an impactor or something.
For feedback questions it is best to ask an expert.
Hey Jimi, where’s the amplifier?
Jimi says
It’s the Sun, stupid.
bobd…”For feedback questions it is best to ask an expert”.
I am an expert and I have given you my input based on decades of working with positive feedback as applied in electronics.
The Sun can amplify nothing. As an amplifier with positive feedback, it would have to accept IR from the surface, amplify it, then output it to the surface at a higher level of solar energy. In other words, IR from the Earth would make the Sun increasingly hotter.
AGW theory suggests heat converted from solar energy by the surface to IR, then radiated to the atmosphere and captured by CO2, can be returned to the surface and added to solar energy so as to make the surface warmer than it is warmed by solar energy alone.
That is one of the most stupid theories I have ever encountered in science. It contradicts the 2nd law and represents perpetual motion.
Furthermore, it’s not a feedback system at all. It could only be a feedback system if CO2 absorbed all surface IR, amplified it, then radiated the IR back to the surface at a significantly elevated intensity that represented CO2 molecules as heat source with a temperature greater than surface temperature.
There is one basic problem with what you say about this
“AGW theory suggests heat converted from solar energy by the surface to IR, then radiated to the atmosphere and captured by CO2, can be returned to the surface and added to solar energy so as to make the surface warmer than it is warmed by solar energy alone.”
This is an observation, not a theory.
So sorry dude.
Nope bob, it has not been observed. It is not a theory, it is pseudoscience. It’s a belief system, devoid of actual science.
JD,
Sit on this and tell me infrared can’t go from a colder body to a hotter body.
https://www.thermotex.com/product/thermotex-platinum/
Sit Ubbu sit
boob, unplug the device and heat can NOT go from “cold” to “hot”.
Plug the device into the required voltage source, and heat can go from “cold” to “hot”. You have added the necessary energy to the process.
Learn some thermodynamics.
https://en.wikipedia.org/wiki/Leap_second
A leap second is a one-second adjustment that is occasionally applied to civil time Coordinated Universal Time (UTC) to keep it close to the mean solar time at Greenwich, in spite of the Earth’s rotation slowdown and irregularities.
UTC was introduced on 1972 January 1st, initially with a 10 second lag behind International Atomic Time (TAI). Since that date, 27 leap seconds have been inserted, the most recent on December 31, 2016 at 23:59:60 UTC, so in 2018, UTC lags behind TAI by an offset of 37 seconds.
binny…”A leap second is a one-second adjustment that is occasionally applied to civil time Coordinated Universal Time (UTC) to keep it close to the mean solar time at Greenwich, in spite of the Earths rotation slowdown and irregularities”.
How does that fit with general relativity theory, which suggested that same leap second can be dilated, to cause space-time anomaly issues in the universe. You have already mentioned the relationship between the second and Earth’s rotational period, GRT theory suggests that period must change to satisfy the theory.
BTW, I just read that Einstein fudged GRT math to move toward an already known Newtonian fact. Don’t know what to make of that but it’s interesting. It’s known that Einstein was seriously frustrated in his attempt to solve the problem, so maybe he got a bit creative, through desperation, as did Planck.
We all do realize that Venus is closer to the sun than the earth? Right?
That’s what my very excellent mother says, but why is Venus hotter than Mercury?
bobd…”Thats what my very excellent mother says, but why is Venus hotter than Mercury?”
Kudos to your mother but who said Venus is hotter than Mercury, based on solar radiation?
I certainly did not say that.
Maybe Pochas94 has a point, but
You do know that due to the high albedo of Venus compared to the Earth, both the Earth and Venus get about the same amount of sunlight below the top of the atmosphere?
Speaking of predictions – While you are bickering here about nonsense The false prophet Al Gore is celebrating ten year anniversary of his epic failed prediction of ice free north pole
https://goo.gl/NNiyLy
Didn’t some guy go swimming up there at the north pole a few years ago?
In ice free water?
But that happened before Al Gore’s prediction, so he predicted something that had already happened.
Go figure
just for giggles here’e the cite
https://www.cbsnews.com/news/man-becomes-first-to-swim-at-north-pole/
bobd…”Didnt some guy go swimming up there at the north pole a few years ago?
In ice free water?”
That would likely have been a woman, Lynne Cox, of the US.
Amazing endurance.
https://en.wikipedia.org/wiki/Lynne_Cox
There’s nothing uncommon about the NP being ice free, especially in the 1 month of summer. It has been known to be ice free, and even above 0C in winter. It’s a weird place and it’s temporary climate can be influenced by wind and ocean currents.
If it is often ice free, as you say why are some of your friends calling it a failed prediction?
An good comment posted at WUWT by Burl Henry:
https://wattsupwiththat.com/2018/12/13/el-nio-is-expected-to-form-and-continue-through-the-northern-hemisphere-winter-2018-19/#comment-2556258
His main statement:
“All La Ninas and El Ninos are caused by increasing or decreasing amounts of SO2 aerosols in the atmosphere. ”
Sounds interesting when sustained by corresponding data, as Henry provides us with.
Look at the preprint of his paper:
https://osf.io/bycj4/
To start, his eruption list prior to Maunder’s minimum is incredible. I know of Samalas in 1257 and of 5 others, but Henry collected a lot more.
His relation between
– volcanic / anthro increases in aerosol levels and La Ninas
– volcanic recovery warming and El Ninos
is interesting.
I think that is called curve-fitting.
A somewhat more scientific evaluation would be welcome.
Bindidon
As yu say “Henry collected a lot more (volcanoes)”.
The problem is that it is difficult to find a date when a volcano is not erupting somewhere, historically and recently.
It is very easy to correalate anything with volcanic activity.
I could produce a graph with a blip for each of your birthdays, and list a volcano which erupted somewhere on each birthday.This does not prove that birthdays cause volcanoes, or indeed that volcanoes cause birthdays.
As gblaikie mentioned there is an accepted link between volcanoes, pollution, SO2. emissions and climate through albedo.
A link between SO2 and ENSO cycles is much less certain. First you have data for only a few cycles; secondly to establish a cause-and-effect relationship you need a mechanism.
“It is very easy to correalate anything with volcanic activity.”
Is that really Entropic Man writing here? Looks rather like a duplicate of Robertson.
“Duplicate of Robinson”
Not quite.
I am sceptical of the paper’s conclusions as I do not regeard the evidence as sufficient.
Gordon Robinson is in denial.
entropic…”Gordon Robinson is in denial”.
I have never been anywhere near Egypt.
The only thing I deny is pseudo-science based on consensus with no hard science to back it.
It just so happen that covers the Big Bang theory, black holes, evolution, General Relativity Theory, GHE theory, and AGW theory.
Not my fault.
Another interesting diversion:
1) Bindidon reports some more pseudoscience.
2) E-man points out it is pseudoscience.
3) Bindidon and E-man join to blame it on Gordon.
4) Gordon comes in to decimate their false logic.
Nothing new.
It seems SO2 injected into high altitudes via Volcanic eruption would be different, than SO2 emission from human sources.
If we concerned about SO2 in tropsphere, it rains a lot in the tropics. And such tropical cloud formation and raining would remove SO2 from the tropical troposphere rather quickly.
And basically, El Nino and La Ninas are tropical events and it seems doubtful China SO2 emission would not have much effect upon El Nino and La Ninas.
Removing SO2 from emission is rather simple to do, and China has a lot modern coal plants which I would guess removes most SO2 emissions- it’s generally minor cost if one building new power plants. Removing other pollution can be more difficult.
” But then the increase of emissions came to an end. In Northern America SO2 emissions peaked in 1970, in Europe in 1980, and in South America a decade later. Since then emissions have been on a downward trend in these regions. At the beginning of the 21st century emissions in Northern America are at a lower level than at any time in the 20th century. The reason for which we will discuss later.
…
The dominance of Europe and North America in total global emissions means that the world’s SO2 emissions peaked in the 1980s, despite a continued increase across the other continents.”
https://ourworldindata.org/air-pollution
And by looking graph the peak was about 140 million tonnes per year for about 2 decades, and dominated by Europe. Or during peak Europe emitted twice as much as China ever did per year.
In terms of volcanoes the amount of SO2 varies:
“Only two years earlier, the major Mt. St. Helens eruption had lowered global temperatures by about 0.1 degree C. The much smaller eruption of El Chichn, in contrast, had three to five times the global cooling effect worldwide. Despite its smaller ash cloud, El Chichn emitted more than 40 times the volume of sulfur-rich gases produced by Mt. St. Helens, which revealed that the formation of atmospheric sulfur aerosols has a more substantial effect on global temperatures than simply the volume of ash produced during an eruption.”
And in terms of LIA, I don’t know if amount SO2 emitted by the various volcanic eruption is known.
Last quote link:
https://www.scientificamerican.com/article/how-do-volcanoes-affect-w/
binny…”His main statement:
All La Ninas and El Ninos are caused by increasing or decreasing amounts of SO2 aerosols in the atmosphere. ”
Highly unlikely since ENSO is driven by ocean currents that begin and end between Australia and the west coast of South America.
gbaikie says:
“The Impact of Heat Islands on Mortality in Paris during the August 2003 Heat Wave
‘Conclusions: Our results support the influence of night temperatures on the health impact of heat waves in urban areas. Urban heat exposure indicators based on satellite imagery have the potential to identify areas with higher risk of death, which could inform intervention decisions by key stakeholders.’
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3279432/”
This is a typically superficial article made by people who think a technical standpoint be of greater importance than considering the social context.
The most deaths due to the heat wave in August 2003 didn’t have anything to do with UHI phenomena.
Look at the map below:
https://www.senat.fr/rap/r03-195/r03-19510.gif
There you see that France’s central region (called ‘Centre-Val de Loire) suffered the most. There are no big cities there: it is a region where, like in the Auvergne and Burgundy, rural aspects clearly dominate. Pollution like in and around Paris is nearly inexistent there. You find only 65 habs/km2!
When you focus at the regions showing the most unexpected deaths in 2003, you see that the departemensts around Paris (the region called ‘Ile-de-France’ which behaves like a giant UHI, as it is the region concentrating bost the most people and the densest industrial and commercial aspects) had the greatest part of these deaths in France (33%). There are 1000 habs/km2 there.
Butt when you look at the percentage of deaths per capita, you see that the Centre region suffered more: 1.1 death per thousand persons compared with 0.7 around Paris.
And within Paris, the most people did not die in August where the heat wave was the strongest: thy died where the number of hospitals per capita was the least, in the south of the town, and where the isolation of people aboe 80 years is the highest.
Moreover, Marseille of course is a smaller town than is Paris, but its UHI factor is by far higher, as
– Marseille’s older parts are by far denser than those in Paris
– Marseille is a hotter place.
But… in Marseille the politicians were well prepared to face a heat wave, better than in Paris and even better than in Nice, where more people died due to the ignorance of Nice’s leaders.
I forgot a detail.
It is evident to me that a GHCN daily station located at Le Bourget airport northeast to Paris will not at all give data fully comparable to what a station inside would provide.
But it is nevertheless interesting to have a look at Le Bourget’s 25 highest absolute temperatures (in C of course):
1947 7 28 40.4
2015 7 1 39.7
1900 7 20 38.6
2012 8 18 38.4
2012 8 19 38.1
1952 7 1 38.0
1947 7 27 37.9
1911 8 9 37.7
1947 6 26 37.6
1911 7 23 37.5
2018 7 27 37.4
2015 7 2 37.3
1998 8 11 37.3
1998 8 10 37.1
1921 7 28 37.1
1904 7 17 37.1
2018 8 7 36.9
2017 6 21 36.9
1946 7 3 36.9
2011 6 27 36.8
1947 6 28 36.8
1945 7 14 36.8
2016 8 24 36.6
2015 7 16 36.6
1990 8 4 36.6
Hum… where is August 2003, you probaly might ask. It is at position 333 (of 1040):
2003 8 7 32.7
The year 1947 appears 9 times within the top 100.
binny…”The most deaths due to the heat wave in August 2003 didnt have anything to do with UHI phenomena”.
What caused temperatures to rise to a level where a heat wave occurred? We know the warming claimed by anthropogenic warming is less than a degree over a century. That can’t explain it.
Neither can it explain record temperatures since the temperature increase is well beyond what one would expect from anthropogenic warming.
CO2 simply cannot account for heat waves or record temperatures.
Robertson
You are so totally fixated in your bloody pseudoscience that you mention CO2 even where it is completely redundant: I did not mention your stupid CO2 blah blah in my comment.
I solely tried to explain gbaikie how useless it is to try to relate heat wave deaths to UHI when a big, big amount of these deaths occured exactly OUTSIDE of any UHI zone, and was due to social problems shared by so many aged and poor persons.
As usual, you do not read comments: you scan them for what excites you for an answer.
You behave like the toros in the spanish corridas: a few cm2 of red stuff somewhere, and you run blindly against it.
Jesus what are you a boring person!
binny…”You are so totally fixated in your bloody pseudoscience that you mention CO2 even where it is completely redundant: I did not mention your stupid CO2 blah blah in my comment”.
Now you are raving like an idiot.
Your MO since you started commenting here has been alarmist in nature. You present home-made graphs pretending to compare the UAH record with that of NOAA and GISS, both fudged. You defend NOAA and GISS when they offer blatant, crooked math, by lowering confidence levels to a ridiculous level like 48% and 38% respectively to move a very ordinary year forward as the hottest year ever.
You obviously support the CO2 warming theory yet here you are raving because I caught you at it.
The UHI effect theory contradicts CO2 warming since it gives a good reason for natural higher global averages. Most surface station data comes from populated regions where UHI is an issue.
As I pointed out to you, the much higher temperatures related to heat waves cannot be explained by CO2 warming but the UHI effect can exacerbate heat waves.
In your petty mind, you likely think somehow that heat waves are caused by climate change, which is scientific nonsense. Climate change is a result of changes in warming, not a cause of it. Something natural is causing the heat waves and the UHI effect is contributing.
Again, Robertson: you are such an incompetent, ignorant and pretentious boaster.
I repeat until you get it:
You are so totally fixated in your bloody pseudoscience that you mention CO2 even where it is completely redundant: I did not mention your stupid CO2 blah blah in my comment.
Gordon Robertson says:
“We know the warming claimed by anthropogenic warming is less than a degree over a century. That can’t explain it.”
That’s right Gordon, it’s not global warming, it’s climate change:
https://tinyurl.com/y7rmdvsr
svante…”Thats right Gordon, its not global warming, its climate change:”
Over the past year or so, you have slowly turned into a blithering idiot. Why would you link to an idiot spreading propaganda?
He talks about water at 32F causing a change in the jet stream.
Duh!!!! Water has to be around 32F or higher. The idiot is likely talking about the Arctic ocean during one month of summer, the rest of the year, there is 10 feet of ice over the Arctic Ocean.
Do you not get the basic reason for that? No solar energy for several months and diminishing solar energy preceding the darkness.
No amount of CO2 will ever change that fact. So, go on raving about climate change, like a blithering idiot, while failing to understand the physics involved in the Arctic.
Roy pointed out a while back that warming in the Arctic moves around and he opined the cause may be conditions in the North Atlantic.
Why can’t you get off you alarmist spiel and try talking some sense, like Roy?
“Why would you link to an idiot spreading propaganda?”
Because it makes sense.
“Water has to be around 32F or higher. The idiot is likely talking about the Arctic ocean during one month of summer, the rest of the year, there is 10 feet of ice over the Arctic Ocean.”
He said the arctic has become warmer. He said there is water where there used to be ice and 10F or cooler.
There is not much left of the 10 feet multi year ice:
https://tinyurl.com/y7rt9t3p
“Do you not get the basic reason for that? No solar energy for several months and diminishing solar energy preceding the darkness.”
That does not explain why the ice is disappearing.
“No amount of CO2 will ever change that fact. So, go on raving about climate change, like a blithering idiot, while failing to understand the physics involved in the Arctic.”
CO2 can reduce the energy loss rate to space.
“Roy pointed out a while back that warming in the Arctic moves around and he opined the cause may be conditions in the North Atlantic.”
Such as warming?
“Why cant you get off you alarmist spiel and try talking some sense, like Roy?”
He says there will be warming. Why can you not learn the 2LOT from him?
For those ignorants living in simple-minded denial of Einstein’s work
In 2005, a guy named Tom van Baak, owner of various atomic clocks and admirator of both Albert Einstein and Louis Essen, wanted to celebrate 100 years of General Relativity and 50 years since Essen’s first atomic clock release.
What he did is amazing, and is described here:
http://leapsecond.com/great2005/
and here
http://leapsecond.com/ptti2006/tvb-project-great-ptti-ppt.pdf
*
Here is a summary of his ‘Project GRE2AT: General Relativity Einstein/Essen Anniversary Test’
The year 2005 was the 100th anniversary of Einstein’s first paper on relativity. It was also the 50th anniversary of Essen’s first cesium atomic clock. Project GREAT (General Relativity Einstein/Essen Anniversary Test) was conceived to celebrate both events in a single experiment.
This paper presents the historical background, implementation details, and surprisingly successful results of Project GREAT, a modern demonstration of relativistic time dilation by carrying multiple portable cesium clocks to high altitude and directly measuring the clock effects of gravitational blueshift as predicted by general relativity. Although several similar, and now classic, traveling clock experiments have been performed since the early 1970s this one is unique for a number of reasons.
This is perhaps the first case where the general theory of relativity is confirmed with a family weekend road trip using surplus atomic timing equipment. The performance of a large collection of vintage and modern cesium standards was measured and the best three clocks pre-selected for the round-trip experiment. With an assortment of sub-nanosecond time interval counters, data logging equipment, AC/D-C power distribution systems, and a large set of batteries the family minivan was converted into a mobile atomic time lab.
The clocks were driven to the highest point accessible by road on Mt Rainier, the iconic volcano near Seattle, Washington, and kept at altitude for 40 hours. Continuous 3-way inter-clock phase measurements were collected during the trip and 5-way rate measurements were made against hydrogen maser references for days both before and after the trip. In addition to practical advice for constructing a mobile time lab, theoretical and actual results, the paper explores multiple algorithms to extract precise time dilation measurements from the raw data.
Einstein’s work predicted for a net altitude difference of 1340 meters a measurement difference of 22 nanoseconds, what was accurately confirmed by two of the three atomic clocks used during the experiment.
*
I propose that the denialists try to repeat van Baak’s experiment, instead of discrediting it in the usual manner.
binny….”In 2005, a guy named Tom van Baak, owner of various atomic clocks and admirator of both Albert Einstein and Louis Essen…”
Essen invented the atomic clock. Don’t you think it a bit odd that Essen thinks Einstein was wrong and van Baak admires Essen?
I am awaiting your explanation of how time can dilate when the second is based on the rotational period of the Earth, which is constant, plus or minus a leap second. How much does GRT claim time dilates, by a leap second?
Robertson
Louis Essen NEVER pretendes Einstein be wrong.
He claimed that Einstein sometimes lacked the precision he would have wished of him, THAT IS ALL.
The rest is, as usual, invention of pseudoskeptics like you.
“I am awaiting your explanation of how time can dilate when the second is based on the rotational period of the Earth, which is constant, plus or minus a leap second.”
Here you show that you understand about time as much as you understand about time series, reference periods, baselines and anomalies: nothing.
The second you mean is that measured at Earth’s surface, you genius!
Time measured 1000 metres or 500 km above Earth’s surface differs due to escape from Earth’s gravitational force.
“Atomic clocks have nothing to do with time.”
That you should tell to those continuing Louis Essen’s work! They would laugh about your endless ignorance.
*
You can deny all that till your death. What does your denial change?
Nothing, Robertson.
binny…”Louis Essen NEVER pretendes Einstein be wrong.
He claimed that Einstein sometimes lacked the precision he would have wished of him, THAT IS ALL”.
No…Essen claimed Einstein did not understand measurement.
No surprise there, Einstein was a theoretical physicist who likely never saw the inside of a lab. He lived in his mind much of the time, in a world of thought-experiments. The only lab he had was the cells in his brain.
Essen argued to have the definition of the second changed from a fraction of the rotational period of the Earth to a multiple of the wavelength of the EM emitted when the Cesium atom changed energy orbital levels. That’s because the Earth’s rotation is not precise to a fraction of a second.
It comes down to the same thing, however. Both references are relatively constant therefore the second cannot change in length unless the basic atomic parameters of the Cesium atom change.
There is nothing proved anywhere that the frequency of the Cesium atom’s radiation changes at velocities approaching the speed of light. So why would Essen agree with Einstein and his GRT, that time can dilate under such conditions?
I think Einstein became trapped in his own mind. That can occur in a world of thought experiment and math unsupported by reality.
binny…”The second you mean is that measured at Earths surface, you genius!”
This statement outdoes any other idiotic statement you have ever made.
Time IS defined at the Earth’s surface. Where else would you measure the rotational period of the Earth wrt the Sun?
This nonsense that time changes with gravity comes from the minds of people unable to think clearly. They are not measuring time for the simple reason there is nothing to measure.
Are you telling me that the Earth’s rotational period changes with altitude and gravitational force?
When measurements are made of time at altitude, and the instrument reads a different time, it is the instrument changing, not time. Handling, heat, and altitude might affect the way atoms vibrate, hence changing the frequency of their emissions. That is not a change in time, it’s a change in atomic parameters.
Besides, any minute changes in atoms has nothing to do with the time dilation suggested by GRT, which is tied to the speed of light.
Einstein clearly messed himself up by basing his theories on acceleration rather than the force that causes acceleration. Acceleration, as we use it, is a definition based on our definition of time.
If you use force and mass, you’d never get caught up in that illusion since it would become very apparent that the only two real entities are the force and the mass.
Whereas acceleration can be claimed to be a real phenomenon, since we can see it happen, the actual measurement of it is based on artificial parameters introduced by humans.
Einstein seems to have forgotten that the time-based ‘second’ element in the human defined acceleration, as in m/s^2, cannot change because it is defined on a constant.
If you cannot see that, you really are an idiot.
binny…”This is perhaps the first case where the general theory of relativity is confirmed with a family weekend road trip using surplus atomic timing equipment.”
What were the clocks measuring? Go into it, examine it closely. I am not kidding.
Atomic clocks have nothing to do with time. The vibrations in an atomic clock are natural vibrations related to atoms processes. They are so regular, however, that we can use them as a very accurate timebase. A timebase is not a clock it is a generator of pulses.
The very short wavelength vibrations must be multiplied bazillions of times to equal one second. The second is based on the angular velocity of the Earth, which is a constant.
There are a lot of researchers who cannot see beyond they ends of their noses.
What would you do if you had to let go of your belief system and your appeal to authority figures?
The usual pseudoscientific nonsense of this site’s most ignorant and most pretentious denialist…
A la poubelle!
binny…”The usual pseudoscientific nonsense of this sites most ignorant and most pretentious denialist”
The true sign of someone in over his head, the liberal usage of ad homs and slurs. At least when I call you an idiot I offer a scientific explanation.
You are nothing more than a Teutonic bag of wind.
And above all, Robertson, I repeat:
I propose that the denialists try to repeat van Baaks experiment, instead of discrediting it in the usual manner.
But never and never would you have even a thin bit of the courage needed to do it!
That is the difference between the van Baaks and the Robertsons!
They do, you discredit and denigrate.
binny…”I propose that the denialists try to repeat van Baaks experiment, instead of discrediting it in the usual manner”.
As Linus Pauling, a leading expert in chemistry of all time, with reference to the need for a double-blind study, once claimed, why is a double-blind study required when an outcome is so obvious?
In a similar manner, why should I read through the an experiment trying to prove GRT correct, when the second upon which it is based is itself based on a constant?
The basis of the GRT presumption, that an observer moving on a platform while measuring motion on another platform, sees time dilate and lengths change, only tells me the human mind is corrupt and prone to illusion. Time does not dilate nor do lengths change, unless a body is heated or frozen.
GRT is about illusions in the human mind. The applicable math works only because Einstein knew the outcome and fudged the math until it worked. I have no problem with the GRT equation about relative motion, but I have a big problem with the conclusion based on GRT that time dilates and lengths change.
That has lead to the idiocy of space-time in which imaginary constructs are created in the universe that have no physical existence.
Sad to say, the Denialist who inhabit this web site aren’t interested in experiments or facts. They are like those religious Fundamentalist who think the Earth is less than 10,000 years old, thus ignore ALL EVIDENCE which refutes this world view.
Take for example their claims that the Moon doesn’t rotate, a well proven fact in astronomy. For months, these guys have ignored the experimental evidence, such as a simple test with a compass which proves HuffingBoy’s analogy of a race horse running around a track. They apparently can’t understand the difference between “translation” and “rotation” in physics, as they continually repeat their nonsense.
I suspect their goal is to post so much garbage that no meaningful scientific discussion can occur.
Oh cry me a river, Swamson.
You’re just trying to escape your past:
http://www.drroyspencer.com/2018/12/can-space-com-teach-us-anything-useful-about-climate/#comment-334086
HuffingBoy, Please explain how the moon can exhibit a cycle in it’s illumination without rotation. Include only objective facts in your reply.
Swamson, your question reveals, again, that you do not even understand the basic motion of the Moon.
The Moon orbits around the Earth. That means the Sun “sees” different sides of the Moon. On Earth, we see “phases” on the Moon, as different areas are illuminated by the Sun.
This is stuff you should have learned in high school, if not before.
Sheesh!
Yes, Huffingboy, we see different “phases” of lunar illumination because the moon rotates WRT the Sun. QED…
As I had to explain to you in my previous comment.
Please continue to make a fool of yourself.
HuffingBoy, So you must agree that the Moon rotates WRT the Sun. If not, please give us your definition of “rotate” in terms of physics.
Swamson, what I “must agree” with is that you don’t even understand the issues you cling to. You can’t think for yourself, except as it applies to grasping onto failed pseudoscience.
You haven’t gotten one thing right, and have exposed your ignorance in almost every comment. You ignorance carries over from your bogus “experiment”.
This issue has nothing to do with the Sun. It has to do with whether or not the Moon rotates on its own axis as it also orbits around Earth.
You have no clue.
Feel free to continue making a fool out of yourself.
HuffingBoy continues with another his usual content free, anti-science, fact free rants. If the Moon doesn’t rotate, it can’t have an axis, since the direction of the axis is determined by the rotation of a body. Of course, you refuse to give us your definition of rotation (or anything else, for that matter), which indicates your gross lack of understanding of physics.
“HuffingBoy continues with another his usual content free, anti-science, fact free rants.”
Inaccurate drivel from a 12-year-old.
“If the Moon doesn’t rotate, it can’t have an axis, since the direction of the axis is determined by the rotation of a body.”
Somehow you stumbled onto somewhat the correct path. The Moon does NOT rotate on its own axis.
“Of course, you refuse to give us your definition of rotation (or anything else, for that matter), which indicates your gross lack of understanding of physics.”
Since you have no background in the relevant physics, perhaps you might understand a simple ceiling fan. A ceiling fan rotates on its own axis. (CAUTION: Ceiling fans can be dangerous. Make sure you always have adult supervision, when near a ceiling fan.)
As expected, HuffingBoy can’t even handle a simple request for the definition of a physical variable called rotation, continuing his mystical denial of physics and disregarding centuries of work in physics and astronomy.
Swamson was finally able to find the definition I’ve linked to numerous times.
See, he can learn. It’s just a slow process.
swannie…”Please explain how the moon can exhibit a cycle in its illumination without rotation. Include only objective facts in your reply.”
That’s a question I’d expect from you after your experiment allegedly disproving the 2nd law. If you think the question through the answer will become obvious.
Try it with two coins. Two coins side by side with a common mark on their perimeters. Now try to rotate one coin around the other while keeping the mark of the one you move on the surface. It is impossible to rotate it by rolling it.
Now slide it around wile keeping the mark on one against the surface of the other.
See how it works? It’s exactly the same as sliding the coin along a flat surface in a straight line without rolling it, only the line is curved into a circle.
Go on, try it on the circle and you’ll see the mark is always pointing in.
Gordo confuses “rotation” with “rate of rotation”. If you move the outer coin around the inner one while keeping the marked point facing the inner one, the outer one will rotate once for each movement from start to finish. The Moon rotates once per orbit
swannie…”Gordo confuses rotation with rate of rotation. If you move the outer coin around the inner one while keeping the marked point facing the inner one, the outer one will rotate once for each movement from start to finish. The Moon rotates once per orbit”
Did you see the mark on the moving coin rotate through 360 degrees, or even one degree?
There is no rate of rotation, it is zero…nada. The mark on the coin is always touching the other coin, it cannot turn with an angular velocity about it’s axis.
What you are seeing is a change of position, which is translation. Because the coin is moving (sliding) on a curved surface, it is constrained to follow that surface. If it follows that surface with the same side always touching the surface, a radial line through the coin touching the marked point will change position since it is part of a rigid body.
If there was rotation, that line would turn around the coin centre, which it does not. That’s because the centre is moving with the line in a concentric circle around the other coin.
All points on the radial line are turning at the same rate, therefore rotation of any part of the line is impossible. It is curvilinear translation.
Gordo, Your version of your coin example is obviously false. Instead of placing a mark on the outer coin, draw a line across it with an arrow at the end instead of a mark. The case with the outer coin moving via “sliding” around the inner one while the marked arrow always touches the inner one (i.e., the arrow pointing toward the center) gives one rotation per circuit of the inner one. The second case with the outer coin “rolling” around the inner one gives two rotations per circuit. The second case would produce a different rotation if the inner coin is not the same diameter as the outer one.
swannie…”The case with the outer coin moving via sliding around the inner one while the marked arrow always touches the inner one (i.e., the arrow pointing toward the center) gives one rotation per circuit of the inner one”.
No it doesn’t. In order to keep one face always facing in, all particles along that line must turn in concentric circles around the stationary coin’s centre. If they are turning in concentric circles, including the axis, they cannot rotate about that axis.
Gordo’s delusional physics continues. Rotation of a free body occurs around it’s center of mass, not some point outside the body. You’re confusing rotation and translation again. Stop trolling and learn some physics.
http://www.mvsrec.edu.in/images/mech_new/DrMM-Notes/dynamicsofrigidbodies.pdf
Page 4 of 28, Fig. 2(b).
The rectangle is NOT rotating about its own center of mass, it is rotating about an off-center axis (point O). All particles of the rectangle move in concentric circles. Replace the rectangle with the moon, the stick attaching it to point O with “gravity”, and there is your “moon’s motion”.
Roy’s bogus EMT, Your reference to Figure 2B shows a body rigidly connected at a pivot to some other body and the resulting motion about that pivot point. Reading a little further, Figure 7 shows a motion which is a combination of rotation and translation. Recall that the motion of the coin is the combination of the instantaneous motions and the effect of the force applied to cause the outer coin to curve around the inner one. The instantaneous motion is a combination of the velocity of the center of mass plus the rotation about that center. This logic is continued in Figures 11, 15 and 17.
Nope, it shows an object rotating about an off-center axis. It made the point so perfectly clearly and simply that you were forced to obfuscate.
swannie…”Rotation of a free body occurs around it’s center of mass, not some point outside the body. You’re confusing rotation and translation again. Stop trolling and learn some physics”.
I said nothing about rotation, I said change of position, which is translation.
Take your straight line with the arrow and point it at a flat surface. Now slide the coin along the flat surface, keeping the arrow pointed at the surface.
No rotation.
Now curve the flat surface slightly and repeat.
Still no rotation.
Keep increasing the curvature until the surface forms a circle.
Still no rotation.
The line is not rotating. The arrow is pointing in different direction because it is being forced by the curved surface to do so. As the line moves around the curved surface, it will have different orientations wrt to a human observer but the notion that it is rotating about it’s axis is sheer illusion.
The same applies to AGW. Sheer illusion.
swannie…keep in mind that the line with the arrow pointed to the surface has all points along the line turning in concentric circles around the circle (ie. parallel to the circle). If all points on the line turn in concentric circles they cannot be turning in concentric circle around their own axis while the arrow stays on the circle.
I have just described curvilinear translation.
Gordo, How do you do it? How do you continue to ignore basic physics? You wrote:
A body in motion continues in that direction unless it it acted on by an external force. The same may be said regarding rotation of the body. The path of the outer coin (and the Moon) is the result of the forces acting on the respective centers of mass which cause the velocity vector to change direction, causing the path to follow the curved trajectory. The fact that the line on the coin changes direction is prima facie evidence that the coin is rotating.
“The fact that the line on the coin changes direction is prima facie evidence that the coin is rotating.”
Not on its own axis.
E. Swanson
At the end of a rather long Google search, I found this document and uploaded it using Google drive:
https://drive.google.com/file/d/1Fiiln0lAnFjPwtYBB1rw1U1x3ynrDegZ/view
It is in German, maybe I manage to translate most parts of it within the next weeks.
Looks interesting at a first glance.
swannie…”They apparently cant understand the difference between translation and rotation in physics, as they continually repeat their nonsense”.
Coming from a pseudo-scientist who claims radiation from a cooler body can increase the temperature of a hotter body that warmed it.
Gordon,
Show me any online university physics/kinematics reference which agrees with your perverted definition of translation. As a matter of fact, you NEVER define translation. You just give stupid examples like the coin nonsense.
skeptic…”Gordon,
Show me any online university physics/kinematics reference which agrees with your perverted definition of translation”.
They all do. Rectilinear translation describes a rigid bodies with all particles in the body moving parallel to each other. The other stipulation is that all points in the body are moving at the same velocity.
Same applies to curvilinear translation with one exception. When a rigid body moves with curvilinear motion, all the particles must move parallel AND their angular velocity is the angular velocity along a radial line through the body, not the individual particle velocity.
A rigid body must turn as a unit, not individual particles. The fact that individual particles on a rigid body can travel at different speeds is due to the fact they must all turn at the same angular velocity. In order to keep up, some of them have to move faster than others.
The Earth has an angular velocity of 360 degrees/24 hours. All particles must meet that requirement, however, different particles travel at different speeds in order maintain that constant angular velocity.
Your problem with CT is that you cannot come to terms with concentric circles being parallel. That’s your hang up.
The definition of both curvilinear and rectilinear translation REQUIRES “every line in the body remains parallel to its original position.” Do you even comprehend simple English?? REMAINS PARALLEL TO ITS ORIGINAL POSITION! Your tangent line is changing position constantly. It does NOT remain parallel to it’s original position. Say for instance your tangent line points east at the top of a circle. Then at the 9:00 o’clock position it points south while moving CCW. It the south line parallel to the north line, which is its original position? Hello! McFly!
The following kinematic reference shows an example of curvilinear translation on page 4. It shows the arrow remaining parallel to its original position as it translates along the curve, and NOT moving tangent to the curve as you stupidly suggest.
https://www.asu.edu/courses/kin335tt/Lectures/Kinematics/Kinematics%20I%20Spring%202005.pdf
And then you have to lie about “all points of the body are moving at the same velocity”. NO WHERE will you find that defined as “angular velocity”. You simply lie to cover your ignorance.
And then you confuse “curvilinear translation” with “curvilinear motion”. Not the same thing. Were you drunk when you allegedly went to engineering school?
You can continue to ignore the truth and make a fool of yourself.
skeptic…”The definition of both curvilinear and rectilinear translation REQUIRES every line in the body remains parallel to its original position. Do you even comprehend simple English?? REMAINS PARALLEL TO ITS ORIGINAL POSITION! Your tangent line is changing position constantly. It does NOT remain parallel to its original position”.
I understand simple English but you seem to have a problem with it.
There is no tangent line on the translating body, it is on the curved surface of the body about which it is translating. The changing angle of that tangent line has nothing to do with it other than as an ‘instantaneous’ reference for a line drawn through the rigid body’s axis with it’s tip always at right angles to the tangent line on the circle.
That means every point on that line is moving in concentric circles around the circle. As long as two lines remain the same distance apart they are parallel, therefore concentric circles are parallel.
That meets the definition of curvilinear translation. It also meets your claim that all points must remain parallel to their original position.
If I start with a tangent line at x = 1, y = 0, on a circle, with a perpendicular line to the tangent line, and I move that tangent line right around the circle, all points on the perpendicular line will remain parallel to the circle.
You seem to have a problem thinking outside the box.
Gordon blubbers:
“That means every point on that line is moving in concentric circles around the circle. As long as two lines remain the same distance apart they are parallel, therefore concentric circles are parallel. That meets the definition of curvilinear translation.”
NO. This is another requirement for curvilinear translation:
“2. Curvilinear Translation: All points move on congruent curves.” [http://kisi.deu.edu.tr/binnur.goren/intermediate%20dynamics/1_kinematics%20of%20rigid%20body(2).pdf]
Not concentric curves like you state! You do know what congruent curves are?
A translating object does NOT change its orientation. If it starts out facing north, it has to face north all throughout its curvilinear motion. If you draw a line through a translating body, that line will not rotate at all. It will always point in the same direction:
“1. TRANSLATION: It is any motion in which every line in the body remains parallel to its original position at all times. In translation, there is no rotation of any line in the body.”
[http://kisi.deu.edu.tr/binnur.goren/intermediate%20dynamics/1_kinematics%20of%20rigid%20body(2).pdf]
And you continue to use the terms “curvilinear motion” and “curvilinear translation” interchangeably. They are not the same. An object can be moving curvilinearly, but not be translating.
Gordon states:
“If I start with a tangent line at x = 1, y = 0, on a circle, with a perpendicular line to the tangent line, and I move that tangent line right around the circle, all points on the perpendicular line will remain parallel to the circle.”
A line drawn through a translating body has to remain parallel to it ORIGINAL position. Your tangent line is not translating. It’s continually changing direction.
You have yet to understand “translation” in ALL your discussions so far. You are way out of your league here if you cannot grasp the simple concept of translation.
tim…” The FACT is that you cant just stretch the circle into an ellipse with the arrows pointing forward. That is NOT how real moons move. You also cannot have one side of the arrow always directly toward the planet that is a DIFFERENT wrong answer”.
Tim, you have obfuscated yourself into a corner.
If you have a circular orbit, with a moon, the moon is always trying to move straight, in a tangential direction to the orbit. The orbit is the resultant between that straight line action and the tug of gravity. The moon’s path in a circular orbit is always tangential (right angles) to a radial line from the circle’s origin to the moon.
As you pull the circle into an ellipse, the moon must still move tangential to the ellipse curve. However, the focal point of the radial line must change. With a circle, you have only one focal point, the origin. With an ellipse, that focal points splits into two focal points, one on either side of the origin.
For any curve, there is a tangent line perpendicular to a focal point for that curve. To find that focal point on an ellipse you must use both focal points. You draw a line from one focal point to a point on the ellipse and another line from the other focal point to the same point.
Then you bisect the angle between line. The tangent is perpendicular to the line bisecting the angle.
https://www3.ul.ie/~rynnet/swconics/TC.htm
About the other line, the one normal to the tangent which defines the condition of the same face always pointing to the Earth. It has to always be pointing in since it is following the tangent line at right angles.
If our Moon had an extreme elliptical orbit, as you have demonstrated, of course, you’d see more of its surface than you would with the current Moon orbit. However, you would NEVER see the far side of the Moon. You would still only see part of one side.
That’s one kind of libration, not rotation. It has nothing to do with a turning motion of the Moon about it’s axis, it is about the orbit.
The Moon still does not rotate, not in the slightest. You see more of the Moon at various points on an elliptical orbit as the ellipse becomes more extreme, but never the far side of the Moon, no matter how elliptical.
I did not think this out well enough when I replied the other day.
Gordon, they are trying to look for ways to discredit reality. I try to be very careful about the way I phrase words, but they still try to misrepresent me.
So, yeah, be more careful, but they are still going to come after you.
You handle their tricks quite well….
Gordon, did you get this:
https://tinyurl.com/y9ag9ex7
We started here but you went wrong in the first sentence of your reply.
svante…”The same physical property can be measured by many units.
Units are irrelevant as long as they measure the same physical property.
1) Power is the property energy divided by the property of time.
2) Time can be measured in the unit s.
3) Energy can be measured by the unit J.
4) Power can be measured by the unit W.
5) W is defined as J/s, or kg*m^2/s^3 like you say.
6) Tk has the unit Watt so the property is power”.
**********
Where did I go wrong in my first sentence?
Using properties of different energies can lead to contradictions if you use them out of context.
1)Power is actually mechanical energy/time = work/time
You can use power to express the work done in an electrical circuit as EI or I^2R. However, that is electrical power expressed in terms of mechanical power. Electric motors are actually expressed in horsepower.
You can convert electrical power to heat and express it in watts using the mechanical equivalent of heat (in watts). However, the natural measure of heat is the calorie. Heat is measured based on the number of degrees it can raise a volume of water.
2)time IS measured in secs, or multiples thereof.
3)Work can be measured by J, energy is far too generic.
4)power can be measured by W in metric terms. In the Imperial system, it is better to use HP.
J is a metric term for work = f.d (in newton-metres). A watt = f.d/sec = N-m/s, but watts are measured in Kg-m/s. In conversion between the Newton in J and the Kg in W, the gravitational acceleration m/s^2 is applied to the Kg, to get the watt in Kg-m^2/s^3
5)W is not ‘defined’ as J/s, it is defined as mechanical work/sec. It so happens a J = work = f.d = N-m.
6)what does Tk mean? Surely T is not temperature?
I think you’re trying to prove that radiation intensity from a body at Thot should be measured in watts, therefore it can be claimed to be power.
I think I proved in my reply to you at the link you provided proves that is not the case.
If you try to visualize EM as it is in reality, an electrical field with a perpendicular magnetic field, you might see what I mean. How can it do work on anything, except maybe an electric charge in a conductor INTERNALLY.
By itself, in space, as a freshly radiated field, it can do no work, therefore it should not be measured in terms of power. It’s not till it encounters mass that it can ’cause’ work to be done but the work is performed by a different form of energy.
svante…I might add that energy is often defined as the capacity to do work. There is a distinction between the capacity to do work, the potential, and the actual work done. That’s why we also define kinetic energy as energy in motion, or perhaps doing work.
You might think of EM as having a potential to cause work to be done. In itself, I would not call it kinetic energy.
Looking at it from a theoretical physical aspect, when a body is radiating to a cooler environment, the atoms in the body are doing work internally through vibration. The degree of vibration is related to the body’s temperature.
As the body radiates, it converts heat to EM and that heat is lost. If the heat is not replaced, the temperature will drop and the internal work done will drop.
So the power expressed at the body’s surface per unit area has to be internal work done by the atoms per unit time. How would that work/unit time get transferred to the EM as power? There would have to be mass in the EM to do work, and there is no mass. So where does the power come from in the radiated EM?
We agree mostly.
I should have said the first message was here:
https://tinyurl.com/yc66ceye with the formula:
Pt = Ph – Pc = esATh^4 – esATc^4 = esA(Th^4 – Tc^4)
You replied time was not a factor. I showed Pt can be measured in the unit Watt which measures the property power and includes the property time.
https://tinyurl.com/y87zx4mx says:
It can be expressed in many other units like you say, but the physical property is power.
The same reference says ‘amount of heat’ is measured in Joule, so you are right about calories, since both are units for the same property, namely energy.
Do we agree on this?
svante….”The same reference says ‘amount of heat’ is measured in Joule, so you are right about calories, since both are units for the same property, namely energy”.
You need be careful when you use the word energy by itself. No one knows what it is. It’s definition says that, the capacity to do work. What the heck does that mean, it does not describe a specific entity? It’s just saying ‘something’ has the capacity to do work.
When you get specific, referring to thermal energy, mechanical energy, chemical energy, nuclear energy, electromagnetic energy, and such, you find all the energies have different forms and different properties. You can relate that specific energy to something.
On the other hand, when you try to lump very different energies like thermal, electromagnetic, and mechanical energy under the generic terms energy, you can make some bad assumptions.
For example, you cannot add mechanical energy and thermal energy because they do not have the same units. First, you have to convert the calorie for heat to joules or watts, or vice-versa.
So, how would you add EM as energy to heat as thermal energy? How do you convert the electron-volts of EM to the calorie of heat, or its mechanical equivalent, the watt? There is a relationship but it’s not as obvious as some might think.
In AGW theory, liberties are taken. It is presumed EM can be absorbed by any body of any temperature and one justification is that the ‘net’ energies can be added.
What net energies?
I think we agree about several points, I’m just saying you need to be aware of the difference between the native measure of a specific energy and the mechanical equivalent.
Joule, the scientist, worked out the relationship between work in joules and heat in calories. When anyone applies watts to heat, he/she has too understand that watts measure mechanical work and that as a measure, the watt represents the mechanical equivalent of heat.
That’s important to understand. When heat is converted to work, there is mechanical work done, either internally or externally. Where is there mechanical work done when a body emits EM, before the EM contacts mass? The only work I can see done is the internal change in work of the atoms. When they radiate heat, they cool, therefore the amount of work they do gets less.
I think the W/m^2 in S-B refers to heat dissipation, not the intensity of radiated EM? It’s fair to claim I am splitting hairs in a way because there is an obvious relationship between the temperature of a body and the intensity of the EM it radiates. I just don’t think it’s a power relationship in watts, as when heat produces work.
One relationship between EM and the electron in atoms is
E = hf. That relates the intensity of EM to the frequency of an electron, which is related to the temperature of the mass in which the electron resides. How do you tie that into power?
Both S and B believed that heat was radiated directly as heat. They knew nothing about EM in their day other than the near-field electric and magnetic fields associated with Faraday’s work with inductors and motors.
Gordon Robertson says:
It is described by the laws of physics, whatever it is.
It can be measured too.
Yes, energy can be converted and measured by the same units.
Yes, like you do in the next paragraph.
Calories and Joules measure the same quantity. Quantities of the same kind can be expressed as multiples of each other. The choice is irrelevant.
Watts are a different quantity, it is called power and is energy per time.
An electron volt is 1.602×10−19 joule [J].
A calorie is 4.2 J.
The relationship is trivial because the are the same quantity.
Again, Watt is different.
Yes, they can be added for black bodies. Multiply by epsilon for real materials.
First thing is to separate units of measurement from the property itself.
Conversion between units is a simple multiplication.
Work and energy are measured by the same units, so the conversion is trivial.
Again, Watt does not measure work, it measures work per time.
Work is commonly used in mechanics, force times distance. Yes, EM carries energy, if you radiate more energy than you receive the object cools. The power is reduced because it depends on temperature.
Power per area is called flux, isn’t it? That’s a different quantity.
S-B multiplies by area, so you get the power.
Trivial. An electron volt is 1.602×10−19 joule [J].
We have learnt a lot since then.
Rob,
“December 3, 2018
Why is the atmosphere so much thicker, and denser than earths? Gravity on Venus is a little less than earths, correct?”
Mainly because Venus never cooled enough to have liquid water on the surface…
Without water to cool the surface, Venus’ crust remained hot and thin, allowing for much greater heat transfer from the interior to the atmosphere. Volcanic activity outgassed much more than on Earth, and its interior is much cooler.
In short, Venus is much more isothermal than the Earth and I expect much closer to cooling to its end state than Earth… When it runs out if water all kinds of interesting changes will occur…
Venus possibly cannot sustain life without any liquid. Do you think Venus has the ability to sustain life underneath the surface since there might be moisture trapped and some liquid might be formed?
I believe everything posted was actually very logical.
However, think on this, suppose you composed a catchier post title?
I mean, I don’t want to tell you how to run your
blog, but what if you added a title to possibly grab a person’s attention? I mean Can Space.com Teach Us Anything Useful about Climate?
« Roy Spencer, PhD is kinda vanilla.
You ought to glance at Yahoo’s home page and note how they create post headlines to grab
people to click. You might try adding a video or a picture
or two to grab readers excited about what you’ve written. In my opinion, it would make your posts a little livelier.
It’s wonderful that you are getting thoughts from this piece of writing
as well as from our dialogue made at this time.
https://comprarcialis5mg.org/
feafes elabora un decleg https://comprarcialis5mg.org/
I am so amaze by how this article was written. It is full of information and I have gain more knowledge because of this, thanks for sharing a wonderful article.
https://eppingelectrical.net.au/residential-electrician-carlingford/