## How to Debunk Climate Deniers and Climate Alarmists

In this video I explain that the accepted debate between the deniers and the alarmists can be debunked by pointing out that the entire paradigm within which the debate is framed is founded on something so ridiculous and easy to refute that it ends up debunking both the main-stream skeptics and the climate alarmists together at the same time!

This entry was posted in Fraud of the Greenhouse Effect and tagged . Bookmark the permalink.

### 240 Responses to How to Debunk Climate Deniers and Climate Alarmists

1. Marshall Rosenthal says:

Earth’s climate has always been changing. Once the climate was inimical to all life. Now, life can thrive, survive. If people left the earth alone, it will continue to be a viable place. The earth responds to complex climate-causing systems. There is no need for alarm.

2. John OSullivan says:
3. Pierre D. Bernier says:

Might as well plug it here for all deniers to point all alarmists to…

The Average Solar Radiation Flux on Earth For Dummies

The Sun’s rays hit the Earth on one hemisphere at a time. I’m not saying it, they are. Look at the picture ! Figure 2 !

And here is how the Clymists interpret that figure…

In the figure the Earth intercepts an amount of solar radiation equivalent to that falling on a disc with the same radius (R) as the Earth, facing the Sun: this comes to (1368 × πR^2 ) W, where πR^2 is the area of the disc (in m^2). However, the Earth is spherical , so the area presented to the incoming solar radiation by the rotating Earth (over a period of 24 hours or more) is 4πR^2; i.e. four times as great. Thus, the solar input per unit area averaged over the surface area of the whole Earth is a quarter of the solar constant; i.e. 1368 W/m^2/4 = 342 W/m^2.

Question ! In that diagram the poles of the disk receive just as much energy as the equator. Therefore the poles should be just as hot as the equator and there should be no seasons ! That’s not what we observe. The poles are cold, the equator is hot and we have seasons ! How do you explain that ? The model is wrong !!! But let’s keep on going !

A part of that solar energy is reflected back into space and never reaches the Earth’s surface. This is called the albedo and it can vary markedly around the world depending on the cloud cover and surface characteristics. The Earth’s average albedo is about 31%, the remainder 69% is absorbed by the atmosphere and the Earth’s surface. So we have 342 W/m^2 * 0,69 = 236 W/m^2 reaching the Earth.

At equilibrium, the Earth should re-emit the same 236 W/m^2 reaching it. The Stefan-Boltzman Law gives us the relationship between temperature and radiative power, a well-established law of physics. The appropriate calculations tell us that, for an Earth-like planet to emit radiation to space at a steady rate of 236 W/m^2, it should have an equilibrium average temperature of -19 °C.

This equilibrium temperature is known as the effective radiating temperature and, were it not for the atmosphere, this would also be the Earth’s global mean surface temperature.

OK Now… This is where all the that GHG stuff has to come in to justify the more temperate average temperature of 15 °C the Earth experiences. They have to invent a fairy tale mechanism to justify that 15 °C. The model is still wrong !!!

How about a more realistic model…

From figure 7, (http://junksciencearchive.com/Greenhouse/Earth-s_Climate_Engine.pdf) we can see that the Sun’s effective rays falling on Earth are weaker as we approach the poles because they fall on a greater surface. The flux at any given location on the hemisphere is a function of the cosine of the solar zenith angle, zero degrees pointing toward the sun, 90° along the periphery of the hemisphere. So, for any point on the hemisphere, from -π/2 to +π/2 in the East-West direction and from -π/2 to + π/2 in the North-South direction, the Sun’s flux density should be…

F(x,y) = 1368 * cos(x) * cos(y).

We have defined what the intensity of each unit beam reaching Earth will have, but not the number of them. Let us imagine the Earth as seen by the Sun as a disk. It is easy to see that a whole lot of unit widths of beam will fit along the equator of the disk but less and less will fit at each latitudes going towards the poles until we get to zero units of flux.

Since this loss of flux is also a function of cos(y), we have to correct the incident Flux density by that factor. So our Sun’s flux density at any point (x,y) on the sphere becomes…

F(x,y) = 1368 * cos(x) * cos2(y).

To find the total flux density on the hemisphere we double integrate that function from –π/2 to + π/2 for both integrals…

F = Intg Intg 1368 * cos(x) * cos2(y) dy dx for (x,y) = -π/2 to +π/2
F = 1368 * sin(x) * Intg cos2(y) dy for (x,y) = -π/2 to +π/2
F = 1368 * [ 1 + 1] * Intg cos2(y) dy for y = -π/2 to +π/2
F = 1368 * 2 * [ (cos(y) * sin(y) + y) / 2 ] for y = -π/2 to +π/2
F = 1368 * 2 * [ (0 + π/2 – 0 + π/2 ) / 2 ]
F = 1368 * 2 * π/2 = 1368 * π

This is the total flux density on the hemisphere. Since the hemisphere has a surface of π/2, the mean flux density on Earth is…

F = 1368 * π / 2π = 1368 / 2 = 684 W/m^2.

Mr. Joe Postma obtained the exact same result using another calculation method (https://principia-scientific.org/how-to-calculate-the-average-projection-factor-onto-a-hemisphere).

If we take into account that same albedo of 31%, 69% of that energy will reach the Earth, which gives us a heating potential of 472 W/m^2. That very same Stefan-Boltzman Law gives us a mean radiative temperature of 302 °K or 29 °C. Warm enough for comfort !!! No more need for that GHG invention here !

We can take it one step further. All around the world there is an East-West band, from latitude 47°S to 47 °N (94°), that is mostly water. Whatever the season, an area from -23,5° to +23,5° E-W by -23,5° to +23,5° N-S, relative to the Sun’s zenith, will always fit in that band. Using…

F =Intg Intg 1368 * cos(x) * cos2(y) dy dx for (x,y) = -23,5° to +23,5°

We come up with a total flux density of 846,4 W/m^2 over that area. Keeping the same x,y co-ordinates, the surface area of a spherical wedge is…

A = 2 * x * r^2

and for a partial wedge centered on the equator (leaving out r^2)…

A = 2 * x * sin(y)

For the designated area (-23,5° to +23,5° by -23,5° to +23,5°)…

A = 2 * (π * 47 / 180) * sin(23,5°) = 0,65419

Therefore, the mean Sun’s flux density on that area is …

F = 846,4 W/m^2 / 0,65419 = 1294 W/m^2.

Allowing for that very same albedo of 31%, 69% of that energy will reach the Earth, which gives us a heating potential of 893 W/m2. The very same Stefan-Boltzman Law gives us a mean radiative temperature of 354 °K or 81 °C. That’s very hot !!! Really no more need for that GHG effect here !

Now we can explain why the equator is hot, why the poles are cold, why we have seasons and why we don’t need back-radiation from the sky to keep the earth warm. In the day time the land and the air in the atmosphere trap some of that enormous amount of heat from the Sun and the water in the equatorial oceans vaporises and traps a lot more of that heat. It takes about 4 times more energy to heat up the same weight of fresh water then of air 1 °C. Taking into account the specific heat capacity and density of sea water and air, 1 cubic meter of sea water cooling 1°C can heat up 3277 cubic meters of air 1°C. That is an enormous amount of energy and we have not even included the latent heat of vaporisation. On the night side of the Earth, the heat in the water is released back to the air, by conduction and convection, as it cools and then released back to outer space. Since this process is not instantaneous but takes some time, the earth stays warm over night, ready for the next day.

A much better model !!!

Pierre D. Bernier,
MSc. Chemistry

4. Herb Roser says:

Hi Joe,
The error is more basic than the flat Earth.The error is the belief that the sun does not heat the nitrogen and oxygen in the atmosphere. Nonsense. Most of the energy coming from the sun does not reach the surface of the Earth but is absorbed in the atmosphere. The problem is that people accept the definition of temperature as bing the mean kinetic energy of the medium being measured. 100 C steam contains 540 calories/gram more energy than 100 C water. 100 C steam and water do not have the same kinetic energy despite what the thermometer says and that is how a steam engine does work. A thermometer does not give an accurate measurement of the kinetic energy of a gas! To get an accurate measurement of the kinetic energy (which is what is being transferred) in the atmosphere you must use the universal gas law which shows that the kinetic energy of the gases in the atmosphere increases with increases altitude. It is the atmosphere that is absorbing and radiating the energy the Earth receives from the sun.
Herb

5. CD Marshall says:

Great video Joseph.

6. Joseph Postma has certainly found a major flaw in AGW climate modeling.

He shows that the total watts of energy hitting the earth’s surface are twice what the AGW proponents say they are. The amount of energy from the Sun at the distance of the Earth is about 1370 watts per square meter. However, because the surface of the Earth is a sphere that energy is reduced by half since the area of a hemisphere is twice the area of a circle of the same diameter giving 685 watts per square meter on the top of the Earth’s atmosphere. The AGW climate models divide 1370 by four instead and without any justification. Even if one takes into account the whole surface of the Earth the amount must be 685 and not half that. Remember that the formula for the area of a sphere is 4 pi r^2, so that of a hemisphere must be 2 pi r2. This does not justify anyone in dividing by four.
Now, the significance of this is that the AGW modelers arrive at a temperature of the Earth from the Sun of negative 18 degrees centigrade. Since the Earth’s temperature is about 15 degrees the difference is about 33 degrees. They then assume this difference must be accounted for by AGW, greatly overestimating climate sensitivity to CO2.
This clearly invalidates their models.

7. Joseph Postma: One criticism of how you do popular science I have, is that the total area of a sphere is not equivalent to speaking of a flat Earth. That is just a cheap shot at your opponents. They are talking about the whole surface of a sphere while you are talking about one hemisphere. To speak of flat Earth just obfuscates the issue and distracts from the point while trying to capitalize on a silly flat Earth craze to sell your book.

8. You continue to miss the point, and I explained it in the OP vid.

Spreading incoming sunshine evenly over the entire surface area of the Earth at once is mathematically equivalent to flat Earth theory, and that is why the diagrams can then be drawn as flat Earth. It is not a cheap shot – IT IS THEIR OWN DIAGRAMS. It is reality. The emperor has no clothes. I am being serious, factual, literal, and I provide references.

9. I know, I know, it would have to be a flat Earth for all the surface to be hit at once. I still think it only puts you in the flat Earth search results.

10. The point is it is not a good way to publicize your ideas.

11. I don’t use “flat Earth” in video titles, or my book title, or my blog titles, and rarely if ever use that phrase in the description blurbs. I do use that phrase with my voice within videos occasionally, and occasionally in blog discussions, and always in a denigrating way.

Thus, I am specifically *NOT* publicizing “my ideas” as being about flat Earth theory, and I only refer to flat Earth theory within discussions in a negative and falsifying way.

12. I think bringing it up at all only hurt your case.

13. STOP MISSING THE POINT!

I bring it up because I factually have to, because their process is factually equivalent to flat Earth theory, and they FACTUALLY draw their diagrams as flat Earth’s because of that.

I’m dealing with facts, not emotions. Yes I know it is emotional for you and others to face the facts that climate science’s starting point is admitted to be flat Earth theory…but this is fact. I am not avoiding a fact because it emotionally affects you. The hurt you feel is your own emotion about a fact. Snap out of it.

14. You forget that I am a long time die hard absolute global warming skeptic since 1995. I am delighted to find that I don’t have to accept luke-warm positions. I am delighted to destroy AGW. You are the one being emotional because “flat Earth theory” means not believing the earth is a sphere. It does not mean dividing the solar input by four when it should be by two. To call it flat Earth theory is to confuse two issues.

15. Just as you misrepresented me just then, you are misrepresenting your opponents, it seems to me. I agree that AGW is totally false through and through. It’s just propaganda.

16. Stop being so obtuse. Your opinion is *NOTED*. Thank you.

I point out JUST HOW FRAUDULENT it is, by pointing out THE FACT that the mathematical starting point is equivalent to flat Earth theory, and that the starting point diagrams are even literally drawn as flat Earth.

It is not confusing two issues – it is clearly, logically, plainly, and simply, CLARIFYING the issue and making sense of the starting point.

17. Zoe Phin says:

2nd sun,

Not only is their earth flat, it’s as flat as flat can be. The flat earth is one-molecule thick, and floating on top of our flat Earth is an atmosphere that only consists of GHGs, N molecules thick, where N is the number of layers needed to make their crackpot math work to seemingly match reality.

18. I just don’t think it helps your case one bit. This just goes to show that unsolicited advice is never welcome. But I’m on the strict skeptic side and you won’t admit or apologize for misrepresenting me. You’re taking a friend as a hostile party. Is that rational? AGW proponents don’t believe the Earth is flat. Sure AGW is a fraud because it was refuted early in the twentieth century and then blithely taken up again later without overturning the criticisms of it.

19. Advice is fine – You’re being belligerent.

You are free to clarify what you think has been misrepresented about you, and I am free to continually re-assess. Your belligerence is hostile, not friendly.

AGW proponents may not believe that Earth is flat, yet, that is what their math and their starting point has done. It is rational to understand this. This is the origin of the error, and the basis of the fraud, because it is from that point from which all errors then flow forth. This is all the easiest, most rational thing and sequence to understand, and it should be understood and appreciated. Sure, it might be difficult for some to face the scale of the error, yet, it is a rational logical fact.

20. I’m sorry that you are so stressed out about this that you imagine I’m hostile to you.

21. I am sorry that you cannot understand that your belligerence can cause stress.

22. Thomas Homer says:

I’m curious about the albedo value used. In Pierre’s example above he’s showing how the angle of incidence changes for sunlight striking the spherical Earth. Why don’t we also employ a gradient for the albedo factor? Using a single value like .3 or .31 seems too simplistic to me. When the sun is directly overhead, it has the strongest impact and (I’m speculating) loses the least amount of sunshine to reflection, while when the sun is either rising or setting it has the least impact but a greater amount of reflection.

If that’s the case, then each term in the cosine calculations should be qualified with a gradient albedo. This means that the strongest sunshine areas would be discounted less while the weakest sunshine is discounted more, resulting in a greater overall average.

23. Julian Fell says:

To a degree the debate here is also following the Chomsky observation. It is aimed at the luke-warmers in a luke-warmer vs no effector context. I am a no-effector but as a local government politician (7 yrs) I had to deal with Joe Public, other politicians and brain washed students from local colleges. Some recent Electees are fanatical “progressives”. All may be characterized by an abysmal lack of knowledge of physics. Try debating with these fellows, especially if they dont want to debate. When a civil discussion occurs, I can point by point explain the consequences of Stefan-Boltzmann, how their anticipated catastrophies cannot happen but after a point they tell me they dont want to debate it any more and flee the discussion. Or they refer you to some IPCC junk diagram. These people want to believe CAGW. They cling to it. They feel threatened if you take away their belief. It is truly a religion in effect.
One of the aids of an ideology is that it can be based on the big lie mechanism. Throw out a simplistic shocking idea… sea level will rise 6 feet in the next 75 years…easy to grasp, to fear and to repeat. Now prove otherwise with numbers like water expands at nine parts per ten thousand per degree C/K and ice melt numbers and their eyes glaze over. They are lost in the complexity of reality. The 15 second sound bite wins.

As Mark Twain noted: It is much easier to fool a man than it is to get him to realize that he has been fooled.

24. Lies are infinite, easy, and don’t require effort, thought, or understanding…

Truth is singular, but often difficult, and required effort, thought, and understanding.

Yes it is the strangest thing ever – people WANT it to be true, and you can demonstrate to them the most basic and obvious facts, and it makes no difference.

25. I want everyone commenting here to know that I agree with most of Postma’s ideas critical of AGW, and I have recently done a very positive Amazon review of his book. I just don’t think the use of the phrase “flat earth theory” s helping.

26. Thank you for the concern trolling.

27. Pierre D. Bernier says:

@Thomas Homer

It is simplistic indeed. I’m just using their own figures to prove them wrong.

http://www.open.edu/openlearn/nature-environment/climate-change/content-section-1.2.1

In the equatorial waters the albedo is indeed about 30% and near the pole about 80% because of angle and snow. It also varies with what’s underneath, water or land or whatever. How can you put that in a simple function ?

28. Pierre D. Bernier says:

@Thomas Homer

Plus what we are really interested in is the Top Of Atmosphere (TOA) Flux where the albedo is not in function yet.

29. geran says:

Joseph, thanks for the insight into the early history of the “Slayers”. I was following the issues over at WUWT, in those days. I couldn’t understand the treatment you gents received there. Later, I figured it out.

I never visit that blog anymore.

30. Zoe Phin says:

Thomas,
Albedo is a simple coefficient which can be placed outside the integral.

ab + ac = a(b+c)

Earth’s albedo ranges from 0.25 to 0.40 throughout the year, with 0.3 being the average.

31. Zoe Phin says:

Julian,
Tell those idiots that oil companies have dug huge caverns into which rising oceans can be dumped. By prevent oil extraction they guarantee serious flooding in the future. But they claim they to want to mitigate disaster …

32. Thomas Homer says:

Thanks Pierre.

“It is simplistic indeed. I’m just using their own figures to prove them wrong.”, and I applaud you for that.

I’m glad the whole [ divide the sun’s energy by 4 then take 70% (1 minus albedo) of that to get 17.5% ] and then claim the Earth is 33C warmer than expected so Greenhouse Gases are introduced to make up the difference, quandary is being disputed here.

Your other comment: “Plus what we are really interested in is the Top Of Atmosphere (TOA) Flux where the albedo is not in function yet.” hits on another issue that half the Earth is intercepting the sun’s energy, even if it’s not reaching the surface. The atmosphere itself is impacted by sunlight, and the difference is like night and day.

33. One criticism of how you do popular science I have, is that the total area of a sphere is not equivalent to speaking of a flat Earth.

Nowhere does JP claim the equivalency that you propose. He does NOT equate total area of a sphere to speaking of a flat Earth. What he does is show the erroneous mathematical CONTEXT within which the total area of a sphere is invoked by greenhouse-theory proponents. Their MATHEMATICAL TREATMENT of total spherical area DEFINES a flat Earth. The sphere of Earth, within their mathematical treatment, COLLAPSES into a flat Earth. That’s what their math means. Their math no longer means a spherical Earth. THEY are the ones referencing a flat Earth, in the way that they do math, and JP simply points this out.

Just because I point out that a person is Christian does not mean that I am a Christian or that I endorse Christianity. I simply point out how other people represent themselves.

It IS shocking that intelligent people today can do simple math in such a way that leads to the inevitable model of a flat Earth. To point this out is not endorsing flat-Earth theory. Rather, it points out how people who would deny believing in a flat Earth, in fact, do their math in such a way as to contradict their own fundamental convictions !

That is just a cheap shot at your opponents.

No. That is an explanation of how cheap their basic mathematical understanding is. The cheapness is in what THEY do with the total spherical area. JP simply points this out.

They are talking about the whole surface of a sphere while you are talking about one hemisphere.

HOW are they talking about the whole surface of a sphere? –they are talking about the whole surface of a sphere incorrectly. They are talking about the whole surface of a sphere in such a way that their talk (their mathematics) collapses it into a flat Earth.

To speak of flat Earth just obfuscates the issue and distracts from the point while trying to capitalize on a silly flat Earth craze to sell your book.

Obfuscates? Not even close. More like blatantly reveals. Distracts? Again, not even close. More like the opposite — gets to the very heart of the matter. Craze? — Yes, a craze certainly exists, but its proper focus is in the greenhouse-proponent camp, where a crazy misuse of mathematics falsely and deceptively fabricates an incorrect understanding that should not require a book to explain. The fact that a book needed to be written about it shows just how crazy this whole thing is.

34. Wonderful clarity Robert. Thanks for that refreshing read!

35. Suppose I use the words, “I do not believe in marriage.” Then I go out, find a suitable person, enlist a minister to join us in matrimony, and live out my life in a domestic partnership. Not once do I use the word, “marriage”, and yet all my actions can lead to no other conclusion than the conclusion that I endorse behavior that leads to marriage.

Some scientists will laugh at the idea of a flat Earth, yet they do mathematics that can lead to no other conclusion than the conclusion that they endorse math that describes a flat Earth.

36. Rosco says:

My 2 cent’s worth.

Roy Spencer denies he uses any model which is equivalent to spreading the sunshine over the day and night.

But he actually does by explicitly stating his support for the ONLY model of the back radiative greenhouse effect that exists !

Clearly somewhere deep down he feels some disquiet and embarrassment ! (As I would without spell checking.)

I have zero problem with their meaningless averages if that’s what you want to do but their failings commence when they claim the average spherical output from an average disk input is the real insolation.

The hottest surface temperature ever measured by NASA’s Landsat weather satellite is 70.7°C or 343.85 Kelvin. If the desert surface in Iran is emitting ~793 W/m2 at 70.7°C in a locality where water vapour is almost entirely absent then how do they explain this ?

This was most decidedly NOT the result of an average input of the order of 239.7 W/m2 or the even more insane figure of 161W/m2 from Trenberth et al 2009 plus the fictitious 333 W/m2 back radiation which has somehow achieved double the heating capacity of the Sun.

That total is ~573 – 220 W/m2 less than the surface is emitting.

And with all that heat trapping from GHG’s just how do satellites actually measure surface temperature ?

It seems both sides either really believe the nonsense that back radiation from a cooler atmosphere can heat the Earth’s surfaces beyond the actual emission temperature of the radiation’s source or they’re some of the least reliable and truthful people on Earth.

The radiation emitted by any source cannot cause a temperature greater than the emission temperature in any object irradiated by it and I don’t care what sort of linguistic gymnastics anyone chooses to indulge in this basic fact is irrefutable.

The surface temperature is not a result of 239.7 from the Sun and 239.7 back radiation as the University of Washington quotes.

“How much energy is received by the earth? Solar radiation incident in the earth’s disk (1368 Watts per square meter) –comparable to energy incident a flat, horizontal surface when the sun is directly overhead on a clear day.

We need to multiply the incoming solar energy by the factor 1/4–the ratio of the area of the earth’s disk (pi r squared) to the earth’s surface area (4 pi r squared)– You can think of this as spreading
out the incident solar radiation uniformly over the earth’s surface (the night side of the earth as well as the day side) 1368 / 4 = 342 watts per square meter.”

37. They can’t understand simple things. What they DO with the math is creating what he denies is reality!

The average of sunshine is NOT sunshine falling over the entire surface at once. This is NOT meaningful!

The average phone number is not meaningful.

The average number of testicles is not meaningful.

Average sunshine is NOT what creates the climate.

They can’t understand simple things.

38. Rosco says:

Another 2 cents worth – the Stefan-Boltzmann equation is NOT the appropriate equation to be performing the ludicrous algebra gymnastics all of these people perform to support their ludicrous claims.

All the Stefan-Boltzmann equation EVER empirically described was the ratio of the total emission from a cavity oven at a fixed temperature. The actual value of the “constant” (which several physicists claim is NOT universal) was only ever a guesstimate until Planck’s equation was derived and accepted.

All of the algebraic manipulations have zero empirical evidence to support them.

I thought the radiative emissions described by the Stefan-Boltzmann equation were as a result of the emitting temperature BUT climate science somehow interprets that an object has to match the input radiation even though that radiation cannot even replace the radiation the hotter object is already emitting.

The relationship between Planck’s equation and the Stefan-Boltzmann equation is that the total power emitted into a hemisphere from an object is equal to Pi times the integral from 0 to infinity of Planck’s equation – the area under the curve. Similarly the relationship between Planck’s and Wien’s equation is that Wien’s relationship is arrived by taking the derivative of Planck’s equation and using the fact that the tangent to the maximum value of any curve is horizontal.

I think it is all first class gobbledygook !

39. Rosco says:

The real danger is that the gullibles, totally convinced of the climate emergency and impending extinction, will support polluting the atmosphere with sulphates to “control” the non polluting CO2 plant food heat trapping. Reflect the Sunlight back into space.

Our public broadcaster – the ABC – has been running this propaganda for some time now :-

https://www.abc.net.au/news/science/2019-07-24/climate-hacking-to-avoid-a-global-warming-apocalypse/11300460

“Stratospheric aerosol injection (SAI) is the most researched of a suite of technologies aimed at cooling the Earth known as “solar geoengineering”.”

“The idea is terrifying, but so is climate change.”

This is all a smokescreen to cover the real motivation – a resurrection of 19th Century Malthusian genocide in the name of global population control. Apparently we are little more than akin to insectivorous parasites awaiting extermination.

After years of hiding from public scrutiny the elitist “experts” have decided the carrying capacity of the Earth is less than 1.9 billion people.

Now to put that into effect will require some impressive genocidal maniacs in charge !

40. CD Marshall says:

“Stratospheric aerosol injection”
They intend on scorching the sky like like in the Matrix. Great, AI rising will be the next step.
Ironic they intend on doing this before a projected maunder minimum. Now they can say they cooled the Earth and the climate change lie lives on. So many people are paid off none will stop this madness until these climate clowns really do destroy the climate.

41. Pierre D. Bernier says:

Addendum (For those math lovers) :

There is a 3rd way to calculate the Sun’s mean flux. Given the above flux equation, if we take into consideration only the meridian on which the Sun’s zenith lies on, our equation becomes…

F(y) = 1368 * cos^2(y)

The total flux density on the meridian is the integral of that function from –π/2 to + π/2…

F = Intg 1368 * cos^2(y) dy for y = -π/2 to +π/2
F = 1368 * [ (cos(y) * sin(y) + y) / 2 ] for y = -π/2 to +π/2
F = 1368 * [ (0 + π/2 – 0 + π/2 ) / 2 ]
F = 1368 * π/2

This is the total flux density on the meridian. Since the meridian has an arc length of π, the mean flux density on the meridian is…

F = 1368 * π/2 / π = 1368 / 2

Let it run for 12 hours, from dawn to dusk, the meridian will receive more flux but it will be a function of cos(x) for x = -π/2 to +π/2 which we already know the answer to be 2. So we have…

F = 1368 / 2 * 2 / 2 = 684 W/m^2.

The same as above.

As for the area from -23,5° to +23,5° E-W by -23,5° to +23,5° N-S, relative to the Sun’s zenith…

F = Intg 1368 * cos^2(y) dy for y = -23,5° to +23,5°

We get 1368 * 0,77583 = 1061

The partial meridian arc’s length is 47 * π / 180, so the mean flux on the partial meridian is…

1061 * 180 / 47 / π = 1293.

And for the whole surface we let the time roll and get…

1293 * (47 * π / 180) / (47 * π / 180) = 1293 W/m^2.

Same as above !

42. If the Earth were flat the total solar input would be 1370 watts per square meter and no one could divide by four, or even by two.

43. Therefore, AGW is not flat Earth theory. The question remains unanswered: How could they divide by four?

44. They divide by four because they think it’s an average, but what is actually does is spread incoming sunshine over the entire surface at once, and this creates a physics identical with flat earth theory which is why the diagrams are then even drawn as flat earths.

45. How could they think it’s an average? Average of what? Four sides of the Earth? A physics identical with flat Earth theory would have 1370 Watts per square meter as I already made clear.

46. Who knows why they think it’s an average – yet they do and there’s a quote and you’ve also seen them say that it is the starting point of climate science which none of them reject. A flat surface at earth’s distance from the sun would have 1370 – they are thus putting the earth at two-times its own distance. Paradox generates endless paradox which is also their GHE…when you start with a paradox. ALL BUNK.

47. If the Earth were flat the total solar input would be 1370 watts per square meter and no one could divide by four, or even by two.

Yes, and so what they DO do is even worse — they collapse a spherical Earth into a shape that is not only flat but also multiple times the correct distance from the sun. This is also what that math means — not only flattening the Earth, but also moving it multiple times its correct distance from the sun

It’s even more emotionally shocking than you are willing to recognize at the moment. Try to look at it for what it is. Don’t give them an inch. It’s just plain ridiculous what they do.

48. Exactly.

49. I certainly never thought they had any justification in dividing by four. AGW is thoroughly pseudo-scientific. I made that comment recently. What I’m particularly interested in is the question of whether solar input requires an amplifier considering that the real input is 1370/2= 685. You know, most solar guys like Willie Soon think they need amplifiers.

50. The entire system is passive and has no amplification abilities. There is no such thing as a heat amplifier. The only active component is sunlight. 1370/2 is also NOT the real input and is NOT the number that drives the physics. The physics is driven locally at every instant with the sunlight that occurs at every position in real time.

51. Pierre D. Bernier says:

Hey guys, look above…

The average flux a meridian receives from dawn to dusk is 1368 / 2. As one meridian emerges from darkness, one dissapears in darkness. So all meridians in the course of 24 hours receive 1368 / 2.

THE DIVIDER IS 2 NOT 4 !!!

TWOOOOOOOOOOOOOOO !!! NOT FOUAAAAAAAAAAAAAAAAAR !!!

52. There is no such thing as a heat amplifier.

(^_^) … Lord (Lordy, Lordy!) Monkton would fastidiously and voluminously disagree — emission temperature somehow feeds back on itself, and somehow gets figured into an elaborate electrical-circuit analogy with painstakingly tedious sophistry-math to counter such a claim.

53. Joseph E Postma says:

With electrons & circuits you’re dealing with fermions, while heat flow is with bosons. These have 100% opposite behaviour. We already have the equations for heat flow. Monkton is a primary gatekeepers. He used to send me emails saying that he was going to sue me for saying that there is no GHE…hahaha! Ooooh you’re so scary “Lord of Full of Shit” Monkton. Where is that asshat anyway these days…someone, we all should, go troll him. He deserves all possible derision and ridicule.

54. That is good to hear as Monckton’s work is not only superficial but entirely beside the point. I am quite comfortable agreeing that amplifiers don’t exist. Then is the solar variation significant without amplifiers? How much difference does it make? Say in the eleven year cycle?

55. “We define emission temperature R0 as the 255 K global mean surface temperature that would obtain on Earth at today’s solar irradiance and albedo but before any greenhouse gases have entered the atmosphere and before any feedback begins to operate; B0 as the feedback response in Kelvin to R0; E0, the sum of R0 and B0, as the equilibrium temperature that would prevail after feedback has responded to emission temperature …”

So, in just these few words alone, I seem to see him claiming that emission temperature produces its own feedback temperature on itself that can be added to itself, or am I missing something.

Lordly, Lordy !

56. Joseph E Postma says:

Ah he’s protected by the mods there. Want his email?

He accepts the GHE and promotes it and has come up with cockamamie ways to protect it. He’s a SHILL, playing for the other side. As I said…why not take a look at the basis of alarmism, which is its greenhouse effect, if you’re actually a skeptic? After-all, what you find there is a flat line, and cold sunshine, and no day & night, and a climate creating itself…all very obvious and very SAFE things to criticize. So then, why not? Because they’re frauds, that’s why not.

“E0, the sum of R0 and B0” [where those terms are temperatures]

WTF!!!??? He’s adding temperatures together! He’s adding the temperature of two cold things to produce a hot thing!!! hahahahah Unreal! Unbelievable! A straight-up COOKOO!!

Note that his approach ENTIRELY leaves out the natural formation of the lapse rate, which by itself makes the bottom of the atmosphere warmer than what it’s average should be. Yes, by leaving that out, as they do with their flat-Earth diagrams, they substitute their fake GHE instead.

57. Pierre D. Bernier says:

Are you sure he’s not Italian ? He has so much spagetti in there !

58. Zoe Phin says:

This tells you everything you need to know:

First image shows very little insolation, and the minimum night time upward flux (a proxy for temperature) is ~450 W/m^2

The second image shows a lot more insolation, and the minimum night time upward flux (a proxy for temperature) is ~450 W/m^2

So let’s divide global insolation by 4, that will be useful! And when it doesn’t … GHGs make up the difference!

59. … adding temperatures, oh yes, … OMG ! — I MUST go over there and comment on that.

Monkton does have an entertaining writing style. Alas, it is wasted on garbage reasoning.

60. Zoe Phin says:

Even if their flat earth assumption was correct,

The area of those 3 co2 bands below 4.8 microns is ~equal in area to the one big 15 micron band. But considering e=hl, their energy content is roughly 5 times greater. Therefore CO2 scatters more energy away from Earth than back down to earth. Therefore CO2 must be a radiative coolant.

They are wrong in so many ways.

61. Zoe Phin says:

Joseph,
I’m curious, do you also believe objects of different mass fall to earth at the same rate like these guys in the comments section:

https://principia-scientific.org/mathematical-analysis-debunks-33-degrees-greenhouse-gas-effect/

Or do you recognize there’s a tiny difference?

62. Joseph Postma: “Who knows why they think it’s an average ” So the reason they think it’s an average is that it’s the average for the incoming solar energy in any one second over the whole earth. That seems like a perfectly justifiable basis for a model. How can that lead to underestimating solar input by a half? We have to have an accurate idea of their position before we can overturn it.

63. Pierre D. Bernier says:

FWIW … About a year or 2 or 4 ago I saw something on TV where they had a big space with a feather and a something big and heavy hanging up in a vacuum. They released both the feather and big object and both arrived on the floor at the same time. They were all hysterical about having proven Galileo right.

64. Joseph E Postma says:

“the reason they think it’s an average is that it’s the average for the incoming solar energy in any one second over the whole earth. That seems like a perfectly justifiable basis for a model.”

Wrong. How is something that doesn’t occur or exist in reality a justifiable basis for a model of reality? By definition it is not. That idea of an average, of spreading solar energy over the whole Earth, is not something that physically occurs or exists, hence is not a model of reality and hence has no justifiable basis. It underestimates solar input because it dilutes solar power over surface area it doesn’t exist upon. We have been over this with you many, many times. We do have an accurate idea of their position – we have quoted them directly, we have looked at their math, and we have identified it as doing something which has nothing to do with reality, as you yourself explain but remain unable to understand.

Sunlight does not spread over the entire surface at once. It is the local, instantaneous heat flow which determines what physics occurs as a result of that heat flow. This false “average” solar input cannot create the climate, such as cumulonimbus clouds, but real-time solar power can. Hence using the false “average” as a starting point will never lead one to a position where they can understand the climate.

Now…let us wait to see how you repeat your position and your question once again, as if it hasn’t ever been addressed in original OP’s or subsequent questions, and just start over right from the beginning again as if nothing has registered in your mind at all.

65. So there’s no difference between the solar input in your models and in the consensus models because you are saying it’s twice as much for the sunny side and they’re saying its half as much for the whole. You’re both saying the same thing except you are claiming a difference that doesn’t exist.

66. That’s some primo sophistry. Yes there is a difference. The difference is that they spread sunshine over the entire surface at once and hence mathematically weaken sunshine to a point where it can’t create the climate, whereas I advocate for sunshine driving physics in real time locally in which case sunshine can create the climate.

See! Told you it would be starting right from the beginning again!

67. The average temperature fallacies that have been legitimately drawn out are about averaging from empirical data. You don’t have much patience with newbies, so I guess you’re running an exclusive club.

68. Joseph E Postma says:

You see this phenomenon you guys? Isn’t this the strangest thing…to see minds work this way? Recall how I’ve explained that when I meet real humans, I can tell because their questions become more subtle, more precise, their comprehension and penetration of a subject increases and their commentary becomes more novel, more unique, etc. But with these other things, zero progress can be made at all, and it is as if they have zero comprehension of even the simplest things. I call them goblins. But see this – maybe an explanation!

False human that actually have no consciousness (just watch it…just listen):

69. I am always starting from the beginning because you haven’t got it right yet. ” hence mathematically weaken sunshine.” This is just you claiming again that your model has more solar input when it doesn’t.

70. Joseph E Postma says:

Your comments make zero sense, and queue off of my comments in no conscious sensible way at all. As if you have no actual real mind at all.

Robert – your model human figure needs to be updated with the reality of having no actual conscious mind at all!

71. Just putting people down wont accomplish anything for you. You make a claim that you cannot substantiate. Your model has the exact same solar input as the AGWers.

72. Joseph E Postma says:

Aaaaannd you’re being re-blocked…because it is a waste of time to talk to a machine. Nice working with you…I’ve gotten everything needed out of this relationship. You can go shut-down now. The stalemate of simply repeating things as if things aren’t ever repeated can cease now – the solution is what the solution is and that’s next 🙂 Truth is a force of nature.

73. I mean hahahahaha…yah…because it’s not substantiation when you directly quote what they say is the starting point of their “science”.

INSANE

74. Pierre D. Bernier says:

@Zoe,

Found the video

abc11.com/science/cool-video-bowling-ball-and-feather-dropped-at-same-time-in-giant-vacuum-chamber/382792/

75. Theory,

So there’s no difference between the solar input in your models and in the consensus models because you are saying it’s twice as much for the sunny side and they’re saying its half as much for the whole.

It’s twice as much, because it physically exists in both space and in time on the sunny side, whereas, for half of the whole it simultaneously does NOT exist — it neither exists in space nor in time for the whole. How can you not see this?

The Earth is constantly rotating in and out of its half-side illumination, and so it is constantly rotating in and out of its half-side darkness. The sun shines on the sunlit side, while the sun does NOT shine on the dark side. On the dark side, there is NO sunlight operating in the space or in the time of that dark side. The effects of the sun having shined on that side, BEFORE it entered darkness, are there, but solar input that caused those risidual effects IS NOT THERE.

There is no average operating EVER in the space or in the time of the dark side that has a physical effect in the time of incidence that an “average flux” demands. A flux exists in a real time in a real place. An average flux NEVER exists anywhere on Earth, let alone on the dark side.

Open your eyes. Open your mind. I know you can do it.

You’re both saying the same thing except you are claiming a difference that doesn’t exist.

Saying that diminished sunshine can be assigned to a space and time where no sunshine exists is NOT the same thing as saying that full sunshine can be assigned to the ONLY space and time where it most certainly does exist in its real-time, real-space amount.

Good God, why are you so resistant to calling these tactics out what they are?

Ah heck, you blocked him, just as I was beginning to try to get through.

76. Joseph E Postma says:

Robert, there is no getting through. We’ve already repeated the same things a dozen times between you and I. We are repeating what has already been said. *Functioning humans do not act this way.* No actual mind, actually operates this way. I even predicted exactly what the response would be, and the response was exactly what I said…because the other thing is an algorithm, a predictable algorithm, hence, NOT an actual information-integrating conscious mind. It is an information-processing algorithm, not an information-integrating mind. Big difference. And we all here have witnessed this exact same algorithmic phenomenon repeated amongst the various people we’ve dealt with over the years…take Eli Rabbet for example. They are all the same. I’ve encountered this behaviour probably *hundreds* of times now. It’s not how a human actually works…it’s not how a real mind actually works. It can be people following an algorithm in order to decieve…sure. But all of the examples I have would require a conspiracy among so many people with such perfectly identical training in the deception algorithm that it becomes unlikely, and what’s more likely is what was in that vid.

77. Pierre D. Bernier says:

Made a little mistake above in Addendum when I say there is a 3rd way to calculate the Sun’s mean flux if we take into consideration only the meridian Up until F = 1368 * π/2 / π = 1368 / 2 is all very fine.
Since we divide by the arc lenght π to get the average on the main meridian and the arc lenght shortens as a function of cos(x) when we move away from that main meridian, we have…

F= Intg cos(x) / π / cos(x) dx = Intg 1/π dx for x=-π/2 to π/2
F= [ π/2 + π/2 ] / π = 1

It’s always 1. So the mean on the meridian is the mean on the hemisphere.
For smaller portions, it’s whatever divided by the same whatever.

78. Zoe Phin says:

Pierre,
That video is wrong. The bowling ball falls faster. Although the difference is negligibly teeny tiny, it is not ZERO.

Yours eyes can’t spot the difference, but it’s there.

79. Flux is a RATE of flow — joules per second, per meter squared.

Note the “second” — that’s a time during which an actual event happens.

There are NO seconds on the dark side of Earth where a flow EVER happens.

Yes, a flow in time DID happen there, when the sun was actually shining in real seconds of the light’s existence in this time, and yes, certain effects of the sun during the seconds described by this flow occurred and continue to have residual effects, when dark falls upon this location.

But flux — energy flow in seconds on a given area in this darkness is NEVER happening. There is NEVER any flux or flow or seconds with associated joules and associated surface area in existence in the dark. An average flux, then, is a complete absurdity, since there are no seconds in darkness where any energy from the sun is falling on any dark surface area.

There is no reduction of real flux to a partial time where it never exists in its specific time-dependent quantity. Flux exists for real time in real locations specific to real surface areas for which the very definition of flux depends. There is no breaking this apart, because to break it apart is to destroy the very definition of what flux is.

80. A piece of art by Mr. Crickets over at Deviant Art (with my added speech bubble) captures the character:

81. CD Marshall says:

The average temperature on the Day side of the Earth’s surface is 15C (correct?) what is the average on the Night side?

When they give an average temperature of the Moon what use does make on anything? In reality the moon’s surface temperature is nowhere near the average. What possible use is an average-to-reality? Astronauts don’t have suits made for an average lunar temperature they have them made to withstand the max cold and heat. I just don’t get it.

82. John says:

This is a comment aimed at Joe, but I appreciate comments from anyone on the answer.
I am fully on-bord with Joe and other Slayer arguments about the reality of how the sun heats the earth – seems total common sense to me. I also laugh at the “back radiation” arguments preached by climate scientisits as this goes against basic physics which I learnt in school when I was 13 years old. My question relates to the often quoted fact that satellite measurements of the average earth emitting temperature to space is -18 Celsius. This corresponds nicely with the flat earth arguments of averaging of solar flux over the entire Earth at any given instant. My question is: How is the average emission temperperature of the earth measured by satellite (in basic terms) and is it indeed -18 Celsius (or is this arrived at by a similar “flat earth” argument)? Is this conmtradictory to Joe’s and the the other Slayer arguments about how solar flux actually falls on the earth?

83. Pierre D. Bernier says:

@Zoe

You are perfectly right. Since a bowling ball and a feather are way lighter then the Earth. you can not see their effect on the Earth. But if you look at the solar system, the Sun is not at the center. It’s the baricenter that is. The Sun wiggles around that center because of the planets moving around the Sun.

Nice day

84. For Theoryofthesecondsun:
Here’s how I’m seeing it:
https://www.dropbox.com/s/qzgdal3fgpaj5oy/EnergyFluxSphere.png
There are two rates going on — (1) a rate of energy flow, which I redundantly call “rate of energy flux”, and (2) a rate of rotation. The rate of rotation exists on the whole sphere all the time. The rate of energy flow, however, exists ONLY on half the sphere, meaning that there NEVER is a flux defined for the side of the Earth where no flux mathematically, physically ever exists.
There is NEVER a rate of energy flow from the sun defined for the hemisphere where no sun shines actively in that moment of observing what a flux might be there. There is flux in the past of a particular point on the dark side. There is flux in the future of a particular point on the dark side. But in the moment, when we speak of flux, where, in that moment, no flux, in fact, exists in that moment of our speaking of it, FLUX DOES NOT EXIST in this darkness. Energy exists as a RESULT of flux, but flux itself that caused the energy dynamics there DOES NOT EXIST and cannot be spoken of as though it exists there as some average all the time.
Flux exists for the time during which it is assessed. If no time exists where, in this time, no flux occurs, then flux does NOT EXIST in this time, because those seconds that define the jouls-per-second are not traversed by any energy flow — the flow happened either in the past or will happen in the future (most likely). Again, in darkness where no energy flow from the sun is happening, energy flow from the sun in that darkness HAS NO MEANING.
Average flux, thus, is an absurdity. It has no real meaning. It has no real application. It is a fiction used in a fictitious mathematical exercise that is meaningless, because the basis of the mathematics is meaningless. The terms of the mathematics are meaningless, falsely, incorrectly fabricated by minds that do not understand the basics of the definitions of the terms that they attempt to use.

85. Test.

86. CD Marshall says:

Someone made this vague claim that NASA said, “satellite measurements show a reduction of earth IR radiation at the CO2 absorption band and an increased absorption there. The only reason for this can be increased absorption CO2 which is the definition of the greenhouse effect. claim that NASA says, ”

Other than the phrase sounding like a contradiction of itself it really makes no sense. Can anyone tell me what he’s trying to say? CO2 absorption band is he claiming the 13-15-18- Micron WL (whatever they are)? Which is also mostly crossed by with water?

The Earth hemisphere in darkness gradually moves into light

But while in darkness, solar flux does NOT exist there in the dark, during those many seconds in dark for which there are NO joules flowing into a dark-hemisphere surface area. Remember, flux is joules per second per meter squared. If the meters-squared are in dark, then there are NO seconds during which joules flow into this dark meters-squared for those seconds. On that area, in those seconds, NO SOLAR FLUX EXISTS. There is no definition for “solar flux” there, because “solar flux” is a time-dependent, area specific concept, for which the relationship of the time element and the space element MUST be connected, in order for the definition to apply.

Yes, the surface area in darkness experiences flux, WHEN this surface area moves into the energy-flow path of sunlight. But when this surface area is NOT in the energy-flow path of sunlight, there is no flow from the sun onto it, and so NO SOLAR FLUX EXISTS in this darkness.

The idea of “average solar flux” ignores this defining requirement of the concept of solar flux, by assigning an energy flow RATE from the sun, DURING times and for areas where NO energy flow from the sun happens during the associated times. This incorrect idea is then used to propel a chain of erroneous calculations, where the resulting numbers further reflect the absurdity of the fundamental error.

88. Zoe Phin says:

Pierre,
You got it. Pretty subtle and neat, right?
But look at the outrage I got at Principia for my blasphemy.

89. Pierre D. Bernier says:

@cdm
“satellite measurements show a reduction of earth IR radiation at the CO2 absorption band and an increased absorption there.”
Makes no sense to me at all If Earth emits less IR then CO2 has less to absorb. How can it absorb more ? Beer-Lambert law has changed since my university years ? Unless it absorbs it from the Sun. ?!?!?!?!?

90. Pierre D. Bernier says:

On July 26 I posted an Addendum (For those math lovers) . Looks like there are not many math lovers around since nobody corrected me on a small mistake nor asked for clarification. So, for the sake of completeness…

Given the above flux equation, if we take into consideration only the meridian on which the Sun’s zenith lies on, our equation becomes…

F(y) = 1368 * cos^2(y)

The total flux density on the meridian is the integral of that function from –π/2 to + π2…

F = Intg 1368 * cos^2(y) dy for y = -π/2 to +π/2
F = 1368 * [ (cos(y) * sin(y) + y) / 2 ] for y = -π/2 to +π/2
F = 1368 * [ (0 + π/2 – 0 + π/2 ) / 2 ]
F = 1368 * π/2

This is the total flux density on the meridian. Since the meridian has an arc length of π, the mean flux density on the meridian is…

F = 1368 * π/2 / π = 1368 / 2

The Sun’s zenith meridian’s flux density average is the same as the hemisphere’s average ? How can that be ?

Consider that as you move East or West, the total solar flux on the meridian diminishes as a function of cos(x). The length of the arc on which you distribute this flux also diminishes as a function of cos(x). So, cos(x) / cos (x) = 1. There is no integral to be had so all meridians have the same average flux density.

Proof. If we sum up all meridians and divide by the x arc length π …

F = 1368 / 2 * Intg 1 / π dx for x = -π/2 to +π/2
F = 1368 / 2 / π * [ x ] for x = -π/2 to +π/2
F = 1368 / 2 / π * [ π/2 + π/2 ]

F = 1368 / 2 * 1 = 684 W/m^2.

The same as above.

As for the area from -23,5° to +23,5° E-W by -23,5° to +23,5° N-S, relative to the Sun’s zenith…

F = Intg 1368 * cos^2(y) dy for y = -23,5° to +23,5°

We get 1368 * 0,77583 = 1061

The partial meridian arc’s length is 47 * π / 180, so the mean flux on the partial meridian is…

F = 1061 * 180 / 47 / π = 1293.

Since the arc’s length is 47 * π / 180 and the same reasoning applies…

F = 1293 * (47 * π / 180 ) / (47 * π / 180 ) = 1293 * 1 = 1293 W/m^2.

Same as above !

91. Hi John. Good question. The -18C is the equivalent temperature of the output flux from the Earth. That is all fine. It’s simply conservation of energy given the input flux. The input flux and output flux aren’t the same value because they occur over different areas, but the total energy is of course the same. What climate science does is use the output flux, equivalent to -18C, as the solar input…and that is of course nonsensical. In that realistic diagram I show sometimes you can see how the input and output fluxes are different, such that the input can create the climate but the output couldn’t, while conserving total energy.

92. Pierre D. Bernier says:

@Zoe,

//en.wikipedia.org/wiki/Barycenter

Barycenter is the center of mass of two or more bodies that orbit one another and is the point about which the bodies orbit. So the Sun orbits something !

93. Rosco says:

My 2 cents worth.

This is as good a statement on general thermodynamics as you’ll ever see (wish I’d said it) –

“There is a general thermodynamic rule which says that you can never focus energy in such a way that the target is hotter than the source.”

Vale every greenhouse effect model – the Sun alone can heat Earth to minus 18°C (despite the ~121°C recorded on the Moon at “noon”), the Earth in response to this input can heat the atmosphere to minus 18°C, the atmosphere in response to this input can emit double the amount of flux associated with minus 18°C because it is a “layer”, and “layers radiate over 2 surfaces, and this is not creating energy out of nothing (LOL) and magically 2 sources of energy from an emission temperature of minus 18°C combine to cause ~30°C.

This nonsense is the result of “flat Earth” models and using 1/4 power of the Sun’s radiation as the input.

The thermodynamic effect of 10 hours @ 239.7 W/m2 (~minus 18°C) input for a total of 2,390 W hours/m2 is NOT the same as 1 hour @ 2390 W/m2 (~180°C) for a total of 2,390 W hours/m2.

Flat Earth physics unequivocally says these are the same thing !

John said – “My question is: How is the average emission temperature of the earth measured by satellite (in basic terms) and is it indeed -18 Celsius (or is this arrived at by a similar “flat earth” argument)?”

This is one of many graphs of measured emission from the tropical Pacific Ocean (from a text by Petty) – you can search for this easily

The emissions from tropical lands are higher and from the poles significantly lower. The average claim is probably a mathematical calculation.

Alarmists put much emphasis on the claim that the peak emission from a Planck curve coincides with the 15 micron absorption band of CO2 – the big bite out of the graph.

But this is only a coincidence of using wavenumber as the independent variable and the vagaries of Planck’s law. Switch to wavelength as the independent variable and the Ozone “bite” at ~10 micron coincides with peak emission while the CO2 “bite” is shifted well away to the right.

If you use Wien’s law to calculate peak emissions you need to be mindful it only applies to the variable chosen – frequency, wavenumber or wavelength. The peak emission value is really meaningless because of this.

94. Pierre D. Bernier says:

@Rosco

I just dont see it ! Wavelenghts and Wavenumbers are a reciprocal of one another. If you change the x axis from one to the other the graph should look the same, and CO2 and ozone still absorb at the same place. Nothing has changed in there chemical structure.Should’nt the cm^-1 change also in the y axis ? Things just dont move aroud in the spectrum, whether it be H, Fe, Ni or what ever.. I just dont see it. Mabe I need a pair of glasses !

95. Pierre D. Bernier says:

@Rosco

Come to think of it. I don’t know where you got that graph but it looks like it’s all messed up. First the bottom X axis from 400 to 1600 is wavelenght, and in nm not micro meters. The top X axis is in wavenumbers not wavelenght. OUTCH !!!

96. That’s excellent bmrgeophyz.

“The GHE is a measure of the bias inherent in the artificial model.”

Diamond! Exactly, that’s what it is a measure of – is a measure of the incorrectness of the model and its foundation. The correct model would be the one with zero bias, i.e., zero GHE!

97. Pierre D. Bernier says:

@Rosco

OK. Got the first part. Near IR on the right. Far IR on the left (inverted).. Still, what absorbs at x wavelenght reemits at the same exact x no matter x is wavelength or wavenumber.

Still missing something

98. Pierre D. Bernier says:

@bmrgeophyz

You really did your homework well. A+

99. bmrgeophyz says:

My attempt at describing the problem is as follows:
The model used to determine the Greenhouse Effect took the incoming Solar constant of 1370 Watts per square metre and spread that across the whole spherical surface of the Earth giving 342.5 W/m^2 as the average irradiance. That is, the model had no night or day and no polar ice caps or Equatorial tropical zone, simply the same irradiance causing the same constant temperature everywhere. This means a non-rotating, non-orbiting Earth receiving equal radiation from all directions of the three dimensional Universe.

This is manifestly different to reality, whereby, at any instant in time there is only one spot on the surface potentially receiving the full irradiance of 1370 W/m^2. That spot circumnavigates the globe every 24 hours along a different path each time but always lies within the Equatorial zone. The remainder of that part of the globe facing the Sun receives the Solar constant reduced by the sine of the angle of inclination of the surface with respect to the incoming radiation. This diminishes to zero along the circumference of the plane facing the Sun and over all of the surface facing away from the Sun. That is, the temperature is always fluctuating back and forth between daily maxima and minima and these constantly change as the Earth orbits the Sun……..

………..In summary, the UN IPCC model defines an isolated sphere in space exhibiting no change in surface temperature whatsoever in marked contrast to the ever-changing temperature both with time and location across the Earth’s surface. The contrived 33 degree Kelvin Greenhouse Effect is not a property of the atmosphere but a measure of the bias inherent in the artificial model used to estimate the average temperature of the surface of an imaginary Earth.
See the Greenhouse Effect page on https://www.climateauditor.com

100. CD Marshall says:

SO if we are having less cloud cover because of less solar activity, but that allows more cosmic rays to reach the Earth’s surface and those rays add to cloud cover, does that increase cloud cover? Does that compensate for the loss of clouds or is it not balancing the equation?

101. CD Marshall says:

Another question is this true or not?
“when you go back millions of the years in record, solar output was much weaker than it is today (as per stellar evolution)”

102. Rosco says:

@Pierre

All I am highlighting is that using the Wavenumber domain for Planck’s equation and the Nimbus satellite records monitored over the tropical Pacific, Nauru I think, you get the graph from Petty’s textbook as shown.

Yes, nothing changes about the absorption bands and how much is absorbed just whether or not this coincides with peak emissions and whether this even matters – alarmists assert this is proof of how dangerous CO2 is but it is nonsense.

In the wavenumber domain the peak emissions of Planck’s curve equates to the bandwidth where CO2 absorbs strongly in the far infra-red – ~600 (~16 micron) to ~800(~13 micron) per cm. Strangely enough the spike in the middle coincides with about 666 per cm – the devil’s number for the superstitious like climatologists.

Here’s one from a NASA site

They make a really big deal over this – typical of all of their alarmist images.

Switch to the wavelength domain and you end up with a graph like this

The reason I mentioned this is that lots of alarmists that should know better make a big deal of the fact that CO2 absorbs strongly in the band of wavenumbers that coincide with the peak of Planck’s curve in a wavenumber curve for ambient Earth temperatures. This band is ~13 to ~16 micron

But in wavelength terms the peak emissions coincide with about 10 microns and the Ozone absorption band impacts this while CO2 is shifted to the right well away from the peak.

Even NASA sites make a big deal of this image – I’m simply pointing out it is irrelevant.

As I said lots of people who should know better make a big deal of this and it is nice to point out their claim is irrelevant – the coincidence with peak emissions is irrelevant.

Just another example of alarmist nonsense like the alarming temperature anomaly graphs etc etc.

Check it out – they always assert CO2 absorbs near the peak emissions of the Earth’s surface temperatures and this is a really big deal. It Isn’t.

103. The graph on the left (Wavenumber) is from NASA, and the graph on the right (Wavelength) is from … Sportisse, B. 2010. Chapter 2: Atmospheric radiative transfer. In: Fundamentals in Air Pollution. From Processes to Modelling. Springer, Dordrecht, The Netherlands, 45–92.

The left graph has slightly different units of measure on the y-axis than the right graph, but I think that they still show Rosco’s point more clearly.

I wish I could find an exact comparison of graphs with all units exactly the same, but I guess keeping this clear comparison obscure helps the alarmists perpetuate their sophistry.

Remember, wavenumber means number of wave crests per centimeter, while wavelength means the length from wave crest to wave crest, where light is conceptualized as waves. The choice of which description to use determines the appearance of the graph.

104. Absorption does NOT mean heat trapping.

105. CD Marshall says:

Someone made this reply to climate science being full of idiots and explaining why other sciences are smarter:
“Geologists and Physicists have a different view of the world. Geologists have a MUCH longer view point, like a billion years. Physicists have a much more math based view point; quantum heat absorption in the IR bands is something they do with their morning coffee, like literature majors do crosswords”.

106. Pierre D. Bernier says:

@ Rosco & RK

OK. I get it ! O3 and CO2 absorb at the same wavelenght (wavenumber), it’s the Plank curve that changes because of the units, making O3 and CO2 look like they absorb somewhere else. Well, one question answered begs another…Why would the Plank curve change if it’s supposed to be measuring the same thing. ? It’s supposed to be showing the spectral distribution of emission of a black body as a function of temperature is’nt it ?

107. Pierre D. Bernier says:

@CDM
/SO if we are having less cloud cover because of less solar activity/
No, we are having less cloud cover because of higher solar activity.

/but that allows more cosmic rays to reach the Earth’s surface and those rays add to cloud cover, does that increase cloud cover?/
Yes, lower solar activity allows more cosmic ray thus more cloud cover.

108. Pierre D. Bernier says:

@Rosco & RK

OK. Got it !
//www.spectralcalc.com/blackbody/units_of_wavelength.html
//www.spectralcalc.com/blackbody/units_of_wavenumber.html

What’s the use of that law if it shoves you all over ?

109. https://www.spectralcalc.com/blackbody/units_of_wavelength.html

https://www.spectralcalc.com/blackbody/units_of_wavenumber.html

What’s the use of that law if it shoves you all over ?

This seems like a good question, but I’m thinking that it shows an underlying expectation that might not be justified, namely that a wavelength framing should mirror a wavenumber framing.

It seems like the Plank curve would visually reflect the manner in which it is framed. Why should framing it in terms of wave-crest-to-wave-crest-measure (wavelength) mirror exactly framing it in terms of wave-crest-quantity-measure (wavenumber)?

The length of something does not equal the quantity of that something. So, why should a length relationship look like a quantity relationship?

A better question might be, “What exactly does the bite focused on by the alarmists mean?” What is that gap?

110. Pierre D. Bernier says:

@RK
/ but I’m thinking that it shows an underlying expectation that might not be justified, namely that a wavelength framing should mirror a wavenumber framing /

Wavelenghts and wavenumbers are reciprocals. Sould’nt one have it’s bulge at the left and the other one at the right ? It does. They just inverted the x axis from right to left for the wavenumbers. The bulge is at the right.

Take your graph above. Take the part in wavenumbers (blue), flip it 180 degress on the Y axis and you have O3 absorbing on the left of CO2 as it xhould be. The bulge changed from left to right. So one or both plank graph are nonesense.

Converting from Wavelenght and wavenumber is a math reciprocal. You are not changing lenght for mass, just the view point. So yes, it shows an underlying expectation that might not be justified.

But I’m still at a loss as to why they use Planks spectrum.

111. Pierre DB,

What else would they use? I’m not understanding why you think that this “shoves you all over.”

If you understand the shifting viewpoints (which you clarified for me), then why is a shift a shove? (^_^)

112. I guess I should get clarification, PDB.

By “they”, you mean the climate alarmists, yes? And your question about why they use Planck’s spectrum? — just mystified, maybe, that they should know what they are doing, but don’t?

They think that they are presenting something scientific — whoo, … Planck, … quantum physics, … that’s supposed show they are speaking from some iconic authority, I suppose. Misusing an iconic authority, and then claiming competence in understanding his contribution to physics. What better person to abuse than Planck?

113. CD Marshall says:

“Richard Lindzen Wins the Frederick Seitz Memorial Award.”
…as long as they toe the greenhouse effect they are part of the Inner Cliche.

114. Pierre D. Bernier says:

Yes, by they I mean the alarmists. They’re shoving all kind of things around in the hope that no one will notice and they are indeed succeeding in fooling a lot of people. Unfortunately.

115. Telling the alarmists exactly what that spectral bite means would go a long way towards adding bite to the skeptic position.

I’ve tried to think this through a bit, but I probably just don’t have the background to do it correctly.

If CO2 absorbs at a certain wave length, then it emits at that wave length. There’s a sort of rhythm going on, right? … and this produces a spectral signature, maybe? … and the alarmists are trying to force an interpretation of “heat trapping” onto something that is merely a signature of some rhythm??

Help me out here, experts.

116. Rosco says:

There is a major difference between the NASA wavenumber graph RK shows and the wavelength graph. There is obviously an explicit simple relationship between wavenumber and wavelength – 1000 wavenumbers per centimetre is equal to 10 microns – the conversion factor is 10,000. 1000 per cm is 100,000 per metre = 10 micron.

Despite the difference in the units as a result of changing variable there is another difference between the graphs RK posted.

For the NASA graph the “y” axis has units of “Flux” – milliwatts per m2 per cm^-1,

The wavelength graph has units of watts per m2 per sr (steradian) per microns. To find the “Flux” values for this graph you need to multiply every value on the curve by Pi. Though there are 2 Pi steradian in a hemisphere the conversion is Pi due to the integral of Planck’s curve – see SpectralCalc PDF if your interested.

The point is if someone tells you that CO2 absorbs at peak emissions you can show them it doesn’t in the wavelength domain. Frequency graphs are similar to wave number which makes sense if you think about it.

The real kicker is that the total area under a Planck curve (The area under any Planck curve is obviously the integral ) is equal to the total power emitted which is the Stefan-Boltzmann equation for that temperature. This applies no matter what variable is chosen – frequency, wavelength or wavenumber.

The integral of Planck’s equation is the Stefan-Boltzmann law, the derivative of Planck’s equation gives an expression for Wien’s displacement law.

This establishes proof that the simple model of the greenhouse effect is not only ludicrous but their algebra is wrong anyway.

117. Rosco says:

Look at the graph from Petty above. It shows a series of Planck curves at various temperatures.

For the region from ~550/cm to ~750/cm the satellite detects radiation emitted by CO2 – absorption = emission at equilibrium.

Look at the black body temperatures associated with those emissions – ~210 K at the lowest point to ~270 K.

This is explicitly what people like Roy Spencer say all the time –

1. greenhouse gases lower the effective emission temperature to space, and
2. the radiation causes surface warming.

But how can the radiation emitted by anything at the associated temperatures cause any warming in a surface 60 or 70 K warmer ?

And their basic premise must mean the atmospheric lapse rate is increasing – cooler at the top and warmer at the bottom. As far as I am aware there is no evidence to support this assertion.

118. Rosco says:

Planck’s equation is the ONLY equation that should be used.

This is because it is Planck’s curve alone that completely defines the emissions fro any temperature.

It explicitly shows a lot of information that is not possible using the SB equation.

My 2 cents worth on this is explained at https://principia-scientific.org/publications/Ross-GHE-Experiment.pdf

119. So, to understand what that spectral bite is, we have to understand your paper and all the relationships and math associated, which would explain why it’s so easy for alarmists to fool people with a substitute explanation that appeals to simple graphic-space subtractions.

A gap on paper is a a blockage, like a rock stuck in a hole that keeps the water from flowing through as fast as it could. Pictorial sophistry, yes?

The picture can mean what they say it means — to hell with all that fancy explanation by those who make so many mistakes that I don’t even know where to begin. (^_^)

Never mind trying to understand the math, in order to discuss WHERE valid disagreement might exist. Just say that whoever does the math doesn’t know some basics that the alarmist cannot really explain. Just claim incompetence, and it shall be so in the eyes of irrational observers.

End of sarcastic rant.

120. Pierre D. Bernier says:

In chemistry, when we want to do a quantitative analysis of a certain chemical, we sometimes us a monochromatic UV, visible or IR light. The transmitted light compared to the original incident light gives us the concentration of the chemical when we compare to known samples. All spectra of light sources we use, unless they are already monochromatic, look pretty much like the graphs above but we use a prism or a diffraction grating to separate the wavelengths or wavenumbers so that we can use the exact one we want. It’s hard to do a real good job on estimating the above graphs because of the slant on the original emission near the max absorption of O3 and CO2 but the results I get at max absorption for O3 are 0.52 and 0.57 and for CO2 0.59 and 0.65. For me, the 2 graphs tell exactly the same story. There is no misleading here. Sorry.

121. Pierre D. Bernier says:

Sorry, I inverted the numbers…

O3 0.48 and 0.44 and CO2 0.41 and 0.36

The conclusion stands.

122. Zoe Phin says:

On Venus, CO2 has absorption bands that don’t exist on earth. Ergo, temperature and pressure are the real “forcings”, and CO2 is just “forced”.

There’s a reason matter (mother) is associated with the feminine, and energy with the masculine.

123. Rosco says:

“Converting from Wavelength and wavenumber is a math reciprocal.”

Pierre this is only true for the variable – 10,000/wave number gives wavelength and vice versa. BUT this change of variable in Planck’s equation requires a differential solution, not just a math reciprocal.

“But I’m still at a loss as to why they use Planks spectrum.”

Only Planck’s equation gives the complete description of the radiation emitted by a black body.

Only Planck’s equation provides the information necessary for remote sensing.

Extracting the temperature expression from Planck’s curve allows for an estimation of temperature from the observed spectrum – this is called the brightness temperature.

Only Planck’s equation allowed for the precise determination of sigma – the SB “constant”.

“For me, the 2 graphs tell exactly the same story. There is no misleading here. Sorry.”

The point I was making is that people who state that CO2 is a problem always cite the left hand graph Robert linked shown below, the wave number graph:-

This shows CO2 strongly absorbing IR in the ~575 to ~750 per cm bandwidth (or ~13 micron to ~17 micron) . They state that this coincides with the maximum energy emissions from Earth’s surfaces and CO2 drastically traps heat.

But this is simply not true or relevant – the coincidence with the peak emissions I mean – because if you switch variable the band of CO2 absorption no longer coincides with the area of peak emissions.

I’m sorry but I disagree – the manner in which alarmists cite the wave number graph gives a false impression to lay people that CO2 stops a lot of the Earth’s radiation escaping to space because it absorbs at a band coinciding with the peak but this is not relevant or even universally true as the wave length graph indisputably shows.

124. Rosco says:

You can’t compare the absorption properties of CO2 to those of Ozone.

CO2 absorbs at near ground level where the vast majority of CO2 is found while Ozone absorbs at ~15 to ~35 kilometres above ground level and none of that re-emitted radiation is coming back to warm the surface.

125. CD Marshall says:

Bill Nye explaining the greenhouse effect…???

Bill Nye: Billy. Billy. Billy. You have chanced on like the most important idea right now in climate science. When you say photons from the sun come in through the atmosphere, that is absolutely true. And they go through out through the atmosphere but not all of them get out at the same energy. Now here’s the strange thing about light and electromagnetism. Now bare in mind we are humans trying to understand nature and if we can’t get our heads around this it’s our problem. But basically, if you do experiments on waves of lights or electricity, electromagnetic waves you will find waves. If you do experiments on photons of light or electromagnetism you will find particles. You can either detect particles or waves. So both of these ideas have helped us in physics understand nature. So here’s what happens. Light from the sun comes in at wavelengths that our eyes detect very well. It hits the earth and is reradiated, the energy is absorbed by the atoms of soil, of bridges, of the ocean, of ice and reradiated or sent back out again at a longer wavelength, it’s a little longer. And I don’t know if you know this but you probably do, what we, you and I call heat is the same thing as light at a wavelength longer than we see with our eyes. There are a lot of animals that see these wavelengths, but that’s not our issue. You’ve seen it with night vision goggles, those cool images. So light from the sun passes through the atmosphere; hits the earth; all these different materials and is reradiated at a longer wavelength that carbon dioxide, methane and some other gases hold in. The visible light at the faster wavelength goes through, the heat at the longer wavelength does not go through to a limited extent, to a significant extent. And that’s how the earth is warm enough for us to live. And because we put so much extra greenhouse gas via various species in the atmosphere, the world is getting warmer faster than it’s ever gotten before. It’s a great question Billy. That is the essence of this. Passes through at one wavelength, starts to go back out at a longer wavelength that is held in by the greenhouse gases. This is the fundamental idea in climate science. Carry-on!

126. Pierre D. Bernier says:

@Rosco

/ the manner in which alarmists cite the wave number graph gives a false impression to lay people that CO2 stops a lot of the Earth’s radiation escaping to space because it absorbs at a band coinciding with the peak /

I completely agree with you. THEY are misrepresenting things by showing only the wavelenght graph. But you and I know that CO2 absorbs just as much in the wavenumber domain, it just does not show as dramatically because the emission is lower giving a sense that absorption is lower. But the ratios are the same.

/ none of that re-emitted radiation is coming back to warm the surface. /

None of it no matter the altitude.

127. A shorter wave length is going to have a higher wave number, right? Something that is a short length is going to have more of them in a given measure.

So, isn’t the Planck curve shifting in accordance to the higher energy of the shorter wave length? Peak energy, thus, is where it needs to be in the right graph?

Might the problem in understanding the variance in the visuals arise from the graphs not using the t same units, because the scale for wave length is lots smaller than the scale for wave number?

128. Extra “t” in my last post … (“the t same units” = “the same units”)

129. My understanding is that the ENTIRE Earth atmosphere radiates at the peak emission of about 10 microns. CO2 is 0.04% of the entire Earth atmosphere — one molecule in 2500 others, ALL of which radiate at the peak of 10 microns, right?

So what if 1 in 2500 molecule absorbs at a wavelength 5 microns more than this. It’s a tiny percentage of the atmosphere doing this absorbing! And isn’t it emitting too?

Am I homing in on the irrelevance of that bite yet?

130. Pierre D. Bernier says:

You’re at a loss. You’re not alone ! CO2 and O3 have their own molecular stucture and absorb at their own natural wavelenght and corresponding wavenumber,15 microns and 666 cm-1 for CO2 and 10 microns and 1000 cm-1 for O3. They don’t change. Why is it that the peak of emission of the black body changes. It’s the same temperature ? The peak emission at 650 cm-1 should have a peak emission at 16.5 ? No ?

I’m at a loss too ! I don’t get it !

131. Now here’s what I’m thinking:

The peak occurs where the waves are a certain number. The peak occurs where an individual wave is a certain length.

A wave-number graph is going to show at what number of waves the peak occurs. A wave-length graph is going to show at what length of waves the peak occurs.

The curve is for the WHOLE atmosphere.

I still can’t put it together yet to my satisfaction.

132. Pierre D. Bernier says:

Did I hear that right… No hurry ? Is he changing his tune ? Afraid of posterity maybe ?

133. CD Marshall says:

Just read an article somewhere that the hottest record in France is off by 2-3 Degrees Celsius compared to all the other stations. The one that set the record, “was meant to be moved to a more neutral location.” (After the flaw was discovered).

134. Zoe Phin says:

Look at this exchange:

Zoe: @EF M
Lame. All of your links can be explained by the real cause of global warming: cloud cover. Earth’s albedo has decreased by 2% since the 1980s. It doesn’t matter that solar irradiance has been reduced by 0.1%, what matters is that the amount of solar radiation reaching the surface has increased by 0.999×0.02 = 1.998%

EF M: @Zoe Phin As CO2 increases, infrared photons increase in the troposphere and decrease in the stratosphere which has been observed for the last 50+ years. This increases the temperature in the troposphere raising absolute humidity but decreasing relative humidity, preventing cloud formation and decreasing albedo by 2%. That cloud loss is primarily at the equator. The primary heat gain is at the poles at twice the rate as the equator which would indicate primarily CO2 temperature gain. You would be in fact partially correct.

Zoe: @EF M
Cute story. You would make a good writer, but a poor scientist. In your fable, CO2 does what ever it “literally” needs to to match observations, physics be damned. You’re a low life.

Solar flux in = Earth flux out

As CO2 blocks near-IR energy (at 2.3 and 4.7 microns) from the sun from reaching the surface, it never gets thermalized into Earth’s IR spectrum. There is less infrared photons to be made.

135. Pierre D. Bernier says:

What does he not understand ?

Solar Flux In = 1368 / 2 * 0.69 = 472 on 1 hemisphere = 29 C (No need for CO2)
Earth Flux Out = 472 on 2 hemispheres = 236 = -19 C

Looks simple to me !

136. Pierre D. Bernier says:

@EFM
Does anyone see CO2 effects here ?

Didn’t think so !

137. Where is that Zoe?

138. Zoe Phin says:

JP,

139. I blocked him from my channel.

140. Hans Schreuder says:

Hah, good on you JP, because these buffoons are the “official face” of climate skepticism, heartily accepting, without ever examining the issue properly, that carbon dioxide causes “some warming” and that the “greenhouse effect” is an indisputable “fact”. These guys, Heartland, CFACT, CEI, etc. etc. are the true enablers of climate alarm. Damn the lot of them!

141. CD Marshall says:

So this fellow Zoe was chatting with who said he was a working engineer with a thorough understanding of thermodynamics, claims the atmosphere can heat itself up more with evaporation because evaporation is work.

That is so far out a child could spot that as being incorrect.

142. Zoe Phin says:

JP,
I think I figured it all out:
Climate scientists deny geothermal energy, except on planets with no GHGs (Jupiter, Saturn).

Geothermal energy alone can heat surface to ~8°C
Sun can heat hemisphere to ~30°C

Thermal fluxes can’t be added, but they can be MAX()ed

During day: Max(8,30) = 30
During night: Max(8,-273) = 8

Average surface temperature: 19°C

Just eyeballing this chart shows 4°C is a good guess for typical conditions.

19-4 = 15°C

And that is exactly what our “surface” thermometers measure!

No greenhouse defect, er uhm, effect. QED

What do you think?

143. Pierre D. Bernier says:

@Zoe

Don’t want to pick a fight with you. I saw what you did to EF M. 🙂

I’m not denying that gepthermal has anything to do but in my view it would be minimal. When you look at the Sun, the real one, the one where you divide the flux by 2 instead of 4, you see that there is plenty of energy coming in to heat what ever during the day and during the night it’s the water that stored that energy that is the heat source to keep everything warm. Water has a tremendously high specific heat capacity plus a very high latent heat of condensation.. I prefer water !

144. Pierre D. Bernier says:

@Zoe

Here is a little table of temperature differences between daily maximums and minimums (average monthly) for differents places on Earth for January-February…

Las Vegas, NV 17
Mexico City, Mexico 17
Niamey, Niger 15
Ulan Bator, Mongolia 15
Wichita, KS 14
Shizuishan, China (Gobi) 13
Los Angeles, CA 10
Montreal, QC 8
Lima, Peru 7
Miami, FL 7
New York, NY 7
Seatle,WA 7
Rio de Janeiro, Brazil 6

The places are in decreasing daily differences. You can see a clear cut between places that are far from big water surfaces (deserts) and those that are near. I dont think geothermal can explain those differences. The presence of water does.

Source : Google search (Mexico city average temperatures)

145. CD Marshall says:

Sorry to add my simple penny to this conversation, but if the water is heated up by geothermal it would add to the increased temperature and therefore on a smaller scale increase temperate around water. Specifically I’m referring to the over one million underwater volcanoes which some are active and semi-active in a constant fluctuation. Just one of those nasty buggers can increase surrounding water temps by up to 5 degrees Celsius. On an average 1-5 thousand are said to be active or semi-active at any given time. What happens when more become active? Very hot oceans that have their normal oscillation cycles upturned , increased weather patterns and naturally stronger, larger and more consistent El Nino effects.

Land based geothermal activity underground can warm water tables, produce pressure and steam below ground that can warm the surface up more. You can’t discount the existence of geothermal on Earth. The Moon yes, apparently it does nothing for the outer crust so I’ve heard, (if it does have a molten core?). How much can it change surface temps I honestly don’t know.

Because of the volume of the oceans we might not see those effects for decades. Stronger El Nino phenomena in the past could have been directly linked to greater geothermal activity 50 years ago or so.

Anyway that’s my penny’s worth which by the way is also the name of one of my cats.

146. Pierre D. Bernier says:

@CDM

I did say / I’m not denying that geothermal has anything to do but in my view it would be minimal./

I dont have any way of measuring the contribution but water still is the recipient of that heat and water is our great thermostat.
By the way, do you remember the name of the owners in the DDT scam ? Really curious !

147. Zoe Phin says:

Pierre,

“I’m not denying that geothermal has anything to do but in my view it would be minimal.”

Minimal? It’s enormous!
The sun can’t penetrate below 10 meters of land and 120 meters of ocean, yet it’s like 10°C and 20°C down there – all from geothermal!

Minimum Upwelling IR is 410 W/m^2

Show me how water vapor can sneak under this equipment.

148. CD Marshall says:

PB I doubt that will ever be easily found on the censored net these days. So I can’t say if it was her or a politician supporting her, it was connected to her in some way just that info is no longer available on the internet. I heard it 10-15 years ago from a guy who likes looking into these things, Seems it was a politician or there spouse that owned the mosquito net company used in Africa. Environmentalist are never innocent.

If I ever come across it I’ll let you know.

149. Zoe Phin says:

Most climate “scientists” believe geothermal can be ignored because it’s measured in milliwatts. The global average is 91 mW/m^2

Formula for conduction:
Q=kA(Thot-Tcold)/d

Rearranging the terms:
Tcold = Thot – Qd/(kA)

Now let’s try this on 10°C @ 10m deep soil.

Thot = 10°C
Q = 0.091
k = 0.837 (from materials database)
A = 1 m^2
d = 10 meters

We’re looking for Tcold, which is at the surface.

10-0.091*10/0.837 = 8.91°C

And what will 8.91°C emit to space by radiation? SB Law!

Emissivity of soil = 0.92

That’s the real flux from geothermal to space!!!

-Zoe

150. Pierre D. Bernier says:

Camp Desert Rock… Still 15 C Max – Min in J-F (desert like difference)

I’m talking about water vapour in the air and from the surface waters of oceans controlling our climate. They get their energy mostly from the Sun. Never said you cant get heat from down there. Ask minors down 1 – 2 or more km down. We also can heat buildings from geothermal, even in Montreal, ask my daughter in law ! My point is I dont see so much heat from geo.

From your above graph, I see kind of a sin wave for 13 hours with a max of about 1000. So, back of the envelope calculation, If I do an integral of cos(x) from -Pi/2 to Pi/2 (12 hrs) I get about 2000. Your blue line is about 500 for 24 hrs for about 2000. Where’s the geo ? My guess is just accumulated heat from the Sun going back up ! Nothing is spontaneous. Convection takes time. Proof, blue and red lines are not perfectly in sync at 13 and 20 hrs. A lag of about 1 hr.

151. Pierre D. Bernier says:

@Zoe

How can soil at 10°C at 10m deep heat soil on the surface already at 40°C ?

152. Zoe Phin says:

Pierre,
“How can soil at 10°C at 10m deep heat soil on the surface already at 40°C ?”

It can’t. That why I wrote:

“Geothermal energy alone can heat surface to ~8°C
Sun can heat hemisphere to ~30°C

Thermal fluxes can’t be added, but they can be MAX()ed

During day: Max(8,30) = 30
During night: Max(8,-273) = 8

Average surface temperature: 19°C … ”

My finance experience (and college classes) made me a decent computer programmer. I wrote a program to analyze the data, and my results:

Red line AVG: 346.4 W/m^2
Blue line AVG: 543.3 W/m^2

Where’s the extra upwelling flux coming from? Geothermal, I must presume.

153. Pierre D. Bernier says:

@Zoe

Don’t think so

You say the global average is 91 mW/m^2 ( not surprised). So low the capacity is rapidly runned over by the scale of the mass. Geo can’t keep up otherwise why temperatures go below 8 ?

As for 346 and 543, how can you get that. 2000 / 24 = 83 and 2000 / 12 = 167 ? Don’t get it. Please explain .

“made me a decent computer programmer.”. Please don’t use such arguments. I red a half 5 1/4 inche floppy seized by police (they never found the other half) before the FBI did. It’s no proof of anything. It just begins a pissing contest.

154. Zoe Phin says:

Pierre,

“You say the global average is 91 mW/m^2 ( not surprised). So low the capacity is rapidly runned over by the scale of the mass.”

I feel like you’re not reading my arguments.

Tcold = Thot – Qd/(kA)

The lower the Q, the higher Tcold.

91 mW/m^2 is large Tcold -> Radiation
0 mW/m^2 is larger Tcold -> Radiation

You’re more than welcome to review my code:

https://www.docdroid.net/5l7KFNK

155. Pierre D. Bernier says:

Little wise girl

Desert rock is right beside a fault/earthquake line. Almost had me fooled !

156. Zoe Phin says:

Pierre,
I downloaded all 2018 data for 6 sites so far, and ran my code:

Bondville 168 374
Desert Rock 236 441
Goodwin Creek 175 404
Penn State 141 366
Sioux Falls 155 352
Table Mtn 188 364

First #: Solar
2nd #: Upwelling IR

I’m not trying to fool anybody.

157. Zoe Phin says:

Pierre,
The problem with water [vapor], is that it can only create a lag effect (slower heating, slower cooling), not additional energy. I can’t think of any reason other than geothermal for why we would have more upwelling IR than the sun allows. -Z

158. Pierre D. Bernier says:

@Zoe,

Perfectly on the same page with you on water only slowing things down enough for comfort. As for geothermal, I’ll have to look it up more closely. Seems odd to me that something in the mW/m^2, on average, could make such big differences as to give in the 50% to 100% more radiation back to the sky. Can’t wrap my head around that. Will come back.

159. Zoe Phin says:

Pierre,

Most climate “scientists” believe geothermal can be ignored because it’s measured in milliwatts. The global average is 91 mW/m^2.

Formula for conduction:
Q=kA(Thot-Tcold)/d

Rearranging the terms:
Tcold = Thot – Qd/(kA)

Now let’s try this on 10°C @ 10m deep soil.

Thot = 10°C
Q = 0.091
k = 0.837 (from materials database)
A = 1 m^2
d = 10 meters

We’re looking for Tcold, which is at the surface.

10-0.091*10/0.837 = 8.91°C

And what will 8.91°C emit to space by radiation? SB Law!

Emissivity of soil = 0.92

That’s the real flux from geothermal to space!!!

The LOWER the Q, the higher the outward radiation!

1 mW “>” 91 mW

-Zoe

160. Zoe Phin says:

Pierre,
In other words, 91 mW/m^2 is heat flux through the medium, not out of the medium. More math is involed for the latter. Have a good night 🙂

161. Pierre D. Bernier says:

@Zoe,

Did some spot checks. For Desert Rock, 22 march 2019, exactly 12 hrs of sunlight topping at about 900. Back of envelope = 1800 / Pi = 573. Blue line eye balled at about 430 – 450 avg.

Seasonality ? Can you do full year ?
Now going to bed !

162. Zoe Phin says:

Pierre,
I told you that I ran full year (2018) data for 6 sites. 6 × 365. See above for results.

163. Pierre D. Bernier says:

@Zoe
The redundancy of three component solar measurements (global, direct and diffuse) provides a useful tool for quality control of the SURFRAD data. In addition, the sum of the diffuse and direct is actually a better measure of total solar than the global measurement alone because when the sun is near the horizon, a change in the cosine response of the global pyranometer’s sensor introduces errors in the global measurement.

I’m missing some of my beautysleep over this !

164. Rocky Squirrel says:

All of this discussion about the average insolation is irrelevant. If you can’t deal with averages then just consider the total energies absorbed and emitted over a suitable time interval.

[JP: THAT IS WHAT WE ARE TALKING ABOUT. THAT IS WHAT AN AVERAGE IS.]

If you do that properly, and make some reasonable assumptions about the effective emissivity of the surface for longwave radiation, then you can prove that without a GHE the average temperature of the surface can be no higher than the effective radiating temperature. Such a mathematical demonstration does not require any assumptions about the temperature of the surface being uniform.

[JP: And yet in real-time the moon gets to +121C, and the surface of the Earth is driven with such intense flux as to generate the weather and the climate…which the AVERAGE 240 W/m^2 can NOT do.]

As for your incorrect solution for the steel greenhouse,

[JP: The solution in my book? I think it was first shown on this blog somewhere. In any case…nice CLAIM. No one has shown where the solution errs…except for me, where I show that the very error of the solution proves the greenhouse effect to be impossible!]

and other incorrect statements about lightbulb filament temperatures, a lightbulb filament will be able to achieve the same temperature with LESS electrical energy input if it is surrounded by a material that reflects the LWR emitted by the filament back onto it. Such a system can then emit the same amount of visible light while consuming less energy than a standard bulb that does not reflect the LWR back onto the filament.

[JP: This is about reducing emissivity. GHG’s aren’t about reflection, but about absorption and re-emission. GHG’s emit half of their energy to space…the material around the filament DOES NOT. Given that GHG’s emit half of their energy to space, whereas non-GHG’s do not…then GHG’s in fact help the amtosphere shed energy to space particularly given that CO2 is collisionally dominated..]

165. Pierre D. Bernier says:

@Zoe,

I know that the Earth’s core is one big ball of molten Iron at 4000K (whatever) with decaying radioactive material. If some of that heat did not come out, the whole thing would blow up. So some heat has to come out. From your figures up here, it looks like half the Earth’s outgoing IR comes from the Sun, the other half from the Earth’s core. Looks ackward to me. Really ackward !

Surfrad measure downwelling IR and upwelling IR. Do they have a way to differenciate what IR is outgoing from conversion at the surface from the Sun’s IR reflected at the surface ?

Also, you are not counting diffuse downwelling solar. Why ?

Until I have more details on each data I can’t pronouce myself. Too many unknowns.

166. Pierre D. Bernier says:

Heck. Just did a test for Desert Rock, 22 and 23 march 2019, and others. Ploting only downdwelling IR and Upwelling IR. Not only they have downwelling IR at night, the bumps in it show up in the upwelling. Some of the upwelling is double counted !

Need something to really seperate.

167. Zoe Phin says:

Pierre,

“From your figures up here, it looks like half the Earth’s outgoing IR comes from the Sun, the other half from the Earth’s core. Looks ackward to me. Really ackward !”

I had the same mental block for a while.
Is it awkward?

Looks to me, if we had no sun and atmosphere, the blue line would go straight to 8°C.

“Also, you are not counting diffuse downwelling solar. Why ?”

I assumed DNI+Diffuse = Global. I looked into max solar, not its parts.

“Not only they have downwelling IR at night, the bumps in it show up in the upwelling. Some of the upwelling is double counted !”

You’re getting close to uncovering the trick.

What happens to the Upwelling IR (heat) as it travels from bottom to top of measuring device?

It teleports into atmosphere and comes down on the device. Get it?

Normal physics:

—>[bottom]———->[top]—–>

Fantasy land:

—>[bottom]———->[top]<—–

There is no Downwelling IR (except negligible inversions).
What they call Downwelling IR, is actually Device Upwelling IR.

So in reality, there's Ground Upwelling and Device Upwelling.

168. Pierre D. Bernier says:

Zoe
/ Looks to me, if we had no sun and atmosphere, the blue line would go straight to 8°C./
Don’t know the relevance. Water, like any other substance contracts and becomes denser with lowering temperature down until 4C. Then the H and O in the water molecules start forming what’s called hydrogen bonds and in some way starts the crystallization process. Then water starts to expand with lower temperatures and therefore rises. That’s why ice form’s on top of water. Salt excluded, the bottom of the oceans can never go below 4C. But I don’t know what that has to do with your blue line ? I know your blue line starts at 4C but the slant can be any. Depends on where you are, pole or equator. Just cant see.

169. Pierre D. Bernier says:

In fact, there is no slant. As soon as the water is at 4C, 20 m or 800 m down don’t matter, it’s a vertical line !

170. Pierre D. Bernier says:

@Zoe

For desert Rock, 22 march 2019, If I add…

Solar Downwelling, 272,66
Solar Upwelling, 53.87
Solar Diffuse, 39,48
I get a total of 366,01

For IR Upwelling I get 365.93

Did I just get lucky or do I win a price ? 🙂

171. Zoe Phin says:

Pierre,
Yes you got lucky.

Why would you add Solar Upwelling to Solar Downwelling?

Solar Upwelling is just solar flux reflected by device, i.e albedo.

What are you trying to do?

Cognitive dissonance?

172. Pierre D. Bernier says:

All I lnow is that the Sun is a full blown fusion nuclear reactor at 6000K. Lots of energy. The Earth’s core might be near 6000K also but it’s only radioactive decay, pressure and friction. The reason it’s so hot is low conductivity to the surface. I just dont see how Earth’s core can give as much energy as the Sun. I know… the energy from the Sun is radiating all over the place but all of Earth’s energy has to radiate to the surface. But still.

That blue line above should be slanted to the left because water at 0C is lighter then water at 4C. It’s a no argument.

Other then that I don’t know.

What the heck do they measure ? I already told you, Every bump in the downwelling IR is seen in the upwelling IR. The upwelling IR seems contaminated. How can we do anything about that ?

Unless you come up with more rational numbers then 2 to 1, I’m not in.

173. CD Marshall says:

Zoe, Pierre

Have any of you been in caves? I’ve been in quite a few caves but never deep Earth caves. In general I never recall a cave near the surface being anything but cool with or without moisture.
When does a cave depth start getting warmer?

Geothermal heat is often apparent to miners working in deep mines, but very few caves extend deep enough into the Earth to have much influence from geothermal heat. For example in mines 5 km deep the temperature approaches 70 degrees Celsius, so cooling equipment is necessary to work in those conditions.

Unless you have an outlet through the Earth’s natural installation of the core I’m not seeing energy getting to the surface as a general rule (meaning heat escapes through the weakest points). That’s why geological hotspots never move only the outer crust moves.

One point of fact is even though the core is far less in magnitude than the Sun it’s heat never ever stops it’s slow ascent to the surface, ever. That heat is always finding an exit.

Certainly some areas allow more warmth to the surface, oceans certainly being an obvious outlet. Heat will always find an escape route at the weakest points in the crust wherever that exists.

So my point is as far I can see heat from the core is selective not general. What is my 2 cents worth? Probably not much in this conversation. Sometimes very intelligent people can’t see the simplest answers.

Disregard all the what ifs and maybes and rip it back down to it’s basest form: Is you basic theory really sound? If you think it is why? Name one simple obvious indisputable point that makes it feasible. Build on that if you can, if you can’t, if hesitation or uncertainty creep in, disregard it and start over.

Answer those questions honestly, and move forward. At this point I think both of you are overthinking it.

174. Pierre D. Bernier says:

@CDM

Very nice points.

I downloaded data in Excel for Desert Rock, 22 march 2019. Had trouble figuring what each colunm represents but finally did it. They have a device for mesuring Downwelling Solar, Upwelling Solar, Downwelling IR and Up welling IR. They have columns with net Solar and net IR (which correspond to the differential of the measured data). Now why would they have a column Net IR if not for the Upwelling IR detector detecting both up and down reflected up.

Net downwelling Solar is 219 W/m2 and Net upwelling IR is 99 with a base line at about 75 in night time when the Sun does not contribute to the Upwelling IR.

Anyway, this whole thing is useless because Net incoming should be about the same as net outgoing which it is far from.

So, unless someone shows me the specifics I’m out !

175. Zoe Phin says:

Pierre,

“They have a device for mesuring Downwelling Solar, Upwelling Solar, Downwelling IR and Up welling IR. They have columns with net Solar and net IR (which correspond to the differential of the measured data).”

Sorry, but in regard to net IR that is incorrect.

A pyrgeometer MEASURES Upwelling IR and Net IR, while Downwelling IR is a derived fantasy name. Downwelling IR is really Upwelling-from-device IR.

Normal physics:

—>[bottom]———->[top]—–>

Fantasy land:

—>[bottom]———->[top]<—–

176. Pierre D. Bernier says:

Total downwelling (global) solar radiation is measured on the main platform by an upward looking broadband pyranometer.
The diffuse component is measured by a shaded pyranometer that rides on the solar tracker.
A third pyranometer is mounted facing downward on a crossarm near the top of the 10-meter tower to measure solar radiation reflected from the surface

An upward looking pyrgeometer on the main platform measures long wave (thermal infrared) radiation emitted downward by clouds and other atmospheric constituents.
Another pyrgeometer, mounted facing downward on the crossarm atop the tower, senses upwelling long wave radiation

These measurements of upwelling and downwelling in the solar and infrared wavebands constitute the complete surface radiation budget

No mention of measuring Net with a pyrgeometer !

If it’s not right, take it up with NOAA not me.

177. Zoe Phin says:

Pierre

“No mention …”
Well duh, they are tricking you!

Do you know how a thermopile works?

Section 2.4

“After completing the installation the pyrgeometer will be ready for operation. The downward atmospheric
long-wave radiation can be calculated with Formula 1″

Ld = Uemf/S * sigma * T^4

Uemf/S = Net IR”

178. Zoe Phin says:

Pierre,

“pyrgeometers are designed to measure the net longwave irradiance between about 3.5 and 50 microns at the detector surface. The net longwave irradiance (Wnet) in Watts per square meter can be determined from pyrgeometer outputs:”

Here they are trying to trick you.

“the incoming longwave irradiance (Win) can be computed in several ways”

Huh, I thought they measure it, but here they admit the truth and give you 4 ways of calculating Down IR.

“Equations 4 and 5 also include approximations to account for possible differences between case temperature (a pyrgeometer output signal) and the effective receiver temperature (the basis for measuring net irradiance).”

The basis for MEASURING net radiance.

A pyrgeometer is just a fancy double thermometer.

Normal physics:

—>[bottom]———->[top]—–>

Fantasy land:

—>[bottom]———->[top]<—–

What happened to the heat traveling up?
In fantasy land it teleports to the sky and comes down.

Take a hot plate, a frying pan, and crack some eggs over it.

In fantasy land, there is no hot plate, it's the sky cooking the eggs.

Thank you Chris.

179. Pierre D. Bernier says:

@Zoe

If you dont trust their pyrgeometer or the way they use them then why are you so hell bent on using their data to prove your theory ?

I know I should not trust them and I dont, no need to remind me. Their data dont make sense and should not be used for anything and nor should you. No way Earth emits as much as the sun.

//principia-scientific.org/how-to-fool-yourself-with-a-pyrgeometer/

CASE CLOSED.. NEXT

180. Zoe Phin says:

Pierre,

On Aug 1st, Desert Rock Upwelling IR starts going up (due to lagged solar) @ 13:05

It reaches a zenith @ 20:55

In that time frame 273558*60 J/m^2 accumulated.

Let’s pretend Aug 2nd is exactly like Aug 1st.

At what time will this energy be depleted (sent to space)?

@ 04:37

The Sun heats Upwelling IR for 7:50 hr
Then for 7:42 hr it cools

Now … what keeps Upwelling IR from falling below 432 W/m^2 for the next 8:28 hrs?

181. CD Marshall says:

Does anyone know what this “The “Green Plate” Effect” is and what’s it all about?

182. Rosco says:

@CD Marshall “Does anyone know what this “The “Green Plate” Effect” is and what’s it all about?”
It is simply another variant on the “Steel Greenhouse” stupidity begun years ago by Willis at WUWT – and they call themselves sceptics ?
This link is the originator of this further ludicrous claim that because you insert a layer between a radiating surface and a “sink” with zero radiation you can double the energy available in true perpetual motion type gobbledygook because a layer radiates over 2 surfaces. He is another PhD qualified loudmouth.
This is the money quote they always cite
“The entire system HAS to heat up to reach the equilibrium condition. T1 and T2 are the equilibrium temps of the plates.”
No increase in energy input but the system has to heat up – stupidity beyond belief !!
http://rabett.blogspot.com/2017/10/an-evergreen-of-denial-is-that-colder.html
It is just another example of the stupidity of the flat earth physics of the greenhouse effect model being debunked by Joe’s videos.

183. Pierre D. Bernier says:

@CDM

C from the green plate cannot heat the blue plate which has a higher temperature.

184. Pierre D. Bernier says:

@Zoe,

Downwelling IR in Mississipi is higher then Desert Rock because of humudity. Got to subtract from Upwelling. Then try to equate Net Solar and Net IR ! Hope you do.

185. CD Marshall says:

Thank you very much gents and or ladies. Sorry I’m not calling you binary even if you want me to. The only binaries I’ll acknowledge are stars…Granted some people are pretty spaced out.

186. Pierre D. Bernier says:

@CDM

Dont know if you ever saw that paper before.

Click to access 1510.02503.pdf

See figure 4. If backradiation existed, the back surface would rise to 297 C for a 1000 W/m2 input. de Saussure never recorded anything above 110 C.

187. CD Marshall says:

So this troll keeps trying to trip me up. This is the conversation so far:
I said:
“ALL gases in the Atmosphere absorb heat, and all heated gases radiate infrared light in close proportion to their temperature. Major gases like nitrogen and oxygen exceeds what minor players like carbon dioxide and water vapor contribute.”

Troll said:
“So show me an infrared spectroscopy spectrum with an absorption band for O-O or N-N vibrational modes to prove me wrong.”

Naturally all gases do heat up. He actually claims Nitrogen and others that aren’t greenhouse gases don’t heat up. I mean this guy is beyond stupid, He’s been trying to trap me since he started the conversation. Unfortunately for him I lived with a mentally ill mother for decades and can avoid mental traps like a chess game.

However I’d like to bring this to closure. If you’d like to join in it’s on Tony Heller’s page. I mean few YTube sites exist that combat global warming. JP I wish you could convert Tony to understanding this game better. He sees it but I’m not sure if he “gets” what he’s seeing.

If you are bored his name is @Romano he comments a lot so def troll. My main conversations with him is on the Fla-Bushcraft Prepper string.

I’d like a critique on how I did if you have the time a few pointers for future references would be appreciated as well.

188. Pierre D. Bernier says:

@CDM

Sorry. IR has to do with chemical bonds between atoms in a molecule. They can stretch, rotate, bend. Since a nitrogen molecule and an Oxigen molecule are perfectly symetrical (N=N , O=O) they can do none. Therefore no IR absoption. The only way Nitrogen and Oxygen absorb heat is by conduction (contact). Sure, when heated it will get all exited and wiggle all over but since there is no dipole moment change, no IR.

//chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Active_Learning/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/03_Text%3A_Molecular_and_Atomic_Spectroscopy/4%3A_Infrared_Spectroscopy/4.1%3A_Introduction_to_Infrared_Spectroscopy

Really sorry

189. CD Marshall says:

Are you saying when heat rises it doesn’t heat up the atmosphere? Or are you just saying it doesn’t create ir. So if ir is being reflected it’s not doing anything to nitrogen and oxygen? Just so we’re on the same page.

190. Pierre D. Bernier says:

@CDM

When air comes in contact with a hot surface on the ground, it picks up the heat by conduction (contact). The air rises and gives up some of it’s heat to the other upper cooler air molecules by contact also, and so on, and so on and so on up.

191. Zoe Phin says:

Pierre,

Earth is a E°K ball.
Sun is a S°K ball.

Sun ball appears for 12 hrs and heats one hemisphere then goes away for 12 hours. Temperatures are averaged for equilibrium.

12 hr Daylight avg: (S+E)/2
12 hr Night avg: E

Total average: ((S+E)/2 + E) / 2

We know S = ((1361/2*(1-0.3))/sigma)^0.25 = 302.75 °K

We know from observations that global near-surface temperature is 288°K.

Total average:
((S+E)/2 + E) / 2 = 288
0.5S + 1.5E = 576
1.5E= 576 – 0.5*302.75
E = 283.08 °K

Earth is a 10°C ball.
Sun is a 30°C ball.

Sun ball appears for 12 hrs and heats one hemisphere to 20°C then goes away for 12 hours.

Earth has a 10°C geothermal source.
(Moon has a 100°K moonthermal source)

You’re welcome.

As far as I know I’m the only one to do this kind of math.

Wise people will come to realize that Kiehl and Trenberth’s 324 W/m^2 backradiation is just a geothermal flipperoo.

192. CD Marshall says:

I find it a little discouraging that almost all physics out out their is off on something. Makes you wonder if these guys should ever be teaching?

193. Pierre D. Bernier says:

@CDM

To be clear… Heat does not rise. It’s the warm molecules that rise giving up some of their heat to cooler molecules by contact. N2 and O2 are completely transparent to IR. No absorption, No emission..

194. Zoe Phin says:

Chris,
N2 and O2 are NOT transparent to IR.

195. CD Marshall says:

Thank you Zoe and Pierre for your input. Zoe that helps a lot.

196. Zoe Phin says:

Chris,

The albedo is 0.3

88% of that is from atmosphere
12% is from surface. [Donohoe,2011]

O2 and N2 are very weak GHGs, but they are still GHGs.

Only the strong GHGs can be responsible for such high albedo.

When alarmists remove albedo (S/4*(1-0.3), they remove the cooling effect of GHGs. After doing so, GHG absorption can only be seen as warming. Get it?

The albedo value stores GHGs’ cooling property.

197. Rosco says:

Not sure I agree totally with Pierre.

Our atmosphere is not predominantly heated by IR from the surface – it is due to contact with the surface and convection. “An example of an adiabatic process is the vertical flow of air in the atmosphere; air expands and cools as it rises, and contracts and grows warmer as it descends.”

NASA says :- “The single most important property of objects that determines the amount of radiation at each wavelength they emit is temperature. Any object that has a temperature above absolute zero (-459.67°F or -273.15°C, the point where atoms and molecules cease to move) radiates in the infrared.”

All of the planetary atmospheres radiate strongly in the infrared whether or not the atmospheres contain any significant IR absorbing gases or not – most of the atmospheres of the outer planets contain virtually no so called greenhouse gases and cannot possibly have a “greenhouse effect”.

I was taught decades ago that electron orbit jumps were involved in electromagnetic absorption and emission.

“Niels Henrik David Bohr was a Danish physicist who made foundational contributions to understanding atomic structure and quantum theory, for which he received the Nobel Prize in Physics in 1922.”

It is also the fundamental principle of the photoelectric effect – “In Einstein’s picture, an individual photon arriving at the surface in Fig. 38.1a or 38.2 is absorbed by a single electron. This energy transfer is an all-or-nothing process, in contrast to the continuous transfer of energy in the wave theory of light; the electron gets all of the photon’s energy or none at all. The electron can escape from the surface only if the energy it acquires is greater than the work function .”

198. CD Marshall says:

So what part of this needs corrected or explained better? Thank for your inputs it really helps. Since I primarily want this about just the atmosphere I prefer to disregard what doesn’t apply to reduce confusion in me and those I’m relaying the information to. Take your time I’m not in a hurry.

“Heated masses always emit light (infrared). ALWAYS. That’s a direct consequence of molecules in motion. And while it’s true that some substances may be transparent to infrared light, it doesn’t follow that they can’t be heated or, if heated, might not emit infrared.

There are three ways for thermal energy to move from one zone to another: by conduction, convection, and radiation. Conductive heat transfer involves direct contact, wherein vibrations spread from molecule to molecule. Convective transfer involves a mass in motion: expanded by heat, a fluid is pushed up and away by the denser fluid that surrounds it. Radiative transfer arises when molecules intercept the light that warmer molecules are emitting, which brings about a resonant molecular vibration — i.e., heating.

All gases absorb heat in the Atmosphere. The Earth first absorbs the visible radiation from the Sun, which is then converted to heat, and this heat radiates out to the atmosphere, where the atmospheric gases absorb some of the heat.

This is around 1% of the atmosphere that can be heated by radiant transfer and ignoring the 99% that is heated by direct contact with the earth’s surface and then by convection. ALL atmospheric gas gets heated mainly by conductive transfer, and that all heated bodies radiate light.

Major gases like nitrogen and oxygen, then, do not just radiate heat, but the total of this radiation vastly exceeds what minor players like carbon dioxide and water vapor contribute. Electromagnetic radiation is produced whenever electric charges accelerate – that is, when they change either the speed or direction of their movement. In a hot object, the molecules are continuously vibrating or bumping into each other (if a liquid or gas), sending each other off in different directions and at different speeds. Each of these collisions produces electromagnetic radiation at frequencies all across the electromagnetic spectrum. Accordingly, any heated gas emits infrared. There’s nothing unique about CO2. It absorbs heat and releases that heat very fast.

Not only will the trace gases need more energy to reach the same temperature as the air that contains them, they will radiate it in all directions instantly and at the speed of light and thus increase the efficiency of the air mass in cooling it, not warming it, in line with the first and second laws of thermodynamics.”

199. There is no meaningful emission of IR coming from the atmosphere, as demonstrated in this article by Alan Siddons:

Further refutation of all issues relating to climate alarm over carbon dioxide emissions can be found here:

Click to access Greenhouse_Effect_Poppycock.pdf

Click to access Greenhouse_Theory_Violates_Thermodynamics.pdf

http://tech-know-group.com/papers/Role_of_CO2-EaE.pdf (peer reviewed)
http://tech-know-group.com/papers/Role_of_GHE-EaE.pdf (peer reviewed)
However, climate alarm will continue unabated until the so-called “official skeptics” stop accepting the existence of that so-called “atmospheric greenhouse effect”.
Oxygen and nitrogen are the true “greenhouse gases” in a real greenhouse as well as in the open atmosphere as those gases are the real “heat-trapping gases” due to their poor IR emissivity. The recent “unprecedented heat-wave” that swept over Europe originated over the Sahara desert where direct insolation heats the sand to over 80 degrees C and which in turn heats the mass of air above it above 50 degrees C. The air then carries this heat across vast distances, even reaching Greenland in the past two weeks, causing wet knickers in the AGW camps.
By now though, even in coastal Kangerlussuaq, a place I visited in 2008, the night temperatures are already falling to zero degrees C despite extensive daylight hours! Having stood on the ice sheet makes one appreciate the vastness of that amount of ice and if there is interest I can repost my pictures of that most amazing visit.

200. Rosco says:

The thermal conductivity of materials is well documented. Air and CO2 are excellent insulators having a very low thermal conductivity. Thermal conductivity includes any radiation component as it is derived by experiment and as far as I know no molecules can be coerced into not radiating while an experiment is undertaken.

If any greenhouse gas had any powerful radiation ability it would have a significantly higher thermal conductivity.

If climate science is right and ~83% of the radiation emitted to space by Earth comes from the atmosphere why would increasing the concentration of the only molecules capable of emitting this IR to space result in less radiation to space – surely logic says more = more ?

201. Pierre D. Bernier says:

@Zoe,

Why do you keep on ignoring the truth ?

//serc.carleton.edu/images/integrate/teaching_materials/energy_sustain/student_materials/geothermal_heat_map_u.s_650.png

IN MILLI WATTS !!!

Also… Yes N2 and O2 absotb some IR but it’s insignificant compared to other molecules. Otherwise we would not be able to make IR spectras of materials other then in vacuum. You dont know what you are talking about. Stop making the Enblers job !!!

You are an economist. Assume that the devises used by surfrad are OK and go do the energy budget. It should balance.I know it can’t be done and I’ll show you in my next post why. For now I dont have the time.

I knew this would turn into a pissing contest !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

202. That’s good CD.

203. p says:

Nitrogen (N=N) and Oxygen (O=O) don’t absorb IR. They don’t have a dipole. They are symmetrical. Look at your own graph above. Littlest of O2 (blue) and no N2 (dark green).

//chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Active_Learning/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/03_Text%3A_Molecular_and_Atomic_Spectroscopy/4%3A_Infrared_Spectroscopy/4.1%3A_Introduction_to_Infrared_Spectroscopy

It’s only after UV rays, high in the atmosphere, have bombarded N=N into N-N+ or N-N- (knocked some electrons around) that they will absorb IR because of the created dipole. Same for O=O. How much of that do you think you are breathing down here on Earth ? How much up there in an aeroplane at 32000 ft ? None. So IR absorption by air is negligible to non existent. Now that that has been cleared up…

I know some people don’t trust the instruments on the SURFRAD stations and I’m not sure I do either. But, just for the sake of argument, let’s suppose that they do measure the right things appropriately (suppose I said). Let’s assume E stands for energy and that the whole system is in equilibrium (other wise everything would freeze down or blow up). So…

E in = E out

E in = Solar in – Solar out (we have both measures)

E out = IR out – IR in + CE out (we have 2 measures)

We have IR out and IR in but not CE out which is conduction (contact) energy lost to space by mere convection.

So the measurement system is incomplete. CE out will be anything necessary to balance the budget.

After the Sun’s ray’s have hit the ground and heated it up, there are 2 ways that that energy can be released to the atmosphere, direct IR emission and conduction by contact with the air. For all PRACTICAL purposes air will get it’s energy from conduction (contact) with the ground, will wiggle up in altitude and give back some of that energy to other colder air molecules by conduction (contact) then move up again. All of that because air does not absorb IR and does not emit either. So according to (Desert Rock, 22 march 2019)…

CE out = Solar in – Solar out + IR in – IR out

CE out = 272,66 – 53,87 – 365,93 + 266,62 = 119,48 W/m^2.

According to SURFRAD, in their mission statement they say…

/ These measurements of upwelling and downwelling in the solar and infrared wavebands constitute the complete surface radiation budget /

Oh ya ? Complete ? And they omit the conductive part of the heat loss ? Almost half the Solar in ?

I wont say what I really feel like saying… I’ll be banned !

And you want to use their data to prove your point ?

204. Pierre D. Bernier says:

205. Pierre D. Bernier says:

@Hans

When I talk about IR downwelling from what ever (water ?) I’m not talking about IR that will heat backup the surface. I know it wont. Whatever that IR is will be reflected back up and double counted by the other IR detector. That’s where their whole scam starts to skid. Plus they are not counting Conduction and convection. If they are please explain.

206. Zoe Phin says:

Pierre,
My comments went in one ear and out the other without processing, I see.

207. Pierre D. Bernier says:

@Zoep

Which one 6 There is so many ! Date and hour stamp please. Starting with what word ?

208. Zoe Phin says:

Pierre,

209. Pierre D. Bernier says:

I don’t read David Copperfield magic !

210. Zoe Phin says:

Pierre,
Really? Could’ve fooled me.

You believe upwelling IR coming up the device gets teleported to the sky and then comes down.

211. Pierre D. Bernier says:

@Zoe

Look at f**KNN diagram. I did not make it. There are 2 IR detectors. One on the ground looking up and the other 10m up a pole looking down. What the hell are they measuring. The speed I piss ?

212. Zoe Phin says:

Pierre,
Just remember, there is no heat plate, the sky cooks the eggs.

213. Pierre D. Bernier says:

Zoe,

You’re not answering the question. Typical leftist troll tactic. They have 2 IR detectors. What the hell are they measuring ? Downward IR ? Upward IR ? .Sideways IR ? Random IR ? What IR ?

214. All that IR detectors do is remotely measure a temperature. All that spectrographs do is measure a spectrum. Although they do it, neither of these should be interpreted as “forcing” or “energy budgets”. There is heat flow, which flows from hot to cold, and cold doesn’t make hot hotter still.

215. Pierre D. Bernier says:

@Joe,

The IR detectors, are they directional or not. If not, why 2 ? One upward looking, the other downward looking ?

216. Detectors are generally pointed in directions. Detecting that something has a temperature doesn’t equate to “forcing” or “energy budgets”.

217. Pierre D. Bernier says:

@Joe

So, IR detectors indicate a temperature. So a temperature at night is just that, a temperature. Not a rate of energy flow. If the minimum temperature in one location is the same night over night over night it does not mean energy accumulation or depletion. It simply means that the system is either in thermal equilibrium at that temperature at that time or that it had not time enough to reach it. It’s therefore a simple question of kinetics, not thermodinamics. If there where more energy in during the day or the system does not have the time to release it at night, then the minimum equilibrium temperature would rise, rising the rate of release, and it would still be a question of kinetics. Am I missing something ?

218. Joseph E Postma says:

You cannot say it is only kinetics and not thermodynamics in a situation which is by reference and definition a thermodynamic situation.

Yes you may be missing something. Your example is not useful. Temperature is always changing and for a temperature to be the same value at some instantaneous time at some location means very, very little, as in between the two days there would have of course been myriad changes of heat flow, inputs, outputs, temperature, physical conditions, etc.

What you’re describing is exactly how the fake GHE of alarmism is described, and so that indicates a pretty fundamental problem with wherever you’re going or whatever you’re trying to say. And as I said, the example you are using is not at all meaningful.

If all physical conditions were constant, and all physical responses were constant, and the scenario was physically trivial, then for a cycling-temperature which had been constant in pattern to now show a change, either to higher or lower temperature, would respectively indicate that either additional heat was input, or heat was taken away.

If however the heating pattern had no change, then a change in temperature could come from a change in emissivity of the material whose temperature is being measured, where lower emissivity would increase temperature, and vice-versa, which of course contradicts GHG-GHE theory given that GHG are emitters which should therefore cause cooling irrespective of “downwelling” or “back” IR since those do not carry heat.

A change in the thermal capacity of the system would also cause the cycling pattern to change.

You seem to be now arguing for the fake GHE of alarmism and its basic tenets about downwelling IR and “kinetics not thermodynamics”, etc.

Try clarifying and simplifying with non-convoluted example what you are trying to communicate or understand.

219. Pierre D. Bernier says:

@Joe
/ You seem to be now arguing for the fake GHE of alarmism and its basic tenets /

No. Not at all. I know that GHE is fake !. I wanted to ask without hurting anyone but since I have to… I’m just trying to understand Zoe’s figures which to me dont make sense…

Bondville 168 374
Desert Rock 236 441
Goodwin Creek 175 404
Penn State 141 366
Sioux Falls 155 352
Table Mtn 188 364

First #: Solar
2nd #: Upwelling IR

The Earth can not be emitting twice the energy it receives from the Sun. I find that you have been especially silent about that subject. So please explain those figures because I sure cant ! If they are real… then so be it (but the geologist’s, or who ever, have been doing a hell of a bad job !

220. The answer is that such averages are meaningless -> this has been the point here for a long time now. Clearly the atmosphere cannot provide twice the energy as the Sun…and so, those averages are physically meaningless.

221. And being meaningless, they are then also, by necessity, by sheer and simple logical and physical and empirical necessity, being interpreted incorrectly – because there is no correct way to interpret a meaningless thing other than simply regarding the meaningless thing as meaningless.

222. Pierre D. Bernier says:

@Joe

Thank you very much. Getting old is one thing. Getting senile is quite another. OUFF !

223. Zoe Phin says:

Pierre,
The Earth can emit infitely more energy than received by the Sun. See South Pole for 6 months.

Earth is a E°K disc
Sun is a S°K disc

Sun disc appears for 12 hrs then goes away for 12 hours. Temperatures are averaged for equilibrium.

12 hr Daylight avg: (S+E)/2
12 hr Night avg: E

Total average: ((S+E)/2 + E) / 2

We know S = ((1361/2*(1-0.3))/sigma)^0.25 = 302.75 °K

We know from observations that global near-surface temperature is 288°K.

Total average:
((S+E)/2 + E) / 2 = 288
0.5S + 1.5E = 576
1.5E= 576 – 0.5*302.75
E = 283.08 °K

Earth is a 10°C disc
Sun is a 30°C disc

224. Pierre D. Bernier says:

@Zoe

Good ! Go talk to Roy Spencer !

225. Zoe Phin says:

Pierre,
Spencer’s rectal walls prevent sound waves from reaching his ears.

Can you solve this problem?:

10°C [ 100mW/^2 ] —> R W/m^2

Solve for R. R is radiation to space.

[ ] : Conductive Material, where
Thot = 10°C
Q = 100 mW/m^2
K = 1
A = 1
d = 10
Absorbtivity & Emissivity = 1

226. Pierre D. Bernier says:

@Zoe

Why is it that you can’t acknowledge that N2 has no significant IR absorption band ? Look at your own spectra. N2 (dark green) has an insignificant band just above 4 microns.

Why is it that you can’t acknowledge that O2 has few and far between IR absorption band (dark blue) which are as thin as Twiggy was in the 1970’s ? Compare then to the water bands (light green). Insignificant !

Why do you pretend that N2 and O2 absorb IR when your own graph shows the absorption band of O3 (purple) when we know very well that O3 is found mainly above the O2 layer in the atmosphere ? Your graph shows the absorption bands for high altitudes where UV rays break everything down (proving my point) !!!

You say / O2 and N2 are very weak GHGs / . They are the strongest because they don’t emit IR (lose energy by IR), for the same reason they don’t absorb IR. Look at Hans’ post above, which I’m completely in agreement with / Oxygen and nitrogen are the true “greenhouse gases” in a real greenhouse as well as in the open atmosphere as those gases are the real “heat-trapping gases” due to their poor IR emissivity /. Why can’t you admit that ? Low emissivity means no IR emission for the same reason that they have no IR absorption.

Why is it that you can’t acknowledge that the Earth has a heat emission in the milli Watts/m2 ? It’s one big Zionist conspiracy I suppose ?

Why is it that you decry the SURFRAD setup and yet are ready to use it to promote your claim that Earth emits more energy then it receives ?

When a molecule in the atmosphere has a need to shed some energy via IR, why should it be shed upwards toward the stars ? Why not towards the Earth, at least some of the time ? Radiation does not mean back-radiation. If that IR is not hot enough it will simply be reflected back just like in Joe’s 5 chambers de Sausure’s set up. Why can’t you admit that ?

Every step of the way you have been posting false things exactly the opposite of what I had posted, whether in chemistry or spectroscopy, which proves to me that you have absolutely no idea of what you are talking about.

Stop pretending. Either you are a plant from the warmist side, a brat who can’t acknowledge anything but herself or someone who refuses to take her medication ! Can you at least answer that one ?

227. Sure seem to have a fixation with Zoe there Pierre. Strange.

228. CD Marshall says:

Joseph. LOL.
Where’s Robert K been? Haven’t seen his comments for a while. Is he well?

229. Zoe Phin says:

Pierre,

“Why is it that you can’t acknowledge that the Earth has a heat emission in the milli Watts/m2 ?”

10°C [ 100mW/^2 ] —> R W/m^2

Solve for R. R is radiation to space.

[ ] : Conductive Material, where
Thot = 10°C
Q = 100 mW/m^2
K = 1
A = 1
d = 10
Absorbtivity & Emissivity = 1

230. CD,

I’m alive and well, and I’ve moved on to JP’s most recent post. I had nothing to contribute to the Zoe-Pierre show. (^_^) — they were doing fine without me.

231. Mack says:

Gidday Joe ,
Here’s a little conversation I’ve had with this bloke David Arthur over at The Conversation. Because this is the No1 govt. funded blog for all the Universities, I have to go my real name, and curb my tongue and be very polite because they’ll cut your comments at the drop of hat….
https://theconversation.com/heatwave-completely-obliterated-the-record-for-europes-hottest-ever-june-119801#comment_1986691
Enjoy.

232. Joseph E Postma says:

What disgusting little weasels these academics are. I stopped reading shortly after he accused you of being condescending…when you simply used the exact phrase he first used on you! What pathetic little creeps these people are. And thanks though for the example of another one of them saying that INDEED sunshine falls spread over the entire surface area of the Earth at once!

These people are such little cowards and creeps. You were doing an excellent job with him! It’s nauseating having to read their replies. I will try to read the whole thread.

233. Joseph E Postma says:

They really don’t understand the simplest of things…for example, that an average is not the actual thing! That there’s a difference between the actual thing driving physics in real time, and an artificial average.

Why can’t they understand such simple things!?

I’m reading more of that thread…you’re excellent Mack! You certainly grasp the physics and the mathematics…more so than any of them!

It must be so embarrassing for them…

234. Joseph E Postma says:

Oh man…I love how he explains that the TOA insolation of 1370 W/m^2 isn’t felt on the dark-side of the Earth(!!!)…but then he thinks it is totally fine to mathematically spread that 1370 TO THE DARK SIDE OF THE EARTH as if this is what occurs “on average”! hahaha

*These people are not conscious.*

235. Zoe Phin says:

Mack,
That guy doesn’t realize that CO2’s extra modes is what raises its specific heat capacity.

http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/shegas.html

N2 – 20.7
O2 – 20.8
CO2 – 29.7

Every molecule of CO2 requires more energy to raise temperature.

236. CD Marshall says:

Let me introduce to you an expert in climate science:

“ahahahahah, solar heat radiation comes through the earths atmosphere, when it hits the earths surface it bounces off as re irradiated heat which co2 and other gases prevent from escaping back into space.”

This followed by a tangent insulting my intelligence.

237. Did you know that heat can bounce!? It’s quite elastic!

238. Many thanks Joe, It really is a nice feeling that this little band is on the right side of history. As I’ve said elsewhere, Einstein was right. human stupidity could be infinite.
Keep up the good work here.
Mack.
Sky Dragon Slayer’s Chief Public Relations Officer.

239. CD Marshall says:

“Chief Public Relations Officer.”

Joseph you could use one of those. Keep you from going Postma on people.
Ha ha…that is so hard to do with these arrogant brain dead trolls.