# Has the greenhouse effect been falsified?

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The greenhouse effect is standard physics and confirmed by observations. |

## Climate Myth...

Greenhouse effect has been falsified

"[T]he influence of so-called greenhouse gases on near-surface temperature - is not yet absolutely proven. In other words, there is as yet no incontrovertible proof either of the greenhouse effect, or its connection with alleged global warming.

This is no surprise, because in fact there is no such thing as the greenhouse effect: it is an impossibility. The statement that so-called greenhouse gases, especially CO2, contribute to near-surface atmospheric warming is in glaring contradiction to well-known physical laws relating to gas and vapour, as well as to general caloric theory.' (Heinz Thieme)

Some climate change skeptics dispute the so-called ‘greenhouse effect’, which keeps the surface temperature of the Earth approximately 33 degrees C warmer than it would be if there were no greenhouse gases in the atmosphere. In other words, without the greenhouse effect, the Earth would be largely uninhabitable.

How do we know for sure this effect is real? The principle is demonstrated through basic physics, because a bare rock orbiting the sun at the distance of the Earth should be far colder than the Earth actually is. The explanation for this observation was based on the work of John Tyndall, who discovered in 1859 that several gases, including carbon dioxide and water vapour, could trap heat. This was the first evidence for what we know now as greenhouse gases. Then, towards the end of the same century, a Swedish scientist named Svante Arrhenius proved the relationship between greenhouse gas concentrations and surface temperatures.

**Empirical Evidence for the Greenhouse Effect**

We only have to look to our moon for evidence of what the Earth might be like without an atmosphere that sustained the greenhouse effect. While the moon’s surface reaches 130 degrees C in direct sunlight at the equator (266 degrees F), when the sun ‘goes down’ on the moon, the temperature drops almost immediately, and plunges in several hours down to minus 110 degrees C (-166F).

Since the moon is virtually the same distance from the sun as we are, it is reasonable to ask why at night the Earth doesn’t get as cold as the moon. The answer is that, unlike the Earth, the moon has no water vapour or other greenhouse gases, because of course it has no atmosphere at all. Without our protective atmosphere and the greenhouse effect, the Earth would be as barren as our lifeless moon; without the heat trapped overnight in the atmosphere (and in the ground and oceans) our nights would be so cold that few plants or animals could survive even a single one.

The most conclusive evidence for the greenhouse effect – and the role CO2 plays – can be seen in data from the surface and from satellites. By comparing the Sun’s heat reaching the Earth with the heat leaving it, we can see that less long-wave radiation (heat) is leaving than arriving (and since the 1970s, that less and less radiation is leaving the Earth, as CO2 and equivalents build up). Since all radiation is measured by its wavelength, we can also see that the frequencies being trapped in the atmosphere are the same frequencies absorbed by greenhouse gases.

**Disputing that the greenhouse effect is real is to attempt to discredit centuries of science, laws of physics and direct observation. Without the greenhouse effect, we would not even be here to argue about it.**

Basic rebuttal written by GPWayne

**Update July 2015**:

Here is a related lecture-video from Denial101x - Making Sense of Climate Science Denial

Last updated on 7 July 2015 by MichaelK. View Archives

fake realityat 07:05 AM on 21 July, 2016@144 HKWhy is it so hard to understand the following implications of the points above?1 + 2: Each square metre of Earth receives 1361 / 4 ≈ 340 w/m2 of solar radiation before the albedo is accounted for.It´s not. But that is the wrong question. The right question is:

What is wrong with your model if it doesn´t replicate reality?

You are assuming that your model is right and reality does something that we can´t describe with radiation laws, so you fill out the gap with stuff you find in the atmosphere, like water, co2, photons etc.

Not one of them is known to have an intensifying effect on a body of higher temperature, but that is exactly what you use it for.

It´s not only your sun with intensity of 240W that is a problem, although it´s really funny, it´s the combination of darkness and stuff that we know as cooling elements in relation to hotter bodies, that is absolutely hilarious. I think you could not make a model that is more dysfunctional, it must be unique as leading theory, in it´s total lack of accuracy.

What you should ask is also: how can anyone buy into that when we know that a sun that doesn´t shine, at ~240K=240W, is not in any way a correct model of reality.

3. When albedo is accounted for, this number drops to 340 x (1-0.3) ≈ 240 w/m2. That radiation flux is what you get from a blackbody at 255 K, or 33 K colder than the real Earth.Why not try to build another model that is more correct. Am Iright in that the number 340 is K?

If so, I think you need to use εσ340^4, where ε=0.7

When you average before albedo(which can be ignored at this stage), you change the outcome pretty much. There is a very big difference between 1370W and 340W but not temperature K.

When we know that we need 1000W/m^2 to get 288K, why not use it?

Because you know better than earth, that radiates at the wrong temperature, and it must be the icecold atmosphere that heats the surface, the sun cannot be enough?

Stupid drunk earth, radiating all wrong so we have to fix it with icecold gas.

4. Even though the Earth is not a blackbody in the IR part of the spectrum, it’s sufficiently close to make the impact from emissivity very minor. An average surface emissivity of 0.96 would only raise the temperature by about 2.5 K on an airless Earth with evenly distributed insolation. The cooling impact from the uneven insolation is probably somewhat larger than that (it’s about 70 K at the Moon’s equator!), so there is at least 33 K of warming that has to be explained in other ways than by insolation, albedo or emissivity.But insulation has certain features that must be in place. One that is very important is som boundary surface or surfacelike structure opposing the warm body, placed at the outer limit of the insulated area, and works by conduction, so that it has to get heated and radiate at it´s own temperature. The gradient between the boundary and body should be as small as possible to get good insulation. Like the opposite of earth´s steep gradients that makes sure that all altitudes above surface is colder, much colder.

Nope, thats not insulation. No blanket, no sleeping bag and no sweater. Just cold air and 3K-vacuum of space.

The 33C is entirely a number that is the ugly child of your model with darkness and cold when the sun heats earth at 260W/m^2

fake realityat 07:24 AM on 21 July, 2016Checking if I am banned yet, last post dissappeared without a trace

Response:[JH] If you were banned, you would not be able to post the above comment.

Tom Curtisat 08:19 AM on 21 July, 2016HK @144, the surface emissivity shown is for IR wavelengths, whereas most of the incoming solar energy is in visible wavelengths. Based on Stephens et al (2015), all sky albedo is 0.293. From the listed flux values, clear sky albedo is 0.154 and clear sky albedo - atmospheric contribution is 0.059. Because of potential overlaps, that means the mean absorptivity of the Earth's surface, weighted for solar energy is between 0.846 and 0.941. 0.9 would be a reasonable approximate estimate. Overall the surface is better at reflecting visible radiation than IR radiation.

false reality's position clearly commits him to the view that the Earth absorbs more energy from the Sun than would be absorbed by a disc perendicular to the Sun placed very near the Earth such that the umbra of the disc at 1 AU equaled the Earth's radius (ie, the Earth was in the full shadow of the disc, and the full shadow of the disc and the full shadow of the Earth coincided). That view clearly contradicts conservation of energy. As false reality refuses to even entertain that fact, there is no more point in discussing the issue with him then there would be in discussing philosophy with a rattlesnake.

MA Rodgerat 10:11 AM on 21 July, 2016Fake Reality Man @150.

You must forgive me but I still find your abilities to comprehend the reality of climate as laughable as the turtle & elephants supporting your fat Earth. Okay, I am sure that engaging with you is something that will be entirely unfruitful but, hey-ho, I enjoy a good laugh.

I shall ignore the nonsense you spout @148, although that said, you do not make demarkation of nonsense entirely easy.

@149 there is in some respects more definition in your assertions You tell us

"that 1000W is the density that has to irradiate 1m^2, 1/2 the sphere. To get 288K surface mean temperature?"This appears to be your obsession. I agree that a thing with a temperature of X degrees Kelvin indeed does require heating by an energy flux of Y from"somewhere"that will allow it to emit the black body emissions (or roughly that amount) commenserate with the thing's temperature of X degrees Kelvin. The problem is that you are very very selective about the"somewhere".You seem to be unable to consider certain alternative"somewheres"such that you proclaim that the only source of heat for the surface of planet Earth has to be the sun. Yet, thinks you, if I put on an overcoat, I do not in any way don an energy source. It is nothing but inert cloth yet I am evidently warmer for the coat. The reason for my feeling less frosty is of course due to insulation. It is a concept you and yourfake realityseem unable to grasp.@149 you insist "

that 1000Wis the density that has to irradiate 1m^2, 1/2 the sphere. To get 288K surface mean temperature?"Assuming my abacus has not lost too many beads, a thing with a temperature of 288K, would be radiating 390Wm^-2. Further, I appreciate you see temperature & external heating as being the definng physics but, as I say, you are being very selective in what you consider to be allowable external heating sources for the surface of planet Earth.And when you state @150m that 1,000W/m^-2 has to be the peak value, I have to ask "Why?" What are you saying? The average has to be roughly (from all sources) 390Wm^-2. That could be 3,900Wm^-2 for 10% of the time, couldn't it? What is this obsession with 1,000Wm^-2?

You say

"We never heat anything up with a colder gas."Indeed. But if the gas is still cool (hey, Fonz gas!!) but warmer than it was, what then is the result? It is no longer a"colder gas!!!!"Even the elephants & turtle know it!! It is warmer."It us a hotter gas!!!!!"The only one here left denying that such cool gas is

"hotter"is our chumfake reality.So chum, defend your position. I look forward to you reply. As I said, I enjoy a good laugh.Response:[DB] Note that, being unable to adhere to the Comments Policy, "fake reality" has recused themselves from further participation in this venue.

Doubtless the physics of the Naetherworld will be more pliable for them.

HKat 22:23 PM on 21 July, 2016Tom @153:

Your numbers seem to agree pretty well with the energy flow chart in @137.

My reason for focusing on the IR part of the spectrum is the fact that an IR emissivity lower than 1 reduces the heat loss to space in a way that resembles the greenhouse effect. In a discussion on another blog (I don’t remember where) someone suggested that the 33 K warming of Earth was caused by a surface emissivity of about 0.61-0.62. If that was true, a surface temp of 288 K would be needed to emit 240 w/m

^{2}.But of course, measurements of many types of terrain show that the IR emissivity in general is > 0.95, so that doesn’t contribute much to the 33 K warming of Earth.

Tom Curtisat 07:01 AM on 25 July, 2016Just as an afterthought, one of the peculiarities of fake realities arguments is the insistence that the standard account of mean global insolation is that it represents a model with four cool suns because the formula for global mean insolation, TSI x (1- albedo)/4, contains a division by four. As he states @116:

The division by four is fully explained as the ratio between the area of a cross section of the Earth defined by the terminators, and the area of the Earth's surface. It requires no further physical explanation ... something so obvious as to not normally require stating.

The peculiar thing, and the reason for this tardy note is that fake realities model requires a division of insolation by two.

By his own reasoning, therefore, he is invoking two hot suns in his explanation of insolation. Not only is he glaringly wrong, but also outragiously inconsistent in his reasoning.C.Sheenat 22:16 PM on 17 September, 2016https://scienceofdoom.com/156. Tom Curtis at 07:01 AM on 25 July, 2016

You are right, it is correct to divide by four for an approximation of earth temperature as a blackbody. If one uses the irradiation combined with albedo, you get a correct energy balance.

That´s how we calculate what a body radiates to space. But I agree with fake reality that it should not be used as a surface temperature for solid mass. It is the temperature that a blackbody had when it is totally isentropic, the same temperature throughout it´s whole body, It should also be absorbing and emitting all radiation at it´s surface, which would be at the top of the atmosphere, or the point where 1370W-albedo is the mean flux.

Since we are dealing with a sphere, and it is only irradiated on half the surface area, and we know that convection and conduction dominates the surface exchange, we should use radiant energy density instead. It has the units J/m^3 and for a sphere we get that by dividing the fluxdensity with 4/3pi*r^2 instead of 4pi*r^2. It is done for all volumes in contact via surfaces, so for the earth surface it is done twice, once for the atmosphere and once for the solid surface.

In wikipedia we can read:

This article is about energy per unit volume.

Energy density is the amount of energy stored in a given system or region of space per unit volume or mass, though the latter is more accurately termed specific energy. Often only the useful or extractable energy is measured, which is to say that chemically inaccessible energy such as rest mass energy is ignored.[1] In cosmological and other general relativistic contexts, however, the energy densities considered are those that correspond to the elements of the stress–energy tensor and therefore do include mass energy as well as energy densities associated with the pressures described in the next paragraph.

Energy per unit volume has the same physical units as pressure, and in many circumstances is a synonym: for example, the energy density of a magnetic field may be expressed as (and behaves as) a physical pressure, and the energy required to compress a compressed gas a little more may be determined by multiplying the difference between the gas pressure and the external pressure by the change in volume. In short, pressure is a measure of the enthalpy per unit volume of a system. A pressure gradient has a potential to perform work on the surroundings by converting enthalpy until equilibrium is reached."

https://en.wikipedia.org/wiki/Energy_density

When circumstances is that the energy exchange is dominated by conduction and convection at a surface, this is the preferred way to address the energycontent of the solid surface, which is the cause of the surface temperature.

When dividing 1370W by 4/3 two times we get an absorbed amount of 770W/m^3. Like I pointed out earlier, earth only gets energy on half the surface, we need to distribute the energy absorbed to twice the volume of the sphere. That can be done in several ways, but dividing by two works good. Half the surfacemass is excited to a level equal to the mean total surface mass constantly. 2m^3 will radiate what is absorbed in 1m^3. That gives us 385W/m^3 radiated through 1m^2. Which is in line with observations.

Then we can find out what earth radiate to it´s outer shell from dividing by four, then we get the atmospheric window of 96W/m^2.

The effective temperature, the energy balance, is then 342W/m^2 for TOA-radiation. And the balance for the surface and atmosphere is 256W/m^2. The TOA-temperature we get from substracting 256W/m^2 from 385W/m^2, which becomes 128W/m^2. This is all done with simple geometry and such a simple solution seems to be correct. Especially considering what my wiki-link says, that it can be connected easily to pressure, magnetic field, enhtalpy etc. as well as W/m^2 and Kelvin. It accounts for mass, volume and the relative energy.

It seems that weather and climate are only products of geometrical functions and solar radiation.

I think your model of radiation, energy and the earth is lacking quite a bit.

Response:[DB] For those wishing to do so, one can also delve into the nitty-gritty details of the mathematics of the physics and chemistry of climate change at the website,

The Science of Doom.Tom Curtisat 08:12 AM on 18 September, 2016C Sheen @157:

1) The TSI is 1370 W/m^2, not 1370 Watts.

2) The formula for the volume of a sphere is 4/3*pi*r^3, not 4/3*pi*r^2.

3) That means the units of TSI/volume/volume (which is your purported formula) is W/m^2/m^3/m^3 which is W/m^8, ie, not the formula for power density at all.

Ergo your formula is in error, and the near coincidence between the purported value of your formula (385 W/m^8) and the surface upward IR flux (390 W/m^2) is coincidental and irrelevant. It provides no evidentiary support for your strange physics at all.4) Further, even were we dividing the solar flux by the volume of two spheres, the formula would not be Solar Flux/Volume

_{1}/Volume_{2}but Solar Flux/(Volume_{1}+Volume_{2}). Preserving your other assumptions that would have a value of 513.75 W/m^5, not 770.625 W/m^8 as is given by your formula. That value is, of course, coincidental to nothing of any note relating to surface flux, so your conclusion that your formula delivers a result "...is in line with observations" is entirely the result of the use of the wrong formula, even if we ignore the disparity of units.5) The only way to make the units to come out correctly is if we assume that you are determining the power density for a column, with a cross section of 1 meter squared at the solar zenith angle, and that the relevant volume is that of two spheres having a radius equal to (1/pi)^(1/3), or approx 0.683 meters. That radius is necessary for the volume to come out as being 4/3 m^3. Even so you will have used the wrong formula as per (4) above. Given this assumption,

6) Were we, despite all these flaws, actually motivated to calculate the power density of insolation in the atmosphere, the relevant volume would be the volume of a sphere with the radius of the Earth plus the height of the atmosphere minus the volume of a sphere with the radius of the Earth. Your two sphere approach is entirely wrong (as should have been evident by now in any event). Further the insolation would be 1370 W/m^2 times the area of a circle with the radius of the Earth.

Given that no climate scientist in the history of the science including several of the founders of the theory of thermodynamics has considered this approach, however, and that you have made such a dogs breakfast of attempting to do so, I see no reason to consider the approach further.

Contrary to the moderator, I see no reason why we should wish onto Science of Doom the refutation of a discussion so jam packed with fundamental errors of geometry.

Response:[DB] Inflammatory snipped.

sjabat 09:04 AM on 25 October, 2016158. Tom Curtis at 08:12 AM on 18 September, 2016

1. I think you understood what he means. It is all simplification and the surface area is just a sign beside a number of importance.

2.Dividing by 4/3pi*r^3 is used when calculating mass-energy density, mostly used for very large masses on a cosmological scale.

It can also be used for radiant energy density:

https://en.wikipedia.org/wiki/Radiant_energy_density

Unit is J/m^3 remembering that Joule per second is equal to Watt. Since radiation moves at light speed TSI can switch units to J/m^3.

I don´t now how you got the very strange unit of W/m^8, as W can just be switched for Joule without problem. It is r^3 that gives the cubic volume the same way that r^2 gives m^2 when calculating effective temperature.

You just extrude the square metre into a cubic metre and switch to Joule. Or, if you are uncomfortable with Joule there are several other names for W/m^3 like spectral exposure or spectral irradiance.

You can view TSI as the bottom of a cubic metre emitting through a square metre, no doubt there is enough energy in solar radiation to fill a cubic metre with an energy density of the same amount as TSI/m^2 in one second, so there is no problem to use W/m^3.

I think that using spherical geometry volume is just the right way to do it. You point at the shrinking volume compared to a cubic metre, and I think that is what makes it work. A squaremetre at the tropopause is represented by a smaller area at the surface, shrinking to a point on the way down to the core.

4. It apparently should be done like that. It seems to work perfectly when accounting for each layer showing what happens at each absorption, first the dry atmosphere and then solid surface and the watercircuit as one body. The number 513 is double the amount of the energy balance between the surface and atmosphere, which also is the mean temperature of the gradient in the troposphere 1027.5/4= 256.875W/m^2. So that number is traceable as well, but to what use? It is of no apparent use and the reason is you not understanding what you were doing.

Interestingly, as I showed, it is not random.

You claim that it is pure coincidence that he gets 770W for the flux of two square metres of the surface. I have looked into this model a bit deeper and found more "coincidences" among other very interesting details. So I will walk you through it and see if you still think that it is pure coincidence.

The effective temperature is 279 at a flux density 343W/m^2 using the whole TSI. 1370/4=342.5

The energy density at the surface is 1027.5J/m^3. 1370-1027.5=342.5(!)

The energy stored in the solid surface mass is 770J/m^3 and the energy balance between the energy density at the surface(1027J) and the following longwave part of the system including the solid surface and atmosphere is 1027/4=257W/m^2. 1027-770=257(!)

Now when absorbed and diffused from half the sphere throughout the entire solid volume, it is more appropriate to use surface flux because we want the surface temperature and the transfer rates through the atmosphere.

Surface flux density is 385W/m^2. Using the energy balance for the irradiation at the surface, 257W/m^2, we get the rate of heat transfer from surface to the troposhere, 385-257=128W/m^2. Which is equal to a flux density of TOA radiation at the temperature 218K(!).

The heat transfer from the atmosphere using energy balance of 257W/m^2 is 257-128=129W/m^2. The exact energy balance for surface irradiation is 256,875 and the exact surface flux is 385.3125.

Using the exact values we get a transfer to the troposphere mean of 256.875W/m^2 of 128,4375 from the surface, exactly half of the energy balance, which is also the intensity equal to TOA longwave flux density.(!!!)

So, when you said that the result 385W/m^2 of surface flux calculated this way by Mr.Sheen is a pure coincidence, exactly what did you mean?

The way he does it, nails the temperature at all relevant points in the surface-atmosphere system. And it balances perfectly leaving nothing left.

An interesting detail is the connection between what is lost in the process of absorption into a new spherical layer and the energy balance between irradiation and that volume. It implies that energy density can be treated

as opposing forces "the old fashioned newtonian way". The force of incoming radiation seems to be balanced exactly on absorption, only transferring the excess into the next layer. It makes me speculate about the absorption process as a macroscopic quantized mechanism.

Now, when presented to a more detailed model, accounting for all of TSI through the process of absorption and emission of both short and longwaves, arriving at values at each layer that is very close to observation and following through all the way out back to the system boundary of TOA, not leaving a single Watt left to be questioned, do you stand by that getting a correct surface flux was "pure coincidence"?

I think this has killed the GH-theory in a single stroke. If using solar radiation as mass-energy and making no distinction between them, treating earth as only a empty spherical multilayer canvas where solar energy is projected, it seems to account for every single watt in the system, putting them in the right places as well. This makes everything above the surface pure solar energy only, which of course is correct since it is the only source of energy present.

Irradiation accounts for both mass and energy at the same time. Making everything above the solid surface a pure product of E=m*c^2. Where all mass above surface is existing in space at squared lightspeed.

The sphere as the only factor explaining the energy inside the system in more detail than the GH-model, accounting for all energy and mass, means that nothing inside the system is acting, it is only reacting. Everything from temperature to albedo or glaciers is a product of energy density in relation to the electromagnetic field. The temperature cannot rise. Unless the sun increase the mass-energy density in the field where earth is positioned.

I think you just lost every bit of credibility and relevance that you imagined that you have in the discussion of climate, climate change and temperature. If you had avoided the aggressive and insulting attitude, and thoughtless throwing of words like "packed with flaws"and "a discussion so jam packed with fundamental errors of geometry", when it actually was you that was incapable of seeing the relationship that I spotted immediately when reading it, you would have spared yourself a lot of shame.

The thing is, the model performed better than GH-theory even when there only was "pure coincidence" that it got the surface flux right. Because that has been the problem all along. Gh-theory don´t even explain anything about the planet, since the effective temperature is the "blackbody", which is an isothermal body with the same temperature throughout, abosrbing and emitting at an infinately thin surface positioned at TOA. That is something that never will exist and the reason for using it as base for the distribution of energy in the system is unclear.

The model of GH-theory is a pure expression of misunderstanding all "science" included. It fails to represent century old concepts of radiation, electromagnetic field, temperature, energy density and heat transfer, claiming to stand on a base of physics.

Next time, remind yourself of carefully analyzing what conclusions that NOT can be drawn from the information you have. That was a big part of physics back when these concepts where discovered. It was probably the key to their great success.

To calculate effective temperature and then discovering the difference to observation, it is a bad idea to use that as a base for new conclusions.

You claimed in bold letters that Mr.Sheen`s formula was in error, but it is you that were in error. Mr.Sheen was exactly right. As you see, that shit is flawless and shiny perfection.

You should have continued to investigate why there was a error in your model instead of assuming that effective temperature is almighty. You only had to read the definition of a blackbody to realise that effective temperature says nothing of how hot the surface should be. It tells you the flux density that would be emitted at the tropopause if the earth had the same temperature through it´s entire volume, absorbing and emitting from a perfectly black infinately thin surface.

A model of what the opposite of earth would emit 10km above the surface was a bad choice for modeling the climate.

The comments about "pure coincidence", the flaws and implying your superior knowledge about geometry and units must seem like a bad idea now. Climate science seems to have kept you in place eating your humbleness and made you blind, leaving you with nothing left but shame.

You should at least have had a second look seeing if there were more "coincidences". I learned a long time ago that I don´t know everything, not even when I know everything.

Response:[JH] Inflamatory & argumentative statements stricken.

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Eclecticat 13:54 PM on 25 October, 2016sjab @159 , there is much in what you say.

The energy density at the

surface fluxinterface, presents a multi-dimensional challenge to conventional climate science ( as well as to physics as a whole ) .Difficulties arise at the flux capacitor site positioning - if the capacitors are placed close to Latitude 80 N , then there is a risk polar sea ice drift could bring

someflux capacitors to the exact positioning where Earth surface rotational speed is at88 mph. With catastrophic consequences. Dimensional distortion would permit heat energy to be projected into the future - with unknowable dangerous effects . . . or far worse, projection into the past, thus producing a runaway positive feedback over the past-present temporal loop. Our planet would become completely fluxed.[ Moderators, please feel very free to delete this post . . . if you should happen to decide to purge all threads of the recent sjabberwocky spam. ]

Tom Curtisat 15:29 PM on 25 October, 2016While part of me thinks that Eclectic's comment @160 is all the comment that sjab @159 deserves, nevertheless, here is the full fisking:

1) I'm sure I understood it to, but

in physics units matter. C. Sheen's cavalier approach to units made it worthwhile pointing out this instance. As we shall see, you are equally cavalier about units (sufficiently so that I suspect you are merely a sock puppet of C. Sheen.) A case in point comes from your response to (2) where you cannot see how TSI/volume/volume leads to units of W/m^8. To spell it out, TSI is measured in W/m^2. Dividing twice by volume is equivalent to multiplying the numerator twice by the units of volume, ie, m^3. Hence W/(m^2 x m^3 x m^3), hence W/m^8. Your claim that, "W can just be switched for Joule without problem" is equally troubling. Watts only become Joules if they are multiplied by a unit of time, or some comlex equation of units that simplifies to a unit of time. This is not just pedantry. It is the bastion against rampant nonsense and pseudoscience.If your units don't work out, your theory is bust(of which more later).2) Dividing by (4*pi*r^3)/3 only applies when you are determining the ratio relative to the volume of a sphere. As demonstrated at points (5) and (6) @158, the two sphere approach of C Sheen as modified in my point (4) also does not work. It was only introduced to demonstrate the irrelevancy of the result (of which more later). If you want to ressurect it, you need to specify what are the two spheres, and why are they important to the discussion. Absent the clear articulation of the reason for dividing by the volue of two spheres, you have no basis for the theory. Worse (and this should be obvious), if your formula is W/m^2 divided by a volume, your result will be in W/m^5, not in Joules/m^3. I did not think this was necessary to point out in point (4) @158 as that was solely to indicate the irrelevance of the number produced, but as you want to take that formula seriously, you skewered by the same logic that brought C Sheen's original formulation undone.

3) Oh, that's right, you had not response to point (3), and indeed repeat an equivalent mistake as noted above.

4) Before beginning on this, I would like to determine the radiant energy density of the incoming solar radiation at the Earth's surface on the sunlit side of the Earth. To begin with, the TSI is 1370 W/m^2. To bring that into units of J/m^3 we need to multiply by some factor having the units of seconds/meter; ie, the inverse of a velocity. That is, we need to divide by a velocity. The obvious velocity to use is that of light.

So, does that work? Imagine we have an incoming solar radiation 1360.5 W/m^2 striking a one square meter plate perpendicular to the incoming radiation. At any instant, the solar radiation that will fall on that plate over the next second is strung out over a one light second, or a 299792458 meter, column extending from that plate to towards the Sun. The energy density of solar radiation of any cubic meter within that column will then be (1360.5 W/m^2)/(299792458 m/s). (Note, for power density we would simply divide by 299792458 meters.) Hence the energy density of incoming solar radiation is 4.54 x 10^-6 Joules/meter cubed. Because the energy density is uniform it would be the same for all sunlit portions of the Earth, and half that averaged over the whole surface. (That, of course, ignores cloud albedo, and atmospheric absorption.)

Armed with this information, let's march through your "coincidences":

a) "1027.5/4= 256.875W/m^2" - "1027.5" is purported to be the energy density of incoming solar radiation, but is very far from it. Further, treated as an energy density, 1027.5/4 = 256.875 J/m^3, which is entirely irrelevant.

b) "The effective temperature is 279 at a flux density 343W/m^2 using the whole TSI. 1370/4=342.5" -

The effective temperature is the temperature of a black body having the same energy flux, and is consequently irrelevant to your example. The surface flux is actually 398 W/m^2, not 343 W/m^2 (see diagram below), and the TSI is 1360.5 W/m^2 at the last solar minimum, and just less than 1361.5 W/m^2 at the peak of the strongest recent solar maximums. Further, for mean energy density you divide by two, not four. Doing so, we find 680.25 W/m^2 is not coincidental with 398 W/m^2 (nor with the 342 W/m^2 back radiation).

c) "The energy density at the surface is 1027.5J/m^3. 1370-1027.5=342.5(!)" - Again, pay attention to units. 1370 W/m^2 - 1027.5 J/m^3 is gobbeldy-gook. You need to introduce a constant with units of m/s or s/m (depending on which side of the substraction it is used), but then it is entirely

ad hoc. That is, ignoring the egregious error in calculating the energy density.d) "Surface flux density is 385W/m^2". No, a density is a value per unit volume. Ergo the surface flux density is 385 398 W/m^2/c = 1.33 x 10^-6 W/m^3.

Given these massive errors in calculating the "coincidences", the rest of your discussion on point (4) is a baseless diatribe, and requires no further response.

Richard Lawsonat 23:44 PM on 9 October, 2018The Moon surface temperature is 130C during the day, falling to -110C in the lunar night, so the average surface temperature would be +20C, would it not? Which is hotter than our +15C, attributed to our GHE.

Is there some characteristic of the Moon temperature cycle that would account for this apparent contradiction?

Thanks for help with this.

MA Rodgerat 01:20 AM on 10 October, 2018Richard Lawson @162,

The average surface temperature of Earth is well known to be 15ºC but an equivalent value for the Moon is not readily available.

The usual statements you will find happily give max & min temperatures for the lunar equator which arguably could be averaged to give a rough lunar equatorial average temperature. The numbers you present (+130ºC -110ºC) appear to be such max min temperatures, although measurements from 2009 gave them as +120ºC to -130ºC making the Moon's equator colder than the Earth's equator which have a range +30ºC to +20ºC.

Since 2009, Williams et al (2017) have published (see their Fig 9a) zonal average temperatures through the lunar 'day'. Williams et al only state average temperatures for the equator (-57ºC) and poleward of 85º of latitude (-170ºC). To calculate an average for the whole Moon, taking all latitudes through a complete lunar 'day', the average for the Moon calculate out at -73ºC, a lot colder than Earth.

Lawrence Tenkmanat 09:35 AM on 10 May, 2020It sounds like the moon equator average temp is a lot colder than the mean of 130ºC and -110ºC (reference in 163). But the mean of those 2 numbers is not, not 20ºC (said in 162), it is 10ºC.

Thank you MA Rodger. The graph in Figure 9a mentioned in 163 shows that you can't just average max hot and max cold to get the average lunar temperature. One can see how fast a planet cools when there is no atmosphere to retain heat nocturnally. The overall temperature average is so much closer to the maximum cold temp, b/c it spends so much more time there. While it takes so much time to heat to max temp during the lunar day (it's only at this max for a short while), max cold temp is reached almost as soon as night falls and it stays at that max cold all night.

Lawrence Tenkmanat 09:48 AM on 10 May, 2020Of course, the moon getting so cold and staying so cold all night has a lot to do with the length of the lunar night too, and many more things I'm not an expert on.

Someone recently tried to tell me the greenhouse effect can't matter b/c the moon which has no CO2 gets hotter than the earth. But that graph was impressive how the temp seems to sit near the Nadir most of the night... and shows that the moon is WAY colder than the earth.

Thank you!