# Do we really have a “33 °C Greenhouse effect”

Written by Jan Zeman, Czech Technical University, Prague

The Wikipedia entry for the Greenhouse Gas Effect states:

“*If an ideal thermally conductive black-body was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% of the incoming sunlight, this idealized planet’s effective temperature (the temperature of a black-body that would emit the same amount of radiation) would be about −18 °C. The surface temperature of this hypothetical planet is 33 °C below Earth’s actual surface temperature of approximately 14 °C. The mechanism that produces this difference between the actual surface temperature and the effective temperature is due to the atmosphere and is known as the greenhouse effect.*”

The statement is almost completely untrue. For instance not even the math adds up: the difference between the two temperatures +14 °C and -18 °C is not 33 °C but 32 °C. But it is not important, what is important here is the fact that there’s not a difference of 33 °C, nor of 32 °C between the hypothetical and real Earth surface temperature. In short, there is clearly a confusion about what is meant scientifically when describing the “surface” of Earth.

I don’t want to rewrite astronomic customs, but for such purposes as a black-body radiation flux equation to and from the planet using the Stefan-Boltzman law, we would think the surface of the Earth should be considered to be “the atmosphere”- not the surfaces of the sea and land. The reason being that it is only the uppermost layer of the planet’s mass that is capable of radiation – in the sense as defined by the Stefan-Boltzman equation – unlike the boundary of the vacuum of space beyond.

This confusion is a result of our human perspective. In the case of big gas planets like Jupiter we observe from the outside and hardly anybody would suggest the immediate exterior of its uncertain small diameter core was the “surface.”

Indeed, there’s an even stranger boundary custom to consider whereby we could discern the “surface” and atmosphere arbitrarily just at the point where Jupiter’s immense atmospheric pressures crosses 10 bar. Nonetheless, when talking about the Stefan-Boltzman law (i.e. about black-body radiation) as applied to Earth and it’s radiation budget, we should consider the gaseous atmosphere as being the Earth’s surface, not the actual surfaces of sea and land below.

**The average temperature of Earth’s surface**

The surface of Earth’s gaseous atmosphere has an ‘average’ temperature of about −18 °C or even a bit lower. The simplified calculation using data from standard ISA atmospheric model you can find here and although it is still rather a very raw estimation – with maybe over one degree uncertainty and given the imperfection of the standard atmospheric model – the average value resulting from it comes out at about 255 Kelvin or lower. The actual value coming from the simplest calculation using averaging over 1,000 meter steps over the lower 50,000 meters of the atmosphere (99.9% of its mass) is 255.024 Kelvin (–18.13 °C) and it looks even lower if we refine the averaging using a larger sample with a 500 meters step – the resulting average temperature value is then 253.86 Kelvin (-19.29 °C).

So 255 Kelvin is the temperature which usually occurs at 5,000-5,500 m above sea level in the standard atmosphere – which is also the point where the atmosphere has about half pressure, which also more or less half the atmospheric mass; half above, half below this altitudes. But remember, it is an ISA atmospheric model average, in the real atmosphere it varies with multiple factors, mainly with latitude.

From the Stefan-Boltzman law we can calculate that for a black-body which has say the <=255 Kelvin average temperature, the radiation of such a black-body would be:

0.0000000567*255^4 <= **239.****76**** W **per square meter [Eq. 1]

Now, let’s compare the figure to the average absorbed solar irradiance (measured since the beginning of the SORCE satellite experiment in 2003 using TIM instrumentation). The total solar irradiance at the top of the atmosphere was about 1361.1 W **±0.47 **per square meter at 1 AU (TIM satellite data here).

Therefore **if we assume that the 30% reflection figure from the Wikipedia statement is exactly true**, then the solar irradiance of one square meter of the Earth surface (atmosphere) which isn’t immediately reflected to space is:

1361.1 / 4 x 0.7 = **238.19 W** (**±0.08) **per square meter [Eq.2]

The resulting value is clearly even lower than what comes as the maximum from Eq. 1. It could suggest that the average incoming solar radiation flux per square meter, since the beginning of the SORCE-TIM operation in 2003, was even a bit lower (1.57 W/m^2) than what the atmosphere would in average radiate according to its average temperature coming from the standard atmospheric model and Stefan-Boltzman law. This would appear consistent with the global warming hiatus during the period.

But not so fast. The value of the resulting incoming radiation is very sensitive to the exactitude of the “30%” reflection (albedo) figure – which very much looks being a rounded rough estimation – so we can’t jump to the above conclusion until we know the value for sure and to several decimal places.*

But what we can anyway conclude with sufficient certainty is that the surface body of the real Earth – the atmosphere – behaves more or less the way Stefan-Boltzman law predicts for a black-body and so it very much looks there’s in fact by far not a “33 °C” difference as purported by the Wikipedia statement. In other words the Earth surface temperature – the average temperature of the atmospheric mass – is relatively close what the Stefan-Boltzman law predicts for such a body.

The 33 °C difference between the so called global surface air temperature and the average atmosphere temperature is most probably given by another physical atmospheric phenomenon rather than a “Greenhouse effect” and this phenomenon is called the moist adiabatic lapse rate. And this graph below, describes the phenomenon more or less confirming the assumption: **Fig. 1**

**The moist adiabatic lapse rate**

The moist adiabatic lapse rate phenomenon is co-caused by considerable atmospheric water content, carrying as water vapor (being significantly lighter than air it is buoyant and rises quickly). It has a considerable amount of latent heat of vaporization, transporting this heat from sea and land surfaces where it is evaporated to the higher levels of the atmosphere. I’ll add that on its way up it is further heated by the sun during the day before condensing. But it is not so important, what is important is the considerable heat leaving the sea and land surfaces.

What needs to be immediately emphasized here is the fact that the latent heat of vaporization doesn’t add the slightest bit to radiation and only when this latent heat is again released during condensation it changes the temperature of the surrounding air at the places where it was released.

The evaporation and evapo-transpiration of heat transport from sea and land surfaces up the atmosphere works much like a compressor-less refrigerator or heat pump (using solar energy and gravity as the energy source and “thermostat”).

It is estimated that at least 505,000 cubic kilometers of water falls as precipitation each year, which is so much water it would cover the whole Earth to a depth of one meter. And this water must be evaporated each year from the sea and land surfaces. Given the water latent heat of vaporization (2.26×10^{6} J/kg) it means that the evaporation of water from sea and land surfaces transports very considerable amount of heat from the sea and land surfaces up to the higher layers of the atmosphere. There the vapor condenses, releases the latent heat, forms clouds and eventually falls back to the ground and into the sea as rain and other forms of condensed liquid or solid (snow) water. We can estimate the amount of heat transported this way by a calculation:

5.05×10^{17} kg x 2.26×10^{6} J = **1.14×10**^{24} Joules [Eq.3]

(~70.8 W.m^{-2} average energy flux from the sea and land surfaces as latent heat – good to note that this figure is still significantly lower than the estimation made in the so called Kiehl-Trenberth energy budget.)

Let’s now compare this figure to an estimation of how much heat Earth’s surface receives yearly on average from the sun according to the Eq.2. Result:

238.19W x 5.10072×10^{14} Joules x 31556926 [seconds in year] = **3.83×10**^{24} Joules [Eq.4]

The comparison shows that about 30% of the total heat converted from the solar shortwave radiation extinction is transported as latent heat via evaporation away from the sea and land surfaces somewhere high in the atmosphere. There it again condenses and releases its latent heat to the atmosphere.

This atmospheric water phenomenon dramatically changes the adiabatic lapse rate from the dry adiabatic lapse rate (~9.8 °C/km) to moist adiabatic lapse rate (varying with air temperature and humidity ~3-9 °C/km) The ISA standard value of adiabatic lapse rate used for aviation is 6.5 °C/km, which is almost exactly the 33 °C difference between sea level mean surface air temperature and the temperature at altitude; where the atmosphere has half of the sea level pressure and where the average temperature of the atmosphere occurs (5 to 5.5 km).

The water vapor as it ascends changes atmospheric temperature profile considerably, and as we have seen in the above graph, it is then more or less consistent with the temperature difference between the average temperature of the atmospheric mass and the average surface air temperature.

**Stefan-Boltzman law and sea**

After showing that the Earth’s atmosphere as a whole behaves close to how the Stefan-Boltzman law predicts for a black-body we can now look how it is with the sea.

It very much looks like the sea surface temperature anomaly is quite intimately linked with the surface air temperature anomaly. Although they seemed to diverge since the 1970s with the warming up to some ~0.15 °C in the mid 2000s and since then they seem again to converge with the cooling by ~0.1 °C. It would seem intuitive to assume the temperature of the air drives the temperature of the sea surface, but it is definitely not the case and the opposite is true.

Why? Because the average absolute sea surface temperature (~17 °C) is considerably higher than the average absolute surface air temperature (~14 °C). Therefore the sea must warm the air and not vice versa.

And not just due to the special 2^{nd} thermodynamic law concerning heat, which says that heat can’t move from the colder body to the warmer one. But simply because the warmer body always releases more energy per section (say the square meter) than a colder one. Athough, as we have already seen, the sea doesn’t release the energy just by the mid-IR radiation resulting from its temperature, but to a considerable extent by evaporation/latent heat transport. **Generally one should assume that the sum of all energy form fluxes released per section, from a warmer body to colder bodies or space surrounding it, should be equal to the radiation energy per section given by the warmer body temperature as the Stefan-Boltzman law predicts. **At least until the Second Law of thermodynamics is falsified, which seems unlikely.

Question: What then warms the sea surface if not air? Answer: mainly the sun.

Question: Can’t the atmosphere warm the sea surface instead of the sun by so called atmospheric backradiation? Answer: No.

Why? Besides the fact that the air is on average colder than the sea, the other reason is that the water is extremely opaque to the long-wave mid-IR spectra (at which only the atmosphere can radiate given its average temperature**); when compared to the shortwave solar spectra. This is in order of million of times as this graph shows (mind the y axis is logarithmic): **Fig 2**

99% of the mid-IR spectra at which the atmosphere radiates is absorbed in the uppermost ~30 micrometers of the sea surface skin and cannot significantly penetrate and warm the water any deeper. Therefore it chiefly contributes not to the sea surface warming, but to the evaporation from the sea surface ‘skin’ we were talking about in the previous section.

On the other hand over 50% of the shortwave solar radiation spectra which penetrated the sea ‘skin’ still gets deeper than 20 meters and still over 5% gets deeper than 100 meters. Unlike for the mid-IR the water is exceptionally transparent to the solar spectra – as we’ve already seen at Fig.2.

How the solar spectra penetrates the sea surface is also suggested by this graph:

The solar radiation still significantly penetrates the sea to the depth of over 100 meters. This is an order of magnitude million of times deeper than the atmospheric mid-IR radiation. It becomes extinct on the way – converted to heat which warms the sea water throughout the so called epipelagic sea surface layer, much, much deeper than it can be warmed by the atmospheric mid-IR radiation. Meanwhile, the ocean, if not covered by ice, has extremely low average albedo of ~0.033 (- value for zero waviness calculated here – for wavy ocean it could be even slightly less due to refractive properties of the air/water interface if wavy, but I omit it). This means that over 96% of the solar radiation reaching the sea surface gets below the surface, becomes extinct/converted to heat in the surface layer and then warms it. It can be estimated that about 90% of the solar radiation reaching sea and land surfaces is combined and converted to heat in the oceans and seas. This is despite the fact the oceans and seas cover only about two thirds of the Earth. But such estimations are not trivial so I will not go into the details as they are not the subject of this article.

**A “Greenhouse effect“ in ocean instead of atmosphere?**

But what happens with the heat in the sea?

This is even more interesting. Because most of the sea surface is warmer than 3.98 °C and the warmer water being lighter stays at the surface, the temperature of the sea surface layer is on average considerably higher (~17 °C) than sea water in the depths (~4 °C). If we go back to the Stefan-Boltzman law the water in the surface layer of the ocean body on average radiates:

0,0000000567*290.15^4 = 401.86 W per square meter [Eq.5.]

Now, this number appears much higher than the number resulting from the Eq.2. But does the ocean really radiate the 401.86 W per square meter out to the atmosphere as mid-IR radiation?

No, because the ocean surface properties are contrary to the atmospheric properties – very considerably different than the properties of a black-body to which the Stefan-Boltzman law applies.

While the ocean is exceptionally transparent to the incoming solar radiation, as we have seen it is much harder for the mid-IR spectra resulting from the sea surface layer temperature to get out. **

But it is not just because the sea water is extremely opaque to the mid-IR spectra allowing the mid-IR photons to travel in it at just several dozens of micrometers. It is chiefly because the refractive properties of the water/air interface raise the reflectivity away from the water dramatically, with the rising temperature and incidence angles. This is depending on the sea’s surface ‘skin’ temperature – which changes the water radiation spectra up to the radiation spectrum with peak wavelength 7.765 μm (water radiation spectrum of near water boiling point) – the water/air interface at average sea surface temperature reflects typically ~30% and with rising surface skin temperature under insolation up to 40%. This calculation is by using standard Fresnel reflectivity model data here. It is of the mid-IR spectra back to the sea, adding to the simultaneous solar irradiance flux and the atmospheric mid-IR radiation flux and facilitating water evaporation from the sea surface ‘skin’ we have already talked about.

But to evaporate water which has on average the 17 °C and sea level pressure one needs not only 2.26×10^6 J/kg but also the energy to heat the water to the temperature at which it evaporates. This energy can be calculated as:

4.1813 x 1000 x (100-17) = 3.47047×10^{5} J/kg [Eq.6.]

and if we go back to the Eq.3 it means at least:

5.05×10^{17} kg x 3.47047×10^{5} J = 1.75×10^{23} Joules of additional heat which must be taken from the water body for the evaporation to take place.

(This then means at least another ~10.9 W.m^{-2} on average is transported from the sea and land surfaces upwards.)

Now we see that evaporation and evapo-transpiration takes from the sea and land up to the atmosphere at least 81.7 W per square meter, which is already 1.7 W.m^{-2} higher number than the Trenberth et. al latent heat estimation. Moreover this value together with mid-IR radiation from the sea would raise the water temperature – which rose according to the sea surface anomaly rise in 20th century ~0.65 °C, most probably due to the rising solar activity.

Can we estimate how much this 0.65 °C sea surface anomaly rise would add to the surface radiation+evaporation energy budget?

[0,00000005670373*290,15^4=]401.885 – [0,00000005670373*(290,15-0.65)^4=] 398.296 = 3.589 W per square meter [Eq.7.]

Which for whole the Earth surface would mean surface air temperature forcing of:

3.589 x (36113200 / 51007200) = 2.54 W.m^{-2} [Eq.8.]

The larger part of it is in direct mid-IR radiation (~~70% – 1.78 W.m^{-2}) – warming first the air immediately above the sea – and in smaller part by latent heat of vaporization (~~30% – 0.76 W.m^{-2}) – transported up warming the atmosphere where the vapor condenses.

Now we see that the rising surface air temperature forcing was due to the raised sea surface temperature, caused chiefly by the rising TSI trend. It is not due to CO2 emissions enhancing any GHE in the atmosphere. And even if something like that was happening, it cannot significantly affect the sea surface temperature. This is because the bulk of the atmospheric mid-IR radiation is absorbed in the very surface ‘skin’ of the sea and immediately effects surface water evaporation, not to the epipelagic sea layer warming.

**Closing remarks: What is more likely in the future: Ice age or a catastrophic global warming?**

Now, we have the Sun, which releases radiation, this then travels through space and then hits our Earth. The minor part is reflected, the larger part absorbed and then re-radiated as it causes the temperature to rise intermittently.

But because the physical body energy exchange with the surrounding bodies or space rises by the fourth power of the body temperature, the radiation of the body rises dramatically. By rule of thumb this is by 5.5W.m^{-2} with every degree of temperature rise. So how on Earth can anyone claim 1.8 to 4 degrees temperature rise in a hundred years? – For such a feat we would need 9.9-22 W.m^{-2} additional forcing to overcome the Stefan-Boltzman law. As we see, such forcing would about equate to the energyneeded to make the water cycle on Earth possible.

So, what makes the average surface air temperature appear ~33°C higher is a forcing of:

[0,00000005670373*288,15^4=] 390.92 – [0,00000005670373*255^4=] 239.76 = 151 W.m^{-2} [Eq.9.]

The vast majority of this forcing originates from Sun heating the epipelagic layer of the ocean. This then heats the atmosphere by direct mid-IR radiation, latent heat transportation system, and then by convection and heat conduction. It is also why rising sea surface temperatures, due to rising insolation, caused the period of late 20^{th} century warming.

It could be estimated that for triggering a runaway glaciation and ice age is sufficient a forcing of minus couple of Watts (up to 10 W.m^{-2}), which will inevitably occur when (northern) summer solstice will get in the phase with Earth perihelion due to the slow Earth axial precession. The ocean absorbing the bulk of the incoming solar energy reaching sea and land surfaces (which wasn’t immediately reflected due to albedo or absorbed by atmosphere), the ocean, bulk of which is at the southern hemisphere, would become less insolated due to the Earth’s tilt/orbit unfavorable phase.

It will inevitably happen some time up to ~10000 years from now. And it will be paradoxically the warm ocean which will help trigger the ice age – it will still produce wast amounts of atmospheric water content by evaporation, which will then precipitate more and more as snow at higher latitudes, more and more will get deposited at lower and lower latitudes and not melted away during the melt seasons, changing the Earth’s albedo considerably and expedite the ice age triggering process with descending solar forcing due to phasing of southern summer with Earth’s aphelion. (The difference between TSI in aphelion and perihelion – most distant and closest points to Sun during the Earth year’s orbit cycle – is almost 100 W per square meter). The snow and ice albedo rise in lower latitudes is a powerful positive feedback for ice-age triggering. There’s nothing like that for warming, because even if the ice recedes in polar regions, it recedes to higher and higher latitudes, where the insolation due to Earth’s curvature is less and less significant. Even if all sea-ice and ice-sheets would melt – which even at current rates (if they would last) is impossible happen sooner than in several dozens of thousands years – it would not trigger any runaway warming, simply because it would not add sufficient forcing not nor any crucial positive feedback.

So what humankind should really once worry about (not now) is a global cooling, not the warming. Due to the Stefan-Boltzman law a runaway global warming which people as James Hansen threaten us with using analogies with Venus (much closer to the sun and receiving almost double TSI) getting so much attention by inept media and politicians is physically patently impossible without major external forcing source (e.g. ever-rising solar activity at the end of stellar life-cycle), because the energy radiation of any physical body rises with fourth power of its temperature and for every degree of the global temperature anomaly rise one would need an additional forcing of ~5.5 W.m^{-2}. Such forcing patently cannot be caused by the minute amounts of CO2 atmospheric content. Do we really believe man is so powerful that we can avoid ice ages just by burning ‘fossil fuels’?

—————————————

* – if the Wikipedia figure would be not 30%, but say 29.54% then the result of equation would be: 1361.1/4*0.703449 = 239.76 W/m^2 – same figure as the theoretical outgoing radiation of the atmosphere based on Stefan-Boltzman law and the average atmosphere temperature, which as we seen is rather the upper bound estimation.

The Earth Bond albedo which in fact the Wikipedia entry figure comes from is listed with values ranging from 28.1 to 33.8 (-in the Trenberth et al. 2009 they apply value 29.8 which would result in 238.87 W/m^2 incoming ASR according to the TSI value measured by SORCE-TIM instrument (surely way better TSI measurement device one generation beyond what Trenberth is referring to) and the resulting planetary energy imbalance would in fact come out as negative – Earth radiating more energy than it receives from sun, which would be consistent with the slightly cooling trend in the last decade, but the uncertainties are so high that we can’t really tell) Given the sensitivity of the Eq. 2 outcome to exactness of the Bond albedo value and the uncertainty of the Earth atmosphere average temperature (which is extremely difficult to really measure with sufficient exactness) we cannot decide whether the Earth radiation budget is positive, negative or neutral until we know the Bond albedo value for sure and with sufficient exactitude.

The current state of our knowledge about this things which is rather poor clearly doesn’t warrant any political action and mitigation spending of any scale and the fact that UN keeps attempting to impose them despite this crucial uncertainties tells more about their desire for power and our money than anything about climate.

** – according to Wien’s displacement law the peak wavelength of atmospheric radiation spectra is:

2897.7685 μm.K / 255 K = 11.36 μm. For such spectrum the the water is order of million of times more opaque than for the solar spectrum as the Fig. 2 shows.

For the sea surface radiation the peak wavelength is 2897.7685 μm.K / 290 K = 9.98 μm

———

Jan Zeman is Associate Professor at the Czech Technical University, Prague

Tags: adiabatic lapse rate, greenhouse gas effect, Jan Zeman, latent heat, solar insolation, Stefan-Boltzman

## A.Rappaport

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[quote name=”Bob Armstrong”](continued)

The atmosphere is of course highly transparent in the visible wavelengths in particular . Thus most heating occurs at the surface — or clouds . However , in all those wavelengths where the atmosphere is opaque , including those where CO2’s [i]AE[/i] is near 1 , it radiates essentially as a black body . It actually contributes to the emissivity in the IR which pulls the temperature as observed from outside down to the 255k range . Otherwise , that portion of the spectrum would be determined by the surface beneath . Were the earth’s radiative balance totally determined by an opaque layer at the top of the atmosphere , there would be no [i]lapse rate[/i] because by the divergence theorem there there would be no temperature gradient. [/quote]

Please clarify your use of the divergence theorem. Lapse rate is a property of the gravitational field and is thermostatic. It hasnothing to do with flux.Uour use of blackbody vs. greybody confuses me. Both implya uniformity of emissivity with wavelength. The Sun, Eart, and space all have emissivities that vaty with wavelength and also vaty with t6he angle from surface normal. There also seems to be a conflict as to just which Kirchhoff’s law of thermal radiation to use for a givem geometry. One states that for any body with no other heat transfer will be in thermodynamic equilibrium only when emission equals absorbtion, with no mention of emissivity. An other law states that surface emissivity equals surface absorptivt at each wavelength and at each direction, with no mention of thermal equilibrium. The Earth is never in thermoddynamic equilibrium because of its thermal mass and its perodic geometry relation to its primaty.

## Bob Armstrong

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With all the heat coming from the top , there would be no source of heat “at the bottom” to cause a misalignment of density with position . I believe this phenomenon has been studied and experimentally verified .

The lack of understanding of why I reference the equations for black and gray bodies indicates to me the distance this field has strayed from the classic analytical methods of physics . These are just the most fundamental relationships needed to construct a planetary thermal model . The expressions presented are necessarily in terms of black and gray bodies be .

cause those are the foundations . The difference between the executable expressions I presented and the SB equations presented by Curt are that mine are for full actual spectra , not some scalar average [i]ae[/i] values .

A couple of years ago , My niece gave me a copy of her undergraduate Griffiths’ Electrodynamics text her dog had chewed up . While somewhat more complicated than heat transfer , such texts go thru the quantitative basics every which way one can imagine requiring and presenting a foundation in vector calculus and functional analysis .

If you know of any such rigorous quantitative analysis of planetary temperature , point me to it . Given the importance the issue has taken on , it needs to be made accessible on the web .

In other branches of applied physics things like the fact that a conservative property like energy equals its equilibrium value over a cycle are proven at the undergraduate level . I get the feeling “climate Scientists” think they can get by with just a social science level of quantitative education .

## Bob Armstrong

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(continued)

The atmosphere is of course highly transparent in the visible wavelengths in particular . Thus most heating occurs at the surface — or clouds . However , in all those wavelengths where the atmosphere is opaque , including those where CO2’s [i]AE[/i] is near 1 , it radiates essentially as a black body . It actually contributes to the emissivity in the IR which pulls the temperature as observed from outside down to the 255k range . Otherwise , that portion of the spectrum would be determined by the surface beneath . Were the earth’s radiative balance totally determined by an opaque layer at the top of the atmosphere , there would be no [i]lapse rate[/i] because by the divergence theorem there there would be no temperature gradient .

I’ve already gone down the path of computing the radiative balance for non uniform gray balls at [url]http://cosy.com/Science/TemperatureOfGrayBalls.htm[/url] . Melding a couple of expressions from there to map these spectral data and functions to the sphere would permit realistic spectral maps of the surface which is necessary to determine the delta created by the atmosphere . The next step is to apply another [i]outer product[/i] to split the atmosphere into voxels covering layers of interest .

So , in half a page or so of definitions , you have the foundation of a pretty serious , yet succinct and well factored enough to be understandable , planetary model .

The entire change in observed temperature since the invention of the steam engine is about 1 part in 300 . I think we’ll only get to the 3rd decimal place understanding of our global equilibrium temperature when into account at least latitudinal difference in surface reflectivity

## Bob Armstrong

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This has turned into a very interesting and informative discussion .

I’m really just an Array Programming Language programmer who only feels he understands something if he can compute it . I got diverted into all this because of the endless non-quantitative word waving and the incredibly amateurish level of understanding of the most basic math and physics I saw displayed on both sides of the debate . The incessant parroting of the 255k value , produced by a step function splitting at the crossing point of the incoming solar spectrum and the earth’s thermal spectrum , [img]http://cosy.com/Science/HypotheticalSpectra.jpg[/img] , as a [i]black body[/i] temperature struck and strikes me as one of the great retardations of the field . It confounds spectrum with total energy density and obscures the fact that any gray body comes to the the SB temperature corresponding to that density . That NASA presents these bogus numbers becomes patently useless when applied to Venus : [url]http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html[/url] .

APLs were originally evolved from the notation for matrix algebra to express applied mathematics for scientific use , but their main market turned out to be complex financial applications . Look at [url]http://kx.com/[/url] the variant I still use for this work to see what I mean .

In general variables can be entire lists or lists of lists of values .

Thus the SB function , written simply as

[i]T2Psb : {[ T ] sb * T ^ 4 }[/i] .

For a gray body we would multiply by an [i]absorptivity=emissivity[/i] constant :

[i]T2Pgray : {[ ae ; T ] ae * T2Psb T }[/i] .

See [url]http://cosy.com/Science/ColoredBalls.html[/url] for the definitions of the handful of [i]verbs[/i] necessary to calculate the heat transfer between pairs of any particular spectra . For these calculations you can’t just multiply two scalar [i]ae[/i] values for source and sink . You need to go beyond T2Psb to the full Planck thermal spectra weighted by the absorptivity=emissivity [i]AE[/i] spectrum between the two bodies . Whichever direction the flow is , the [i]AE[/i] act like symmetric filters . There respective [i]ae[/i] at each wavelength multiply .

Thus for 2 bodies , eg , the earth and sun , to be in radiative balance

[i]( +/ ( AEs * AEe ) * Planck[ WL ; Ts ] ) = ( +/ Planck[ WL ; Te ] )[/i]

or

[i] 0 = +/ ( AEs * Planck[ WL ; Ts ] ) – ( AEe * Planck[ WL ; Te ] )[/i]

where [i]WL[/i] is a regular sample of the interval over which the Planck functions for either of the 2 temperatures have significant values . That’s essentially all the code that is necessary for the calculation for uniformly colored opaque balls where

[i] reflectivity : {[ AE ] 1 – AE } [/i]

(continued)

## Don Tonaronton

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Sunsettommy, thos os so weak and revealing.

Are you really that incompetent?

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## Don Tonaronton

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Sunsettommy, thos os so weak and revealing.

Are you really that incompetent?

## Sunsettommy

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[quote name=”Don Tonaronton”]Sunsettommy, thos os so weak and revealing.

Are you really that incompetent?[/quote]

There was a bug in the software that was preventing me from banning you successfully but now that the problem has been identified…………….

## Curt

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The idea that the top-of-atmosphere radiation to space matches (on average) the incoming solar radiation, and the surface temperature is then warmer than this due to the lapse rate is an excellent presentation of the standard explanation of the greenhouse effect.

Well done!

(If the atmosphere were not radiatively active, it would be the surface that is radiating to balance the incoming solar radiation, and it would be at about 255K.)

## Bob Armstrong

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“(If the atmosphere were not radiatively active, it would be the surface that is radiating to balance the incoming solar radiation, and it would be at about 255K.)”

nonscience .

## Curt

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Conservation of energy (1st law of thermodynamics) is nonscience? Wow!

And you just did the same calculations, just with a different albedo…

## Bob Armstrong

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Show me your computations .

As I have repeated numerous times , for any gray ( flat spectrum ) ball , however black or white , its equilibrium temperature , as can be calculated thru several different approaches , is about 279k .

## Curt

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Your assumption of a “flat spectrum” ball, with the same absorptivity/emissivity over all wavelengths, does not come close to modeling the real world.

The earth’s “bond albedo”, that is, over the wavelengths of the sun’s spectrum, is about 0.3, yielding the 0.7 value for absorptivity/emissivity you use in your calculations. But most of that albedo is due to clouds reflecting a significant fraction of the sun’s radiation back to space.

The emissivity of the earth’s surface over the wavelengths corresponding to its thermal radiation is about 0.95. You mock “alarmists” for (sometimes) rounding this up to 1.0, but that is a far better approximation than the value of 0.7 that you use.

## Bob Armstrong

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Did you have [i]any[/i] quantitative physics in school ?

Is a [i]Black Body[/i] which definitionally has an absorptivity=emissivity coefficient of 1 [i]across the entire spectrum[/i] “come close to modeling the real world.”

I find your post incoherent . Work thru some of the links I’ve presented and then you may have something rational to contribute . But I doubt your brain is capable of tying together the many threads required to grok the math .

Work thru my

## Curt

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Your objections make no sense. I was simply pointing out that your “flat spectrum” model, with constant absorptivity/emissivity across all wavelengths, whether a blackbody (1.0) or your graybody (0.7) is not a good model for the earth’s behavior. I have been through your analysis, and found it wanting.

I will repeat what I said before. The earth’s Bond albedo is approximately 0.3, which says the earth absorbs about 70% of the sun’s radiant energy that it intercepts. We know this to with a few percent from real measurements.

The earth’s surface emissivity in the wavelengths of the thermal radiation that it emits is about 0.95 – that is 95% of what a blackbody of this temperature would emit. We also know this to within a few percent from real measurements.

So, quantitatively speaking, using separate values for the two ranges of wavelengths, solar and terrestrial, yields a much more accurate result than your simplified “flat spectrum” model.

## Jan

| #

” will repeat what I said before. The earth’s Bond albedo is approximately 0.3, which says the earth absorbs about 70% of the sun’s radiant energy that it intercepts. We know this to with a few percent from real measurements.”

We don’t.

The Bond albedo values you can find in recent literature range from 0.27 to 0.352 – which clearly isn’t “few percent”, but 8.2% – which I would definitely not call “few”.

Moreover the exactitude of the Bond albedo we would need to be able to say anything serious about surface radiative budget for the purpose of climatology would need to be exact not to few percent, but to 1 promile (-which is equivalent of ~0.34W/m^2 for purpose of calculating surface incoming solar energy, which is equivalent of ~0.06 degree of surface anomaly change – more or less equivalent to decadal trend for whole 20th century warming)! To put it in the perspective: the range of 0.27-0352 Bond albedo value is equivalent of ~27,9W/m2 surface solar forcing range, which is equivalent of ~5 degree of surface temperature change!

As you see the “few percent” really by far isn’t the exactness of the Bond albedo value which would allow any conclusions in climatology whatsoever.

## Curt

| #

Several points – all of the satellite based measurements of Bond albedo, rather than earthbound estimates, cluster tightly around 0.29 and 0.30. But even if the range is what you say, it still does not negate my point, that the earth’s absorptivity to the sun’s radiation is significantly different from its emissivity/absorptivity in the far infrared range of its own thermal radiation.

Your post is about whether there exists a significant atmospheric greenhouse effect or not – whether you can account for the earth’s average surface temperature of about +15C without it. This is a very different issue from how much further additions of radiatively active gases add to that effect, which is what this most recent comment of yours concerns. This is a constant point of confusion at PSI, and it provides ample ammunition for alarmists to claim that skeptics have no clue as to what they are talking about.

## Jan

| #

I don’t deny there is about 15C average surface temperature which is some 33C higher than e.g. average surface temperature of the Moon virtually without atmosphere but at basically same orbit around Sun, I also don’t deny that all bodies which have temperature above absolute zero radiate – including our atmosphere, moreover I emphasize, that if such radiation is affected by emissivity change or temperature change it will change heat dissipation rate not only towards surface but to space too – moreover more, because the aperture to space is significantly wider than aperture back to surface due to atmospheric body spherical curvature and descending volumetric absorbtivity with altitude..

What I question is why the atmosphere causes the temperature difference and especially whether the elevated CO2 levels in the atmosphere due to human emissions (I don’t deny it) can have the effect of the recent observed warming magnitude.

I have serious reasons to doubt already the basic premises of the underlying “greenhouse” hypothesis, because it in my opinion violates especially Stefan-Boltzman law for radiative transfer.

Warmists themselves continually provide ammunition against themselves, for examples: by the notorious ignoring of the 2nd law of thermodynamics, postulating net positive radiative transfer from colder body of atmosphere to warmer bodies of the land and ocean surfaces – which in principle can’t happen and therefore I deem such concept superfluous to say it mildly (instead of concerning themselves by the question how much the changes of the atmospheric properties due to “GHE” content change can change the temperature of the atmosphere and therefore allow for change of the net radiative loss from surface -which would be the correct approach), by ignoring big net radiative cooling effect of the clouds blocking the solar irradiation and the net negative cloud feedback to rising surface temperature – considerably higher than the proposed CO2 forcing, by ignoring the completely broken correlation between CO2 and surface temperature anomalies lasting more than decade, by ignoring the solar activity/surface temperature anomaly correlation and the distinct signal of the solar cycles in the temperature anomalies data (while the CO2 signal is clearly completely missing there, moreover in the decade of the highest antropogenic CO2 emissions in history), by doing outright impossible warming predictions based on outright impossible anthropogenic CO2 emissions predictions and outright impossible CO2 forcing enhanced by outright impossible positive feedbacks estimations, moreover fitted into models using real data trends – and coming to results which in principle hindcast simmilar things as observed in the real data, but without any forecast predictive value whatsoever, by outright lies that recent warming period is “unprecedented”, while 90% of just holocene was warmer than last decade and there were at least 15 periods in it which had 5-15 times faster warming rate than the last warming period…etc.

It is absolutely unimportant what ammunition maybe PSI provides, because it doesn’t affect any policies – unlike the warmist overshots which cost hundreds of billions yearly.

## Ben Wouters

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[quote name=”Jan”]I don’t deny there is about 15C average surface temperature which is some 33C higher than e.g. average surface temperature of the Moon virtually without atmosphere but at basically same orbit around Sun.[/quote]

Average surface temperature of our Moon is ~197K (Diviner project)

288K – 197K = ~33K ??????????????????????

## Jan

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Finally interesting information and correction. I don’t know where I red it the Moon has 255K, but I checked the Diviner and you’re right. So now we have the Earth with 289 K (according to Trenberth) average surface temperature and Moon at the same distance from Sun (even with considerably lower albedo) at 197 K – it indeed looks like 92K difference, not 33K.

I must think about what are the implications, but anyway thanks to point me to the Diviner.

## Ben Wouters

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Implications are that a grey body that reflects ~30% of incoming solar at our distance from the sun and does one rotation in every orbit will have a radiative balance temperature around 151K.

So the Climaclowns who believe our atmosphere with a thermal mass equal to ~3 meter water can warm the surface (even including the deep oceans), have to explain a temperature difference of ~140K iso 33K.

see [urlhttp://principia-scientific.org/supportnews/latest-news/123-moons-hidden-message.html]eg. this article.[/url]

## Ben Wouters

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Again the link:

http://principia-scientific.org/supportnews/latest-news/123-moons-hidden-message.html

## Jan

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Just one quick question which falls into my mind – where on Earth we find so low temperature as 152K? – even the lowest temperature in atmosphere is listed like ~180K in Mesopause – but it is true no significant albedo above – so 681W – 331 K – 167K if I use the 2.7K -still too low, but true that 750K+ in thermosphere. I’m puzzled, as I said I don’t understand much from tropopause up.

## Ben Wouters

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[quote name=”Jan”]Just one quick question which falls into my mind – where on Earth we find so low temperature as 152K?[/quote]

Nowhere. Earth is basically a ball consisting of moilten rock with temperatures of 1000K and higher.

The thin crust is barely able to contain this heat.

The temperatures in the Thermosphere are high, but this is only for a few lonely molecules, no thermal mass involved.

## Jan

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high temperature particles though scattered through vacuum in substantially thick layer could be relatively good insulator.

## Jan

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Very interesting ideas. Thank you for the link.

To me also looks unlikely the atmosphere, having so small heat content, and the subpromile fraction of this heat content in CO2, could somehow alone change the surface temperature multiple K (as the CAGW hypees still prophesize -after decade+ of non-warming, yet highest CO2 emissions in history) and somehow prevent the tremendous heat content surplus be radiated away due the multiple K higher temperature. -It looks to me from realm of pure tales.

Now, what could explain the huge temperature discrepance? I have maybe an partial suspicion. As I look into the ocean, it surprises me, how much and how deep it is able to absorb the solar spectrum.

Some interesting numbers:

In tropical region: 74.8% of ocean (73% NH, 76.5% SH)

Tropics-polar circle still 70.5% ocean (48.3% NH, but 92.7% SH)

Polar circle-pole 48.5% (NH 73%, SH 24%)

Equator-65° 72.15% (NH56% SH 87.2%) (-working version of ocean/land stratification: tumetuestumefaisdubien1.sweb.cz/GLOBAL-INSOLATION-GRID-1DEG.xls)

I’ve looked how much the ocean reflects the solar spectrum, I’m still not quite sure with the result, the estimation is by matrix of Fresnel equations which still doesn’t have sufficient resolution, but as it looks to me now, it reflects in average less than 5% of the solar spectrum at the latitudes up to the polar circle. Rest gets undersurface and extinct there results in accumulation (Reason: way higher reflectivity of the water-air interface for mid-IR from inside than for solar spectrum from outside – a “greenhouse effect”, somewhere else, than CAGW herd would like to have it) of heat content several hundred times higher than that of the atmosphere and average surface temperature ~13C higher than bottom.

I can roughly estimate ocean intercepts over 90% of all energy Earth surface absorbs from the solar irradiaton. So it is at hand to assume, that what chiefly heats the air close to surface is indeed the ocean. The next will be of course the land surface, also significantly heated by Sun to temperatures at comparable latitudes sometimes far exceeding that of sea surface, but to much shalower depth, given by very low transparence of the landmass surface materials for solar spectrum and also less is accumulated, given by considerably lower volumetric heat capacity of land surface materials than that of water (except that of biologic nature, which is again mainly water anyway).

I can also assume what chiefly heats the surface is clearly the Sun and what heats the air above is clearly the surface -less above land, more above the ocean -not the colder atmosphere.

What we can also assume is that water plays crucial role in troposphere too, changing its lapse rate several °C per kilometer. And this effect arguably itself makes difference of ~33K between average surface air temperature and average boundary+troposphere+tropopause layers (0-20km -pressure weighted, 255K occurs at 5.1km in ISA model) temperature.

What mechanism(s) makes the paradox temperature gradient above in the remaining ~4.6% of atmospheric mass with top at 750+K I don’t well know -but sources of heat are sure: both Sun and surface.

## Ben Wouters

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Sun light doesn’t reach much deeper than `100m into the oceans. The sun only warms a shallow layer, the Surface Layer.

Below that the Thermocline starts, the separation between surface layer and deep oceans. Solar heat will not pas to below the thermocline.

The temprature of the deep oceans is set by geothermal heat.

see [url]http://principia-scientific.org/supportnews/latest-news/124-real-global-warming.html[/url]

## Jan

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It is even much less than 100m, most of the solar energy (now I don’t talk about spectrum) is absorbed in upper ~6 meters. But that’s pretty deep in comparison with usual materials at the surface of the land. all that mixed down by surface evaporation leaving salt behind, waviness and turbulences and only in surface layer (funny it is deepest at about lat40°) we get heat content 50+ atmospheres.

I would think the deep ocean temperature is given by water dilatometric anomaly – all water with temperature significantly colder or heated (say by geothermal heat) significantly above the point would buoyantly ascend up.

## Shooter

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Yeah, but the Earth’s not gray or black or white.

## Bob Armstrong

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Such comments strike me as revealing a lack of any hard science , physics or engineering education . Years are spent on spheres and planes and cylinders and sines and cosines and other simple abstractions . That’s how physics is done . And you learn such things as inner products and orthogonal functions and how to analyze observed data in terms of them .

A gray , being a flat spectrum , is “orthogonal” to color , non-flat spectra . So it separates out the total energy density , eg at a point in our orbit , from the effect of the non-flat spectra of , for instance CO2 or H2O or chlorophyll , The equations allow the calculation of the equilibrium temperature of any colored body in terms of the spectra of any set of source .

Consider my example of calculating the equilibrium temperature of deep water . Here’s a slide from the power point I’m working on which shows the calculation for water . http://cosy.com/Science/WaterEquilibriumTemp.jpg . ( I’m not happy with the slide and hope to make the calculations more concrete . )

See also the other material on my http://CoSy.com . There clearly is an unconscionable gap in the teaching of this most basic physics , and indeed the classical rigorous analytical methods of physics in the “climate science” branch . It’s as if the field doesn’t recognize that it is just a branch of applied physics .

## Jan

| #

The atmosphere is indeed the surface of the Earth (at least for the purpose of emissive ballancing solar irradiation) and it indeed has average temperature of ~255 K.

But it doesn’t radiate just from the top, but from whole the gaseous atmospheric body, which is gradually thinning towards vaccuum so it “reabsorbs” the outgoing radiation less and less towards the space, while the opposite is true back towards the land and sea surfaces below. The resulting tropospheric temperature gradient called dry adiabatic lapse rate is therefore primarily caused by the gravity.

The variating water vapour content, carrying the latent heat in the air – which is anyway in principle outside realm of radiative heatloss – is causing then shift from the dry adiabatic lapse rate to moist adiabatic lapse rate, so the tropospheric temperature gradient is in reality considerably less steep then it would be if the air doesn’t contain any water vapour. However this effect doesn’t significantly reach beyond troposphere, which is bulk of the atmospheric mass.

Correct me if I’m wrong but I thought the “greenhouse effect” is not that atmosphere radiates to the space, but that “greenhouse gases” in the atmosphere absorb some of the IR radiation from surface and “backradiate” it back to surface causing warming of it -In my opinion superfluous concept circumventing much simpler (at least in the number of unknowns) implications of laws of thermodynamics as well as Stefan-Boltzman law of radiative transfer [P=epsilon.sigma.area(Th^4-Tc^4)] prima facie excluding possibility of radiative transfer from a colder body to a warmer one.

To explain: The CO2 holds less than 0.0003 heat content in the troposphere so it can’t have significant effect on its temperature and therefore it can’t have significant effect on the (much higher) sea and land surfaces heatloss and therefore their temperature. It is so simple as that – the radiative transfer in open system is dependent on the emissivity and temperature of the body which is warmer and temperature of the colder body the warmer loses heat to, not on the emissivity of the colder body. So the slight emissivity changes to air caused by CO2 content change are irrelevant and in principle have no effect on the radiative loss from the warmer body – the sea and land surfaces (warmed by sun, because they have way higher absorbtivity for solar spectrum than air) losing heat to colder air.

Higher emissivity of air caused by higher CO2 content in principle causes higher emission in all directions -the same (or even higher, because there is slightly wider aperture caused by spherical curvature of the atmospheric mass) towards space as towards the land and sea surfaces. So only effect additional CO2 causes is that it adds C to atmospheric mass, making the atmospheric column slightly taller (several centimeters) and that the plants thrive on it.

## Tim Folkerts

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1. A critical word in ‘adiabatic lapse rate’ is ‘adiabatic’ = without exchanging heat with the surroundings. In this particular case, if a parcel of air rises without exchanging heat with surrounding air, then as it expands, it will cool. Convection is the ’cause’ of the adiabatic lapse rate, not radiation. In fact, since thermal radiation from one part of the atmosphere to another is by definition NOT adiabatic, then radiation within the atmosphere works to eliminate the lapse rate. Only continued convection can maintain the adiabatic lapse rate.

2. I would say that the “greenhouse effect” is the simply the warming. The [i]cause[/i] of that warming is thermal radiation. Period. Since you can’t have radiation to space without radiation downward to the ground, these are simply two aspects of the same explanation.

Its kind of like asking what makes a car accelerate forward. Some would say it is the tires pushing back agasint the ground. Others might say no, it is the ground pushing forward on the tires. But Newton’s 3rd law say you can’t have one without the other, so the two answers are equivalent if you understand the underlying connections.

## Jan

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The cause of tires pushing against the ground is clearly the torque originating in the engine of that car, not the reaction of the ground – which for that is called “reaction”, not “action” or “M. Jackson”.

And with it colapses your inappropriately relativistic argument.

The cause of the warming was rising solar activity and not the atmospheric thermal radiation due to rising atmospheric CO2 content. I could write also “Period”. But unlike you I’ll explain why:

The CO2 carries less than 0.5 PROMILE of atmospheric heat content, so it can’t much influence its temperature on which depends the radiative transfer from the (warmer) surface to the (colder) atmosphere –

P=epsilon.sigma.area(Th^4-Tc^4)

– which clearly doesn’t depend on emissivity of the atmosphere, nor its change due to rising CO2 content)!

What depends on atmospheric emissivity is the radiation transfer from (warmer)

atmosphere to (colder) space, but it rises with emissivity in all directions including that towards the space, tending to equilibrium, so the changes of emissivity given by rising minute CO2 content can’t significantly rise the temperature of the atmosphere (because the additional heat is lost to space exactly due to the higher emissivity) and therefore can’t significantly change the radiative transfer from surface to the atmosphere and therefore can’t significantly change the surface temperatures as was observed!

That’s why I say the “greenhouse effect” surface warming hypothesis relying on so called “backradiation” (the thermal radiation back to surface) is superfluous, circumventing the Stefan-Boltzman law for radiative transfer – violating it and generally the 2nd law of thermodynamics too.

-So if there was a factor causing significant surface warming from say the 1970’s to 2000’s it in principle couldn’t be the rising minute CO2 (and other “GHE”) content in the atmosphere, because its temperature can’t significantly rise due to minute changes of CO2 content and cause lowering of radiative transfer from the surface, resulting in its rising temperature.

-That’s also why the observed correlation of the CO2 atmospheric content with surface temperature anomalies clearly floped to negative last decade despite all that rising emissions and IPCC unscientific blabber, while there still rests the much better correlation with solar activity – and the correlation not only is way better than with CO2, but the temperature anomalies data clearly contain also the distinct solar cycle signal (unlike a signal of CO2) – determining the causality – see my article here:

http://wattsupwiththat.com/2013/04/19/another-escalator/

So I have really serious reasons to believe the factor causing the recent warming was chiefly the cummulation of the surface heat content caused by rising solar activity trend. Which, contrary to common belief and according to all solar activity indices measured sufficiently long to the past (SSN, F10.7, GCR) ROSE well until 2000’s – I’m ready to prove it beyond any reasonable doubt – here e.g. from SSN:

http://tumetuestumefaisdubien1.sweb.cz/SSN1890minima-2012+1964minima2012Trendsv2.gif , http://tinyurl.com/qa8xpyz

## Tim Folkerts

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See my comments @ 2013-10-03 17:58

## Curt

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Jan – You are misinterpreting both the Stefan-Boltzmann equation and the 2nd Law.

In the S-B equation you cite, the second term (epsilon*sigma*[-Tc^4]) IS the back-radiation term. The first term is the “forward” radiation term.

As for the 2nd Law, none of the standard greenhouse models show the cooler atmosphere radiating more power to the earth than the warmer earth radiates to the atmosphere. This means that the resultant heat transfer is always from higher temperature to lower temperature. No 2nd Law violation there.

## Jan

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Tell me please, you mean this equation:

P=epsilon*sigma*area(Th^4-Tc^4)

And if the answer is yes, then please tell me, the epsilon there, is it emissivity?

And if the answer is yes, then let’s imagine we use the equation for determining the “forward” radiative heat transfer from Earth surface to atmosphere.

Can you please tell me – for me to correctly put numbers into the equation:

1. the emissivity in the equation is for what wavelength (as you surely know emissivity is wavelength dependent – and quite untrivially for IR spectra..) – of the “forward” surface radiation or for the wavelength of the “backradiation”?

2. I maybe want too much, and both “forward” and “backward” radiations anyway look like consisting from whole spectrum of wavelengths, so it will maybe suffice if you please tell me (because there is only one epsilon in the equation) – is it emissivity of A. the surface, B. the atmosphere, or C. somehow both?

3. If somehow the answer to the previous question is the C. then please tell me where I find the values and for which wavelength from the x dozens micrometer wide significant spectra I should chose the corresponding emissivity value to get the result right.

Many thanks for correct answers in advance.

Oh, not to forget, the answers are extremely important to me, I have even a feeling, that at the time I’m not much interested in anything else .. so please spare me from potential OT.

## W5OVF

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[b]Correct me if I’m wrong but I thought the “greenhouse effect” is not that atmosphere radiates to the space, but that “greenhouse gases” in the atmosphere absorb some of the IR radiation from surface and “backradiate” it back to surface causing warming of it -In my opinion superfluous concept circumventing much simpler (at least in the number of unknowns) implications of laws of thermodynamics[/b]I cannot correct, Your Math is correct, Your explainations fine, however you use the same mistaken assumptions:

!.Radiative heat transfer from the surface, to the atmosphere, affect the surface temperature. (It is to small to do anything.)

2. One can use the S-B equation to determine the effective radiation temperature of a planet at a distance from the primary! This requires a suitable geometry and emissivity such as planets 2,3,4 do from the Sun.

3. Please consider the measurements.

http://coen.boisestate.edu%2Fvsridhar%2Ffiles%2F2011%2F10%2FS_2JGR_lsm_2002.pdf

I think or:

Validation of the NOAH-OSU land surface model using surface flux measurements in Oklahoma V. Sridhar,1 Ronald L. Elliott,2 Fei Chen,3 and Jerald A. Brotzge.

Figures 13,14,15 show the thermal flux both measured and computed. I still cannot get the real measured numbers.

Please review, and see if your theory matches the measurements.

:

:

## Jan

| #

Unfortunately the link doesn’t work for me. Try please repost it.

Radiative transfer from land and sea surfaces to atmosphere is given by the difference of fourth powers of their average temperatures

P = sigma*emissivity*(average temperature of the surface^4 – average temperature of the atmosphere^4)

and is chief way the surface dissipates heat content resulting from solar irradiation absorbtion.

## Tim Folkerts

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For the link, go to wikipedia and look for “thermal radiation”.

Here is another website:

http://www.efunda.com/formulae/heat_transfer/radiation/view_factors.cfm

## W5OVF

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[quote name=”Jan”]Unfortunately the link doesn’t work for me. Try please repost it.

Radiative transfer from land and sea surfaces to atmosphere is given by the difference of fourth powers of their average temperatures

P = sigma*emissivity*(average temperature of the surface^4 – average temperature of the atmosphere^4) and is chief way the surface dissipates heat content resulting from solar irradiation absorbtion.[/quote]

Please Google: Validation of the NOAH-OSU land surface model Then download the PDF. The figures 13,14,15 show the “measured” heat flux for both sensible and latent heat, as a plot of surface flux vs time of day.

All heat is transfered to the atmosphere via convection not radiation. Radiative transfer between the surface and atmosphere is so small it can be ignored. Except for the 8-14 micron band, to space, thermal radiation from the surface is absorbed within 35 meters, with a delta T of

## W5OVF

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[quote name=”Jan”]Unfortunately the link doesn’t work for me. Try please repost it.

Radiative transfer from land and sea surfaces to atmosphere is given by the difference of fourth powers of their average temperatures

P = sigma*emissivity*(average temperature of the surface^4 – average temperature of the atmosphere^4)

and is chief way the surface dissipates heat content resulting from solar irradiation absorbtion.[/quote]

All heat is transfered to the atmosphere via convection not radiation. Radiative transfer between the surface and atmosphere is so small it can be ignored. Except for the 8-14 micron band, to space, thermal radiation from the surface is absorbed within 35 meters, with a delta T of < 0.3 degree Celsius. You figure the flux. This flux never increases on its way to space.

The use of the S-B equation, backward, to attempt to determine the temperature of the absorber, then emitter, at a lower temperature,

without detailed measurement of spectral emissivity of all surfaces, is the most outrageous scientific fraud ever perpetuated

on innocent earthlings.

## Tim Folkerts

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[quote]All heat is transfered to the atmosphere via convection not radiation.[/quote]

More heat is transferred to the atmosphere by latent heat of water than by either of these.

[quote]Except for the 8-14 micron band, to space, thermal radiation from the surface is absorbed within 35 meters, with a delta T of < 0.3 degree Celsius.[/quote]

I would enjoy seeing your references for this statement. For one thing, the edges of the absorption bands are ‘ragged’ so for some wavelengths the distances should be longer than 35 m.

More importantly, the 8-14 um band gets absorbed by clouds. Since this band is ~ 40% of all the thermal radiation from the surface, and clouds cover ~ 65 % of the earth, this amounts to ~ 0.4*0.64*396 something around 100 W/m^2 into the atmosphere. The clouds, are not merely 35 m up with a delta T < 0.3, but potentially several km high. This will result in a transfer on the order of 10 W/m^2 (perhaps a bit higher or lower - I don't have the expertise to get really good numbers). IN any case, it is definitely NOT zero. ************************ In any case, I think this is a bit of a red herring. The question is not how much energy is transferred specifically to the atmosphere by radiation. The more interesting question is “how much MORE energy would be transferred away form the surface if not for the IR gases?”. Now, instead of delta T

## Curt

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Jan – You are getting close, but you must realize how all of these things tie together. You realize that the “top of atmosphere” must radiate as much to space (on average, of course) as the earth receives from the sun, which sets an effective TOA temperature of about 255K, and due to the lapse rate, a significantly warmer surface temperature.

While the radiation to space certainly is not from a definite surface as with a solid opaque object, it is quite opaque to many of the wavelengths that dominate Planck radiation at these temperatures (which is why we use space telescopes to look at these wavelengths astronomically), so the radiation to space for these wavelengths comes from high up in the atmosphere. You can think of the atmosphere with CO2 and water vapor as appearing in these wavelengths like fog does to us in the visible wavelengths.

Now for the necessary corollaries to this: A substance that emits at a given wavelength also absorbs at that wavelength. The better it emits at the wavelength, the better it absorbs. The emissivity at any wavelength for a substance is exactly equal to the absorptivity. So if the atmosphere can radiate in these wavelengths, it can absorb in these wavelengths.

Next, the atmosphere can have no directional preference in how it radiates. A location in the atmosphere will radiate evenly in all directions, which means it radiates as much down as it does up. If you accept that it radiates “up” to space, you must also accept that it radiates “down” toward the earth’s surface.

Also, with the earth’s surface significantly warmer than 255K, it radiates at a higher power flux density than the TOA. Roughly speaking, it radiates an average of about 390 W/m2. (It also transfers power to the atmosphere through sensible and latent heat transfer.) Since it only receives on average about 240 W/m2 of solar radiation, it needs an additional 150 W/m2, plus whatever is needed to counteract sensible and latent heat losses, to be in even rough energy balance.

So where does this additional continual power flux come from? Many PSI folk think it comes from the force of the weight of the atmosphere. But for this weight to transfer power, it must fall (as with hydroelectric power generation), and it has nowhere to fall, as it has already fallen as far as it can. Some credit moving downdrafts for this, but this must be fully balanced by any updrafts, so that effect cannot provide ongoing power either.

(To be continued…)

## Curt

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(Continued from above…)

But the radiatively active atmosphere can. It is continually radiating power back toward the earth’s surface. Let’s look at the version of the power transfer equation you give:

P = epsilon * sigma * area * (Th^4 – Tc^4) (in Watts)

It is derived from the Planck radiation flux density “p” of the two objects:

Object “h”: p[h] = epsilon[h] * sigma * Th^4 (in Watts/m2)

Object “c”: p[c] = epsilon[c] * sigma * Tc^4 (in Watts/m2)

where “epsilon” is the emissivity of each object (not necessarily the same).

So the power that object “h” radiates toward object “c” is:

P[h->c] = p[h] * area = epsilon[h] * sigma * area * Th^4 (in Watts)

and the power that object “c” radiates toward object “h” is:

P[c->h] = p[c] * area = epsilon[c] * sigma * area * Tc^4 (in Watts)

where “area” is the effective cross-sectional area through which the two objects can radiate toward each other.

Now, “sigma” is a physical constant that is common to both equations, and “area” also must be common to both equations.

It is tempting to say that the power transferred between the two objects can be calculated as:

P = P[h->c] – P[c->h]

It would be true for blackbodies (epsilon = 1.0) that absorb all of the radiative power they receive. It works pretty well for graybodies with the same emissivity (epsilon[h] = epsilon[c]) and constant over all of the relevant wavelengths, and if the radiative power not absorbed is reflected back to the original object, not passed through, and not reflected elsewhere.

It is important to realize that the equation you give is really a simplification that is often given in introductory courses to make problems solvable for beginning students.

And your assertion that the emissivity of the colder object does not matter is simply not true. While I used the letters “h” and “c” in the above analysis, there was nothing in the analysis that depended on object “h” having a higher temperature than object “c”. The emissivities/absorptivities of both are important.

In the case of the radiative power transfer between the earth and the atmosphere, let’s use the simplified equation you present. With a radiatively active atmosphere, Tc can be taken as ~255K. If the atmosphere were not radiatively active in these wavelengths, as would be the case without CO2 and H20, the earth’s surface would be in radiative exchange with deep space, with a Tc of ~3K.

Therefore, with a radiatively active atmosphere, the earth’s surface temperature must be significantly higher to radiate away the power it receives from the sun than it would be if the atmosphere were transparent to its radiation. This difference is the “greenhouse effect”, and it is a necessary consequence of the analysis you present.

## A.Rappaport

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[b]Therefore, with a radiatively active atmosphere, the earth’s surface temperature must be significantly higher to radiate away the power it receives from the sun than it would be if the atmosphere were transparent to its radiation. This difference is the “greenhouse effect”, and it is a necessary consequence of the analysis you present.[/b]

This “may” be true, but only if the “radiative”heat transfer from the surface to the atmosphere determines the temperature of that part of the atmosphere.

This is not true on this earth. Latent of evaporation and the convection thereof, completely dominate the the physical temperature of any part of the atmosphere. Radiation from the surface can be ignored. All required radiative heat transfer originates in the atmosphere. All is radiated into the two PI steradians of cold space. Your increase in atmospheric CO2 can only result in a decrease in temperature of the atmosphere and surface.

## Tim Folkerts

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You have this so completely backwards!

[i]”Radiation from the surface can be ignored” [/i] BECAUSE it is balanced by the radiation downward from the sky. IF the sky were not radiating back, then your radiation from the surface could NOT be ignored becasue it would suddenly be MUCH LARGER!. Which of course would cause the surface to cool significantly until conduction, convection and evaporation decreased enough to restore the balance.

[i]”Your increase in atmospheric CO2 can only result in a decrease in temperature of the atmosphere”[/i] AT THE TOP. The top will a) get higher and b) get cooler as more CO2 is added. The result at the ground level will be an increase as required by a) the lapse rate and/or b) conservation of energy.

## A.Rappaport

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[quote name=”Tim Folkerts”]You have this so completely backwards![/quote]

[i]”Radiation from the surface can be ignored” [/i] BECAUSE it is balanced by the radiation downward from the sky. IF the sky were not radiating back, then your radiation from the surface could NOT be ignored becasue it would suddenly be MUCH LARGER!. Which of course would cause the surface to cool significantly until conduction, convection and evaporation decreased enough to restore the balance.

From the surface, All thermal radiative heat transfer from the surface to the air is complete within 35 meters with a delta T of less than 0.3 degree Celsius This radiative flux as calculated never increases on its way to cold space.

[b][i]”Your increase in atmospheric CO2 can only result in a decrease in temperature of the atmosphere”[/i] AT THE TOP. The top will a) get higher and b) get cooler as more CO2 is added. The result at the ground level will be an increase as required by a) the lapse rate and/or b) conservation of energy.[/b]

Your so called TOP (effective radiative temperature) does not get higher, nor change temperature) it always stays at the effective

radiative temperature of the atmosphere. The increase in emissivity of CO2 can only decrease temperature of TOA, and decrease the temperature of the surface

## Tim Folkerts

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[quote]Your so called TOP (effective radiative temperature) does not get higher[/quote]

My “top of atmosphere” is around 220 K near the tropopause. This is the “last layer before the CO2 gets so thin that the IR passes off to space”. The altitude of this will depend on the amount of CO2.

[quote]… always stays at the effective

radiative temperature of the atmosphere[/quote]

I agree that the effective radiating temperature of the earth will always be 255 K.

[quote]The increase in emissivity of CO2 can only decrease temperature of TOA, and decrease the temperature of the surface[/quote]

If the top and bottom decreased in temperature, then the effective radiating temperature would DROP, so you have just contradicted what you had said in the previous sentence!

The first half is right — the ToA would get cooler. To conserve energy and to keep the effective radiating temperature at 255 K, the bottom must warm up!

## Curt

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[quote name=”A.Rappaport”] Radiation from the surface can be ignored. [/quote]

How ridiculous can you get? You cannot ignore the typically almost 400 W/m2 radiating from the surface in calculating the earth’s energy budget any more than you can ignore $400/month car payments in calculating your household financial budget. Yes, you would have income sources and other expenses as well, but if you ignored your car payments, your results would be completely wrong.

If you can ignore the radiation from the surface, how do you explain that the earth’s surface gets colder than the air at the surface many nights, and that this effect is particularly pronounced on clear nights?

Have you ever really looked around on a morning after a clear night when the air temperature got almost, but not quite, down to freezing? You will see frost on upward facing surfaces that have a clear shot to the sky, but not under trees, etc. You can often see half a roof with frost and the other half (under a tree) without frost.

There are many more examples that are easy to find, if you would just look.

## A.Rappaport

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[quote name=”Curt”][quote name=”A.Rappaport”] Radiation from the surface can be ignored. [/quote]

How ridiculous can you get? You cannot ignore the typically almost 400 W/m2 radiating from the surface in calculating the earth’s energy budget any more than you can ignore $400/month car payments in calculating your household financial budget. Yes, you would have income sources and other expenses as well, but if you ignored your car payments, your results would be completely wrong.

If you can ignore the radiation from the surface, how do you explain that the earth’s surface gets colder than the air at the surface many nights, and that this effect is particularly pronounced on clear nights?

Have you ever really looked around on a morning after a clear night when the air temperature got almost, but not quite, down to freezing? You will see frost on upward facing surfaces that have a clear shot to the sky, but not under trees, etc. You can often see half a roof with frost and the other half (under a tree) without frost.

There are many more examples that are easy to find, if you would just look.[/quote]

Radiation sfom the surface to space in the 8-14 micron band is less than 20 Watts/m^2.

Radiation fromthe surface to the atmosphere is less than 1 Watt/m^2 this small amount does not change the temperature of the atmosphere at any altitude. The latent heat of evaporation of water adds 78 Watts/m^2 of heat energy to the atmosphere continiously, and sets the temperature of the atmosphere. It is the atmosphere itself not the surface that radiatively cools this planet.

## Tim Folkerts

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[quote] It is the atmosphere itself not the surface that radiatively cools this planet.[/quote]

That is one legitimate way to think about it. But this line of thought says that this cooling by the atmosphere (due in large part to thermal radiation from GHGs to space) comes from high up in the atmosphere. And then the lapse rate guarantees that the surface will be warmer than 255 K.

Personally, I like explaining the GHE this way, rather than specifically using the term ‘backradiation’, because I think it is more intuitive.

But whether you want to discuss the ideas in terms of ‘backradiation’ or ‘thermal resistance’ or ‘the atmosphere itself radiatively cools this planet’, all of these require the atmosphere to have thermal IR properties, and all lead to a surface that is warmer than 255 K.

## Jan

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Because estimations how much m2 of Earth in average radiates and the emissivity – intimately linked in SB law – were mentioned here several times, I’ll note that emissivity is strongly angle dependent and tends dramatically to zero at high emission angles (and also changes with temperature and wavelength – both determine the actual refractive indexes of the adjacent materials).

I’ll not do here all the vast multidimensional matrix of fresnel equations for all angles, possible materials, their surface stratification, possible temperatures and their stratification and resulting spectra to show the total surface ε is not 1, nor very close to 1 and how much it really is. I’m not paid the billions in grants to calculate – for those who are – the multibillion dollar number.

Just simple example with water – ocean, covering more than 2/3 of this planet: When ocean (which is liquid and with fairly horizontal surface) radiates the IR, it (despite the directions of the original emissions are random) radiates dramatically more of it at angles closer to straight up towards the space (because that part of the random direction IR emissions which passes the interface at low incidence angles is not much impeded by the high water-air interface reflectance) than at high angles to the sides. (And I’ll rather not go into details what ocean waviness would do with such side radiation…)

Therefore even roughly speaking the average 390W/m2 Curt mentioned before, or “almost 400W/m2” now, and which are clearly not really measured, but are instead derived using SB law assuming ε=1 or very close to 1 (for example Kiehl and Trenberth use value 396.4W/m2 obtained using SB law with exactly 16°C temperature and ε=1, while apparently seriously discussing Wilber1999 water ε=0.9907 value -not even a laboratory blackbody has such high emissivity!!) is unrealistic and is very apparently a result of false assumption that surface ε=1 or is very close to 1.

Just for very instant example:

(Basic premise I cite from Wikipedia: http://en.wikipedia.org/wiki/Planck%27s_law_of_black-body_radiation “For any interface the sum of the emissivity and reflectivity is 1. An ideally perfectly reflecting interface has emissivity zero, reflectivity one. An ideally perfectly transmitting interface has emissivity one, reflectivity zero.”)

What if (and I say what if – because it is just example by far not taking into account whole radiation spectrum) the total emissivity of the ocean would be not ε=1 but say ε=0.7 (rounded value estimated using fresnel equations from water-air interface angular spectral reflectances for 9-15μm IR radiation, with optical depth and waviness effects omited)

the P for the ocean would be 0.7×0,00000005670373×290^4 = 280 W/m^2

-It is by far not result of any real value – it is just example to demonstrate what a different ε value does with the SB law equation result (- but likely still more realistic than “almost 400W/m2” derived assuming ε=1 which is utterly impossible for any real materials interface). Think about it.

## Bob Armstrong

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Of course one of the most immediate elaborations of a model is to add a Lambertian cosine function to each surface element .

My main point is that in an APL language , each such elaboration requires at most a few additional succinct definitions . As an open project , it can counter all the billions the watermelons have to create inscrutable million line models , all of which are wrong .

My main interest is evolving programming language itself , and that’s what I have to get back to now .

## Jan

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[quote name=”Bob Armstrong”]Of course one of the most immediate elaborations of a model is to add a Lambertian cosine function to each surface element .

[/quote]

I’m not sure, most of the Earth surface is covered by ocean, which looks to behave extremely non-Lambertian way and specular reflectivity looks like it absolutely dominates on its interface with air. So I’m not sure what for would be to use Lambertian function for surface which looks like way much better described by Fresnel equations and sonic spectral waviness models.

Can you please elaborate?

## Tim Folkerts

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I agree that getting a good estimate of “the emissivity” of the entire earth is a challenge, for the reasons you gave.

I suspect that the “correct” number is somewhere between Hanson’s original estimate of 1.0 and your estimate and your estimate of 0.7. I suspect that it is closer to 1.0 than to 0.7. You are welcome to try to get a better estimate of this number (either from the literature or your own experimentation).

## Jan

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[quote name=”Tim Folkerts”]

I suspect that the “correct” number is somewhere between Hanson’s original estimate of 1.0[/quote]

(ε=1.0 “estimate” for common material is not a valid estimation, it is an utter nonsense – even laboratory blackbodies have usually emissivity lower than 0.99 – and that’s moreover the value for close to 0° emission angle.)

[quote name=”Tim Folkerts”]and your estimate of 0.7. I suspect that it is closer to 1.0 than to 0.7.[/quote]

I suspect it will not be the case and I suspect the estimated value 0.7 is still overestimation and that the total average total water ε will be even lower than 0.7 for possible significant IR spectra. I have serious reasons why I think so. I’ll try to describe them for possible discussion:

What I tryied so far was calculating matrix of water-air interface reflectances for all angles 0-90° by step 1° for wavelengths 6-16μm by 1μm step, then averaged them and got total average water-air interface reflectance Rs+Rp/2=0.29354, subtracted it from 1 and so I got the raw total emissivity estimation value ε=0.70646. (I note, that I completely omited the extinction coefficient – the imaginary part of the refractive index and I also didn’t corrected the result for waviness -both would presumably further lower the emissivity)

I’m aware it is still by far not whole possible range of spectra for all possible temperatures of the ocean surface, but the point is that the lowest real part refractive index has water at ~12μm, (complex at ~11μm) the real part of refractive index average for 200nm-6μm (the imaginary part plays some role in UV region) is n=1.33 and for 16-30μm it is n=1.47 (and the imaginary part plays already very significant role in whole IR region), while for the range 6-16μm (which I did the emissivity raw estimation for) the n=1.27 (and lower n means lower reflectivity of the water-air interface -higher emissivity for the range) – so it very much looks to me so far that if I go further to further widen the spectrum, refine the resolution and use imaginary part, the total average reflectivity likely will go further up in both directions from the center wavelength 10μm (which is peak wavelength of 17°C- average SST spectrum and the water-air interface has ε=0.69 value coming from my estimation for this separate wavelength)

So it very much looks to me, that if I indeed do the estimation for whole possible significant IR spectra range, the ocean emissivity estimation will come out even lower than the ε=0.7 raw estimation. (I of course still don’t know for sure, because the n does funny things between ~3 and 6μm and the 1μm resolution would definitely be not enough there for meaningful average even for raw estimation).

But even if not, and you’ll be right the water total ε is somewhere between 0.7 and 1 then if it would be on the end just say 0.9 for the most significant spectra (which would mean global surface radiation budget change of ~30+W/m2), I still can well imagine, what havoc it anyway could wreak in whole the arduously build AGW hypotheses complex – it would anyway crumble like a cardhouse.

## Don Tonoranton

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Just checking, tommy, if you are using names or IP adress to discriminate. You are such a fool You have deleated many comments of others .Keep on deleating!

## Sunsettommy

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[quote name=”Don Tonoranton”]Just checking, tommy, if you are using names or IP adress to discriminate. You are such a fool You have deleated many comments of others .Keep on deleating![/quote]

You have been moderated,Your questions answered several times then banned when you couldn’t stay reasonably civil here as requested numerous times.

I have been UNPUBLISHING your recent comments.They are still there for Administrators to see.

## Don Tonaronton

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[quote name=”Sunsettommy”][quote name=”Don Tonoranton”]Just checking, tommy, if you are using names or IP adress to discriminate. You are such a fool You have deleated many comments of others .Keep on deleating![/quote]

You have been moderated,Your questions answered several times

Whew Tommy please let the banned admisistratords decide! They they must ageree with you, as you are “MODERATOR”. They MUST disaGREEW WITH YOU to the greater public as that is aLL THEY HAVE. Please go away fool s.

I have been UNPUBLISHING your recent comments.They are still there for Administrators to see.[/quote]

Ok Tommy, let us see if the administration

does agree with your haphazardd “moderation”, they must, as you are the moderator, they must not, if they are to have any creditability

with the larger population.

Suck it in Tommy!

## Bob Armstrong

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For me to understand something is to be able to compute it . Understanding mean planetary temperature is impossible if there is a lack of understanding of even how to calculate the temperature of a radiantly heated colored ball .

That “climate science” claims that label while such discussions as this go on shows how far it has been retarded relative to any other branch of applied physics . The basic question of the equilibrium temperature of a radiantly heated colored ball would have been a homework assignment 80 years ago .

There are a number of approaches to calculating the temperature of a [i]gray[/i] ( flat spectrum ) ball in our orbit . The simplest is just adding up the energy impinging on a point in our orbit and applying the Stefan-Boltzmann 4th root relation . This works out to about 279k given an effective solar temperature of about 5780k .

By the divergence theorem this will apply to any gray ball of any thermal conductivity whether rotating or not . These claims can be experimentally tested with relatively simple apparatus .

I am working to make these computations understandable , ie : calculable , at the “talented teen” level . See [url]http://cosy.com/Science/RadiativeBalanceGraphSummary.html[/url] and referenced pages .

For the .ppt on these non-optional calculations I’ve put together , I approximated the spectrum of deep water as a crude approximation of the spectrum of the earth’s surface . Here’s the relevant slide : [url]http://cosy.com/Science/WaterEquilibriumTemp.jpg[/url] . While a very crude approximation , it shows the Earth , if totally ocean , would be much closer to 0c ( 273k ) than 255k .

## Curt

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Your calculations that result in 273K for a water planet ignore the albedo of clouds – that’s what would bring it down to 255K.

## Bob Armstrong

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Look , all I am concerned with first is getting the most basic classical physics correct . FIRST one has to know how to calculate the temperature of a radiantly heated opaque colored ball . I see little indication that even that is well understood at the level which would be required in any other field of physics or engineering , eg , electrodynamics – which is much more complicated because of curl .

I am not interested in anything other than equations and their computation . That endlessly parroted 255k BS has no chance of explaining anything because the total phenomenon being considered is on the order of the 3rd decimal place – tenths of a percent .

I am just ever more convinced that “climate scientists” are people that never had a more rigorous analytical course than “physics for poets” .

Frankly , all I am concerned with initially is computing the [i]radiative balance[/i] for any arbitrarily “painted” sphere . The divergence theorem then puts massive constraints on any variations in temperature within that sphere . But if you don’t know how to compute that external radiative balance to an accuracy of a part in 1000 , you are just word waving with no quantitative , experimentally testable understanding .

## W5OVF

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[quote name=”Jan”]Spherical surface of Earth has only 2x cross-sectional area for solar irradiation, than it would have a circle of same diameter – because the sun is only one in our solar system and always insolates only half of that Earth sphere, from one side. Only because Earth rotates around its axe once a day makes the insolation distribute over both sides.

Most of the heat in the atmosphere comes not from convection, nor latent heat of water evaporation. – most of the heat (>65%) is created by the absorbtion of EM radiation coming both from sun and from the warmer surface of the Earth. To claim otherwise is clearly a with others.nly flagrant denial of basic knowledge.

-It’s good to think faster than write.. [/quote]

Only when trying to communicate with others! Not recording nascent thoughts is a good way to loose them.

Please explain your >65% by radiation? Most convective heat transfer with latent heat (mass transport) of the atmosphere is lateral not vertical. It is powered by the latent heat and creates “all’ of the wind power in this atmosphere. Most of 77777the heat in the atmosphere is latent heat not sensible heat.

## Jan

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I already explained why the heat loss from the surface via latent heat of evaporation is about 30% (when I compared the amount of latent heat yearly needed to evaporate the water for precipitation cycle and the amount of heat resulting from solar radiation absorbtion -in the second section of the article.)

The rest comprises of radiative loss and a bit also the direct heat conduction to the atmosphere (but the usual surface materials -mainly water – nor air are good heat conductors, so the dissipation this way is just a minor one).

The chief way how the heat dissipates from the Earth surfaces is the radiant loss and in case of dissipation of the heat back to space from the atmosphere in fact the ONLY way (heat as a form of energy is always bound to mass – so if you would lose heat directly, not via radiation, you would literally lose the atmosphere – which fortunately the gravity doesn’t allow much).

Also, contrary to your claim, most of the heat in the atmosphere is not the latent heat of water evaporation, but specific heat of the air.

It can be easily proved by calculation:

At ANY TIME the specific heat of atmosphere approximately equates to 1.0012 [J/g/K – heat capacity of air] multiplied by mean temperature of the atmosphere [~ 255K] and by atmospheric mass [5.148×10^21 g] = 1.31×10^24 J. While the latent heat circulated by water cycle is only 1.14×10^24 J in whole year – it clearly can’t be there at any time, because it all the time dissipates via conversion to specific heat and subsequent radiative loss to space (all bodies – including body of atmosphere – radiate if they have over absolute zero temperature).

The latent heat also doesn’t intervene in all the atmosphere – whenever the water condenses on its way up by convection the latent heat is released and adds to the specific heat of the air surrounding it. The air then radiates the heat to space – in such case clearly not from the land and sea surfaces, but up from the altitude where the water condensed. (Which is one of crucial mechanisms for cooling the surface, because latent heat doesn’t add a slightest bit to the radiation loss from its carrier until it is released – so until the vapor condenses it doesn’t in any way heat the air around, and when it eventually does it is way up from the surface.)

## W5OVF

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[b]Also contrary to your claim, most of the heat in the atmosphere is not the latent heat of water evaporation, but specific heat of the air.[/b]

Humm! lets check that!

[b]It can be easily proved by calculation:

At ANY TIME the specific heat of atmosphere approximately equates to 1.0012 [J/g/K – heat capacity of air] multiplied by mean temperature of the atmosphere [~ 255K] and by atmospheric mass [5.148×10^21 g] = 1.31×10^24 J. While the latent heat circulated by water cycle is only 1.14×10^24 J in whole year – [/b]

Great! someone willing to d the arithmetic!!

I agree with your numbers! I should have said available sensible heat, vs avalable latent heat in the atmosphere. My fault! Sensible heat heat at a temperature lower than any temperature about the earth, to me, is already entropy and can do nothing except be radiated to cold space. Sorry for any confusion.

[b]It clearly can’t be there at any time, because it all the time dissipates via conversion to specific heat and subsequent radiative loss to space (all bodies – including body of atmosphere – radiate if they have over absolute zero temperature).

[/b]

The latent heat 1.14×10^24 J is also subject to conjecture The actual amount of latent heat in the atmosphere is 2500 Joules/gram of water vapor times the number of grams of water vapor in the atmosphere at that time.

I am glad for the opportunity for discussing differences in POV.

[b]The latent heat also doesn’t intervene in all the atmosphere – whenever the water condenses on its way up by convection the latent heat is released and adds to the specific heat of the air surrounding it. The air then radiates the heat to space – in such case clearly not from the land and sea surfaces, but up from the altitude where the water condensed. (Which is one of crucial mechanisms for cooling the surface, because latent heat doesn’t add a slightest bit to the radiation loss from its carrier until it is released – so until the vapor condenses it doesn’t in any way heat the air around, and when it eventually does it is way up from the surface.) [/b]

All of this is true, thank you.

This latent heat is essentially the only player in the atmospheric thermodynamic processes not radiative heat from the surface. The small temperature difference between the surface and atmosphere greatly

limits any radiative heat transfer between them.

The rate of heat transfer in the atmosphere with convection can be 10^4 higher then that of diffusion, especially with 300 kph wind speed. The surface area is vast,and the conductive distance is two molecules at a 4 degree Celsius diference,. Run the numbers.

As the warm air mass from the tropics (moves) transports to the poles, the only thing that can maintain the temperature difference is the latent heat of water vapor. Radiation from the surface is nonsense!

## W5OVF

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[b]Also, contrary to your claim, most of the heat in the atmosphere is not the latent heat of water evaporation, but specific heat of the air.[/b]

Ahh let us check this out.

[b]It can be easily proved by calculation:

At ANY TIME the specific heat of atmosphere approximately equates to 1.0012 [J/g/K – heat capacity of air] multiplied by mean temperature of the atmosphere [~ 255K] and by atmospheric mass [5.148×10^21 g] = 1.31×10^24 J. While the latent heat circulated by water cycle is only 1.14×10^24 J in whole year – it clearly can’t be there at any time, because it all the time dissipates via conversion to specific heat and subsequent radiative loss to space (all bodies – including body of atmosphere – radiate if they have over absolute zero temperature).[/b]

Good! Some one that is willing to run the numbers.

I agree! Why is sunsettommy screwing with my posts? John?

## Jan

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The heat dissipation from the atmosphere to space via radiative loss is more or less equal to the heat gain from absorbtion of the solar radiation by Earth (+the approx. 0.09 W/m^2 geothermal heat) they always tend to equilibrium – and it can’t be otherwise, because the radiative loss rises with fourth power of temperature – so if surface temperature rises (for example because the solar activity rises), the heat dissipation rate rises accordingly – that’s what the full expression of the Stefan-Boltzman for radiative loss [P=epsilon.sigma.Area(T^4-Tc^4)] in fact determines – if the Tc is always same [near absolute zero], the sigma is constant, the Earth surface area also stays same, the emissivity stays same and only the air temperature rises (because the land and especially sea surface temperatures rise heating the air), than the radiative loss rises with temperature accordingly.

That’s why a runaway surface warming (the Hansen scares us about for decades, never realizing what a fool he makes from himself …especially given his education in physics he should understand basic principles of thermodynamics) is physically impossible – because you would need to get somewhere ~5.5W/m^2 forcing for every degree of the surface temperature rise. -The figure direcly comes from Stefan-Boltzman law:

To demonstrate: if you say have the average global surface air

temperature 288 K, the surface in average radiates:

0,00000005670373*288^4 = 390.1 W per square meter

if the surface temperature rises one degree then the surface would radiate:

0,00000005670373*289^4 = 395.6 W per square meter.

395.6 W/m2 – 390.1 W/m2 = 5.5 W/m2

(which you would need to add to the system, for making the surface temperature in average rise the 1C)

– this also shows that the recycled IPCC prediction from the brand new AR5 -of the “most likely” 3C temperature rise for CO2 atmospheric content doubling is a nonsense – The Stefan-Boltzman law and what the IPCC itself says about CO2 forcing prima facie exclude such possibility – the CO2 doubling from preindustrial levels (even if it would be possible by anthropogenic emissions – which isn’t – we simply don’t have enough carbon available for such feat) clearly wouldn’t be able to add the ~16,5 W/m^2 forcing inevitably needed for rising the global average surface air temperature the 3C.

…I only wonder how such nonsenses as the “most likely” “3C” IPCC prediction could gain so much attention and nobody sweeps it from the table as prima facie utterly physically impossible once for all.

## Bob Armstrong

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The final slide in my .ppt , save one of a quote from Lenin relevant to this global statist stupidity , shows what utter nonscience Hansen’s claim that Venus’s surface temperature is an example of a [i]runaway greenhouse effect[/i] : http://cosy.com/Science/VenusCalc.jpg . It should have never gotten past pal review .

Good to see you come to the same conclusion by a different path .

## W5OVF

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test

…I only wonder how such nonsenses as the “most likely” “3C” IPCC prediction could gain so much attention and nobody sweeps it from the table as prima facie utterly physically impossible once for all

## Greg House

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[i]the solar irradiance of one square meter of the Earth surface (atmosphere) which isn’t immediately reflected to space is:

1361.1 / 4 x 0.7 = 238.19 W (±0.08) per square meter [Eq.2][/i]

Jan, why exactly do you divide 1361.1 by 4?

Let me guess: is it because the dark side of the Earth “gets nothing”, right?

If yes, would you agree that “getting nothing” is equivalent to the temperature of the dark side being -273°C (absolute zero)?

## Jan

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Yeas I divide it by 4 because sphere surface area is four times bigger than surface of a circle of same diameter.

To your second question: No, surface materials have usually quite high heat capacity, so it takes time for them to cool. It cant happen overnight.

## Greg House

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Jan, there are circles, triangles and many other geometrical figures, so what? Your explanation does not make sense at all.

What you effectively have done in your calculation is that you implicitly assumed that the dark side of the Earth “gets nothing”, as warmists usually put it, so you divided what one side “gets” between 2 sides.

If you did not committed the error of deriving average temperature from average radiation, which is false because the dependency is not linear, so if you derived your “average temperature of the earth” from average temperatures on both sides assuming that the dark side “gets nothing”, then the temperature on the dark side in your calculation would have been -273°C (absolute zero).

Maybe you did not understand all that and just reproduced this well known warmists” calculation, I do not know, but this a total crap, sorry.

## Bob Armstrong

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The calculation of 279 or so is precisely correct for a flat spectrum ball in our orbit . See http://climatewiki.org/wiki/Category:Essential_Physics for a derivation from a somewhat different approach . Our temperature is linear with the temperature of the sun .

## Jan

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This is most probably a mere misunderstanding.

It’s high-school school mathematics:

I divide the average TSI (all SORCE-TIM era) by four to get average TOA TSI per square meter of WHOLE Earth (not just one side) and multiplied it by Bond albedo, to get average surface TSI per square meter of WHOLE Earth (not just one side).

I don’t postulate flat Earth top side insolated, nor steady Earth with other side dark, nor with absolute zero temperature.

There’s in principle NOTHING WRONG with the equation, nor it in any way implies absolute zero temperature anywhere. It copes with radiant power per square meter in the first place, not in any way with temperature. The resault of the equation IS more or less correct (if the Bond albedo value is sufficiently correct – which I discuss in notes).

But I’m very well aware the result from equations like this is just a global average simplification, which doesn’t say anything about effective insolation and the imbalance between northern and southern hemisphere – which I tryied to note about in the article – nor about differences between different latitudes. It’s just a global average, nothing more.

-That’s why I’m working on global insolation grid, which will in 1-degree resolution cope with uneven land/sea stratification (-preliminary data fully describing the global land/sea stratification with 1 degree resolution you can find here: tumetuestumefaisdubien1.sweb.cz/GLOBAL-INSOLATION-GRID-1DEG.ods)

– which is important, because sea surface has generally very significantly lower albedo than land surfaces and therefore effectively absorbs considerably more irradiation, converting it to heat than land surfaces.

## Greg House

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[quote name=”Jan”]This is most probably a mere misunderstanding.

It’s high-school school mathematics:

I divide the average TSI (all SORCE-TIM era) by four to get average TOA TSI per square meter of WHOLE Earth (not just one side) and multiplied it by Bond albedo, to get average surface TSI per square meter of WHOLE Earth (not just one side).[/quote]

You can perfectly mathematically divide whatever by whatever but this does not mean that your result is what you claim it is [b]physically[/b], in this case the “average temperature of Earth”, see my next comment.

## Jan

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Unfortunately I don’t see any next comment from you. But I have somehow strong feeling that what you cite is primarily about average TOA TSI, not about average temperature of Earth.

To clarify: When I was talking about “average temperature of Earth’s surface”, it was about average temperature of atmosphere and the way I estimated it was not using TOA TSI, but using ISA model, I came to the average temperature of atmosphere the ~ -18°, determined how much blackbody would radiate at such temperature and only then I compared the result to the average 1AU TOA TSI to find out the results are very simmilar and their slight difference could well be a result of Bond albedo uncertainty.

It was just a omparison showing startling simmilarity between the theoretical radiation of a blackbody at 255K and the average TOA TSI, which I for purpose of comparative demonstration determined exactly the way the warmist do.

## Bob Armstrong

| #

Jan ,

I appreciate what you are doing , and like I say , I want to look at your analysis of the vertical structure of atmospheric energy balances if I ever am motivated to implement more detailed computations .

But , it should be made clear that 255k is not the temperature of a black ( or gray ) body in our obit it is the temperature of a body with a weighted average absorption with respect to the Sun’s spectrum of 0.7 and an absorption=emission spectrum with respect to longer wavelengths of 1.0 . The temperature for a truly flat spectrum , however dark or light , is about 279k . That is the black body temperature .

## Jan

| #

I don’t know what has the blackbody temperature 279K and I don’t know how you got the figure, but the average temperature of our atmosphere coming from the standard atmospheric model used globally for aviation purposes is around 255K – Which is figure well agreeing with the temperature figure resulting from Stefan-Boltzman law for a surface of a planet with Bond albedo value 0.298 used by Trenberth. That’s all I know for sure in this respect.

## Bob Armstrong

| #

This is a problem .

A black body is defined as one with an absorption=emission coefficient , [i]ae[/i] perhaps it should be called a Kirchhoff coefficient of exactly 1 across the entire spectrum .

Simply integrating the energy impinging on a point in our orbit , almost all of which comes from the 5.4 millionths of the celestial sphere covered by the sun’s disk . That comes out to am energy flux of about 342 w%m^2 which by Stefan-Boltzmann 4th power relationship works out to about 279k .

Note that any level of gray ( constant [i]ae[/i] ) other than the 1.0 of a black body will drop out of the equation . Thus this is the temperature for any gray ball . Thus this value is orthogonal to any variations [i]ae[/i] across the spectrum .

I’ve put together power point summarizing the analyses on my http://CoSy.com : http://cosy.com/Science/AGW.ppt . It goes thru the calculation for the assumption of an average reflectivity with respect to the solar spectrum of 0.7 and 1.0 in the longer wavelengths which produces the ubiquitous 255k assertion . It goes thru the calculation for any arbitrary [i]ae[/i] spectrum .

Note that 0.7 assertion has little to do with surface reflectivity ( dislike the term [i]albedo[/i] for various reasons ) . My guess is , it’s largely due to clouds . As a first approximation to the surface , I approximated the reflection spectrum of 10 meter deep water . That has a reflectivity of about .08 with respect to the solar spectrum giving an equilibrium temperature of about 273.4 .

That would imply a “greenhouse effect” of around 15c rather than 33c .

Curt mentioned an estimate of about [i]ae[/i] 0.95 in the long wavelengths and considered rounding that to 1.0 “good enough for government work” . No it’s not . The total change in temperature since the invention of the steam engine is only on to order of 0.3% , an order of magnitude smaller . For the record , 0.95 works out as follows :

.7 % .95 />/ 0.737

r ^ % 4 />/ 0.926

r * 278.7 />/ 258.2

a difference several times greater than the less than 1 degree this whole fuss is about . I hope my notation is self explanatory .

I don’t have time to delve in to it now . As long as calculating the temperature of a simple colored ball is not firmly understood , I see little point in going further .

But I’m curious about where the 255k value for atmospheric temperature is derived . One thing that is frequently forgotten in such energy balance calculations is that the sun’s energy is virtually “undiluted” in at higher altitudes and cannot be ignored . Living at 2500 meters above Colorado Springs , we can feel it .

## Greg House

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[quote name=”Jan”]Unfortunately I don’t see any next comment from you.[/quote]

A few comments have disappeared, including yours. Maybe a software bug. I’ll repost mine.

## Greg House

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[quote name=”Jan”]Spherical surface of Earth has only 2x cross-sectional area for solar irradiation, than it would have a circle of same diameter – because the sun is only one in our solar system and always insolates only half of that Earth sphere, from one side. Only because Earth rotates around its axe once a day makes the insolation distribute over both sides.[/quote]

This is so blatantly wrong.

There is no “distribution” of a certain amount in your calculation, what you have done is cutting the solar [b]power[/b] in half (referring to 2 sides) and this is exactly what is completely false in your calculation. The heating power is not “distributed” at all, it only results in different temperatures depending on some parameters. Trying to distribute [b]power [/b]is simply false.

To illustrate that, let’s consider a simplified case, where the Earth is flat on both sides and instead of rotating just switches the sides twice a day. So, the Sun heats one side [b]by it’s full power[/b] and then it heats the other side [b]by it’s full power as well[/b], not by half of it. The dark side cools then, of course, for 12 hours. The side facing the sun is heated for 12 hours. The average temperature would depend on the rates of warming and cooling. If radiative cooling was slow, both sides would stay at [b]almost maximum [/b]temperature the solar power can produce according to the SB law. There is no room here for your ridiculous dividing the solar power by 2.

## Jan

| #

The Earth rotation distributes the solar radiation going through the Earth diameter cross-section over the whole surface of Earth.

It is so obvious that to engage in discussion with somebody who repeatedly denies it and instead comes with flat-earth examples doesn’t look to me being much seminal and I would think it should mark end of discussion with you. But I’m still patient enough.

It is also obvious even the insolated side is not insolated evenly, and by full power – because Earth is sphere – e.g. near terminators the insolation of the surface is much lower than in the middle of the cross-section (noon), while it is in turn the atmosphere there which absorbs bulk of the solar power going through the crossection there. (Just to imagine the magnitude only the terminator tropospheric crossection of the atmosphere constantly absorbs several times more energy from Sun than is the difference between solar minima and solar maxima TSI per whole Earth and some part of this light is scattered by the atmosphere [always colimating part of the light towards surface as the result of air having higher refraction index than vacuum] towards the surface beyond the terminator – that’s why there is still not dark after sunset.)

The insolation of the bright side surface goes down with cosinus of the longitude from the top of the bright side hemisphere, which constantly moves with both rotation (day) and ecliptic (year) and with (sinus(atmospheric optical depth) + atmospheric optical depth)/2. The vertical atmospheric optical depth moreover goes down with latitude, but the function is not simple, there are different atmospheric conditios between tropics, between tropics and polar circle and beyond polar circle, developing the characteristic Polar, Ferrel and Hadley cells. The effective surface insolation at the top of the insolated hemisphere (and everywhere else) also depends on whether it is land or ocean what is insolated there, because they have considerably different albedos.

This all shows that to imagine Earth as a discworld is same way worthless as to imagine that the side which is insolated at the moment is insolated everywhere by same power. It is not the case – even if we average over day the higher latitudes are still less insolated than lower ones, even if we average over the year the effective insolation of southern hemisphere is still higher than the northen one, because at the southern hemisphere is generaly more ocean (having considerably lower albedo than land).

It is possible to develop an effective insolation model of the Earth’s surface based fully on real data and real sea/land stratification (not on arbitrary global constants fudging to make the flat-earther models at least seemingly work …but obviously without any predictive value whatsoever) to get more realistic numbers for the surface heat budget than from the global averaging as Trenberth does (and I used the alarmist overtly-simplistic method for the sake of comparison), but it is definitely not a simple task.

## Greg House

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[quote name=”Jan”]The Earth rotation distributes the solar radiation going through the Earth diameter cross-section over the whole surface of Earth.

It is so obvious that to engage in discussion with somebody who repeatedly denies it and instead comes with flat-earth examples doesn’t look to me being much seminal and I would think it should mark end of discussion with you. But I’m still patient enough.[/quote]

I am patient, too. I re-posted 2 lost comments and I guess you wrote your answer before reading the 2nd one, where I demonstrated by a simple example that your approach contains an error resulting in too low average temperature.

I suggest you read the 2nd comment: http://principia-scientific.org/latest-news/320-do-we-really-have-a-33-c-greenhouse-effect.html#comment-2046

## Greg House

| #

[quote name=”Jan”]I actually don’t cut the solar power in half, I cut it in quarter. (-I divide the TOA 1AU 10 year average TSI by four not by 2, because the total surface area of a sphere equals four times the circular crosssection – surface area of a circle having Earth radius, not 2 times).[/quote]

Again, it is wrong to cut it in quarter, and I illustrated that by an example of a flat earth, where in accordance to your approach you would cut it in half. That is why I was talking about cutting in half and I am surprised that you can not follow such a simplification. I simplified the case to make the error in your approach obvious to the readers and to you.

Let me try it a slightly different way. At any given moment only a half of Earth (a hemisphere) is being heated by the full power of the Sun. The other half [b]has already been heated by the full power of the Sun[/b] and is cooling. Get it? So, to determine the average temperature you must take this into consideration, which you failed to do.

Let’s come back to my simplified scenario with a flat earth and compare your approach to mine. So, we have a flat planet of 2m² area (1m² each side) switching sides twice a day, the Sun heating by 1000 W/m² perpendicularly to the area. There is no cooling in your scenario, so let’s ignore cooling here as well. Let’s also assume instant warming of the side facing the Sun and the emissivity=1.

The temperature of the side facing the Sun would be according to the SB law 364K. The temperature of the dark side would be the same 364K, because it has already been warmed [b]by the same full power[/b] of the Sun (when it faced it). The average temperature of the whole surface would be 364K then.

According to your and the warmists approach, since, as you put it, the total surface area of the planet equals 2 times the cross section (which is correct), the solar power should be cut in half (yielding 1000/2=500) and then the average temperature should be derived from this 500 using the SB law, so the result would be 306°K. This way you get a 48K [b]lower temperature[/b] as a result [b]for no physical reason[/b]. The Sun, however, just warms one side and then the other side, and the temperature of each side has nothing to do with the temperature of the other one, these temperatures are unrelated to each other. If there were 100 sides, the average result would be the same, each side would be heated by the full power, not by 1/100 of it. It becomes obvious if one averages the temperatures.

## Bob Armstrong

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The fact that the simple calculation of even a radiantly heated gray ball is in dispute shows the retardation of the understanding of the most basic physic in the AGW debate and the cause of its utter stagnation over decades .

I’ve given links to the classic quantitative expressions needed to calculate any of these arrangements and indeed much more general cases . I won’t bother repeating them .

## Jan

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Again, I didn’t determined the average surface temperature from the insolation in my article at all.

What I did there was that I determined the average temperature of the atmosphere (which is indeed the surface of Earth for purposes of global radiative equilibrium), nit from SB law but from ISA model (and compared resulting blackbody radiation to solar input) – this average atmospheric temperature (~255K) in itself determines the cooling (so it is untrue to claim I failed to consider the cooling) – all bodies which have nonzero temperature, including our atmosphere, radiate according to SB and Wien’s law. In fact the IR radiation to space is ONLY way the surface of Earth dissipates heat to space-cools. Capito?

Sun continuously warms the Earth surface, more at equator, much less at the poles, more when in zenith, much less at the terminator and beyond (in fact the atmospheric colimation causes Sun warm Earth surface even beyond the terminator).

To divide Earth surface to insolated and uninsolated halves makes not much sense, because Earth rotates fast, so in just 1 day the longitudinal differences are wiped out as are wiped out the differences given by the obliquity in just 1 tropical year. Only differences in effective insolation which rest then are the differences given by latitude and land/sea uneven stratification.

The average TOA insolation per square meter of the Earth surface in SORCE-TIM era (since 2003) was 1361.1/4 = 340.275 W/m^2, a part (according to Trenberth 0.298 – but I note Bond albedo significantly varies with cloudiness dependent on extraterestrial GCR seeding factor, partially depending on solar activity, but let’s pretend the Trenberth value is more or less correct average) of that got reflected and the bulk got absorbed by atmosphere and sea and land surfaces – 1361.1/4*(1-0.298) = 238.87 W/m^2 in average – of course much more at equator, much less at poles.

No significant surface temperature anomalies trend was observed, so we can assume the decade being more or less at the thermal equilibrium which generally means also Kirchhof equilibrium. Therefore the average Earth surface (atmospheric) temperature I determine from insolation (now for you first time I ever did something like that here) is slightly higher than (238.87/0.00000005670373)^0.25 = 254.76 K (-18.4 °C) (I say slightly higher because air emissivity is slightly lower than 1).

The result relatively well agrees with the average atmospheric temperature derived from the ISA model (-17.91 °C – not -18.13 °C – which is error I made in the article nobody pointed out – so I do).

And BTW the total surface area of a planet (including Earth) equals 4 times the insolation cross-section of same radius, not 2 times – which you now deem “correct”, adding another absurdity to this discussion. – When I was talking about the 2 times it was obviously about the actually insolated hemisphere.

The rest I’ll rather not comment, because your reasoning collapses with this blatant error.

## Greg House

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[quote name=”Jan”]And BTW the total surface area of a planet (including Earth) equals 4 times the insolation cross-section of same radius,[/quote]

Again, even if the total surface area of a very very big planet was warmed only by the solar radiation going through just a tiny tiny cross-section, this circumstance alone would not affect the average temperature of the planet at all. If we ignore rate of cooling, of course, which you did.

The point is that this solar radiation warms one spot, then another and so on [b]by its full power[/b]. If there are 1,000,000 spots warmed [b]in succession[/b], then each of them will be warmed [b]by the same power[/b] and you still can not cut the solar power by 1,000,000 and then derive the average temperature from the resulting value. Get it?

You really really need to make an effort here.

## Jan

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It is always some unrealistic examples you use. And I definitely will not make any effort to pursue such nonsenses.

The reality is that in just one day (one Earth revolution) the average TOA insolation of the whole Earth is already ~340 W/m^2 and due to partial reflection and partial absorbtion of the irradiance by the atmosphere then the average insolation of surface comes out about ~239 W/m^2 – much more in tropical regions, much less in polar regions, absorbed more by the sea, less by the land.

The effective (absorbed) insolation directly determines the temperatures – because it is the absorbed Joules which add to the heat content, rising temperature due to the limited specific heat capacity of the materials.

Again, this temperatures itself directly determine also the radiative transfer – cooling – for adjacent bodies as the planet and its atmosphere under this law: P=epsilon.sigma.area(Th^4-Tc^4) – where epsilon is average emissivity of the radiating planet, sigma is Stefan-Boltzman constant, area is the interface area between the planet and the atmosphere, Th is the absolute average surface temperature of the planet, Tc is the absolute average temperature of the atmosphere (the same works then for atmosphere and adjacent space – just the epsilon then is the emissivity of the air, Th is then the absolute average temperature of the atmosphere and the Tc is then the absolute temperature of space.

So it is clearly impossible to ignore cooling when we talk about couple of hundred K temperatures on the planet in the near absolute zero temperature space – the temperatures itself imply cooling via radiative transfer. Your argument that I ignore cooling is therefore moot.

## Greg House

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[quote name=”Jan”]So it is clearly impossible to ignore cooling when we talk about couple of hundred K temperatures on the planet in the near absolute zero temperature space – the temperatures itself imply cooling via radiative transfer. Your argument that I ignore cooling is therefore moot.[/quote]

I suggest you not distort my point.

Yes, you ignored the rate of cooling when cutting the solar power by 4 only based on the relationship between the cross-section and the total area. No rate of cooling was involved in that action of yours. You just lowered the average temperature deriving it from a false value of “average power”.

## Jan

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When one calculates incoming power (- and one must do it separately, because the incoming energy is in form of shortwave spectrum with peak wavelength 502nm in visible region, based on the temperature of solar surface, part reflected by albedo for such spectrum, while the outgoing power is in form of longwave mid-IR spectrum based on temperature of the planet) it IN PRINCIPLE is not calculation of cooling, but calculation of power purely causing heat content rise, causing warming. It doesn’t mean one ignores the simultaneous heat dissipation by the longwave radiation – the cooling, going on all around the planet 24/7. It is just not what one is calculating, when calculating the incoming solar radiation power – which surely never causes any heat content decrease – cooling. So as you hopefully now see I don’t ignore it, it is just not at all the subject of the incoming power calculation.

## Greg House

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[quote name=”Jan”]It is always some unrealistic examples you use. And I definitely will not make any effort to pursue such nonsenses.[/quote]

Unrealistic?

The example I used was not designed to be “realistic”, it was a simplified example to demonstrate how absurd the approach you used was.

It was specifically designed that way to make it possible for lay persons to understand, what kind of unscientific crap your calculation and similar ones of warmists’ was.

## Bob Armstrong

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Greg ,

Just exactly what are the values you calculate ? I have not seen what you contend is the correct answer for anything or how you calculate it .

## Greg House

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Here is my answer, Bob.

I imagine me acting as a sort of defense attorney on behalf of Mr.CO2 falsely accused of “warming the world”. As you possibly know, this accusation is partly based on a certain bogus calculation warmists and some useful idiots like so much.

I find it sufficient to demonstrate by what sort of trick they made the world colder than it is and I have absolutely no interest in finding out how exactly warm the world is.

Your position is similar to the one of a judge who asks the defense “Who killed the victim then?” after it has been proven that the accused has an alibi.

## Bob Armstrong

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Sorry , I only respond to computations which produce testable quantitative values . It appears you cannot do that . So whatever you are saying macht nichts .

On the other hand , the calculations I’ve presented are basic geometry and 19th century insights which are tested every time someone uses an IR thermometer .

## Jan

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I don’t believe Mr. CO2 is a culprit either. Here’s why:

The CO2 in atmosphere is well mixed and carries only

## Jan

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The discussion here really sucks, most of my reply disappeared immediately by posting.

So again: I don’t believe Mr. CO2 is culprit either – here’s why:

The CO2 carries only

## Jan

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Because this system is unable to post my reply in whole, here it is in txt:

http://tumetuestumefaisdubien1.sweb.cz/reply.txt

## Tim Folkerts

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[quote]P=epsilon.sigma.area(Th^4-Tc^4 )

– it clearly doesn’t depend on emissivity of the atmosphere![/quote]

This is the “freshman” level version of the equation, designed only for very simple cases. A more complete version can be found, for example, here: http://en.wikipedia.org/wiki/Thermal_radiation#Radiative_heat_transfer

This DOES include the emissivity of both materials.

[quote]… but it [thermal IR] rises with emissivity in all directions including that towards the space[/quote]

This is subtle. The IR to space from the cold atmosphere increases with more GHGs. But the IR from the warm surface decrease (because more gets blocked). The net effect of MORE GHG is LESS IR to space (at least until everything warms up and/or other changes occur to restore balance).

******************************

Other factor, like the sun or oceans that you mention later in this reply, will indeed also be important players in the climate. It is certainly possible that internal feedbacks (eg evaporation) or external changes (eg the sun or cosmic rays) can party overpower the changes due to CO2, but that is VERY different from your argument that CO2 is intrinsically incapable of warming the earth.

## Greg House

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[quote name=”Tim Folkerts”]This is the “freshman” level version of the equation, designed only for very simple cases. A more complete version can be found, for example, here: http://en.wikipedia.org/wiki/Thermal_radiation#Radiative_heat_transfer

This DOES include the emissivity of both materials.[/quote]

What has this formula to do with gases?

## Jan

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Tell me, where you see a non-“freshman level” equation at the said link which does include emissivity of both emiting and receiving materials?

..I hope you don’t confuse it with formula for the two surfaces greybody…

What I see there is a formula determining “radiative heat transfer from one surface to another as equal to the radiation entering the first surface from the other, minus the radiation leaving the first surface” -which could be interpereted using emissivities on both bodies -if only the emissivity wouldn’t be so intimately linked to absorbtivity by Kirchhof law -as if the mother nature prefered 2nd law of thermodynamics over the IPCC agenda. 😉

How much the “GHG’s”, which add to the absorbtivity of the air and as you acknowledge add also to its emissivity, “block” the IR from surface and purportedly cause the surface warm?

The changed emissivity of the atmosphere, if ever, in my opinion would intervene only intermittently until equilibrium in it is (quickly – IR moves at speeds close to c in air) established, after that it will not much affect the radiative transfer from surface to space through the atmosphere, because it will not much change its average temperature – at least not by the minute amounts of the added CO2, which in reality could have changed the atmospheric temperature only like in order of 0.1C (a temperature change equivalent to the atmospheric heat content change caused by the carbon content rise in it since preindustrial level – and so no “intrinsic incapability”), which would change the radiative transfer from surface to atmosphere ~0.38W/m^2 and the same it will change the radiative transfer from atmosphere to space (so no net atmospheric temperature change due to emissivity change results ever on more than subsecond scale and only what rests is the atmospheric temperature change due to changed heat content, lowering radiative transfer from surface 0.38W/m2, causing its ~0.1C warming).

I would think the subtle changes in the atmosphere due to the rising minute CO2 content can’t much add to, nor offset the observed global surface temperature anomaly changes of almost order higher magnitude than could be caused by the CO2 rise, and so caused presumably by other factors – as the observation, especially of the more than decade long non-correlation of the CO2 atmospheric content rise with surface temperature anomalies would suggest, and while we still find distinct solar activity signal in the temperature data and while solar activity indices show rising solar trend exactly up until beginning of 2000’s -links in my previous reply.

Yes, a correlation doesn’t prove causation, but where no correlation is present whatsoever for N=100 and higher, there a statistician hardly will expect a causation – whatever hypothesis supports it.

Wouldn’t you think that if the solar activity changes or whatever factor as you say overpower the temperature rise purportedly due to CO2 -moreover in the time of its fastest atmospheric content rise in history, it simply means the CO2 is not the chief factor causing the temperature change?

## Tim Folkerts

| #

Yes, that equation for the radiative heat transfer is the one. The heat depends on …

* A1 the area of object 1

* A2 the area of object 2

* F12 the geometry

* ε1 the emissivity of object 1

* ε2 the emissivity of object 2

That pretty clearly involves both emissivities. The other link has basically the same result.

[quote]if only the emissivity wouldn’t be so intimately linked to absorbtivity by Kirchhof law [/quote]

I’m not sure what your objection is. Kirchhoff says (ε1 = a1) and (ε2 = a2).

It says nothing about (ε1 = ε2). Different materials can and do have different emissivities.

## Jan

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Please look at the equation, try it! You’ll hopefully find out that if you leave ε1 same and go up with ε2 the resulting radiative transfer will go UP with it – while for say rising emissivity of the air (caused by CO2) where say ocean radiates, in the paradigm of the “greenhouse effect” hypothesis you would expect exact opposite.

The equation is for ONE gray body with two surfaces of different emissivity – could be useful say for the planet when you’ll have one average emissivity for ocean, say ε1, the other ε2 for land – then A1 would be area of ocean interface with air, the A2 area of land interface with air and result will be radiative transfer P for the planet surfaces to atmosphere.

It has clearly still nothing to do with emissivity of the receiving body whatsoever.

Also you could try what it will do if T1 will be lower than T2 – you will see that regardless what the emissivities are the result will be – in accordance with 2nd law of thermodynamics – a negative figure (but nonsensical, because you would apply emmisivities of receiving material to the radiating material). So please try equations, understand what they are intended for and how they work before you claim something, which if I would use the way you think it could be used, I would get result that enhanced emissivity of the atmosphere caused by CO2 will cause cooling of the surface – which I think is not exactly what you want me to come to.

My objection with Kirchhof for the atmosphere was that if its absorbtivity rises (say due to CO2) it rises also its emissivity – so more heat will get dissipated to space OR the temperature of air will descend if more radiation to absorb and dissipate not available at the spectral regions of the added absorbtivity – facilitating so heat dissipation from the surface interfacing the atmosphere from other side.

There is IR radiation of the atmosphere but as long as the atmosphere is colder (because it constantly radiates to near zero temperature space, not insulated from it by anything) than the surface it is unable to transfer single Joule there, you must heat the atmosphere for making heat dissipation rate from surface to atmosphere lower (and it will be warmed by sun) – an IR absorbtion would do, IR emission however would cool it back. Therefore what are factors which could cause the surface warm is either warmer atmosphere, or higher TSI (which will also warm the atmosphere), or less clouds (possibly caused by less GCR, caused by higher solar activity), or lower surface albedo, resulting in warmer surface which will warm the atmosphere until new equilibrium established. I’m afraid you can’t much warm atmosphere by adding still minute amounts of CO2 there. (I wish we could avoid iceage just by burning fossils, but I’m afraid it is not the way which will work – the natural sequestration is too high, the reserves of fossils too low and the CO2 effect too small).

## Tim Folkerts

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Please look at the equation more carefuly, try it! You are correct that if you leave ε1 same and go up with ε2 the resulting radiative transfer will go UP with it. However, in the paradigm of the “greenhouse effect”, t[i]his is exactly what we expect! [/i]

Read through the description at wikipedia:

http://en.wikipedia.org/wiki/Idealized_greenhouse_model#The_model.

They show that increasing the emissivity of the atmosphere does indeed increase the temperature of the ground.

## Jan

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No, it isn’t what you’ll expect! The temperature of the surface, nor atmosphere is what is changing in your equation (-unlike as in the “idealized” model at the page you link now, which really scandalously assumes atmosphere absolutely transparent, but operates with albedo, planet surface being a blackbody -which it by far isn’t -while absurdly operating with emissivity, and makes whole chain of incorrect conclusions, which literally (self)”evidence” themselves by circle because the changes made to Ts and Ta are exactly solely attributed to the “greenhouse” effect emissivity changes using the tiny ε/2 in the Ts equation – while the Ts and Ta changes on real Earth can of course have whole bunch of other causes…) .

I thought you’re brighter and you will try your equation thoroughly before you teach eagle fly.

If you did, you would find that both ε1

and ε2 drive the resulting Q absolute value UP, without any changing the T1 and T2 whatsoever. (-as one would indeed expect with equation intended to calculate two different emissivity surfaces greybody radiation – which is what the equation clearly is intended for – it is even quite clearly stated there)

-So if you would e.g. use (incorrectly) the equation for radiative transfer from surface, thinking that for ε2 you use the elevated emissivity of atmosphere (and for ε1 the unchanged emissivity of surface), then the elevated ε2 would drive the resulting Q from surface to atmosphere !UP! -exact opposite than what you would like it to do in the “greenhouse effect” paradigm – because while the surface temperature didn’t change a bit nor atmospheric temperature, the radiative transfer from the surface would go anyway up with ε2 – and should therefore cause COOLING of the surface. (-but the equation is intended for something else and therefore the described result is not coming from it.)

Same will happen if you use the equation for radiative transfer from atmosphere back to surface and to ε1 you will put the elevated atmospheric emissivity, to ε2 the surface emissivity, to T1 the temperature of the atmosphere (lower) to T2 the temperature of the surface (higher) – the resulting Q will be a negative number, which nevertheless will get more negative with rising both ε1 or ε2, suggesting exactly same result as if you used it for radiative transfer from the surface, a number of exactly same absolute value, just with negative sign caused by swapping unchanged T1 and T2 values, still suggesting the surface radiation go up with elevated ε, without change of its temperature. (but again I note, the equation is intended for calculation of something else)

I’m really astounded to what lenghts of absurdity the “greenhouse” groupthink leads people, that they apparently stop thinking, and even somebody patiently tryies to explain them why they are mistaken, they don’t listen and keep making fools from themselves, claiming something, which the reality contradicts prima facie. (and not just by this equation, which is clearly not even for purpose this groupthink people apparently firmly believe and which works exactly opposite way they believe it works, but on many other instances)

## W5OVF

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[quote name=”Jan”]Unfortunately I don’t see any next comment from you. But I have somehow strong feeling that what you cite is primarily about average TOA TSI, not about average temperature of Earth.

To clarify: When I was talking about “average temperature of Earth’s surface”, it was about average temperature of atmosphere and the way I estimated it was not using TOA TSI, but using ISA model, I came to the average temperature of atmosphere the ~ -18°, determined how much blackbody would radiate at such temperature and only then I compared the result to the average 1AU TOA TSI to find out the results are very simmilar and their slight difference could well be a result of Bond albedo uncertainty.

It was just a omparison showing startling simmilarity between the theoretical radiation of a blackbody at 255K and the average TOA TSI, which I for purpose of comparative demonstration determined exactly the way the warmist do.[/quote]

Yes indeed, you use the warmest technique

average everything until there is no understanding left!. Your math is very good. Glad to see someone do that, rather than handwave. The actual thermodynamic

processes on this planet change from second to second. These changes determine what “is”. The averages only describe what

“might have been likely”, but is not!

## W5OVF

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[quote name=”Jan”]Yeas I divide it by 4 because sphere surface area is four times bigger than surface of a circle of same diameter.

To your second question: No, surface materials have usually quite high heat capacity, so it takes time for them to cool. It cant happen overnight.[/quote]

Hi Jan

Indeed the spherical surface has 4x the cross-sectional area. So what? Think of what does happen. Surface + atmosphere gains sensible heat when the Sun is in view. The atmosphere radiates this heat to space at all times in all outward directions to a lower temperature. Most of the heat in the atmosphere comes from convection with he tremendous latent heat of evaporation of water. If you wish to understand, you must think, hard and difficult to do, but that thinking is all you have that is truly yours!

## Greg House

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We can forget the atmosphere to simplify the matter and focus on this dividing by 4.

What surprises me is that reasonable people understand intuitively that the AGW thing is a hoax, but still uncritically accept some of their fictions, like this calculation above and also the “global temperatures”, “warm periods” or “ice ages” or even “global temperatures” from millions of years ago.

## josullivan

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I can advise that Jan has written in to PSI to advise that he is looking to re-visit this and make corrections/updates as necessary, and taking into account feedback provided herein. Thanks for all your comments, they help inform the ongoing peer-review process.

## Jan

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Spherical surface of Earth has only 2x cross-sectional area for solar irradiation, than it would have a circle of same diameter – because the sun is only one in our solar system and always insolates only half of that Earth sphere, from one side. Only because Earth rotates around its axe once a day makes the insolation distribute over both sides.

Most of the heat in the atmosphere comes not from convection, nor latent heat of water evaporation. – most of the heat (>65%) is created by the absorbtion of EM radiation coming both from sun and from the warmer surface of the Earth. To claim otherwise is clearly a flagrant denial of basic knowledge.

-It’s good to think faster than write…

## Greg House

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[quote name=”Jan”]Spherical surface of Earth has only 2x cross-sectional area for solar irradiation, than it would have a circle of same diameter – because the sun is only one in our solar system and always insolates only half of that Earth sphere, from one side. Only because Earth rotates around its axe once a day makes the insolation distribute over both sides.[/quote]

This is so blatantly wrong.

There is no “distribution” of a certain amount in your calculation, what you have done is cutting the solar [b]power[/b] in half (referring to 2 sides) and this is exactly what is completely false in your calculation. The heating power is not “distributed” at all, it only results in different temperatures depending on some parameters. Trying to distribute [b]power [/b]is simply false.

To illustrate that, let’s consider a simplified case, where the Earth is flat on both sides and instead of rotating just switches the sides twice a day. So, the Sun heats one side [b]by it’s full power[/b] and then it heats the other side [b]by it’s full power as well[/b], not by half of it. The dark side cools then, of course, for 12 hours. The side facing the sun is heated for 12 hours. The average temperature would depend on the rates of warming and cooling. If radiative cooling was slow, both sides would stay at [b]almost maximum [/b]temperature the solar power can produce according to the SB law. There is no room here for your ridiculous dividing the solar power by 2.

## Jan

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I actually don’t cut the solar power in half, I cut it in quarter. (-I divide the TOA 1AU 10 year average TSI by four not by 2, because the total surface area of a sphere equals four times the circular crosssection – surface area of a circle having Earth radius, not 2 times).

It is just a simplification. It doesn’t imply a random square meter of Earth really receives the resulting average radiant power from the Sun, nor that it receives it steadily, nor that it determines a temperature. -I actually didn’t make such link (TSI->temperature) in the article at all!

The net radiative loss (of the land and sea surfaces) in purely physical sense depends on their temperatures and the temperature of the surroundings (air) [P=epsilon.sigma.Area(T^4-Tc^4)].

It is true that the surface temperatures somehow depend on TOA TSI [per square meter of insolated Earth surface] – which varies in time and with location from almost zero at night to ~1405W/m^2 at the time of southern solstice/perihelion phase (when – ~ a noon around the end of December) around the tropic of Capricorn (where), but I actually don’t define their relationship in the article.

To define such relationship (TOA TSI -> surface temperatures) is not an easy task, because there are several factors which play role and some of them not very well understood (for example the cloudiness).

I hope now it is clear.

## Greg House

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[quote name=”Jan”]I actually don’t cut the solar power in half, I cut it in quarter. (-I divide the TOA 1AU 10 year average TSI by four not by 2, because the total surface area of a sphere equals four times the circular crosssection – surface area of a circle having Earth radius, not 2 times).[/quote]

Again, it is wrong to cut it in quarter, and I illustrated that by an example of a flat earth, where in accordance to your approach you would cut it in half. That is why I was talking about cutting in half and I am surprised that you can not follow such a simplification. I simplified the case to make the error in your approach obvious to the readers and to you.

Let me try it a slightly different way. At any given moment only a half of Earth (a hemisphere) is being heated by the full power of the Sun. The other half [b]has already been heated by the full power of the Sun[/b] and is cooling. Get it? So, to determine the average temperature you must take this into consideration, which you failed to do.

Let’s come back to my simplified scenario with a flat earth and compare your approach to mine. So, we have a flat planet of 2m² area (1m² each side) switching sides twice a day, the Sun heating by 1000 W/m² perpendicularly to the area. There is no cooling in your scenario, so let’s ignore cooling here as well. Let’s also assume instant warming of the side facing the Sun and the emissivity=1.

The temperature of the side facing the Sun would be according to the SB law 364K. The temperature of the dark side would be the same 364K, because it has already been warmed [b]by the same full power[/b] of the Sun (when it faced it). The average temperature of the whole surface would be 364K then.

According to your and the warmists approach, since, as you put it, the total surface area of the planet equals 2 times the cross section (which is correct), the solar power should be cut in half (yielding 1000/2=500) and then the average temperature should be derived from this 500 using the SB law, so the result would be 306°K. This way you get a 48K [b]lower temperature[/b] as a result [b]for no physical reason[/b]. The Sun, however, just warms one side and then the other side, and the temperature of each side has nothing to do with the temperature of the other one, these temperatures are unrelated to each other. If there were 100 sides, the average result would be the same, each side would be heated by the full power, not by 1/100 of it. It becomes obvious if one averages the temperatures.

## Greg House

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[quote name=”Jan”]Most of the heat in the atmosphere comes not from convection, nor latent heat of water evaporation. – most of the heat (>65%) is created by the absorbtion of EM radiation coming both from sun and from the warmer surface of the Earth. To claim otherwise is clearly a flagrant denial of basic knowledge.[/quote]

Really? Who proved that and how exactly?

I suggest you’d better not throw around with expressions like “denial of basic knowledge”, given your ridiculous calculation of “average temperature”.

## Jan

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I actually was replying W5OVF (not you), who claimed “Most of the heat in the atmosphere comes from convection with he tremendous latent heat of evaporation of water.”

I showed already in the article it is not the case (the exactness of the proof depends on global precipitation estimation exactness), but that anyway the surface loss of heat via evaporation is at least 30% of the total heat loss from there – and is very important (because also the evaporation rate positively depends on the sea surface temperature, positively depending on solar activity trend – which was rising in the 20th century and from the mid term perspective it was rising up until mid-2000s – – as you can see here: http://tumetuestumefaisdubien1.sweb.cz/SSN1890minima-2012+1964minima2012Trendsv2.gif).

Just to tell the important: I have serious reasons to believe that the solar activity (together with Milankowich cycles) is the chief driver of the Earth climate (and the surface temperatures) on the short and mid-term scale – direcly (by changing TSI) and indirectly (by deflecting more cloud seeding GCR in periods of higher activity). On the really long-term scale however the overal GCR level takes over as the solar system passes regions of higher and lower GCR activity on its way around the center of our galaxy. That’s the core of reasons why I question the CO2 GHE hypothesis.

Who proved that? – we know quite well how much precipitation is there yearly on global scale, so ve can estimate the latent heat (Joules) needed to evaporate the water and compare the result to the estimation of energy (Joules) yearly absorbed by Earth surfaces from Sun (-and presume it must be all dissipated, because heatloss of a body is dependent on forth power of its temperature, so the heat can’t stay there, bacause the temperature of space is near absolute zero). I actually did such comparison in the second section of the article. I even showed the surface heat dissipation via evaporation figure ~30% based on estimation of heat needed for water cycle.

I challenge you: disprove it (that’s what the scientific method is all about). Before you do it, don’t tell me suggestions what not to write ..and avoid throwing around of expressions as “total crap” – when is quite very clear from multiple your subsequent replies you don’t much even understand why I divided the TOA TSI by 4 – while the reason for that is clearly a ground-school geometry and you repeatedly even didn’t noticed I didn’t any way determined any temperature such way.

(and btw, even if the Earth wouldn’t rotate and would have always one side dark, the temperature of the surface there anyway wouldn’t be absolute zero. There are multiple reasons for it.)

And btw: What exactly is “ridiculous” on calculating the average temperature of the atmosphere estimation from the ISA model used in aviation for decades?

## Bob Armstrong

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I take a more basic approach to demonstrating the falsehood of the 255k calculation at http://cosy.com/Science/RadiativeBalanceGraphSummary.html . For the .ppt I’m working on , I calculate the equilibrium for 10m deep water to be ~ slightly above 0c .

I’ll have to look at your analysis of the vertical structure of the atmosphere if I am ever motivated to elaborate my implementation of the essential physics .

## Jan

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I actually don’t demonstrate falsehood of the 255K, what I do is I show using respected atmospheric model widely used for aviation that the 255K or even bit lower is indeed the average surface temperature of Earth – only the “surface” of the Earth in the thermodynamic sense is not the land or sea surfaces, but the atmosphere as whole – the average temperature 255K which works for Moon without atmosphere works also for Earth with atmosphere – of course elsewhere.