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New Climate Model Trumps Flat Earthers of Greenhouse Gas ‘Science’

Written by PSI Members

A team of experts from the “hard” sciences working with climate researchers at Principia Scientific International (PSI) have devised what they believe is an important new energy model of our planet that turns conventional “flat earth” climate thinking on its head.

Published below, the diagram deftly accounts for all the energy Earth receives from our sun without the need to factor in the hotly disputed “greenhouse gas theory.” The diagram serves as a simplified version of an earlier PSI model produced in answer to a “put up or shut up” challenge (May 10, 2013) by climatologist, Dr Roy Spencer that appears to have the now subdued Spencer stumped.

Pointedly, PSI’s model depicts our planet in three-dimensions, unlike the preferred flat earth two-dimensional model favored by Spencer and other climatologists (the Kiehl-Trenberth model). PSI believes it is crass and contrary to the advancement of science that promoters of the “greenhouse gas theory” (GHE) should insist on relying on the outmoded flat-Earth model. The GHE is increasingly discredited because despite its core claim that more atmospheric carbon dioxide means higher temperatures the hard evidence proves this has not happened.

 Miatello EARTH ENERGY BUDGET

COMMENTS – EARTH’s SOLAR ENERGY BUDGET

This diagram (‘Miatello Model’), in our opinion as members of Principia Scientific International (PSI), is a precise, simple and visually immediate way of presenting and depicting the Earth’s solar energy budget.

 Moreover, and more importantly, this diagram links the actual solar energy input more accurately to the “average temperature of Earth” (about 288K = 15°C, 59°F), without needing to include the questionable aspects of a “greenhouse effect” and “warming by back-radiation”.

The diagram indicates a perceived error made by Kiehl & Trenberth (K-T) in their widely accepted energy budget diagram (shown below) with regard to the geometrical shape of the Earth – as a rotating spherical planet – and its rotational speed as the main factors influencing the % of solar energy’s absorption and emission.

The K-T diagram depicts our Earth as a “static, flat” surface without rotation, no difference between day and night, no summer or winter and a solar irradiance that does not melt ice nor generate water vapour.

But Earth is not a “flat, static” surface, it is a spherical rotating planet, whose absorption of solar irradiance is ruled by its geometrical shape (according to Lambert’s sine/cosine law and therefore absorption and emission of solar energy is greater at the Equator and the Tropical areas and lower in near-Polar areas and – no less important – the rotational speed of Earth around its axis, are all important factors dictating the amount of actual solar energy absorbed by our planet.

If the Earth was a planet without rotation, there would just be averaged total solar energy absorption of 32-33% – that is, an average of 64-66% energy absorption in the irradiated hemisphere and 0% absorption in the dark hemisphere.

Conversely, if the Earth’s rotational speed higher than the present, e.g. 30 rpm (revolutions per minute), like a disk on an old gramophone, there would be – in almost any time span – the same amount of solar irradiance both in the illuminated and in the dark hemisphere and the Earth would receive a total 64-66% of solar energy.

Being that Earth’s rotational speed is much lower, its total energy absorption is around 50% (51% as per NASA).

In this diagram, the average emissivity of Earth’s surface is calculated to be around 82-83%.

Earth is not a black-body; therefore, its emissivity cannot be either 1, as for a black-body, nor 0.93 as for a tarmac surface.

Laboratory black-body enclosures at 373K (100°C) are not behaving like perfect black-bodies, having an emissivity of 0.98. (Encyclopedia Britannica, 1963 edition, Volume 11, item: “Heat”, p. 335)

Therefore it is reasonable to calculate an emissivity of about 0.84 – 0.86 for ocean surfaces (covering 7/10 of the Earth’s surface and having a high reflectivity of solar radiation) and less than 0.8 for the many different ground surfaces (covering 3/10 of the Earth’s surface).

Thus, the product 0.5 (mean absorption of solar energy for geometry/rotational speed) and 0.82 (emissivity) gives the total average emissivity of Earth, namely around 0.41.

Multiplying the solar constant (= 1367 W/m^2) x 0.7 (= “filtering” effect of the atmosphere owing to albedo, scattering of solar radiation by aerosols and clouds) x 0.41, an average distribution of solar radiation on the surface, i.e. 390 W/m^2, is obtained. That correlates, through applying the Stefan-Boltzmann equation, to nearly 15°C, which is the “average temperature” of Earth as evidenced by weather stations.

It is interesting to note that this result could have also been observed by Kiehl & Trenberth had they not disregarded the Earth’s rotation (= actual absorption) and emissivity (= grey-body) of the Earth.

In the K-T diagram the average solar irradiance on Earth’s surface is given as nearly 161 W/m^2. Dividing 161 by 0.41 for total emissivity as per above arrives at a figure of 392 W/m^2, corresponding – according to the same S-B equation – to 15.2°C, closely similar to our calculation.

KT Earth Energy Budget Fig3

Thus, the perceived error in the K-T diagram – dividing by 4 the average irradiation from solar constant and impinging on Earth’s surface (rather than using the actual solar constant) based on a portrayal of a flat, motionless, cold Earth, with a wrong emissivity of 1 (as a black-body – rather than the grey-body that it is) – arose perhaps from a prescriptive requirement to include a debatable Greenhouse Effect and heating by back-radiation.

We regard this diagram as the most visually precise way to explain how Earth’s solar energy budget works and how it is linked to the “average temperature” of Earth (nearly 15°C).

Compiled by members of Principia Scientific International, July 2013

With special thanks to Alberto Miatello, Alan Siddons, Pierre Latour, Joe Postma, Martin Hertzberg, Nasif Nahle and Hans Schreuder and John O’Sullivan.

 

ADDENDA (August 01, 2013)

 

The correctness of the new PSI Solar Energy Model, with its  use of solar irradiance and emissivity to work out the expected average temperature of the Earth’s surface,  is confirmed by another radiative model, mentioned by Wikipedia, although less precise than ours: http://en.wikipedia.org/wiki/Climate_models

Wikipedia is calling it the “Zero dimensional model” and their calculation is as follows:

A) ( 1367 x 0.7 x 1/4 x 0.6 (emissivity of Earth) x σ ) ^ 0.25 = nearly 288 K

The PSI Model calculation is as follows:

B) (1367 x 0.7 x 0.5 (absorbed energy) x 0.8 (emissivity of Earth surface) x σ ) ^ 0.25 = nearly 288K

The Wikipedia radiative/emissivity model 1/(4 x 0.6) (emissivity of Earth Including clouds, water vapor, etc.) gives 1/2.4 = 0.41, which  is exactly the same as the PSI radiative/emissivity model: 0.5 (absorption) x 0.82 (emissivity of Earth surface) = 0.41

It therefore appears correct to multiply the radiation received and filtered by our planet x its emissivity to find the expected surface temperature.

Now, Readers could ask: “ are there any differences between the Wikipedia model and the PSI model?”

Yes there are!

1) Wikipedia’s model is wrongly calling a “greenhouse effect” what would correctly be called the “water cycle”, including the emissivity of clouds and water vapor, in the total emissivity of Earth.

The PSI Model on the other hand is just and more precisely focusing on the emissivity of Earth’s surface ONLY (nearly between 0.82 and 0.88) WITHOUT including the emissivity of the atmosphere, clouds, water vapor, etc., (difficult to calculate) in order to show that it is possible to find an average temperature of Earth’s surface up to about 20°C – WITHOUT any greenhouse effect – and then it is the COOLING WATER CYCLE that reduces the average temperatures (it doesn’t heat, it cools!) down to 15° C, as evidenced by weather stations from around the world.

2) Wikipedia’s model is too “rough”, it is like a map showing only the city of departure (irradiance 1367 W/m^2) and the city of arrival (total emissivity of Earth 0.6), without showing the itinerary.

The PSI model is showing HOW the solar irradiance is heating the surface of Earth, introducing the important concept of ROTATIONAL SPEED of the Earth, which is a crucial parameter in order to find the actual solar energy ABSORBED (50%) by our planet, before calculating the emissivity.

Rotational speed of a planet is of vital importance, because the faster a planet rotates, the higher is the percentage of solar energy it absorbs.

The PSI model is not a “zero-dimensional” model because it is showing the result of the correct rotational speed of a 3D planet.

3) The PSI Model clearly distinguishes between the temperature of the Earth surface (up to 20° C), BEFORE the cooling via the water cooling cycle, whereas the Wikipedia model shows only a total “rough” emissivity of Earth, including the emissivity of clouds, atmosphere, etc. which is difficult to accurately evaluate due to the dynamics of the atmosphere.

Finally, just an important remark regarding an incorrect comparison between Venus and the PSI model.

The PSI model is valid and applicable ONLY for planets having no atmosphere or a semi-transparent or a rarefied atmosphere, like Mercury, our Moon, Earth, Mars and elsewhere where a star like our Sun is the main and driving factor of temperatures and climate changes.

 

******

 

Due to reader comments below that are in error, Alberto Miatello has offered the following addenda:

@ Geraint Hughes and Anon

Geraint Hughes and Anon are in error, whereas  Miatello is correct.

I didn’t forget to divide per 4 the solar irradiance, whereas you clearly forgot to read the full article.

I said clearly that it was possible to use the “model” by Kiehl and Trenberth and divide by 4 the solar radiation and even use the amount of solar energy of 161 w/m^2, that Kiel and Trenberth themselves show, to indicate the amount of solar energy reaching the Earth surface.

And  now, if you divide 161/0.41 (total average emissivity of Earth) you get again 392 W/m^2, corresponding with SB to 288.4 K = 15.3° C

So, K-T simply “forgot” to take into account the full emissivity of Earth (0.41), depending upon the energy actually absorbed in 1 year by our Earth (nearly 0.5 of solar energy reaching the surface day/night and in the 4 seasons) x emissivity of surfaces of Earth (oceans, deserts, rocky mountains, forests, clay soils, etc., etc), which is nearly 0.82.

So, what you wrote is incorrect, because I did not forget that the surface of the sphere has four times the area of a flat disc, as K-T did, but – in a different way than K-T – I correctly remembered to calculate the actual average  EMISSIVITY of Earth!

Moreover, this PSI model is fully in accordance with the KIRCHOFF LAW of thermal radiation, stating that:

e = E*a

the total emissivity of a body, at any temperatures, is proportional to the coefficient of absorbance.

Therefore, it is correct to multiply the average absorbance of Earth in 1 year (0.5 of solar energy received) x emissivity of the surfaces of Earth (0.82).

No matter whether a body is spherical, or cubic, or anything else, it is fully correct to multiply the actual energy it absorbs x its average emissivity, and that’s what I’ve done!

So, please, study the Kirchoff Law, before posting!

But it is not over yet!

I can easily prove that the PSI model is fully correct and totally in agreement with our positions, as PSI members, against the bogus theories of GHE/AGW, because someone yesterday told me that maybe the emissivity of Earth surface can be higher than 0.82.

That’s a correct remark; I actually took an approximate value 0.82, since it is difficult to work out a precise number of average emissivity for millions different surfaces here on Earth.

But I can easily prove that the PSI model also works with a greater emissivity.

So, let’s take an average  emissivity of Earth surface for 0.88 (very high).

Then we get:

1367 x 0.7 x 0.5 x 0.88 = 421 W/m^2

( 421/5.67 x 10^ -8) ^0.25 = 293.5 K = 20.4°C

So, even assuming a higher emissivity of Earth’s surface, corresponding for instance, to 20.4°C average temperature, then the lower average temperature of Earth recorded by weather stations  can easily be explained by taking into account the COOLING EFFECT OF WATER CYCLE! (rains, snows, hail, cold winds, etc.

The PSI model explains only the effect on the surface temperature of the average solar radiation reaching a rotating planet as our Earth is, in 1 year, without taking into account other factors (in the model we don’t take the emissivity of air, clouds, etc.).

But the water cooling effect on Earth is very powerful:

2.4 x 10^6 x 5 x 10^17 = heat vapor x quantity of vaporized water

This corresponds to a huge amount of energy: 3.8 x 10 ^16 W

Namely, 1/3 of the energy from the Sun to the Earth is moving the water cooling cycle and that’s completely disregarded by AGW/GHE supporters!

So, the average “solar” temperature of Earth = 20.4 ° C less 4°- 5° of cooling by the water cycle = 15°-16° C of average temperature of Earth!

To summarize:

The PSI model is totally correct because it explains clearly that the average temperature of Earth (nearly 15°C) can be worked out by taking into account the average solar irradiation absorbed by a rotating body such as our Earth in one year, with a surface average emissivity and WITHOUT any sort of mysterious “backradiation”, GHE, etc.

The PSI model is totally correct because it shows only the average temperature of the surface, BEFORE the cooling convective effect of the WATER CYCLE.

The PSI model can also foresee an average surface temperature of Earth of – say – 20°C and then the difference with the average temperature of Earth (15°C)  is due to the WATER COOLING CYCLE.

So, everything is fully working and without the need for a GHE to explain the differences! Earth is heated by our Sun (to 15-20°C as an average) and then the atmosphere is COOLING (and not heating) through the water cycle.

But, please, I would appreciate that those who are posting comments read better, before posting.

Thanks.

Alberto Miatello

 

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Comments (71)

  • Avatar

    Mohammad

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  • Avatar

    bwebster

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    Evidently “Guest” is a troll using the familiar tactics of the wide-eyed “true believers” – so would it be possible to just delete the space wasted by these posts?

    One cannot engage those who refuse to deal in the real world. Evidently, “Guest” is one of those whose denial of reality must be preserved because he/she doesn’t have the maturity to challenge his own premises.

    Real scientific inquiry demands skepticism until hypotheses are confirmed by experimentation or observation. Computer models are tools for fools who have no other refuge for their failed hypotheses underlying their flawed theories.

    • Avatar

      Sunsettommy

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      [quote name=”bwebster”]Evidently “Guest” is a troll using the familiar tactics of the wide-eyed “true believers” – so would it be possible to just delete the space wasted by these posts?

      One cannot engage those who refuse to deal in the real world. Evidently, “Guest” is one of those whose denial of reality must be preserved because he/she doesn’t have the maturity to challenge his own premises.

      Real scientific inquiry demands skepticism until hypotheses are confirmed by experimentation or observation. Computer models are tools for fools who have no other refuge for their failed hypotheses underlying their flawed theories.[/quote]

      I am the one who deleted the comments but it was not a guest who made the comments because when I deleted the the username [b]coolist[/b] reverts to the default word guest.

  • Avatar

    JeffC

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    PSI has embarked upon a program of deleting comments by those who expose the errors in articles such as this. Such comments will be preserved on my website where anyone interested can read screen captures before they were deleted. This puts PSI among the ranks of SkS, SoD and some other blogs where comments are deleted when no answers can be given, as is the case here. Miatello cannot refute the fact that he failed to divide by 4, and also he used 35% (namely 0.7 x 0.5) instead of 51% or 55% with emissivity. The link is http://www.climate-change-theory.com/slayers.html

  • Avatar

    JeffC

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    Miatello writes …

    [i]”In the K-T diagram the average solar irradiance on Earth’s surface is given as nearly 161 W/m^2. Dividing 161 by 0.41 for total emissivity as per above arrives at a figure of 392 W/m^2, corresponding – according to the same S-B equation – to 15.2°C, closely similar to our calculation.”[/i]

    Here he is using his “effective emissivity” of 0.41 (which he modified because of absorption and albedo) to then boost the K-T value of 161 back up to his 392W/m^2. But the 161 figure has already been adjusted for albedo and absorption by the atmosphere, and so too has his “effective emissivity.” Hence he is just getting back to an approximate estimate of mean TOA radiative flux, for which he should have used a quarter of 1367W/m^2 because the surface area is 4 times the cross-sectional area.

    K-T (and NASA) do not assume absorptivity of 1.0 for the surface. For example, NASA shows 4% reflected by surface and 51% absorbed. Hence this implies absorptivity of 51/55 which is about 0.927 and it’s not too far out.

    Miatello does all this fiddling with figures in an attempt to avoid dividing by four. So in fact it is his model which is a flat Earth model supposedly received all solar radiation perpendicular to its surface, whereas K-T have allowed for the fact that the surface is spherical and has an area 4 times that of the cross-sectional disc through which radiation is defined and measured.

  • Avatar

    JeffC

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    You say [i]”Being that Earth’s rotational speed is much lower, its total energy absorption is around 50% (51% as per NASA).”[/i]

    It has nothing whatsoever to do with rotational speed.

    The reason NASA says 51% is because 26% is reflected by the atmosphere and clouds, 4% is reflected by the surface and 19% is absorbed by the atmosphere and clouds. So 26+4+19 = 49% does not enter the surface. But the other 51% does. We don’t need to make further adjustments to emissivity. We just multiply the 51% by 0.25 (to allow for the spreading of the energy over 4 times the area) and then we use the correct emissivity (between 0.8 and 0.9) but we don’t get anything like the observed temperatures – simply because the surface is not a grey body.

    And that’s why radiation calculations don’t tell you any planet’s surface temperature if that planet has a significant atmosphere.

    When are you going to be honest, admit your mistakes and withdraw you article which is a disgrace to PSI. Even Christopher Monckton recognises that, and soon Anthony Watts and he are sure to make a laughing stock of PSI.

  • Avatar

    GHEbreaker

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    @Geraint Hughes

    Please, be honest and admit your mistake.
    You wrote that
    “It is not possible for a sphere in space with an emissivity of 0.4 and an absorptivity of 0.5 to attain a temperature of 288k with only 390 w/m2 energy input over the area of its absorption a much lower temperature results.”

    So, at first you mistake input for output, and you said that Earth was too cold, according to my diagram, then you’re saying that it would be too hot!
    (and you disregard the water cycle!)

    Please, try and reach an agreement with YOURSELF before posting! ah ah ah!

  • Avatar

    GHEbreaker

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    @jeffC

    You’re another guy who is normally twisting my words, and becomes ridicolous…

    I never said that 33% of solar energy “strikes” the dark side in the night.

    I said that 33% of energy STORED by surface is RELEASED IN THE NIGHT BY SUB-SOIL LAYERS.

    So, please, find other persons to cheat by twisting someone’s else words, you found the wrong person!

    Alberto Miatello

    • Avatar

      JeffC

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      I didn’t twist your words – I quoted them exactly: [i]”Being that Earth’s rotational speed is much lower, its total energy absorption is around 50% (51% as per NASA).”[/i]

      You are clearly implying that the reason Earth’s surface absorbs about 51% of the insolation which originally entered the atmosphere has something to do with its rotational speed. Does that “rotational speed” as you call it cause more or less reflection and/or absorption of incident Solar radiation by clouds and atmosphere? Does it give you a reason for not dividing by 4 (and thus multiplying the actual mean Solar radiation by 4) or inventing an “effective emissivity” which I’ve not heard any physicist talk about, and for which I can see no need, other than to try to baffle readers with incomprehensive fictitious fissics.

    • Avatar

      JeffC

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      I’ll quote you again: You said [i]”33% of energy STORED by surface is RELEASED IN THE NIGHT BY SUB-SOIL LAYERS.”[/i] Well, 70% of Earth’s surface doesn’t have any “sub-soil layers” because it’s mostly salt water. So what does it do? Release the other 67% on top of the 66% released during the day? What does any such release by evaporative cooling or conduction (followed by convection) have to do with radiative flux? I suppose you will try to convince readers that it will change the emissivity of the oceans from about 0.85 to whatever gives you the right answer. /sarc

      PS Don’t forget to divide the incident Solar flux by four to get the mean because it is spread over four times the area and is, after all, measured in Watts per square metre. And you can just use NASA’s 51% for absorption because that will take care of all the reflection and absorption by the atmosphere so you won’t need to double count some of it to get that fictitious 35% (0.7 x 0.5) which is in your calculations which I quote …

      [i]”(1367 x [b]0.7 x 0.5[/b] (absorbed energy) x 0.8 (emissivity of Earth surface) x σ ) ^ 0.25″[/i]

      • Avatar

        JeffC

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        One further note. If you are going to multiply by emissivity (which should be closer to 0.9 than 0.8) then you should use the total flux reaching the surface, including the amount reflected.

        Hence you should have 1367 x 0.25 (spreading over 4 times the area) x 0.55 (NASA absorption 51% plus reflection 4%) x 0.88 (or your best estimate of emissivity)

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    Rosco

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    Kiehl& Trenberth’s energy budget has a serious mistake in that it adds fluxes without consideration of ambient conditions.

    It shares this error with all the models which show insolation and backradiation adding up as a simple sum instead of combining as the Stefan-Boltzmann equation requires – that is a “net” energy flow not a simple sum.

    Surely this is obvious – the air or the surface always has some temperature and any energy associated with a lower temperature does not heat the warmer object.

    The SB equation involving radiation exchange between 2 objects is a “net” relationship. If net energy is flowing from an object it is cooling not heating.

    Plug some of the temperature equivalents into the SB equation from K & Tand you’ll see the surface as they have it is never heated by the solar radiation or the back radiation.

    The diagram simply makes no sense.

    • Avatar

      D Cotton

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      Yes you are right about the fact that the Sun can never heat the surfaces of Earth or Venus etc to the observed temperatures with direct solar radiation. That is why Miatello must be wrong in assuming it can be. First he effectively uses 4 times as much radiative flux, because there is no allowance (dividing by 4) for the fact that the energy has to be spread over 4 times the area. So that leaves him with far to much flux. Hence he tries to cut that down by double counting the effect of albedo. His “0.7 x 0.5” is (reduction due to reflection by atmosphere and surface) x (reduction due to reflection by atmosphere and surface plus reduction due to absorption by atmosphere and clouds.) So, even though he knows NASA says 51% is absorbed by the surface, Miatello is assuming only 35%. Then he wrongly assumes the surface acts like a grey body, and he even invents an “effective emissivity” which allows for 50% loss, even though that loss is already more than accounted for in his 35% figure.

  • Avatar

    GHEbreaker

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    @JeffC

    Sorry, but again you confuse the emissivity of Earth surface THAT I CLEARLY AND CORRECTLY indicated as 0.82 (it could be up 0.9 as a grey body), and the TOTAL EMISSIVITY, which is the result of the energy received (0.5 of solar energy) x emissivity.

    And that’s totally in Agreement with Kirchhoff’s Law, stating that the emissivity of a body is e = E * a, e is the total emissivity, a is the coefficient of absorptivity, and E a number proportional to the absolute temperature.

    So, the higher is the energy a body receives, the bigger becomes its absolute temperature, and then the higher is the total emissivity (up to 6000K, as Plank discovered).

    I don’t invent, it’s you who don’t know Kirchhoff Law!

    A body receiving 50% of energy emits less energy than the same body receiving 100% of energy, are you questioning this, by chance

    • Avatar

      JeffC

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      I have answered this in a deleted comment of which a screen capture may be read [url=http://www.climate-change-theory.com/slayers.html]here[/url].

      Show me any physics text which defines total emissivity that way. Emissivity (total or otherwise) is an independent variable just as Christopher Monckton pointed out to you. Look up Wikipedia.

      Your calculations are wrong because you leave out the 1/4 which has to be there to allow for the fact that the energy in the incident solar radiation that would have passed through a disc of size equal to the cross-section of the Earth is in fact spread over four times the area.

      When are you going to acknowledge your error in this regard, which is of course a critical error making all your conclusions wrong.

    • Avatar

      JeffC

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      No, the Earth’s surface is not a grey body, so you can’t use SBL to calculate its temperature. Again I say, look up a bit of physics in Wikipedia or elsewhere before you make such blunders in public.

    • Avatar

      JeffC

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      Kirchhoff’s Law (which I teach my students) is relating to bodies which are heated (and which cool) only by radiation, which is not what the surface does.

    • Avatar

      JeffC

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      What I am questioning is your 50%, because you left out the 1/4 and so your 50% (or 35%) should have been more like one-eighth of 1367W/m^2.

      The mean solar radiative flux absorbed by Earth’s surface is of the order of 170W/m^2. You are obviously disputing this value which has been confirmed with empirical measurement and cited by NASA for years.

      Go and edit the Wikipedia article if you think you’re right, and take out that 1/4 in the Wikipedia article which you quoted in the article above.

    • Avatar

      D Cotton

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      You continue to imply that Earth’s surface acts like a grey body. It doesn’t because it is not perfectly insulated against transfers by non-radiative processes – not by a long shot. Almost as fast as the Sun tries to heat it in the morning it loses energy by conduction, convection and evaporative cooling. You cannot estimate what its temperature ought to be using SBL, as you would have to acknowledge if you used the correct figures, rather than 35% (0.7 x 0.5) absorbed by the surface instead of NASA’s 51% after 4% reflection by the surface. So even though your comment here says 0.5 of Solar energy, your calculations in the article use 0.7 x 0.5. Make up your mind.

      And by the way, I have been teaching Kirchhoff’s Law for decades – it also requires a perfectly insulated body that only receives and loses thermal (kinetic) energy by radiation. Learn the pre-requisites for these laws to apply – that’s what I stress to my tertiary physics students.

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    GHEbreaker

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    @coolist

    read better…

    50% of energy absorbed by rotational speed comes from the average 66% of energy absorbed by the radiated side + 33% of energy released by the dark hemisphere in the night, and that’s absolutely correct!

    And it is correct to consider the average “filtering” action of atmosphere, and albedo 0.3 , therefore 1367 x (1 – 0.3 = 0.7), before the energy reaches the surface.

    Moreover, you’re in error, because in my calculation I include the emissivity 0.8-0.9 of the Earth surface, that K-T diagram doesn’t include at all!

    I repeat, the error by K-T was in non considering the EMISSIVITY (0.8-0.9) of the Earth surface.

    And that’s the reason why, at the end, if you divide 161 w/m^2/0.41 (my total emissivity) you get again 390 W/m^2 = 288K = 15° C

    According to your statement, I should find a cooler surface than K-T, but I find a WARMER surface!

    Alberto Miatello

  • Avatar

    Coolist

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    Where does Miatello get his [i]”0.5 (mean absorption of solar energy for geometry/rotational speed)”[/i] ????

    It sounds suspiciously like the 51% figure which NASA net energy diagrams show as being absorbed by the surface. But Miatello wants a different figure, namely 0.7 x 0.5 namely 35%. He calls 0.5 “absorbed energy” but he double counts the reflected energy loss by then multiplying 0.5 by 0.7. The 0.7 comes from 30% albedo (reflection.)

    Let’s be clear about what the NASA figures say:

    Total energy reaching the Earth: 100%

    Less …

    Reflected by atmosphere: 6%
    Reflected by clouds: 20%
    Reflected by surface: 4%
    Absorbed by atmosphere: 16%
    Absorbed by clouds: 3%

    Total absorption and reflection: 49%

    Hence, absorbed by surface: 51%

    NOT 35% Miatello!

  • Avatar

    Geraint Hughes

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    This is going to be my last comment cos I cant be bothered any more. I am not convinced.

    If the output of the Earth is 390 w/m2.
    And the Earths diameter is 12,756 km, assuming its a perfect sphere area is 511,185,933 km2. This gives an output of 199,362 Terrawatts.

    If we take absorption side Earth as a single sphere with the same diameter has a maximum absorption area of 127,796,483 km2 and assumed its a energy hungry black body eating all energy in its path, the max input figure is 177,697 Terrawatts.

    How do you account for the difference in energy between input and output? How does the fact its spinning increase energy levels?

    • Avatar

      Coolist

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      You are right Geraint. What trumps Miatello’s “new model” is the fact that he does not realise the very basic fact that the measurement of 1367W/m^2 is the flux that passes through a plane that is perpendicular to the radiation path. What strikes the Earth is that which passes through its cross-sectional circular disk. But that energy gets spread over four times the area during a 24 hour period, and it is also attenuated by about 50% due to reflection and atmospheric absorption. NASA says only 51% of effective TOA flux is absorbed by the surface. Hence what is absorbed by the surface is about one-eighth of 1367W/m^2. That is a well known fact confirmed empirically. What a blunder!

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    JeffC

    |

    This would have to be the most extraordinary claim imaginable. Miatello says 33% of the Sun’s energy strikes the dark side of Earth – at night! Can’t say I’d noticed.

    But seriously, does Miatello not realise that incident solar radiation is only 1367W/m^2 passing through a plane which is orthogonal to the radiation vector? During the course of each 24 hours it is spread over an area about four times the area of the disk through which it passes. Hence its mean intensity is about the equivalent of 25% of 1367W/m^2 namely about 342W/m^2 (orthogonal to the surface) at TOA. Then that mean figure is of course reduced to about 51% of its TOA value by the time it reaches the surface. Hence the mean intensity over the whole surface is only about 170W/m^2 as is well known and confirmed empirically. But Miatello wants to use a value 50% of 1367!!!

    As has been said in other comments, you can’t just alter emissivity to fudge the right result. Miatello appears to agree that the emissivity of the oceans and water surfaces is about 0.85 and they make up about 70% of surface area, so the mathematically the weighted mean emissivity is highly unlikely to be less than 0.70 or greater than 0.95. I’m happy with the value of 0.88 that I read in some comment.

    So try working with 170W/m^2 and 0.88 emissivity and I’ll donate $1,000 to PSI if you get over 287K. My trusty on line SBL calculator gives me 241.6K.

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      Greg House

      |

      [quote name=”JeffC”]incident solar radiation is only 1367W/m^2 passing through a plane which is orthogonal to the radiation vector? During the course of each 24 hours it is spread over an area about four times the area of the disk through which it passes. Hence its mean intensity is about the equivalent of 25% of 1367W/m^2[/quote]

      This is so wrong, Doug.

      You can only correct for the difference between a disc and a half-sphere, not a disc and a sphere.

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        JeffC

        |

        Fine – We can look at the instantaneous situation rather than a 24 hour period if you wish, so the hemisphere which is sunlit at any one moment is “corrected” to receiving half as much, namely half of 1367W/m^2. And at that instant the other hemisphere is receiving zero. Hence the mean for the two hemispheres is a quarter of 1367 as I said.

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          Greg House

          |

          [quote name=”JeffC”]And at that instant the other hemisphere is receiving zero. Hence the mean for the two hemispheres is a quarter of 1367 as I said.[/quote]

          Doug, at that instant there are a lot of other hemispheres in the universe that are receiving zero. What would be the mean then? Right, almost zero. Get it now, why your approach is so wrong?

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            JeffC

            |

            Uranus receives about 2.7W/m^2 of incident Solar radiation, being 29 times further from the Sun than Earth. No solar radiation reaches its small solid core (55% the size of Earth) about 20,000Km below the top of the Uranus atmosphere. Yet the Sun’s energy is maintaining temperatures of thousands of degrees down there. I understand why, but maybe you don’t.

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    GHEbreaker

    |

    @Rojclague

    ah ah ah! You wrote that Kirchoff Law and Stefan Boltzmann only apply to blackbodies. (?????)

    Ah ah ah, please, go to high school, maybe they can suggest you a good handbook!

    • Avatar

      Coolist

      |

      SBL can be applied to a gray body of course using the appropriate emissivity and/or absorptivity. BUT, even a gray body transfers energy to its environment only by radiation. That is true of the whole Earth plus atmosphere system. BUT you are trying to apply SBL to an internal interface, namely the surface-atmosphere boundary. At that interface perhaps more than half the energy transfers are by non-radiative processes. So these remove energy which is no longer then available for radiation. Hence, under an overhead Sun at the Equator, nearly 690W/m^2 is received and, using 0.88 absorptivity, SBL calculations say the surface could be raised to about 70C. So the results of SBL calculations (everywhere on the surface) are very inaccurate because the surface is cooling by non-radiative processes while the Sun is trying to heat it. You can’t work out surface temperatures this way, especially on other planets, now can you?

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      D Cotton

      |

      The SBL applies to black and grey bodies. Such bodies must (by definition) be perfectly insulated against any transfer of heat by non-radiative processes to or from their surrounds. Earth’s surface is thus not a grey body and so SBL calculations will not give the observed temperatures. This is because, while the Sun is trying to heat the surface to such temperatures the surface is simultaneously losing heat by non-radiative processes.

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    GHEbreaker

    |

    @Monckton

    Please, don’t twist my words!

    In the diagram it is not written that albedo alters emissivity, it reduces the % of energy absorbed, not emissivity

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      D Cotton

      |

      But in the text you claim that absorption cuts emissivity in half to your “effective emissivity” of 0.41 and you then use that 0.41 to adjust the K-T result. This is illogical, because the K-T result is already adjusted for about 50% absorption and reflection. So you are adjusting twice.

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    Geraint Hughes

    |

    Look at my equation the input and output sides are clearly identified.

    If you are now saying you used a 0.7 albedo you are confusing yourself. Your diagram says absorption 50%. Your absorption is either 50% or 70% which is it? It cant be both.

    With an input of 956.9 and emissivity of 0.41 the steady state temp would be 318k.

    And on the emissivity side it is either 0.8 or 0.4 which is it? It cant be both.

    Absorptivity and emissivity doesn’t change just because something is turning, it is determined by the spectral properties of the materials used, NOTHING ELSE.

    If you are trying to prove back radiation doesn’t exist with your model, you will fail, because I can prove it does exist with two flat plates in a vacuum compared to 1 flat plate in a vacuum.

    The model you need to use, is a sphere inside a sphere, this gives a clearer picture of energy transfer. If you are trying to resolve it with maths for just 1 sphere, you will fail.

    That doesn’t mean I think CO2 cause global warming cos it doesn’t nor does it mean I think that Kiel & Trenberth is right because its not.

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    Geraint Hughes

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    I didn’t say 390 w/m2 was reaching the surface. YOU DID ITS ON YOUR DIAGRAM AND IN YOUR ARTICLE.

    I said that’s wrong.

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    GHEbreaker

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    @Geraint Hughes

    33% of solar energy in the dark side surface, of course comes from sub-soil and sub-ocean layers, in the night.

    Alberto Miatello

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      Coolist

      |

      What comes from the sub-soil is not radiation, so I wouldn’t suggest including that figure in the radiative flux used in your SBL calculations. Besides, such energy is merely returning the energy which was absorbed by the soil (and water surfaces) during the sunlit hours.

      Your radiation figure is wrong because it is not converted to the equivalent of direct radiation perpendicular to the surface at every point. You are in effect multiplying solar flux energy by four, because you are not diminishing its value when it strikes the surface at an angle and is spread over four times the area. Thus everything you conclude is invalid.

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      D Cotton

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      And 70% of the dark side doesn’t have sub-soil – it has salt water in case you hadn’t noticed. And evaporative cooling is not radiation. Nor are conduction and convection. If you talk about absorption of insolation (which is what sets temperatures) then the mean inward flux per square metre is about an eighth of the 1367W/m^2 because it gets spread over 4 times the area and then gets halved by absorption and reflection.

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        Greg House

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        [quote name=”D Cotton”]the mean inward flux per square metre is about an eighth of the 1367W/m^2 because it gets spread over 4 times the area and then gets halved by absorption and reflection.[/quote]

        No Doug, in reality both sides of the Earth are warmed by the Sun and the dark side cools very slowly and never reaches an absolute zero. You can not reasonably say “the other side receives nothing” and assume it’s temperature 0K nor can you halve the solar power.

        Or just think about barbecuing things, you know, maybe this example can help.

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    GHEbreaker

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    @Geraint Hughes

    All you wrote above is WRONG!

    You clearly don’t distinguish between INPUT energy and OUTPUT energy!

    You wrote

    quote

    “It is not possible for a sphere in space with an emissivity of 0.4 and an absorptivity of 0.5 to attain a temperature of 288k with only 390 w/m2 energy input over the area of its absorption a much lower temperature results.”

    unquote

    Are you kidding???

    Please, the energy INPUT over the surface of Earth is NOT 390 W/m^2, that’s the OUTPUT!!!

    The input energy is 1367 W/m^2, better, 1367 x 0.7 = 956.9 W/m^2 spread all over a rotating sphere, receiving 66% on the radiated side, and 33% on the dark side.
    66 + 33/2 = 50% average in 1 year.

    Kindly, stop twisting our words!

    Alberto Miatello

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    Monckton of Brenchley

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    This article incorrectly assumes that albedo alters emissivity, when the two are distinct; and that emissivity

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      josullivan

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      [quote name=”Monckton of Brenchley”]This article incorrectly assumes that albedo alters emissivity, when the two are distinct; and that emissivity[/quote]

      You can tell Chris Monckton is rattled when he has to resort to “accidentally” misinterpreting what the article says.
      In the diagram it is not written that albedo alters emissivity, it reduces the % of energy absorbed, [u]NOT[/u] emissivity, your lordship!

      • Avatar

        Coolist

        |

        Josullivan

        Miatello wrote [i]”Thus, the product 0.5 (mean absorption of solar energy for geometry/rotational speed) and 0.82 (emissivity) gives the total average emissivity of Earth, namely around 0.41.”[/i]

        In physics emissivity is the ratio of what a gray body emits compared to a perfect blackbody. The fact that some incident energy is reflected before the radiation reaches the body has nothing to do with the body’s intrinsic emissivity. You do not alter the emissivity of the oceans (about 0.85) by varying the albedo or the absorption levels in the atmosphere. His Lordship is right on this point.

        The article is also wrong for the reasons I have given in other comments. The surface-atmosphere boundary does not act like a gray body because, by definition, a gray body only receives and transfers energy by radiation. The surface passes more energy to the atmosphere by non-radiative processes than it does by radiation. It also passes energy each morning by conduction to regions in the outer crust beneath it and into the upper layers of the oceans. Hence the Sun cannot warm it as much as it would if the surface were perfectly insulated as are blackbodies and gray bodies. You cannot apply SBL to the surface because it is not a gray body.

      • Avatar

        Coolist

        |

        The article starts ..

        “Wikipedia is calling it the “Zero dimensional model” and their calculation is as follows:

        [i]A) ( 1367 x 0.7 x 1/4 x 0.6 (emissivity of Earth) x σ ) ^ 0.25 = nearly 288 K

        The PSI Model calculation is as follows:

        B) (1367 x 0.7 x 0.5 (absorbed energy) x 0.8 (emissivity of Earth surface) x σ ) ^ 0.25 = nearly 288K”[/i]

        Neither is correct. Your (B) omits the 1/4 for example. It should be ..

        (1367 x 0.25 (distributing over 4 ties the area) x 0.51 (fraction absorbed by surface) x 0.88 (realistic emissivity of surface) x σ ) ^ 0.25

        But nothing like this using Stefan-Boltzmann will ever give the observed temperatures simply because the surface is not a gray body and it is not radiative balance which determines its temperature. Try your calculations on Venus, for example.

      • Avatar

        D Cotton

        |

        Monckton was perfectly correct in saying that emissivity is an independent variable. Miatello’s “effective emissivity” has no place in physics, and in any event he used 35% instead of 51% or 55% for total absorption and reflection. And he failed to divide by 4 to allow for the spreading of insolation over 4 times the surface area.

    • Avatar

      Coolist

      |

      Yes your Lordship. The conclusions in the article is indeed totally incorrect. Miatello has assumed that the energy which passes through an imaginary disk with the radius of Earth then strikes the top of the atmosphere with the same vertical intensity that it does when it strikes orthogonal to the surface, such as when the Sun is directly overhead in the tropics. He only reduces it 50% for night and day, forgetting that the Sun does not warm things quite as effectively at 75 degrees latitude for example.

    • Avatar

      Coolist

      |

      Your Lordship:

      It seems PSI has resorted to SkS tactics in that they now delete comments which exhibit errors in their articles, including my reply to yourself.

      Miatello did indeed think that he could reduce emissivity by 50% somehow. After all he had to do something to fudge the “right” answer, because he made a huge mistake and failed to allow for the fact that the radiation passing through the cross-sectional area of the Earth is actually spread out over four times the area of that cross-section during each 24 hour period.

      In any event, no one can calculate what temperatures the Earth or Venus surfaces ought to be using the Stefan Boltzmann Law, because neither is a gray body and so the calculations are irrelevant.

      I am keeping screen captures of all my comments that are deleted, so, when I have time I will post links to the deleted entries and perhaps email the links to you in case you’d like to write some more about this atrocious PSI article which will be their downfall.

    • Avatar

      JeffC

      |

      I don’t often support you, your Lordship, but you are close enough with this comment. Miatello does invent a term “effective emissivity” but in so doing he abuses the normal definition of emissivity and starts talking about the emissivity being about 0.41 which is about half its real value. He writes, for example [i]”Thus, the product 0.5 (mean absorption of solar energy for geometry/rotational speed) and 0.82 (emissivity) gives the total average emissivity of Earth, namely around 0.41.” [/i] This at the very least is nonsense to a physicist.

      These are other key errors which were explained in some of the deleted comments which can still be read [url=http://www.climate-change-theory.com/slayers.html]here[/url].

  • Avatar

    Guest

    |

    Hi! Can my comment be deleted, just for fun?

  • Avatar

    JeffC

    |

    I see my comment was deleted, but fortunately I saved a copy and posted it on [url=http://wattsupwiththat.com/2013/07/31/dueling-desktops-anthony-watts-versus-al-gore/#comment-1376973]WUWT[/url], [url=http://www.drroyspencer.com/2013/07/senate-epw-hearing-climate-change-its-happened-before/#comment-85589]Roy Spencer[/url], [url=http://judithcurry.com/2013/07/30/denier-blogs/#more-12367]Judith Curry[/url] and [url=http://joannenova.com.au/2013/07/unthreaded-weekend-12/#comment-1302132]JoNova[/url].

  • Avatar

    JeffC

    |

    There is an obvious error in the calculations in this article, because the mean Solar radiation over the whole surface of Earth is most certainly not over 400W/m^2.

    It seems the author has overlooked the fact that, when the Sun’s rays strike the surface at an acute angle, then the intensity is reduced because the energy which passes through a 1m^2 cross-section of the radiation then falls over a larger surface area. For example, if the angle is 45 degrees then the surface area is 2m^2 based on the square of the cosine.

    The overall effect is that the mean intensity is reduced, not by 50% due to half the globe being in darkness, but by about 75% due to the additional fact explained above.

    PSI is barking up the wrong tree in trying to explain surface temperatures using calculations based on incident solar radiation, because the Earth’s surface does not act like a gray body with emissivity 0.88 or whatever. If it did, the Sun could heat the equatorial regions to nearly the boiling point of water, as happens on the Moon.

    I agree with Rojclague below. I trust this comment will not be deleted as I have seen several others with opposing views to the author have been without providing any reason. In my view they were valid comments.
    .

  • Avatar

    Geraint Hughes

    |

    Just so you know, in my calc, I get an emission rate of 160.62 w/m2 for the surface temp of 288k with an e of 0.42.

    If we assumed a really small sphere, say 4m2 that gives a total output of 642 watts. That means across the circle area of absorption this much must be being absorbed.

    In your explanation you also seem to confuse emissivity with emission rates.

    It is possible to objects to have the same temperature but be emitting different rates of heat. I.e. A blackbody spehere at 279k has emissivity and absorptivity of 1 and emits at a rate of 340 w/m2 but a grey body sphere at 279k with emissivity of 0.5 and absorb of 0.5 will also be at 279k but will be emitting just 170 w/m2.

    If you think that the emissivity of earth is 0.41 you need to prove this. Say, by saying that the Oceans of this planet are absorbing energy but not emitting it, because the energy is being locked up in the deep sea and because deeper waters are cooler than higher waters heat absorbed under the surface the direction of heat transfer is downwards in the oceans and so subsurface heat never gets emitted outwardly. In effect the oceans are acting like a multi-layered absorber, with only surface emitting. (most of incoming energy transmits straight through the top) IR from below, never makes it out because water absorbs all IR after a couple of metres.

    • Avatar

      JeffC

      |

      Yes Geraint, Miatello does not use the term emissivity in accord with its definition, but he doesn’t use much real physics anyway. The deleted comments may be read [url=I don’t often support you, your Lordship, but you are close enough with this comment. Miatello does invent a term “effective emissivity” but in so doing he abuses the normal definition of emissivity and starts talking about the emissivity being about 0.41 which is about half its real value. He writes, for example [i]”Thus, the product 0.5 (mean absorption of solar energy for geometry/rotational speed) and 0.82 (emissivity) gives the total average emissivity of Earth, namely around 0.41.” [/i] This at the very least is nonsense to a physicist.

      These are other key errors which were explained in some of the deleted comments which can still be read [url=http://www.climate-change-theory.com/slayers.html]here[/url].
      ]here[/url].

  • Avatar

    Anon

    |

    So now that someone has told you (AM) that the emissivity should be 0.88 not 0.82 you calculate a higher mean surface temperature above 20C and suddenly remember that the water cycle cools the surface through the process of evaporation. Good, you’re some of the way there. Now you should also take into account the effect of other non-radiative cooling processes (like convection and even conduction of heat down below the surface) which also prevent the Sun from warming the surface as much as it would a perfect black or grey body.

    PSI member Joe Postma (whose name is listed under this article) calculated (in his “Model Atmosphere” paper) that the Sun should heat the surface to 87.5C when it is directly overhead. It doesn’t do so because non-radiative cooling competes with it simultaneously, because the surface acts nothing like a grey body with emissivity 0.88 or whatever. All calculations of surface temperatures, including your mean temperature, are prone to similar very significant errors which can be as large as 40 degrees because of non-radiative heat loss. Your claim of accuracy to three significant figures is thus absurd.

    [b]Solar radiation reaching the surface does not explain Venus surface temperatures, so your conjecture is dismissed because it only takes one planet to prove it wrong.[/b]

    • Avatar

      Anon

      |

      Postma’s figure of 87.5C can be calculated using NASA’s estimate that 51% of incident Solar radiation gets through the atmosphere and into the surface. So, where the Sun is at its zenith we get nearly 700W/m^2 heating the surface.

      But, if there is only 700W/m^2 during the day where the Sun is directly overhead, how could there possibly be a mean of over 400W/m^2 for the whole Earth’s surface, night and day?

      Go and argue with Postma about your estimates of Solar radiation reaching Earth’s surface. Then try to explain Venus temperatures of 730K supposedly resulting from a mean of about 10 to 20W/m^2 getting through its atmosphere.

      • Avatar

        Anon

        |

        Well, let’s use the K&T value of actual mean radiation reaching the Earth’s surface, namely 161W/m^2 because it’s close enough to my 170W/m^2, but a long way below your now revised 421W/m^2 which is of course impossible. You don’t just multiply by 0.5 to get the mean. You have to take into account the angles of incidence over the whole hemisphere. I don’t see any integration in your calculations. The fact that Solar radiation strikes most of the hemisphere at an angle less than 90 degrees results in a need for an additional halving approximately. That of course is why the K&T value is about a quarter of the 700W/m^2 (in my comment above) for an overhead Sun.

        Now you tell me why you then make a further adjustment for night and day in reducing the emissivity, when that is already taken into account in adjusting the radiative flux. The surface is not much cooler at night and still keeps radiating with similar emissivity. The 0.88 or 0.82 or 0.73 or whatever estimates of emissivity are in the correct ball park – not your alternative 0.41. You just don’t like the fact that using 161W/m^2 and realistic emissivity gives about 255K. And that’s before water cooling, convection etc. The whole point is that, just as on Venus, and as Rojclague says below, it’s not all about Solar radiative flux reaching planetary surfaces. Read his comment and the prerequisites for the Kirchoff Law to apply (as I make a point of teaching my students) before replying.

    • Avatar

      Anon

      |

      I am aware of your paper about Greenhouse refutation in which you have three pages about Venus. Yet strangely you don’t make any effort to explain Venus surface temperatures as being due to radiative balance like you try to explain Earth’s temperatures. That’s not very consistent now is it? .

      In fact in your paper you recognise the very point which Rojclague makes below when he concludes [i]”Surface temperatures are set by the lapse rate and the height of the convective atmosphere ( the troposphere “[/i]

      Indeed he is onto the right track about all planetary surface temperatures including Earth’s. After all, why should physics operate in a totally different and unique way on Earth? You yourself came so close in your paper, but you don’t correctly explain how the energy gets into the surface of Venus, because pressure doesn’t actually transfer energy from the top of the Venus atmosphere (where it is absorb from solar radiation) down into the surface. Maybe you’ve read somewhere else what does..

  • Avatar

    Geraint Hughes

    |

    It actually needs an alpha of 0.47 (not 0.53) to attain 288k.

    EQ1: 0 = ά E A(a) – έ A (e) σ T4

    This is the equation for determining temperature of any object in space.

    EQ 2: T = [(A (a)/A(e)) * (άE / έ σ)] 1/4

    Equation is re-arranged to find T.

    Spheres are easy because parts of the equation cancel each other out.

    Absorptivity is 0.47 emissivity is 0.41.

    EQ 3 : T = [ ( (1/4)/1) * (0.47*1367 W/m2) / (0.41 * 5.67 x 10-8 W / (m2K)) ] ¼

    EQ 4 : T = 0.25 * (642.49 W/m2/2.3247 x 10-8 W/(m2K) ) ¼
    EQ 5: T = 0.25 * 27,637,544,629¼
    EQ 6: T = 6,909,386,157¼
    EQ 7 : T = 288k

    This works best with hollow objects and for a hollow sphere perfectly as every part of the absorbing side can transmit via radiation to every part of the sphere to achieve isothermic temps.

    Solid objects with specific heat capacities will experience different peaks and troughs across the body.

    High SHC objects experience small swings, low SHC objects big swings.

    It is not possible for an object to emit more energy than it absorbs and using super high SHC gets exactly same answer as hollow model.

    So for a sphere in space with an alpha of 0.47 and an epsilon of 0.41 an absorption input of 642.49 w/m2 is required otherwise different steady state temperature results.

  • Avatar

    Geraint Hughes

    |

    It is not possible for a sphere in space with an emissivity of 0.4 and an absorptivity of 0.5 to attain a temperature of 288k with only 390 w/m2 energy input over the area of its absorption a much lower temperature results.

    I am not in error, its a physical impossibility.

    If you don’t believe it, go make one as a satellite and send it in space.

    I will see if the maths cuts and pastes onto this page and stays within char limit.

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    Rojclague

    |

    Miatello says

    “The PSI model is totally correct because it shows only the average temperature of the surface, BEFORE the cooling convective effect of the WATER CYCLE.”

    A radiative model corrected for “convective effect”. That is the AGW theory.

    Your fudge factor, emissity of the surface is criticised and you have added another. You seem not to be convinced by your own theory. Occam’s razor suggests you are not going in the right direction.

    Reflection from the surface is included in the albedo factor. So is counted twice.

    Kirchoff’s Law of Radiation and the S-B Law only apply to a black-body. A black body gains and loses energy by radiation only . The surface of the earth is losing energy by radiation and convection. It is not a black body. It is wrong to treat the surface of the earth as a black body and then correct for convection.

    Kirchoff’s Law can only be applied at the top of the atmosphere.

    Surface temperatures are set by the lapse rate and the height of the convective atmosphere ( the troposphere ).

    • Avatar

      Joseph A Olson

      |

      In regards to ‘the cooling convective effect of the WATER CYCLE’

      The average thunderstorm is 15 miles in diameter….
      weighs 10 x 10^8 kg….
      releases 10^15 joules of energy at altitude…and…
      THERE ARE 16 MILLION thunderstorms world-wide per year.

      This energy, removed from the surface, released with NO radiative component at altitude is a massive COOLING system. To claim that ‘water vapor’ warms the planet is the most apparent FRUAD in all recorded history.

      There is, in addition DRY convective heat, that is also dissipated with no radiative component. ‘Radiative Balance’ is a false metric….advanced for a political agenda.

      • Avatar

        bwebster

        |

        [i][b]Excellent observation[/b][/i] Joe.

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    P. Tuvnes

    |

    PSI Members have made a considerably improved and more correct energy budget compared to the flat-earthers Kiel & Trenberth. In the PSI spirit of open peer review I hope PSI Members will respond to the critics expressed in comments and revise the energy budget accordingly. (The Q&A post of J. Postma is a good model for that).

  • Avatar

    CleanEnergyPundit

    |

    High time this AGW proposition is shown up as a new ‘Phlogiston’ though having more sinister implications

  • Avatar

    Geraint Hughes

    |

    Something wrong.

    Using your absorptivity and emissivity figures, I can arrive at 288K with 0.53 absorptivity and 0.41 emissivity, except I show an energy input of 642 w/m2 on the absorption area. If I use 390 w/m2 as you indicate I get only 254k.

    I thought perhaps maybe you were assuming the energy was spread fully over the whole surface meaning max obtainable 341 w/m2.

    I agree with you, the flat earth model used by greenies is totally wrong. But to get a sphere to 288k with 0.53 solar alpha with 0.4 epsilon needs 648 w/m2.

    The other thing I tell greenies, that is wrong about their flat earth model, is that their maths only represents a single sphere and not a sphere within a sphere.

    Rock inside atmosphere.

    The area of the earth including the atmosphere is much bigger than just the rock and water portion of Earth itself and when you include the increased area of absorption of the atmosphere an extra 10% is available were it to be fully absorbed. I have a xl spreadsheet that can show a sphere within a sphere and can adjust trasmisivities, alphas and epsilons of both spheres and if you insert sensible figures 288k is shown for the inner sphere.

    Greenies are just bad at maths.

    CO2 doesn’t cause global warming.

    • Avatar

      Anon

      |

      Yes Geraint. The calculations are full of errors in this article. Miatello forgot to divide by four (due to surface area of a sphere being four times the area of a circle with the same radius) and then, instead of using NASA’s 51% for the percentage of solar radiation reaching the surface, he uses the product of two non-independent factors, namely 0.7 and 0.41. So he gets 1367 x 0.7 x 0.41 = 392. Then, even though he says emissivity of 70% of the surface (the oceans) is about 0.85, he somehow thinks the weighted mean can be reduced to 0.41, which is mathematically impossible. The fact of the matter is that planetary surface temperatures cannot be determined purely from the incident insolation. Such is IPCC thinking from which PSI needs to divorce itself. No such calculations will ever “work” for other planets, and nor will they work for Earth. PSI had a good paper on planetary surface temperatures in the PROM menu, but removed it.

  • Avatar

    andersnordenfelt

    |

    To my taste this is a bit too “surface oriented” and “earth centered”. Fiddling around with several quasi-arbitrary parameter only to explain one average temperature will not be convincing.

    I suggest you take a look at the planetary temperatures in the pressure rage 100-300 mB and see what you find.

  • Avatar

    ExposingTravestiesOfPhysics

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    Three of my comments have now been deleted because they demonstrate that the author of the above article has failed to divide the incident Solar flux by four. This has to be done simply because the area of the cross-sectional disc (circle) is only about a quarter of the surface area of the (roughly) spherical surface over which the radiation has to be distributed. The mean solar radiative flux reaching Earth’s surface is not 390W/m^2 but, according to NASA, is more like 170W/m^2. Hence all the above calculations and conclusions drawn from those calculations are not valid.

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    WhatsUpWithPSI

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    The article reads [i]”Multiplying the solar constant (= 1367 W/m^2) x 0.7 (= “filtering” effect of the atmosphere owing to albedo, scattering of solar radiation by aerosols and clouds) x 0.41, [b]an average distribution of solar radiation on the surface, i.e. 390 W/m^2, is obtained[/b].”[/i]

    Well, even if there were no atmosphere the Solar radiation passing through an imaginary disc above the Earth (with the same radius as the Earth) would then have to be distributed over a surface area which is four (4) times as large as the surface area of the disc. Hence at the most, without an atmosphere reflecting and absorbing about half the radiation, there would be a quarter of 1367 W/m^2 and with an atmosphere about one eighth, namely about 171W/m^2.

    Now use a realistic emissivity of about 0.7 (given that the oceans dominate) then your “reliable” Steffan Boltzmann Law gives a mean surface temperature of 256K. If you prefer 0.6 emissivity then you get 266K. You can’t go much lower than 0.6 if the ocean is 0.85.

    [b]So there is a glaring error in the calculations in this article which will no doubt spark an article on WUWT if I draw Anthony’s attention to it.[/b]
    So even

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    PhysicsResearch

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    There is no justification for dividing the emissivity in half because of day/night considerations. The ocean is virtually the same temperature at night as during the day, and so it will emit about the same – certainly not zero which is implied when multiplying 0.82 x 0.5 to get the 0.41 figure used.

    What makes the whole calculation even more irrelevant is the fact that about two-thirds of the energy that transfers from the surface to the atmosphere does so by non-radiative processes.

    Nowhere near all of the radiation from the surface is actually transferring surface energy to the atmosphere, but the calculations imply that all of it is doing so.

    Hence, just like Trenberth, Miatello makes the same mistake of ignoring non-radiative heat transfer from the surface to the atmosphere when using the Stefan-Boltzmann Law to calculate surface temperature. It cannot be done that way because radiation is only doing about a third of the energy transferring.

    You need to understand that the supporting temperature at the base of a planet’s atmosphere is slowing down the cooling in the pre-dawn hours. You cannot explain this slowing with your model because you ignore the gravity effect which is the primary determinant of planetary temperatures, even beneath the surface.

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    PhysicsResearch

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    The model is wrong because, in general, planetary surface temperatures cannot be determined from knowledge of direct radiative flux from the Sun. Venus receives at its surface only about 10% as much direct insolation as does Earth’s surface. Yet the Venus surface is about 730K even at the poles. And Uranus receives none at the base of its theoretical troposphere where it is hotter than Earth’s surface.

    Planetary surface temperatures are determined by the supporting temperatures at the base of their tropospheres, and those temperatures are supported by energy absorbed in the atmosphere which is then distributed in all directions as thermodynamic equilibrium evolves. Such a state must have an autonomous temperature gradient in a gravitational field, for otherwise there would be unbalanced potentials.

    You cannot make a model such as the above “work” for Venus, let alone Uranus. And it only appears to work for Earth because of incorrect assumptions about radiative heat transfer and emissivity.

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