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.
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.
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
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!
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.