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## Fixing the basic AGW calculations

Posted by Jeff Id on December 19, 2010

Derek Alker, has kindly offered what is definitely the most skeptical thing I’ve ever posted here.  Please don’t interpret this as my work in any way as my views on these things differ, but as I have no time  at least it will make for interesting discussion.  Derek has obviously put a  huge amount of effort into this so, lets see if anyone can find problems with his work or dramatic conclusions.

Instead of doing the full reformatting which would take hours, I’ll just post a link to the discussion and spreadsheet.

The article:

Derek – Free to all pdf to end AGW scam Saturday 18_12 version.

The spreadsheet:

Derek – Excel sheet calculator, disc, sphere, hemisphere temps 18-12

1. ### mrpkwsaid

Interesting.
Not completely sure as to what it means (even after reading the summary).

Can someone dumb it down for me???
(Too many equations !!!)

2. ### Edsaid

Why cant we get to the nub of the problem ? with NUMBERS NOT ADJECTIVES

It is utterly futile to think that mankind can affect climate.

The numbers are very simple and are backed up by acceptance from a renowned UK government advisor, The US Department of Energy and many reputable scientific minds.

To understand the FUTILITY of Man-made Climate Control by limiting CO2 emissions, watch:

On average world temperature is ~+15 deg C. This is sustained by the atmospheric Greenhouse Effect ~33 deg C. Without the Greenhouse Effect the planet would be un-inhabitable at ~-18 deg C.

Running the numbers by translating the agents causing the Greenhouse Effect into degrees centigrade:
• Water Vapour and Clouds account for about 95% of the Greenhouse Effect = ~ 31.35 deg C
• Other Greenhouse Gases GHGs account for ~5% = ~1.65 deg C
• CO2 is about 75% of the remaining effect accounting for the enhanced effects of Methane, Nitrous Oxide and other GHGs = ~1.24 deg C
• Most CO2 in the atmosphere is natural, more than ~93%
• Man-made CO2 is less than 7% of total atmospheric CO2 = ~0.09 deg C
• the USA contribution to CO2 is ~20% equals = 17.6 thousandths deg C
• UK’s contribution to World CO2 emissions is ~1.8% = 1.6 thousandths deg C.

Closing the whole UK Carbon economy could only affect world temperature by this minuscule and immeasurable amount.

These figures only have to be just close to the right ball park to show how futile the whole emissions control effort could ever be.

And closing the carbon economies of the Whole World could only ever achieve a virtually undetectable less than -0.09 deg C. How can the Green movement and their supporting politicians ever imagine that their remedial actions and draconian taxes to control CO2 emmissions are able to limit warming to only + 2.00 deg C?

The UN IPCC, have changed their nomenclature from “Global Warming” to “Climate Change”. It now looks as if that too is morphing into “Global Climate Disruption”. Thus “Climate Change Believers” now back the horse whichever way it runs. Nonetheless all their policy recommendations are only ever intended to control excessive Global Warming by the reduction of Man-made CO2 emissions.

This is the BLINDING PARADOX of the Warmist position.

So any current global warming is not man-made and in any case such warming could be not be influenced by any remedial action taken by mankind however drastic. If the numbers above are even close to the right ballpark, they should be greeted with UNMITIGATED JOY:
• concern over CO2 as a man-made pollutant can be discounted.
• it is not necessary to destroy the world’s western economies to no purpose.
• if warming were happening, it would lead to a more benign and healthy climate for all mankind.
• any extra CO2 is already increasing the fertility and reducing water needs of all plant life and thus enhancing world food production.
• a warmer climate, within natural variation, would provide a future of greater prosperity for human development and much more food for the growing world population. This has been well proven in the past and would now especially benefit the third world.

Nonetheless, this is not to say that the world should not be seeking more efficient ways of generating its energy, conserving its energy use and stopping damaging its environments. It remains absolutely clear that our planet is vastly damaged by many human activities such as:
• environmental pollution.
• over fishing.
• forest clearance.
• industrial farming.
• farming for bio-fuels .
• and other habitat destruction.
And there is a real need to wean the world off the continued use of fossil fuels simply on the grounds of:
• security of supply
• increasing scarcity
• rising costs
• their use as the feedstock for industry rather than simply burning them.

The French long-term energy strategy with its massive commitment to nuclear power is impressive, (85% of electricity generation). Even if one is concerned about CO2, Nuclear Energy pays off, French electricity prices and CO2 emissions / head are the lowest in the developed world.
However in the light of the state of the current solar cycle, there is a real prospect of damaging cooling occurring in the near future and lasting for several decades. And as UK power stations face closure the lights may well go out in the winter 2015 if not before.

All because CO2 based Catastrophic Man-made Global Warming has become a state sponsored religion, supported by classic NOBLE CAUSE CORRUPTION. So any misrepresentation is valid in the Cause and any opposition however cogent or well qualified is routinely denigrated, publically ridiculed and as the world now sees literally terminated thanks to the Splattergate 10:10 film.

3. ### amabosaid

If I may make a formatting suggestion:
Don’t use middle-alignment when you have a large amount of text. It makes it harder, for me at least, to stay focused on reading.
If you use left-aligned text the eye will “naturally” know where to move when you’ve read until the end of a line.

4. ### mrpkwsaid

# 2
I disagree
“It is utterly futile to think that mankind can affect climate”

Without a doubt mankind can/does affect local environments but to affect climate, the entire planet with all of mankind being involved with the sole purpose of changing the climate would be needed.

But, also it may be possible that we are currently affecting the climate with man made C02. (certainly not the the absurdly catastrophic level that the CAGW crowd tells us)

I just don’t know

5. ### Jeff Idsaid

I’ve not sent any real time but believe that the divide by 2 vs 4 argument to have problems. The incoming energy of the earth is correct enough but the emission is over the entire surface area of the sphere. This is why we wouldn’t use half a sphere for emission. I may have missed some point, however my guess is that the calculations are in error at that point.

6. ### M. Simonsaid

industrial farming

If we end industrial farming we will have to put a lot more acres into production or starve.

I wonder which would be the best choice?

I vote for clear cutting forests and using the land for agriculture. Although slash and burn farming may be the better choice.

And for those of you too thick to get it: {/sarc} (love them curly brackets).

7. ### Oliver K. Manuelsaid

Thank you, Jeff, for this message.

There are many factual flaws in the AGW propaganda.

In my opinon, the most basic flaw is this: Changes in Earth’s climate cannot be understood without considering the empirical fact that:

a.) Earth’s heat source is a variable star, the unstable remains of a supernova that violently exploded and gave birth to the solar system five billion years (5 Gyr) ago [“Strange xenon, extinct superheavy elements and the solar neutrino puzzle”, Science 195 (1977) 208-209; “Isotopes of tellurium, xenon, and krypton in Allende meteorite retain record of nucleosynthesis”, Nature 277 (1979) 615-620].

b.) Neutron repulsion in an energetic but tiny (~10 km), compact neutron core powers the Sun [“Attraction and repulsion of nucleons: Sources of stellar energy”, Journal of Fusion Energy 19 (2001) 93-98; ) “The Sun’s origin, composition, and source of energy”, 32nd Lunar & Planetary Science Conference (2001) paper #1041].

c.) Cyclic changes in solar inertial motion (SIM) induce solar cycles of solar surface activity (sunspots) and changes in Earth’s climate [“Earth’s heat source – the Sun”, Energy & Environment 20 (2009) 131-144]

http://arxiv.org/pdf/0905.0704

With kind regards,
Oliver K. Manuel

8. ### M. Simonsaid

Re: #5 Jeff,

Global average temperature would be OK if radiation was a first order function. Instead it is a 4th order function. A 1 degree rise (temperature scale is not an issue for this comment) gives a (roughly) 4X increase of radiation over a linear model.

To get “real” radiation you need to take the temperature at multiple points (more is better for accuracy) and sum (integrate). You also have to account for the fact that the radiation sphere for a planet with an atmosphere is not the same as the surface sphere.

Of course you know this. I’m commenting for our less mathematically inclined readers.

Question: is the radiation sphere 33 degrees colder than the surface sphere?

9. ### Dereksaid

Thank you Jeff for posting this. Indeed amongst the many issues I try to clearly raise the most obvious one is of energy received by a sphere from a single source.

How can more than half the sphere be lit? It can not, so in the context of studying earth’s planetary climate system dividing by anything more than 2 for received energy at the surface from the (variable) sun is plainly unphysical non-sense.
But why do “they” do it, there must be a reason? There is a reason, it is a mathematical “quirk”.

The excel workbook clearly shows all the calculations in the K&T type budgets are for a black body. Earth IS NOT A BLACK BODY.

The more subtle issue I try to describe (and refer to above) is the 33 degrees mathematical “quirk”,
created by incorrectly comparing disc, hemisphere and sphere (that are really a disc) calculations.
The calculations and the sums I have done below each K&T type plots illustrate this most clearly I think.
I also illustrate at least five other ways (with the W/m2 range they would most likely be used in) to “create”
a “33 degrees effect” by incorrect, and plain wrong comparisons / calculations / sums.

Therein (the mathematical “quirk”) lies the rotten, pseudo science basis of AGW.

I am hoping someone with better maths training than me can describe the “quirk” I have noticed in better mathematical terms,
but I think I have illustrated well enough that it IS there, and being used.
The use of the “quirk” explains the adherence to black body, the difference between disc computer models, and hemisphere K&T budgets,
and almost all the other “strange” things AGW will not allowed to be questioned, such as,
” where is night??? “

10. ### Ryan Osaid

Derek,

First, echo Jeff and Simon . . . you must use the entire sphere for emission and integrate.

While there are certainly geometric simplifications used by GCMs (though I do not know if the simplification you discuss is actually used by any GCM), the problem with this as an explanation is that the geometry does not change between unforced (no CO2 effects) and forced (CO2 effects included) calculations – either using simple, 1-D models that can be done by hand or using GCMs.

In order for a geometric issue to explain away the difference between unforced and forced scenarios, the geometry must change between the scenarios. Unfortunately for your explanation, the geometry does not change, and cannot explain why different behavior is seen in the unforced and forced scenarios.

So while it is true that GCMs and smaller analytical models make some geometric simplifications (and may even get the geometry wrong), this has nothing to do with why they find a difference when CO2 and other GHGs are added.

11. ### Dereksaid

Ryan O,

With respect, the piece is not looking at emission,
the piece is looking at, in this respect, energy received with regard to “shape” calculated for.
The piece is looking at what is received at the earth’s surface, otherwise
we would not be looking at the increase supposedly due to a so called “greenhouse effect”.

ie, the first K&T plot in the piece (page 9), 161, and 494 are what are received at the earth’s surface.
161 from the sun, 333 from “back radiation” supposedly, hence
161 and 494 (161 + 333 = 494) are calculated to see the difference at the earth’s surface, as depicted,
supposedly due to the “greenhouse effect.

Looking at the minus 18 degrees lie “section” within the piece may also help.

12. ### scienceofdoomsaid

The article is extremely confused.

According to AGW unproven hypothesis and black body the earth’s surface temperature should be very, very cold at night

According to basic physics the earth’s surface temperature should not be very very cold at night. The oceans have vast heat capacity. It’s trivially easy to demonstrate that the ocean temperatures will not drop much at night. Why does the author make such a ridiculous claim? The only answer can be that the author has almost no understanding of the theory he is trying to debunk.

There is stuff in the article about averaging temperatures and geometry which somehow disproves something about “AGW”.

I can’t claim to understand the author’s thoughts, but maybe it is worth explaining the correct calculation. This is nothing to do with “AGW”. It’s just very basic physics.

1. From satellite measurements the energy received from the sun is about 1367 W/m^2. The sun is (from our perspective) a point source a long way away. Therefore, the sun’s rays are intercepted by the area of a disc with radius 6,360 km. So the total energy we receive from the sun every second = 1367 J x surface area of a disc (with radius 6,360 km). But for now let’s just call the area A1.

2. Some of the sun’s energy is reflected. About 30% is reflected (measured by satellites). Therefore the total energy absorbed is only 70% of the value in section 1.

3. The earth’s climate system radiates to space. This is measured by satellite. The globally annual average measured is about 239 W/m^2. So the total energy radiated from the earth every second = 239 x A2. A2 is the area that emits this radiation.

4. We would expect that the earth’s climate system is approximately in energy balance with the sun. Over any given year there is likely to be a small imbalance but because we don’t see strong warming or cooling the energy received must be approximately equal to the energy lost.

Question. What is the relationship between the energy absorbed from the sun (in 1 and 2) and the energy radiated to space?

A factor of 4.

Check – Solar radiation absorbed = 1367 x 0.7 = 957 x A1
Radiation lost to space = 239 x A2.

If these two are in balance then:

957 x A1 = 239 x A2
A2/A1 = 957/239 = 4.0

Why?

The energy received from the sun is intercepted by an area equal to the “disc” area. See the 2nd graphic in The Earth’s Energy Budget – Part One.

The energy radiated by the earth’s climate system is – of course – radiated by the whole surface area of a sphere.

The area of a disc = pi x r^2
The area of a sphere = 4 x pi x r^2

The radio between them is 4.

It’s very simple and doesn’t need any averaging of temperatures. We are only considering energy per second. We could multiply up both sides by the number of seconds in the year and the total surface area in both cases. Then we would have total energy in Joules over a year.

The ratio would still demonstrate that solar radiation needs to be divided by a factor of 4 to equate it to radiation emitted by the climate system (or that radiation emitted by the climate system needs to be multiplied by a factor of 4 to equate it to the radiation absorbed from the sun).

Hopefully this demonstrates clearly that the ratio of 4 is the right ratio.
The satellite measurement averages include days, nights, weekends and public holidays..

Now when we consider the relationship between the radiation from the earth’s surface to solar radiation absorbed it is just the same geometrical relationship.

The calculated emission of radiation from the earth’s surface is around 390 W/m^2. Note that this is much higher than the radiation emitted from the earth’s climate system to space (which is 239 W/m^2). This in itself tells you that there is an effect which needs to be accounted for.

Generally, this is described in temperature terms, which is where the 33’C “greenhouse” effect comes from. If the surface of the earth emitted only 239 W/m^2 (globally annually averaged) then the maximum average surface temperature would be -18’C. Instead it is around +15’C. And regardless of whether “average temperature” means anything, the “greenhouse effect” can be quantified in terms of emission of radiation, which can be averaged (or totaled to calculate for total energy).

Note: The last statement “calculated emission” in the penultimate last paragraph has a fascinating effect on many people. “Ah ha” – it’s made up.. So as not to hog too much space here, without having a huge number of expensive measuring devices all around the globe, including the oceans it isn’t possible to measure this surface radiation at a high enough spatial resolution to provide the average calculation. (You can’t measure it from space). However, whenever we do measure it, it always matches the Stefan Boltzmann equation. And the emissivity of the earth’s different surface types has been also measured and examined. The Stefan Boltzmann equation has 150 years of experimental proof behind it. Etc, etc. Basic physics and nothing to do with “AGW”.

13. ### Steve Fitzpatricksaid

Derek,

First, let me third what Ryan O, Jeff, and Simon said.

“How can more than half the sphere be lit? It can not, so in the context of studying earth’s planetary climate system dividing by anything more than 2 for received energy at the surface from the (variable) sun is plainly unphysical non-sense.”

I don’t think so. The illuminated cross-section (the area of the shadow cast by the earth, the area of the Sun’s light continuously striking the Earth) is Pi * R^2, but the total area of a sphere is 4*Pi*R^2, a factor of four larger. The average solar intensity for the whole of the Earth has to be 1/4 of the intensity of the Sun on a perpendicular flat disc with a radius equal to the Earth’s radius; ~1365/4 = 341.25 watts per square meter. Radiation lost to space is also based on the total Earth surface.. 4*Pi*R^2, the same area as you need to use for calculating average surface radiance.

14. ### Steve Fitzpatricksaid

#12,
“The article is extremely confused.”

I think that sums it up accurately.

15. ### Dereksaid

Jeff – Just to illustrate a little more clearlyin regards of the divide by 2 or 4 argument you mention.
If we look at the K&T type plot on page 9 again.
The solar input is 341, which is obviously divided by 4, so it should be 682, as planet earth does have a dark side.

I would of thought the other figures taken off the solar input are from the lit hemisphere (already divided by two in effect),
23 reflected by surface,
79 reflected by clouds and atmosphere, and
78 absorbed by atmosphere.

OK, so to compare more realistically the 78 and 79 figures might be slightly larger, but never the less, this leaves us with
a far more reasonable sum of,

682 solar input, minus 23 + 78 + 79 = 502.

Given 78 and 79 might be a little larger, and I’ not sure of the arcuracy of the 23 figure,
Albedo measurements.
http://www.globalwarmingskeptics.info/forums/thread-628-post-7506.html#pid7506
” How accurate is the earthshine method in determining global albedo?
The earthshine method doesn’t give a global albedo estimate.
It covers about one third of the Earth at each observation occasion and
certain areas can never be ‘‘seen’’ from the measurement site.
Furthermore the measurements are sparsely sampled in time, and
only made in a narrow wavelength band of 0.4 to 0.7 µm (Bender 2006). ”

Citation: Bender, F. A-M. (2006), Comment on ‘‘A multi-data
comparison of shortwave climate forcing changes’’ by Palle´ et al.,
Geophys. Res. Lett., 33, L15812, doi:10.1029/2006GL025745.

then,
this is very close to, and could well be, the 494 depicted as being received by the surface.

Mixing hemisphere and sphere figures it seems can have quite an (unrealistic) effect,
not usually noticed.

16. ### Oxbridge Pratsaid

#14, absolutely.

17. ### Steve Fitzpatricksaid

Derek,

I would think you were joking here, but you seem too serious for that to be the case. Really, you are simply wrong. You need to think about the geometry involved and about what it means when someone says “average solar intensity”. The Earth is not a flat disc. The average surface intensity is for the total surface area, and that total surface area is 4 times the area of one side of a flat disc. The average solar intensity is defined as total watts of sunlight divided by total surface area of the Earth.

18. ### Dereksaid

*13 – Err, yes, a disc of the same diameter has 4 times less surface area than a sphere, that’s the point.
A hemisphere has twice the area of the same diameter disc.

” The illuminated cross-section (the area of the shadow cast by the earth, the area of the Sun’s light continuously striking the Earth)”

Steve, would it not of been easier just to write DISC, which does equal Pi * R^2, as you note.

19. ### Dereksaid

*17 “The Earth is not a flat disc. The average surface intensity is for the total surface area, and that total surface area is 4 times the area of one side of a flat disc. ”

What, no night. ?

What, no hemisphere. ?

Who is confused. ?

20. ### Steve Fitzpatricksaid

Derek,

You have a bunch of engineers and scientists telling you that you are mistaken about a very basic calculation. I would go back and think this through again if I were you.

21. ### Steve Fitzpatricksaid

19.

Night hemisphere ==> area = 2*Pi*R^2
Day hemisphere ===> area = 2*Pi*R^2

Total area = 4*Pi*R^2

Total sunlight energy blocked by the Earth = Io*Pi*R^2

Where Io is the solar intensity in space (watts per sq meter)

Average solar intensity = total energy /total area = (Io*Pi*R^2)/(4*Pi*R^2) = Io/4

22. ### Jeff Idsaid

Derek,

I still haven’t read your entire work in detail. I’m a little slow about these things but typically, the calculation is energy in = energy out. From that you get temperature. The emission is from the surface of a sphere at average measured temp. I completely agree with M. Simon on the non-linearity which makes the problem more complex than this but if you have energy in – which can be calulated from a disk or sphere, it’s still the light blocked by our blue sphere from the sun (energy in) vs energy out, again from our little blue sphere in all directions. They must add, temperature which doesn’t match these calcs is attributed to greenhouse effect. So when you replied to Ryan above,

“With respect, the piece is not looking at emission,”

it is quite difficult to understand where your calculation is not an in vs out. Perhaps you can clarify that?

23. ### JTsaid

I agree with Jeff, et al, but there is a subtle shape effect due to the difference between the shape of a flat disc and a convex hemisphere, being the change in intensity of the incident radiation as you move from the center of the surface of the hemisphere toward the edges. At the center the angle of incidence of the incoming solar radiation is 90 degreesto the surface so a flux of 957 watts per square meter produces an intensity of 957 watts per square meter on the surface, but at the edge of the hemisphere the angle of incidence is 0 degrees so the intensity produced there is also 0. This effect can be given mathematical expression by multiplying magnitude of the incoming flux by the cosine of the angle between the flux vector and the vector normal to tangent to the surface at the point of absorption. Using a disc for calculation purposes reduces the surface area over which the absorption is occurring and creates a uniform intensity of flux across the entire surface instead of the non-uniform flux which actually exists. This correctly keeps track of the global averages but ignores the spatial heterogeneity of the flux absorption and thus the spatial heterogeneity of the surface heating. Treating the explanation as a boundary conditions problem as 12. does not sort out the details by correctly attaching the fluxes to the surfaces into which they are absorbed and from which they are emitted, nor does it explicate the time lags associated with the temperature variations and the energy stores which develop and subside over the 24 hour rotational period. Might I suggestthat a re-working of the Trenbarth energy budget diagram showing a spherical surface and top of atmosphere, and the heated layer of land and sea from which stored energy is released at night might reduce the confusion which the diagram in its present for seems to be causing.

24. ### Mark Fsaid

The solar intensity reaching the earth will be a function of cosine of angle away from noon and from the “tilted” equator, modified by bending due to refraction and diffraction as rays pass thru the atmosphere, with some critical angle considerations, integrated over the hemisphere. Obviously, the path through the atmosphere to the edge of the “disk” is greater than that along a radial axis of the earth aligned with the sun. I don’t think I’d have gotten through first year Engineering with a pi-r^2 or ^3 simplification. But I’m open to correction.

25. ### Jeff Idsaid

#24, My guess is that the difference between disk and by integrated angle is less than 10%. I’m wrong often though 🙂

26. ### Steve Fitzpatricksaid

#24,

The details could make some difference (refraction angles, etc.) but the first approximation is the ratio of the area of a disc to the area of a sphere. Derek needs to get that far before worrying about the nuances.

27. ### Ryan Osaid

Derek,

Hopefully you don’t take all of this too personally. It can be quite a shock to see what happens when you first post ideas on an open blog. The result is not always an ego boost! 🙂

Though it may take some time to realize, our goal here is not to criticize you, but rather to share our different knowledge and experience to try to come to a better understanding of what is really happening with the climate. We are (hopefully) just as hard on each other from a factual standpoint as we are on the warmist crowd.

I think it is clear that most of us do not feel your explanation is sufficient . . . and I hope that you take this as a learning experience rather than a rejection. We’ve all posted stuff that others thought was rather ridiculous. Besides, real learning only occurs when you realize you are wrong. Nothing is learned if you realize you are right (especially when you are actually wrong).

Science progresses through falsification, not confirmation. Almost every theory is eventually disproved. That doesn’t mean the theory was crap . . . it just means that new information is available to demonstrate that the theory was wrong, or, at least, incomplete. 😉

28. ### Steve Fitzpatricksaid

Ryan O #27,

You are a true gentleman.

29. ### Mark Fsaid

25 Jeff – 10 percent? What does that do to the error bars? Isn’t 10 percent a lot of solar radiation?

30. ### Jeff Idsaid

#29 Yup, it is. I recently left the comment at Judith Curry’s blog that I wouldn’t be surprised that the 33 C was in error – either way. I haven’t seen a good calculation yet. She used it in a thread which had almost 300 comments, I still haven’t had time to read it yet. Keep in mind thought that the error we should expect from high angles of incidence and specular reflection tends toward greater than 33C.

31. ### AMacsaid

Science of Doom, Thanks for the clear explanation of disc vs. sphere with respect to energy balance, at #12.

32. ### AusieDansaid

I agree with much of what has been said, but I’m glad I’m not doing the calcualtions or even expressing the physics in any formal manner.

I have only read this post quite quickly, so maybe I’ve missed the following.
But it seems that you all have not talked about the effect of the circulating atmosphere and ocean currents, which add a lot more to the mix.

If we assume that the earth is in approximately an equilibrium energy balance, at least in our lifetimes (the small percentage of average temperature increase, when expressed in degrees Kelvin over 100 years, makes me bold enough to claim that).

Well, there must be too much heat ariving in the tropics, relative to the global average, and too little reaching the poles. Heat must therefore be redistributed from the equator to the poles.
However, the ocean and atmospheric currents only part way compensate for this, so the globe must be in some sort of rough, temporaty dynamic equilibrium, in order to keep global, as well as local temperatures all in (seperate) equilibrium.

A model that could explain this complexity would be a marvelous thing.
Then we could add further complexity to test the various hypotheses proposed to explain the 0.7 degree per century warming – recovering from ice ages, CO2, sun spot activity (solar magnetic flux variations), poor thermomter siting, wrong data adjustments and so forth.

I find it all fascinating but perhaps governments would be better advised to get back to worrying about more urgent matters and stop trying to tax and regulate the most effecient forms of energy generation and propulsion out of existance.

33. ### AusieDansaid

What I was trying to say is that there seems to be an equilibrium energy balance for the earth as a whole and there must be an approximate balance of heat being transferred from the tropics to the poles as their temperature difference does not vary dramatically, and that all the systems must be in a dynamic balance too.
Magic!

Why do the circluating systems not carry enough heat around so that everywhere on earth is at the same temperature?
I conclude that there is a balance between the system maintaining overall global balance with those regional systems moving the heat about.

Surely some engineer reading this, can tear this last post apart and show me, quite simply, where I have gone wrong.

34. ### Brian Hsaid

I personally think Derek is having a problem with the time aspect. His numbers relate to the instantaneous incident flux — but that happens for each hemisphere for only half the day.

As for emissions, the daylit hemisphere must emit more than the night side, as it is warmer, and e-radiation is a function (4th power) of temperature. The net balance is net influx during the day, and net output at night, of course.

35. ### scienceofdoomsaid

Jeff Id said on December 19, 2010 at 9:38 pm:

#29 Yup, it is. I recently left the comment at Judith Curry’s blog that I wouldn’t be surprised that the 33 C was in error – either way. I haven’t seen a good calculation yet. She used it in a thread which had almost 300 comments, I still haven’t had time to read it yet. Keep in mind thought that the error we should expect from high angles of incidence and specular reflection tends toward greater than 33C.

The 33’C isn’t a number that can really be expressed with precision. It’s more a headline number like the rate of inflation or the increase in house prices last year in Kentucky.

It’s a number to help people visualize what the “greenhouse” effect means in number of extra coats they would need.

What can be expressed with a much higher degree of precision is the difference in total (or average) energy radiated from the surface and the total (or average) energy radiated from the climate system.

Converting this energy to temperature depends on the range of radiated values across time and across the surface of the earth. If the temperature was constant (in time and space) then 239 W/m^2 would correspond to a temperature of 255K (-18’C).

The more the temperature varies (in time and space) the lower the “average temperature” to still be radiating an average 239 W/m^2. See The Dull Case of Emissivity and Average Temperatures. For the mathematically minded, this is because radiation is proportional to the 4th power of temperature, so we don’t have a linear relationship.

I think the “high angles of incidence and specular reflection” are probably irrelevant to the inappropriately-named “greenhouse” effect.

36. ### scienceofdoomsaid

Another curiously confused comment from the article:

Earth is a grey body, because it has volume, a black body does not have volume, simply that is the end of “the controversy”.

I have seen similar inaccurate comments before but can’t fathom where they came from.

The difference between a “grey body” and a “black body” is this:

1. A blackbody has an emissivity of 1. So energy radiated, E = εσT^4, where ε=1.
2. A grey body has an emissivity >0 and <1. So energy radiated, E = εσT^4, where ε<1.

From the article it appears that "volume" is equated to "thermal mass". Yet in both 1. and 2. above, the body will have thermal mass. Thermal mass is totally independent of the emissivity of the body. They are not related in any way. In any physics textbook they are completely independent.

It’s like saying color is related to weight. Or texture is related to height..

Net change in energy = energy absorbed – energy radiated.

Net change in energy = mc.ΔT
where m = mass, c = specific heat capacity, ΔT = change in temperature

These equations are still completely true regardless of whether the body is a black body (no 1 above) or a grey body (no 2 above).

The quotation from the article fails the most basic test.

I welcome the author of the article explaining why he believes a blackbody has no thermal mass. What thermodynamics or physics textbooks states this? It’s a bizarre statement.

Why did the blog owner post this article??

37. ### Carricksaid

SOD:

Why did the blog owner post this article??

I think that was obvious: To give it public scrutiny, and because Derek requested that it be posted.

We learn as much from wrong results as we do from correct ones, as long as we understand what is wrong with them.

One of the latent functions of anonymous, confidential reviews is to protect the authors from the embarrassment of publishing “howlers”. That is one of the dangers of “puffball” reviews that established scientists in a field sometimes receive.

38. ### Jeff Idsaid

I don’t think we should be too hard on Derek. He presented his analysis openly with all calculations and he put a lot of effort into it. Sure it wasn’t correct but some of you might remember that this blog owner has made a few mistakes over the years. You guys have probably forgotten about the correlation analysis at CA or the global sea ice post which went on WUWT! I have a lot of respect those who don’t fear messing with the data or speaking their minds. When he asked to post it, my warning was that tAV readers can be a little rough sometimes, he said no problem.

Good on him.

39. ### Carricksaid

I agree with Jeff. Derek was willing to hang his work out here for the world to see. That takes guts and he deserves some credit for that.

And as I said, we can learn from examining the errors in his submission just as we learn from reading correct results.

(I’ll point out that Science of Doom has posted several blog articles featuring errors in articles from other sites. Apparently he agrees that deconstruction wrong arguments can be educational too.)

40. ### Shub Niggurathsaid

There is something wrong in explanation #12 but I cant put my finger on it. (however, this does not mean what #12 is trying to explain is wrong)

41. ### Steve Fitzpatricksaid

Carrick #37,

“One of the latent functions of anonymous, confidential reviews is to protect the authors from the embarrassment of publishing “howlers”. ”

Yes, but simply asking someone with basic understanding to read it over for obvious errors is probably a good first step in avoiding embarrassments.

42. ### PaulMsaid

Shub, you could start by questioning

This is measured by satellite. The globally annual average measured is about 239 W/m^2.

43. ### Ryan Osaid

You know, I just visited SoD for the first time. I got sidetracked by the link to “Why ‘Science of Doom’ Doesn’t Understand the 1st Law of Thermodynamics” and it made for some quite entertaining reading. Poor Gord. In the end, though, I came back to SoD, read the articles on Trenberth & Kiehl understanding thermodynamics. Much to recommend.

I think I will make SoD one of my regular visits!

44. ### Dereksaid

To Jeff, and ALL,

I have taken no offense at all from the comments here so far, although I have not read them through properly as of yet,
I’ve just skimmed through them.
First of four 11 hour factory floor shifts this week just completed.

It would appear that all that has happened here by most people posting comments is the quick skimming of my piece and excel workbook.
No one appears to have read, or even skimmed the articles and papers to read first that I state at the start of my piece.
Hands up, did anyone read even one of them, particularly Alan Siddons, and Claes Johnson. ?
I doubt anyone did read, even one, let alone consider the contents.

So, where does that leave us all, back at square one, the confusion of shape calculated for is still KING here at least.
Let alone the consideration it may cause problems, and that those problems do “create” an imaginary “33 degrees effect”….

I thought my post 15 in the comments here may have helped shed some light, but apparently not.

ie,

1) – AGW says, Solar input 341 (sphere figure) minus (23 + 78 + 79 hemisphere figures) = 161 received at surface, plus 333 “back radiation” (sphere figure) = 33 degrees greenhouse effect.

I say 341 (sphere figure) is unphysical, this planet has day and night, so that IS a lit and a unlit hemisphere = 2 hemispheres = 1 sphere,
NOT A 4 X DISC – that is just a big unphysical disc, that causes MUCH confusion, and miscalculation.

So,
2) – My piece says, 682 solar input to lit hemisphere, minus (23 + 78 + 79 (lit hemisphere)) = 502 [roughly] received at earth’s surface on the lit hemisphere.
My piece continues with night, but is too long to go into in this comment here, other than surface heat retention, and varying later release, because earth IS a grey body. AND, of course my daily water jacket plot.

Basically the above calculations get the same answer, but one has no imaginary “33 degree effect” conjured up by the incorrect comparison and calculation of different “shapes”.
I think the fact night follows day is a fairly strong indicator which of the two above calculations is more correct.
In fact 1) is fatally flawed, by it’s misuse of “shape”.

I also showed there are many other ways to incorrectly combine, calculate shape and incorrectly get a “33 degrees effect” that does not exist, except in the calculations.

Some here have even tried to argue whether earth is a grey body or a black body as calculated by AGW…Unbelievable.

I sincerely thank Jeff for posting my piece here, and the “airing” it has received,
you largely seemed to have lived up to my expectations…
Patiently though, I still await actually considered comments.

45. ### Dereksaid

Re “howlers”

Which is the howler in regards of received at the earth’s surface. ?

1) 341 – (23 +78 +79) + 333 = 494 W/m2 for the entire globe (includes, or ignores night).
4XDisc – (H, H, H) + 4XDisc = Sphere.

or,
2) 682 – (23 +78 +79) = 502 lit hemisphere only. ie, night not included.
All Lit Hemisphere

Time will tell.

46. ### scienceofdoomsaid

Derek,

I’ve read Claes Johnson. He doesn’t really believe Planck’s law, although he is quite obscure about it.
You can read our exchange on this subject.

Planck’s law is very well established and one of the foundations of modern thermodynamics.

If you rely on Claes Johnson then ask yourself if it is because you are sure than 100 years of physics is wrong, or because Claes says something you like to hear.

And I look forward to your comments on why you claim:
“Earth is a grey body, because it has volume, a black body does not have volume, simply that is the end of “the controversy”.
See # 36.

Your statement is completely wrong. Thermal mass & volume are completely independent of the emissivity of a body.

47. ### scienceofdoomsaid

Curious comment from PaulM – #42.

Why would you question the satellite measurement?
You can check it for yourself with the Ceres data.

I don’t think I can insert an image so I uploaded the 2009 monthly global mean to science of doom.

You can see that the value is around 239 W/m^2. This is day time and night time averaged.

This is the total radiation emitted from the earth’s climate system per unit surface area of the planet. Which also tells you that the solar radiation absorbed per unit surface area of the planet is around 239 W/m^2. But not exactly, because there is no reason why the planet has to be in perfect balance over a 12 month period.

48. ### scienceofdoomsaid

The image in #47 didn’t work.
Here is another attempt at displaying the monthly global mean OLR.
Sorry if it doesn’t work.

49. ### Andresaid

Well now,… as a Norseman occationally visiting southern Europe during summer vacations, I’ve always enjoyed the afternoons more then the midday’s sun.

50. ### Steve Fitzpatricksaid

Derek,

Despite the best efforts of a number of people (Jeff, Ryan O, SiOfDoom, and others), you seem immune to learning. Too bad.

51. ### Jeff Idsaid

Derek, here is the main error.

“My piece says, 682 solar input to lit hemisphere, minus (23 + 78 + 79 (lit hemisphere)) = 502 [roughly] received at earth’s surface on the lit hemisphere.”

You have 682 incoming from a hemisphere and have subtracted sphere numbers it should be 682 minus (46+156+158) because this is the incoming/reflected/absorbed energy on the hemisphere. Then you need to compare that to the outgoing and backradiation energy on the sphere. If you do it your way correctly, you will get exactly the same answer either way.

52. ### curioussaid

Does anybody have a ref. for the methodological process used for the original calculation (inc. error bands) of the annually and spatially averaged measure of outgoing radiation? I’ve had a look but can’t find it – I think it is perhaps in a Kiehl et al 1994 paper as referenced in Kiehl and Trenberth 1997:

“Kiehl, J.T., J. J. Hack, and B. P. Briegleb, 1994: The simulated earth radiation budget of the National Center for Atmospheric Research community climate model CCM2 and comparisons with the Earth Radiation Budget Experiment (ERBE).
J. Geophys. Res., 99, 20 815–20 827”

Thanks

53. ### CNY Rogersaid

Derek,

I think a point Science of Doom made in his first post is the key problem: “We are only considering energy per second. We could multiply up both sides by the number of seconds in the year and the total surface area in both cases. Then we would have total energy in Joules over a year.”

It looks like you are taking the snapshot approach (the references to night) but when you total everything over the entire year the result will be different.

Thank you for putting all the information out for review.

54. ### Dereksaid

Thank you All, I will digest the comments and points raised ASAP, but that will be in a few days time.
I hope to get round to them over or just after Xmas, and will reply if it seems sensible to here.

Two small points,
1) just slightly more than a hemisphere will be lit will it not as the sun is not a “point”, I know it’s a long way away,
and my 1M beach ball and marrow fat pea a football pitch apart simile helps imagine this.
Presumably earth’s shadow is a “cone” shape.

2) Presumably the “disc” scenarios have all their surface perpendicular to the sun’s surface, so they will not be exactly flat.
I imagine they will actually be “saucers” rather than flat discs.

Apols, just trying to get the “shapes” sorted out in my head correctly.

55. ### Steve Fitzpatricksaid

Curious #52,

See this thread, with DeWitt’s and my comments: https://noconsensus.wordpress.com/2010/12/20/fixing-the-basic-agw-calcs-ii/#comments

DeWitt was kind enough to make a scale change which makes the basic data more useful.

56. ### curioussaid

55 Thanks Steve – I had seen those and I also recall DeWitt’s original post some time ago. The paper I think I’m after is the one in my comment 52. I have the Kiehl and Trenberth 1997 paper but the Kiehl at al 1994 seems to be paywall only. If anybody has a link to a freebie that would be appreciated as would any other link to a discussion of how they (K et al or other source) arrived at the planet wide, annually averaged, equivalent radiation figure that SoD quoted in point 3) of comment 12 above. Thanks

57. ### kuhnkatsaid

SOD,

“Your statement is completely wrong. Thermal mass & volume are completely independent of the emissivity of a body.”

Then when I heat a rock it should emit all the energy immediately and I cannot put it in my pocket to keep my hands warm for a while?

This is the concept that APPEARS to be at issue for ignorant people like me.

I have no doubt that the CONCEPT of a black body has been very useful as many other CONCEPTS. I DOUBT that it really encompasses the earth system and like many other concepts needs other CONCEPTS to be added or subtracted to make a reasonably workable model. If you could EXPLAIN to us ignorant types how the math and models that are obviously above our heads takes this into account you could probably win at least a couple of battles if not the war.

58. ### kuhnkatsaid

SOD,

“You can see that the value is around 239 W/m^2. This is day time and night time averaged.”

Is this m2 surface area, emission altitude area, or TOA area?

59. ### curioussaid

SoD re: 12.3 – I see you are now posting on this topic at you site:

http://scienceofdoom.com/2010/12/23/understanding-atmospheric-radiation-and-the-greenhouse-effect-part-one/

I’ve had a read but I still can’t see the reference for the algorithm that produces the monthly average global figure which you plot in your first black and white figure. The graphic isn’t numbered but you introduce it with the line:

“If we summarize this data into monthly global averages:”

and you follow it with

“The average for 2009 is 239 W/m². This average includes days, nights and weekends. The average can be converted to the total energy emitted from the climate system over a year like this:

Total energy radiated by the climate system into space in one year = 239 x number of seconds in a year x area of the earth in meters squared

= 239 x 60 x 60 x 24 x 365 x 4 x 3.14 x (6.37 x 106 )²

= 239 x 3.15 x 107 x 5.10 x 1014

ETOA= 3.8 x 1024 J

The reason for calculating the total energy in 2009 is because many people have realized that there is a problem with average temperatures and imagine that this problem is carried over to average radiation. Not true. We can take average radiation and convert it into total energy with no problem.”

As this is taken as a key component in the debate about the Earth’s energy budget which is the fundamental concept behind the current argument of AGW please can you supply a reference for the derivation methodolgy of this monthly value? I’ve had a look at the CERES site and found these docs of which 2.0 and 3.0 look promising but have not had chance to absorb them:

http://ceres.larc.nasa.gov/atbd.php

I’m hoping to get a feel for the possible error bands on the calc which produces the 239W/m2 figure. Thanks.

60. ### Dereksaid

Post 1.

mrpkw said
December 19, 2010 at 1:31 pm

Interesting.
Not completely sure as to what it means (even after reading the summary).

Can someone dumb it down for me???
(Too many equations !!!)

I will happily admit that I do not like equations, yes, they are “short hand”, but they can also be very, very misleading.
My usual “defense” is that I try to “walk” myself verbally through an equation. So,
in this case what would my verbal walk through the equation be?

The equation.

(P divided by 5.6704) raised to the power 0.25, and then multiplied by 100.

Verbal “Walk through”
P is the power of a beam in Watts per meter squared received at a black body’s surface,
this is divided by 5.6704, and raised to the power 0.25, the resulting figure is then multiplied by 100.
The end figure is the temperature expected at the black body’s surface for the power of beam it is receiving in degrees Kelvin.

OK, so what does it mean? Well firstly the equation does not infer any shape or area whatsoever.
P is merely a rate, not a shape. Yes, the answer applies to all the surface area receiving the same rate, but at the end of the day,
the equation is (merely) a “point” calculation.
This in itself I suggest gives credence to the points raised in my piece,
which is about the effect of incorrectly applying “shape” using the equation.
There is a mathematical “quirk” between the “shapes” as incorrectly applied, and
this mathematical “quirk” does produce a “33 degrees effect” that
does not necessarily exist in reality upon planet earth.

What does dividing P actually do.?
In the case of comparing it to earth, or a black body receiving energy from a single source, dividing P moves the object away from the source.
This must be the case as that is the only way the W/m2 received by the black body could be reduced, with a single and constant source of P.
If we agree that earth is nominally 93.5 million miles from the sun, then dividing P by two doubles the distance between the, and if P is divided by 4, then it quadruples the distance between them (187 million miles, and 374 million miles).
So, I agree (at least partially) with Jeff Id, when he posted in comment 51 that my use of 682 is incorrect, but that also means that the 341 figure is also incorrect, and incorrectly used.
Yes, Jeff you are correct (and I had realized, but had not had time to post, to be honest) in noting I should also have doubled the other figures mentioned. But as I do not see the point in considering a disc 187 million miles from the sun, or a disc 374 million miles from the sun, when earth is only 93.5 million miles from the sun, I suggest we are all in error, and it is not a small error.

The other point I would like to raise is that the often quoted “disc” is not a disc at all really, and neither is it flat.
The equation is for an area receiving the same rate of input, so
from a point source of P the shape would have to be a saucer to maintain a constant R and therefore input.
Also, the sun is not a point, nor is it’s effect surface of emission either flat or constant.
One look at any image of the sun and it’s solar flares confirms this.
So, the often quoted flat disc is a) more of a saucer shape “mirroring” the sun’s surface, and
b) is the surface of the saucer is constantly in a flux moving to and fro to maintain a constant W/mm2 input fro the source.
In short, when the equation is used to compare to the earth, then the resulting “shape” is more like a very wobbly saucer shape that is most commonly 374 million miles from the sun…

I think it is safe to say “we” need a better “model”….

My tentative, first suggestion.
Image a meter square at eye level about ten feet in front of you. The square is vertical, or perpendicular to your view of it.
If the square is tilted away from you (vertically) does it reduce in apparent height to you, yes.
If the square is then also tilted away from you (horizontally), it’s visible area to you is further reduced.
Your “view” would be of the effective area exposed to P.
Given a rate and time (W/m2 for example) for P, and an effective area exposed (Latitude and longitude “corrected”), then a volume (Joules) per square would be quite easy to calculate.

To apply this to a lit hemisphere of known size (earth), a known number of miles from a source of P (the sun),
would only require each “square” to be corrected for actual area on a hemisphere’s surface.
So, this would be a grid of degree squares (180 x 180) corrected for latitude and longitude inclination (from source of P),
and then corrected for actual area on a hemisphere’s surface.
Given the rate, time and area are thus calculable, then a volume (Joules) answer is relatively easily obtainable.
This would be a “model” for a result for a hemisphere “shape” that was it’s actual distance from the source of P.
A vast improvement over the present “disc” “calculations” that are apparently (when P is divided by 4, 374 million miles from the sun….).

This “hemisphere model” approach would be quite versatile, and it would seem to me to be capable of far more than merely a “snap shot” view,
it could easily be doubled into a “sphere”, and / or “rotated” for example, and
incorporate many things not presently included in modeling, such as
surface heating, retention and varying later release….Oh, and night of course…..

Given the title of the thread I hope I have remained on topic..

To the regulars here – I had to listen again to the stupid claims of AGW on the news this morning.
Some environmental group or other is now claiming polar bears have an 80% chance of extinction within 40 years…
(I wonder if THAT will make it past “peer review” into the AR5 report….Give me “blog review” ANY AND EVERY DAY.)
– that is why I sometimes may appear a bit “controversial” – AGW is baloney, plain and simple,
please start questioning it’s principles,
not merely restricting yourselves to quibbling the AGW figures.
I hope the example of “shape” with regards to K&T type plots, and modeling I have been looking at in this piece / thread
at least partially helps, to move us all forward to a better understanding.

61. ### Dereksaid

Clarification –

Please note in my comment at post 60 above, by “point” calculation, I mean for an area equivalent to a square meter.
This does not I believe change the basic thrust of what I have posted, the equation says nothing whatsoever about “shape”.

I dislike this type of blog format because you can not correct your own mistakes afterwards, that said,
at least you do get my thoughts in the raw so to speak.
There are some minor typos in the above comment which I hope people see as such, rather than misinterpret.

I will also check the distances involved for P /2 and P /4 “disc” “shape” and post them here if there is any change in distance.

62. ### Dereksaid

Just to be polite, and above board so to speak,
I also have the below threads in regard of the pdf and excel workbook/s that you have all kindly “blog reviewed” here,
at the GWS (my “home”) forum.

Open thread for considered questions.
http://www.globalwarmingskeptics.info/forums/thread-1039.html

63. ### DeWitt Paynesaid

Then when I heat a rock it should emit all the energy immediately and I cannot put it in my pocket to keep my hands warm for a while?

That’s not it at all. If the rock had no heat capacity, which is the coefficient relating energy temperature, then it would indeed assume the ambient temperature immediately. But it does have significant heat capacity so it can’t change temperature instantly. Emissivity is a separate property of the surface of a solid object and is completely unrelated to its heat capacity. Emissivity = absorptivity and has a range from zero to one. A black body has an emissivity of 1. A perfect reflector or a perfectly transparent object has an emissivity of 0. Everything in the real world is somewhere in between. A polished aluminum surface has a much lower emissivity than a black anodized aluminum surface. You can see this with an IR thermometer (no home should be without one, ~\$50 at auto parts or kitchen supply stores). Boil water in a shiny steel or aluminum pot and in a black anodized aluminum or cast iron pot. Point your IR thermometer at the two pots and compare the readings. The dark pot will be close to 100 C while the shiny pot will be much lower.

64. ### Dereksaid

I have not replied to Science of Doom as of yet, particularly his comment 46 including,

“Planck’s law is very well established and one of the foundations of modern thermodynamics.”

Yes, SoD, I am aware, and have skimmed your comments / exchanges with Claes Johnson.
In my opinion, and from my everyday observations, you have lost comprehensively.

Why, is explained best to date, in this short piece by Frank Davis, titled Maxwell vs Planck.
http://frank-davis.livejournal.com/132165.html

I understand my views are best described as “Maxwellian” and your views as, well, those of a “Planck”.

I do have distinct reservations with one of Frank Davis’s comments though, namely,

“If so, it will only be physicists – and not politicians or anybody else – who will ultimately be able to resolve the dispute”

My reservations are,

i) Professor Claes Johnson is a respected professor of applied mathematics, not a physicist.

ii) It was physicists arguing between themselves that got us into this mess in the first place,
if we are to learn just one thing from this, it is surely the essential requirement for multi-disiplinary discussion / peer review.
(Particularly where paradigms within one area clash – astronomy most likely next….)

65. ### scienceofdoomsaid

I stopped checking on this thread after it looked like people had stopped posting. I see some comments and questions that I will try to respond to.

First, Kuhnkat said on December 22, 2010 at 12:12 am

Citing my original comment of December 20, 2010 at 6:03 pm:

Your statement is completely wrong. Thermal mass & volume are completely independent of the emissivity of a body.

Kuhnkat said:

Then when I heat a rock it should emit all the energy immediately and I cannot put it in my pocket to keep my hands warm for a while?

This is the concept that APPEARS to be at issue for ignorant people like me.

I have no doubt that the CONCEPT of a black body has been very useful as many other CONCEPTS. I DOUBT that it really encompasses the earth system and like many other concepts needs other CONCEPTS to be added or subtracted to make a reasonably workable model. If you could EXPLAIN to us ignorant types how the math and models that are obviously above our heads takes this into account you could probably win at least a couple of battles if not the war.

No. Thermal mass & volume are still completely independent of the emissivity of a body. I think you haven’t understood what that the independence of these parameters means.

Here is an example, slightly simpler than putting something in your pocket:

Let’s take a body of 1kg with specific heat capacity of 1000J/kg.K. Therefore a heat capacity, C, = 1000J/K.
Surface area = 2m^2.
Emissivity = 1 (a blackbody).
Temperature (time=0) = 300K
Background temperature = 0K (lost in the vastness of space)

After time =0 the body radiates energy, E=2 x 1.0 x 5.67×10^-8 x T^4 [eq 1 from Stefan-Boltzmann’s law, E = area x emissivity x stefan-boltzmanns constant x T^4]
Change in temperature, T, dT = E/C [eq 2]

We can calculate this change in temperature T vs time based on the above equations.

I can plot a graph and post it on my blog if you are interested to see it.

Now let’s take an example of a body with all of the above characteristics and starting temperature but an emissivity = 0.6 – i.e., not a blackbody.
Guess what? Same equations with one changed parameter:

After time =0 the body radiates energy, E=0.6 x 5.67×10^-8 x T^4 [eq 1]
Change in temperature, T, dT = E/C [eq 2]

The resulting dynamic temperature change will be different.

The fact that the body does OR DOES NOT not emit with an emissivity of 1 is completely INDEPENDENT of its specific heat capacity, mass, surface area, thermal conductivity.. and every other parameter you might think of.

It doesn’t mean that a rock will emit all of its energy immediately before you put it in your pocket. I have no idea why you think this is something that anyone would believe, or what its possible connection is with “blackbody” or “non-blackbody”.

I would appreciate you explaining what your thinking is here. I would like to understand it.

66. ### scienceofdoomsaid

On my earlier comment #65, I missed the surface area in the second set of equations. Should be (see bolded “2”):

Guess what? Same equations with one changed parameter:

After time =0 the body radiates energy, E=2 x 0.6 x 5.67×10^-8 x T^4 [eq 1]
Change in temperature, T, dT = E/C [eq 2]

67. ### scienceofdoomsaid

#58: Kuhnkat on December 22, 2010 at 12:21 am:

Citing my comment: “You can see that the value is around 239 W/m^2. This is day time and night time averaged.

Asked:
Is this m2 surface area, emission altitude area, or TOA area?

That’s an interesting question. What surface area does the satellite measurement convert to?

The difference between the surface area of the earth and the surface area of TOA is small. There is about a 1.5% difference assuming a 50km TOA.

But I don’t know the answer to the question although I will see if I can dig it out.

68. ### scienceofdoomsaid

#59 Curious said, referencing my CERES graphs, on December 23, 2010 at 1:12 pm:

I’ve had a read but I still can’t see the reference for the algorithm that produces the monthly average global figure which you plot in your first black and white figure. The graphic isn’t numbered but you introduce it with the line: ‘If we summarize this data into monthly global averages’..

The monthly averages are produced by the CERES website. I can’t post graphics here, so I have posted the link and the screen grab of what the selector tool looks like to obtain monthly averages – in the comments for the article here.

As this is taken as a key component in the debate about the Earth’s energy budget which is the fundamental concept behind the current argument of AGW please can you supply a reference for the derivation methodolgy of this monthly value?

It’s a fundamental component of basic climate science. What is so special about “AGW” when we are interested in the top of atmosphere outgoing longwave radiation?

I don’t have a reference for it. The CERES website says it is the average, so I believe them and expect it to be the mean of all the values for the month. The values are area-weighted to start with. What else could “average” be?

I recommend reading http://ceres.larc.nasa.gov/documents/ATBD/pdf/r2_2/ceres-atbd2.2-s8.0.pdf and reporting back if you find that instead of “the mean” they calculate something different.

Previous studies from the ERBE 1985-1989 values also reported similar, but not identical global annual averages. However, CERES is the most accurate measurement system so far – for measuring OLR.

I’m hoping to get a feel for the possible error bands on the calc which produces the 239W/m2 figure. Thanks.

You can see the error figures cited from Dessler et al (2007):

The uncertainty of an individual top-of-atmosphere OLR measurement is 5 W/m², while the uncertainty of average OLR over a 1°-latitude x 1°-longitude box, which contains many viewing angles, is ≈1.5 W/m²

69. ### scienceofdoomsaid

Derek said on December 29, 2010 at 4:09 am:

I have not replied to Science of Doom as of yet, particularly his comment 46 including,

“Planck’s law is very well established and one of the foundations of modern thermodynamics.”

Yes, SoD, I am aware, and have skimmed your comments / exchanges with Claes Johnson.
In my opinion, and from my everyday observations, you have lost comprehensively.

As and when you can explain what is flawed about Planck’s law you will have everyone’s attention.

Note my response on Frank Davis’ blog:

I ask the people who accept Claes Johnson’s ideas – what has he actually demonstrated is wrong with the “greenhouse” effect? “Demonstrated” doesn’t mean “stated”.

And if you can’t understand his paper, then why accept his conclusions?

His paper is “Computational Blackbody Radiation” dated September 16, 2010

what is the connection between parts 1-7 full of maths and part 8? – where he says:

“A main lesson of this note is that “backradiation” is unphysical because it is unstable and serves no role, and thus should be removed from climate science..”

In fact, Johnson’s paper appears to be trying to deal with the theoretical rationale behind Planck’s law rather than the actual results. The measured results are still the measured results. And his paper doesn’t demonstrate anything at all about climate physics.

Still, if you think “I have lost comprehensively” then you will have no trouble explaining the connection between the statement in part 8 (cited above) and the balance of his paper.

How does he demonstrate this claim?

70. ### Curious Confusion about “Blackbodies” « The Science of Doomsaid

[…] another blog, there was a confused article posted about the earth’s energy balance. Within the article was […]

71. ### scienceofdoomsaid

I posted an article Curious Confusion about “Blackbodies” hoping for enlightenment about what so many people believe is “the problem” with “blackbodies”.

72. ### Dereksaid

SoD (I genuinely hope that is the appropriate abbreviation), quoted himself as saying,

“I ask the people who accept Claes Johnson’s ideas – what has he actually demonstrated is wrong with the “greenhouse” effect?

I think you have missed the basis here, at a very basic level, that Frank Davis also missed,
when referring to “Maxwellians” and “Plancks” (as I put it), of the terms and their “definitions”.

A “Maxwellian” in this sense is someone who does not, infact sees as fanciful, and / or a gross and misleading (and wrong) oversimplification, the whole notion of AGW as presently proposed / advocated (most people actually).

A “Planck” in this respect is someone who either believes in AGW “wholesale”, or merely quibbles the figures, WITHOUT questioning the principles (“theories”) of AGW, usually because it HAS BEEN ACCEPTED FOR A LONG TIME …Is that science ?

Most people I know and meet do not “believe” in AGW because they can not “see it” – this is the point.
This is the real “Maxwellian” / “Planck” issue. Can it be seen in everyday observation – No.

“We” can not see, or feel “back radiation, when “we” should be able to.
“We” can not see, or feel the “greenhouse effect”.
As for, “all radiation is positively absorbed” – in everyday life, in all other forms of heat / energy movements – PULL THE OTHER ONE.
What makes IR so different in AGW “explanations”… Most conclude “bulldung”, hence most are “Maxwellian”,
not “Plancks”..

Frank Davis, and SoD missed the point when it was said that “Maxwellians” see what they want to see in the maths,
even if they do not understand the maths.
The point is, that Claes Johnson’s maths agrees far more closely with most people’s everyday OBSERVATIONS.

I think it was Karl Popper who said that effectively none of our science is correct, most understand this, and
most understand that Claes Johnson is far nearer (apparently) the (apparent by observation) truth than AGW “explanations” so far.
THAT is the “Maxwellian” point in these “discussions” to date.
Whether Claes Johnson is excatly right or not, is actually a “secondary point”.

SoD, with respect, it is not that people accept Claes Johnson’s maths “wholesale”, it is that people “see” his points, made mathematically,
make much more sense than AGW’s, and most main stream sceptics points do at present.
Principally because the principles of AGW are not being questioned by the “Plancks”.

To make the point that how “things” are being questioned is the issue a bit better, I will refer to you this Dr. Spencer thread,
nearly referred to by Frank Davis (it was a follow up thread / experiment to the “Virginia” thread),
http://www.drroyspencer.com/2010/07/first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature/

and my “unanswered” as of yet question….

Derek says:
July 30, 2010 at 2:56 PM

Errr, I assume the air temperature sensor is getting some regolith (and vegetation) released heat, particularly at night
(relative cooling by day I would also assume), and the cavity sensor is getting none, or very little (day and night).
Would this be bound to produce a higher air temp than cavity temp at night, and slightly lower (air temp) during (at least the early part of) the day. ?

If you have allowed for this, and I had missed it on my quick skim through so far, then please accept my apologies.

yours,
Derek.

*
Roy W. Spencer, Ph. D. says:
August 2, 2010 at 6:03 AM

the box is meant to minimize the rate of heat flow from *everything* except IR radiative loss by the cavity, and IR radiative gain from the sky.
End of quote.

I do not see, as usual, that the point I raised was even addressed, let alone answered..

“Plancks”.. Therein lies the “problem”.

73. ### scienceofdoomsaid

Derek,

Re #72:

I believe in the scientific method which means even things we can’t “feel” and “experience” are still true.

I can’t feel that the earth is rotating but I believe it is from the evidence.
I can’t feel that the earth follows around the sun in an ellipse but I believe it does from the evidence.
I can’t experience the intensity vs wavelength graph of EM radiation but I believe it is true from the evidence.
I can’t experience most things that are clearly demonstrated by science.

Anyway, now that you have clarified your fundamental position on the scientific method I don’t think we will find new areas of agreement.

74. ### Dereksaid

Nice misquote SoD. I said see and feel, not feel and experience..

I do understand we can see the earth rotating, it is called day and night..
I do understand we can feel earth’s elliptical orbit around the sun, it is called summer and winter…
I would of thought a filter of a certain wavelength works at all intensities of that wavelength, but not at all wavelengths,
a pair of sunglasses (anti UV tinted for instance) is probably a good enough example..
Arrh well, that is the world “Plancks” live in, a virtual world, if it works for you, so be it,
I prefer real world observation/s.

Talking of which can you answer my observation / question of Dr. Spencer’s experiment’s regolith effect ignored problem.??
AND,
Can you offer a better explanation, or find fault in my explanation and interpretation of
what a thermal image of a greenhouse AND it’s surroundings actually shows,
as I described in my pdf..
ie,
” Question, have you ever looked at a thermal camera picture of a greenhouse and it’s surroundings. ?

The greenhouse will be warmer than it’s surroundings, so it will be a brighter image than it’s surroundings.
AGW “theory” says the greenhouse is warmer because it traps radiation,
YET the thermal image will clearly show beyond doubt that the greenhouse IS radiating MORE than it’s surroundings
(with the same solar input).
What does this show. ?

It shows whatever is cooling the surroundings is far more powerful than radiation, AND
that it is not radiation doing most of the cooling (otherwise the higher radiating greenhouse would be cooler than it’s surroundings).

If one opened the doors and windows in the greenhouse, especially on a breezy day,
the greenhouse would soon reduce it’s temperature to that of it’s surroundings.
Logically the temperature difference was removed from the greenhouse to the surroundings or aloft by air.
Air transporting sensible, and latent heat (of water vapourisation), I would suggest.

When the doors and windows are closed it logically follows that the increase in temperature inside the greenhouse is due to
the reduced transport of sensible and latent heat (of water vapourisation) from inside the greenhouse to the surroundings.
Obviously even with the doors and windows of the greenhouse are shut,
some conduction and convection at the greenhouse glass surfaces occurs, and
this explains why a greenhouse remains warmer than it’s surroundings for some time after sunset.

Hopefully this has illustrated that a greenhouse works by reducing conduction and convection,
NOT by “trapping” radiation, as AGW pseudo climate science, “physics”, and “theory” suggest. ”

SoD wrote,
” Anyway, now that you have clarified your fundamental position on the scientific method”
Yes, I have indeed, real world observation is the basis of my approach, and fundamental physics and effects..

” I don’t think we will find new areas of agreement. ”
Whilst you adhere to (seemingly unquestioningly) “Planck” based pseudo / playstation AGW “climate science” “principles”,
(against real world observations) you are correct in this statement.

75. ### Carricksaid

Derek:

I do understand we can see the earth rotating, it is called day and night..

Not true. Since you can’t feel the rotation of the Earth, the obvious conclusion is the universe is revolving around us. That was the explanation the ancient gave, and it took until Galileo for it to be totally overthrown. The correct explanation requires inference, because experience doesn’t help.

I do understand we can feel earth’s elliptical orbit around the sun, it is called summer and winter…

Again a modern scientific explanation. What you “feel” of winter and summer cannot be used to infer this. All you know is the sun moves north in the summer, and south in the winter.

Certainly not the elliptical orbit part… that was established by Kepler. (Even the original heliocentric explanation assumed circular orbits.) What you can see directly is that the sun moves to higher latitudes in the summer and lower ones in the winter. Again the correct explanation requires inference, because experience doesn’t help.

I prefer real world observation/s.

Here’s some questions:

How old do you think the Earth is (and why)?

How old do you think the Universe is?

What is the distance from the Sun to the Earth?

You won’t get very far at all in explaining how your universe works without inference.

And what is the distance from the Sun to the Andromeda Galaxy?

76. ### Dereksaid

Carrick Post 75 – Answering your first “point”, your “obvious conclusion” is only if you ignore the perfectly observable curvature of the earth.

The obvious conclusions you describe are best kept for the flat earth society, where they will be considered no doubt in isolation,
of other real world observations that are inconvenient..

77. ### Carricksaid

Derek, a spherical earth doesn’t force us to a heliocentric picture of the solar system. The Greeks knew the Earth was curved (and even measured its curvature) but still assumed a geocentric universe.

78. ### Carricksaid

One of the biggest “ah ha” moments was the realization that by Newton that the force that attracts an apple to the Earth is the same force that attracts the Moon to the Earth. This was entirely inferential, as was his inverse square law of gravitation, it’s not based on “real world” observations. (Observations support it, just as they do the radiative physics that you seem to be in denial about.)_

79. ### Dereksaid

I think you have the who is in denial confused between us Carrick.

I have shown, and you have not even attempted to answer that “accepted” interpretations (at quite a few levels now) “maybe” in error..

I think Newton’s apple was in the real world, presumably he based his calculations upon this observation from the real world…

Carrick, why not try to deal with my problem with Dr Spencer’s experiment, where he ignores the regolith effect,
or,
my interpretation of what does a thermal image of a greenhouse AND it’s surroundings ACTUALLY show. ??

80. ### Carricksaid

Derek, I think you are very confused, or at least your arguments are.

You’ve claimed that day and night proves the Earth is rotating (it does not).
You’ve claimed that winter/summer proves an elliptic orbit of the Earth around the Sun (it does not).
You’ve claimed that the observed curvature of the Earth precludes the Sun rotating around the Earth (it does not).

my interpretation of what does a thermal image of a greenhouse AND it’s surroundings ACTUALLY show. ??

Dealt with that on Jeff Id’s thread, includes “real world” data. If you add IR trapping, the temperature of the interior of the greenhouse is increased.

And of course as I mentioned on that thread the “atmospheric greenhouse gas effect” works differently than the “real greenhouse effect”. Unfortunate nomenclature doesn’t change the underlying meaning.

81. ### Steve Fitzpatricksaid

Derek,

You have some very unusual ideas about how the world works. A few days ago, over at Lucia’s Blackboard, Julio (who is a professor of physics, specializing in quantum mechanics) commented as follows:

“For a scientific conversation (as any other conversation) to get anywhere, the people involved have to first agree implicitly on certain basic beliefs that will serve as the foundation of everything that follows.”

I think that sums up the problem here pretty well. People like Julio, Carrick, Jeff, Lucia, Ryan O, Science of Doom, Gavin (at Real Climate), and many others (including me) all share a basic understanding of how the world works, even though we do sometimes strenuously disagree about the details. That understanding is based on the science and engineering taught at colleges and universities all over the world: basic physics, chemistry, thermodynamics, radiative transfer, etc. I do not know why you reject most all of that basic science as mistaken, but clearly that is the case. Under these circumstances, it is just not productive for most people here to discuss science or engineering subjects with you.

I wish you well, but urge you to read some basic texts on physics, and especially radiative transfer.

82. ### Jeff Idsaid

“even though we do sometimes strenuously disagree about the details”

I thought we always agreed on everything?!

83. ### Steve Fitzpatricksaid

Jeff,
“I thought we always agreed on everything?!”
Nah..

84. ### Carricksaid

Jeff ID:

I thought we always agreed on everything?!

Yeah, we’re a real echo chamber!

85. ### Eli Rabettsaid

OK, If you start to read Derek’s slides, you soon come to the point where he goes off track

1)The complete ignoring of planet Earth’s permanently unlit side(hence the “disc world” and “evenly lit sphere” approaches) This avoids “explaining” why the earth is no where near as cold as it should be (according to AGW and black body) on it’s dark side. The computer models and the K&T type plots BOTH ignore night.”>

And if you carefully go through the rest of it, and the conversation here, you see that Derek, unlike Keven Trenberth, is ignoring day and night. The Earth does not have a permanently unlit side. Mercury does. The Earth does turn, daily, and this is accounted for in energy balance arguments (2 of the factor of 4*).

This is pretty simple, and really does not require very much, if any, physics or math to find and point out. FWIW, SoD pointed out that the surface does not cool to 2.7K at night because of the heat capacity of the surface and oceans.
———————————–
*by which the average intensity of sunlight striking the surface is reduced from the intensity perpendicular to the direction between the sun and the earth at the earth’s orbit. The other factor of 2 is accounted for by the obliquity, or the angle, at which sunlight strikes the surface other than at the equator discussed above. The average amount reflected is 30%, but this is more complex, being a function of where (angle, cloudiness and surface structure – water, ice land, etc.).

Oh yeah, if Eli wanted to be really picky, he would point out that a disc has two sides. . .

86. ### Eli Rabettsaid

OK, forgot to close the blockquote. happy new year.

87. ### Dereksaid

Eli Rabett said
January 1, 2011 at 8:49 pm

” Oh yeah, if Eli wanted to be really picky, he would point out that a disc has two sides. . .”

and if I felt like correcting you Eli, I’d point out a 4 times surface area diameter disc, 374 million miles from the sun,
would have twice the surface area of the earth’s sphere, AND be too damned far away.
Which is WHY AGW only uses ONE SIDE OF THE DISC, AND HENCE IGNORES NIGHT.

But hey, that’s the real world I observe from…….

To all others, have any of you yet worked out dividing P DOES NOT change the shape of the area K is calculated for.
Dividing P ONLY increases the distance between P and the surface K is calculated for.
You all seem so preoccupied with echoing each other you have not seen
the shape of P determines the shape of the surface K is calculated for.

88. ### Dereksaid

Eli Rabett said
January 1, 2011 at 8:49 pm

” This is pretty simple, and really does not require very much, if any, physics or math to find and point out. FWIW, SoD pointed out that the surface does not cool to 2.7K at night because of the heat capacity of the surface and oceans. ”

As I have been saying for a long, long time now. Surface heating, retention and varying later release must be taken into account.

Which is pretty difficult to “square” with the K&T type plots that treat earth’s surface as a black body over the period of a year.
How many oceanic phases do we know of that last longer than a year……………..

SoD agrees with me, in that the surface does not cool to some ridiculously cold temp because of surface heat retention, and varying later release,
it is just that he, and you forget to mention AGW / K&T / modeling does not mention it, nor quantify it. However you try to dress that up.
Even Dr. Spencer “missed” it as I also pointed to an example of earlier in this thread, even when I prodded him about it, he still “missed” it.

89. ### Richard T. Fowlersaid

Derek said,

“You all seem so preoccupied with echoing each other you have not seen the shape of P determines the shape of the surface K is calculated for.”

Nonetheless, Derek, would you agree that the total incoming radiation is the same for either shape, within the existing models you have described — that is to say, within the “disc” model, within the hemispheric model, and within the sphere model?

To be clear, in my question, by “total incoming radiation” I mean, “total incoming radiation for the whole, entire planet” not “total incoming radiation for a certain subset of it”.

RTF

90. ### Dereksaid

I am not certain what you mean by your question Richard T Fowler,
that said, I will try to answer what I think you are asking.

The equation does not describe a model, it is to all intents and purposes a “point” calculation.
Dividing P does not alter the shape of the area K is calculated for,
dividing P merely increases the distance between P and the surface K is calculated for.

Put another way, the effective shape of the surface that radiates P has to be “mirrored” by the surface k is calculated for,
otherwise a constant input over the area K is calculated for would not happen.

Dividing P is not a model, whether it is portrayed as “disc”, “hemisphere”, “sphere”, or any other shape incorrectly.
(It just so happens that the average hemisphere and disc temps are the same as dividing P by 2 and 4 respectively, but that is a different matter, and one that confused me considerably whilst writing the pdf. I have now altered the introduction to the pdf at the GWS forum to hopefully at least partially offer a better explanation of what the pdf attempts to cover.
http://www.globalwarmingskeptics.info/forums/thread-1028.html
I hope you find that useful – and I have repeated it below, as almost all here also see to have missed this very basic point and differences. )

Total incoming radiation (rate) for the point that the resulting K is calculated for would be a better and more accurate description I think.
BTW – I would say such “models” can not apply to our planet because it ignores many other known variables, geothermal inputs being just one, and no one has yet shown how a single source of P can illuminate all of a globe. Hence I have described such an illuminated globe as unphysical.

Repeated from GWS forum thread introduction to pdf.
Later addition – The pdf I have put here is best read AFTER reading the articles I refer the reader to at the start of the piece. Particularly the Alan Siddons articles.
However for clarity it may be worth me briefly stating some of the relevant points here
just in case the reader does not get time, or have immediate access to the referred to articles.

The formula that the pdf is based upon AGW’s, K&T type plots, and climate modeling’s use and interpretation of,
is the below formula.

(P / 5.6704)^0.25 * 100 = K.

In this thread at Jeff Id’s the Air Vent blog where the pdf and excel sheet is kindly being “blog reviewed”, I have posted (slightly edited and corrected here) the below.

I will happily admit that I do not like equations, yes, they are “short hand”, but they can also be very, very misleading.
My usual “defense” is that I try to “walk” myself verbally through an equation.
So, in this case what would my verbal walk through the equation be?

The equation.

(P divided by 5.6704) raised to the power 0.25, and then multiplied by 100 = K.

Verbal “Walk through”
P is the power of a beam in Watts per meter squared received at a black body’s surface,
this is divided by 5.6704, and raised to the power 0.25, the resulting figure is then multiplied by 100.
The end figure is the temperature expected at the black body’s surface for the power of beam it is receiving in degrees Kelvin.

OK, so what does it mean? Well firstly the equation does not infer any shape whatsoever.
P is merely a rate for an area equivalent to a square meter, not a shape as such.
Yes, the answer applies to all the surface area receiving the same rate, but at the end of the day,
the equation is (merely) a “point” calculation.
This in itself I suggest gives credence to the points raised in my piece,
which is about the effect of incorrectly applying “shape” using the equation.
There is a mathematical “quirk” between the “shapes” as incorrectly applied, and
this mathematical “quirk” does produce a “33 degrees effect” that
does not necessarily exist in reality upon planet earth.

What does dividing P actually do.?
In the case of the equations use when comparing the equations results to earth, or a black body receiving energy from a single source,
dividing P moves the object away from the source.
This must be the case as that is the only way the W/m2 received by the black body could be reduced, with a single and constant source of P.

If we agree that earth is nominally 93.5 million miles from the sun, then dividing P by two doubles the distance between the sun and earth, and
if P is divided by 4, then the distance between the earth and the sun is quadrupled (187 million miles, and 374 million miles respectively).
So, I agree (at least partially) with Jeff Id, when he posted in comment 51 that my use of 682 is incorrect, but that also means that the 341 figure is also incorrect, and incorrectly used.

I do not see the point in considering a disc 187 million miles from the sun, or a disc 374 million miles from the sun, when
earth is only 93.5 million miles from the sun, I suggest we are all in error, and it is not a small error.

The other point I would like to raise is that the often quoted “disc” is not a disc at all really, and neither is it flat.
The equation is for an area receiving the same rate of input, so
from a point source of P the shape would have to be a saucer to maintain a constant R and therefore input.
Also, the sun is not a point, nor is it’s effective surface of emission either flat or constant.
One look at any image of the sun and it’s solar flares confirms this.
So, the often quoted flat disc is
a) more of a saucer shape “mirroring” the sun’s surface, and
b) the surface of the saucer is constantly in a flux moving to and fro to maintain a constant W/mm2 input from the source.
In short, when the equation is used to compare to the earth, then
the resulting “shape” is more like a very wobbly saucer shape that is most commonly [b]374 million miles from the sun…

I have recently posted this elsewhere which maybe of help.
In short, with regard to the equation,

(P/5.6704)0.25*100=K

1) The shape of the source (or combined sources of) P determines the shape of the area for which K is calculated by the equation.

2) Dividing P merely increases the distance between the source of P and the black body surface area temperature calculated.

3) Misinterpreting dividing P to assigning to the surface area that the temperature is being calculated for to another “shape”
is the error by which an imaginary “33 degrees greenhouse effect” has been “created”.

I hope the above helps people understand the pdf better.

91. ### Richard T. Fowlersaid

Reply to Derek @ January 3, 2011 at 4:42 pm (#90).

Derek, please try to remain calm and take this a little more slowly. Now I invite you to have a very careful look at the following.

You have described three models which you claim are in use by others. I am in no position to dispute that, and indeed it would appear from the debate here that neither are any of your critics.

So when you tell me, “The equation does not describe a model”, I can only view this as avoidance of my question. Perhaps it is not a deliberate avoidance, but I don’t see how it cannot be an avoidance.

There are three models described in your article, true or false?

Assuming you will answer “true” to this question, my next question is:

What is the total incoming radiation into each model?

If, and only if, you can answer this question with a straight answer, then we have grounds to continue exploring this matter together. If you cannot, I will have no choice but to conclude that you do not know the most basic facts about those models which you have claimed knowledge of.

I respectfully request that you limit your response to a maximum of three issues at one time. Attempts to cover more than this are completely unproductive, because I will not respond to them, and I doubt anyone else who doesn’t already agree with you will read completely through them.

I look forward to seeing your answers to the two questions I have posed in this comment.

RTF

92. ### Dereksaid

In response to Richard T. Fowler Post 91 –
Remain Calm ? I don’t think you know me at all well, if at all.

You say,
” If, and only if, you can answer this question with a straight answer, ”

” If you cannot, I will have no choice but to conclude that you do not know the most basic facts about those models which you have claimed knowledge of ”

Please reread Post 90, visit the GWS forum thread, and read the intro and then the pdf.
If it is also possible please also read the four authors articles I refer to in the Sky Dragon book (links in pdf) at the start of the pdf.

I hope this gives you the appropriate directions to find the answers to all your questions in regards to the pdf, and my travels of discovery in the process of writing it.

I think you may well have missed that I say quite clearly and openly in the pdf that I noticed a mathematical quirk I hope others of better maths training will expand upon or, explain better.
This has not happened, but I have come to a better understanding myself, and as is my way have shared it in the order I came across it, adittedly a bit back to front. But that is the way it happened, so yes, I would write the pdf differently and in a different order now, in particular the intro on the GWS forum thread would be expanded and be the first part of the pdf.

93. ### Richard T. Fowlersaid

Derek,

Respectfully, I will not make the effort to acquire and read the books or articles you referenced prior to receiving your answer about incoming radiation. You need to be able to tell me how much radiation is entering. Otherwise, you know absolutely nothing relevant about the three “worlds” you claim to describe in great detail.

First, prove that you understand the most basic fact about what the modelers are doing with these three worlds. Then you will have my continued attention.

I am looking for three figures in watts. If you don’t know, but you believe it to be in one of these books, please look it up and type it into a comment, and post it here. Thank you very much.

RTF

94. ### Dereksaid

Richard,
Looking in the excel workbook will be a good start.
I see no need for me to repeat here just for you what is already in that.

” I am looking for three figures in watts. ”
They are in the excel book, plus
there are calculators you can put whatever W/m2 figure you want in, it gives the answers for P, P/2, and P/4…

” please look it up and type it into a comment, and post it here. ”

No, download the excel workbook and use it with my compliments.

95. ### Richard T. Fowlersaid

Derek,

I already had downloaded and reviewed it. Please, there is no need to be difficult here. I am an agnostic about back-radiation, at the present time.

I am trying to show you how I perceive the three models in question, based on your article and the responses to it. But because of the disagreements that have occurred with others, in order to do this, I have to engage in Socratic dialog: question, and answer, and question, and answer. Also, I may choose to answer questions if that is what you want; but I do want an answer to my present question.

I cannot do it the way you want me to, i.e., just go and look it up. The reason I cannot is because the answer to my question has to be deduced from other facts, some of which may be in your workbook, but not explicitly stated, and others which (relating to geometry) are not clearly stated in either the workbook, or the article, or your comments. They are hinted at, but in what I consider to be an ambiguous way. Therefore, I ask that you please consider taking my question seriously.

I am assuming good faith on your part, meaning that I assume your purpose here is simply to convince me that your stated beliefs are true. If you have solid evidence, I want to understand that evidence. And if you have solid evidence, it should stand up to a few elementary questions. I’m not trying to play “gotcha” here, or “Who’s the smartest?” I’m trying to put your assertions to the test. If my questions are easily answered, then I ask that you kindly answer them. If the answer is not clear from the contents of the workbook, then I ask that you provide additional information.

If the problem is that you do not know the exact figures used for terrestrial surface area in the models, then I ask that you just express the watts as a multiple of A, where A is the figure used for surface area of the earth in the “disc” model. In such case, if you are saying that the “disc” model has incoming radiation of 1366A, then, in essence, my question to you is, how much radiation as a multiple of A is the hemispheric model receiving, and how much, as a multiple of A, is the spherical model receiving? If you are unsure whether all three use the same figure for A, then we can agree to assume that they do, for the purpose of engaging in an approximate comparison of the models.

The greater intent here, with the question I am posing, is to test the soundness of the idea that an alleged energy imbalance results (in whole or in part) from an apples-to-oranges comparison of different geometries. My original question was phrased as “would you agree”, because I thought you probably *would* agree. I am still trying to find out for sure if you do.

Thank you for your consideration of my question.

RTF

96. ### Dereksaid

Thank you for explaining Richard, it is greatly appreciated. I shall try to answer your question.
You are correct to assume my good faith, but my purpose is not to convince, it is to see if others agree with what I think I have found.
I have tried to show I have solid evidence and do indeed want to know if what I suggest does stand up to questioning.
I will of course do my best to provide any additional information if I can.

” I ask that you just express the watts as a multiple of A, where A is the figure used for surface area of the earth in the “disc” model. In such case, if you are saying that the “disc” model has incoming radiation of 1366A, then, in essence, my question to you is, how much radiation as a multiple of A is the hemispheric model receiving, and how much, as a multiple of A, is the spherical model receiving? If you are unsure whether all three use the same figure for A, then we can agree to assume that they do, for the purpose of engaging in an approximate comparison of the models. ”

As I understand this you are asking about the volume of energy received per model type, because
we have rate (W/m2) and area (m2) and time (per second).
I have not done this calculation, but presumably it is rate (W/m2/per second) times the number of square meters surface area.
I realise the models should all use the same figure for A (in total), but not for how much is lit.
The equation used is in effect a point calculation of “induced” temperature for a given rate.
As I understand the equation it will produce the same volume but it will say nothing of the varying distribution of the received energy over anything other than a uniformly receiving of P shape.

For a (saucer like 4 x diameter of earth) “disc” of appropriate surface area A, (374 million miles from the sun)
1366/4 times A = Volume X

For a “Hemisphere”, which is really a 2 x diameter “disc” (187 million miles from the sun)
1366/2 times 1/2 A = Volume X

There is no actual diameter of the earth disc model as such (93.5 million miles from the sun),
this figure is used only to give the maximum “point” temperature for a “hemisphere” at 93.5 million miles distance from the sun.
1366 times 1/4 A = Volume X

The total volume of received energy for a shape is a different calculation, that I have not done.
This was the thinking behind my tentative first suggestion in post 60 earlier on this thread.
If that is the model you are asking about, then I have not started on that yet, as I was putting it up here for consideration,
particularly as it would show the “distribution” of the received energy that the presently used equation can not do.

I hope this helps, and at least partially answers your questions.

97. ### Richard T. Fowlersaid

Derek says, “As I understand the equation it will produce the same volume but it will say nothing of the varying distribution of the received energy over anything other than a uniformly receiving of P shape.”

This is what I was getting at.

I agree to an extent, but please note it has nothing to do with seconds of time. P is either watts per square meter, or, in aggregrate for the whole model, just watts. Adding in the figure of “per second” cannot possibly be right, and if you didn’t know this, then I caution you that there may be much about this that you are in grave error about.

That said, I agree that the lack of proper accounting for heat flow (and what Scienceofdoom has called heat capacity, which I have no reason to doubt), is potentially a serious problem. But that having been said, I think your geometric argument carries much less weight than your original article has suggested, given equivalent total incoming radiation and may best be overlooked for now. From what I am reading from the physicists, both here and at Scienceofdoom.com, I think this is a cul-de-sac, and may best be left alone for now.

I will have some more comments after I have digested the entirety of your most recent comment.

RTF

98. ### Richard T. Fowlersaid

Derek says, “The total volume of received energy for a shape is a different calculation, that I have not done.”

I respectfully submit that the calculation is simple:

Let D be the radiation per m^2 for the “disc” model.

“Disc” incoming radiation = DA
Hemispheric incoming radiation = DA
Sphere incoming radation = DA

In other words, if the “disc” rate is 1366A, then the rates for the other two are also 1366A.

My next question:

Can you see how your article and some of your comments repeatedly suggest that in the hemispheric model, only half of the world is receiving any radiation at all?

If so, can you see that this is not consistent with a model in which total incoming radiation = 1366A? If your suggestion were true, then the total would be only 683A. This would be ridiculous of course, and it is unimaginable that it could be slipped past the entire physical community without their noticing it.

Since you seem to agree that that is NOT what happened, you would be well advised to avoid suggesting that that IS what happened. You have repeatedly (and at times, forcefully) suggested that you believe it did happen. If that is not your belief, you need to change your language so as not to suggest it.

If you’re still interested, I will have more questions and maybe some comments about other aspects within the next 12 hours.

RTF

99. ### Eli Rabettsaid

FWIW, Derek doesn’t believe the earth rotates

1)The complete ignoring of planet earth’s permanently unlit side(hence the “disc world” and “evenly lit sphere” approaches) This avoids “explaining” why the earth is no where near as cold as it should be (according to AGW and black body) on it’s dark side. The computer models and the K&T type plots BOTH ignore night.

which accounts for half of the factor of 4 in figuring out the difference in the amount of energy conveyed to the earth per unit area from the sun as opposed to the sun’s hitting one side of a disc. The other factor of two comes from the fact the angle at which the sun strikes various parts of the surface.

A nice example of the last factor is that the intensity of the sun hitting the surface other than at the equator is less than the intensity of the sun hitting a solar array set so its surface is perpendicular to the sun.

Derek also does not believe that the oceans have a significant heat capacity.

Oh yes, Derek also doesn’t believe that the disc has two sides. . .

100. ### Richard T. Fowlersaid

“FWIW, Derek doesn’t believe the earth rotates”

Dr. Rabett,

That may have been a fair characterization of his comments at the time he made them, but I would prefer to give him a chance to “revise and extend his remarks”, as they say in the U.S. Congress….

It is my belief that Derek does indeed know that the earth rotates, but may perhaps have been unaware that the hemispheric model contains the entire surface of Earth on one of its sides.

I think I understand the other issues you are raising. But I thought I might try to address the most important thing first.

Unlike yourself, it is important to me that such details be straightened out, because I do not have ready access to all the details about what the climate scientists and others are doing with these models. I am just a regular guy who, when it comes to these models, only knows what he has read in the blogosphere.

Thanks for helping me to understand.

RTF

101. ### Dereksaid

OK, let us just compare notes here, in reply to Richard T Fowler I said that W/m2 is a per second rate.
Does anyone disagree with this.?

Eli seems to think I do not realise the earth rotates, I assume he has not read my post 60 in this thread, if he has, he has obviously not understood it.
I would further inquire of Eli, please explain how a single source of P can illuminate a whole globe, as far as I understand this is unphysical,
so the “globe” must be a saucer like disc to maintain a constant W/m2 input over it’s whole surface. The “globe” model is NOT a globe at all (unless it can be shown a single source of P can illuminate a whole globe…),
it is a “disc” that is 4 times further away than it should be to the source of P.

Richard T Fowler writes,
Let D be the radiation per m^2 for the “disc” model.

“Disc” incoming radiation = DA
Hemispheric incoming radiation = DA
Sphere incoming radiation = DA

I had earlier stated, I thought the way to do this calculation is, where A equals the surface area of the globe, that,

For a (saucer like 4 x diameter of earth) “disc” of appropriate surface area A, (374 million miles from the sun)
1366/4 times A = Volume X

For a “Hemisphere”, which is really a 2 x diameter “disc” (187 million miles from the sun)
1366/2 times 1/2 A = Volume X

There is no actual diameter of the earth disc model as such (93.5 million miles from the sun),
this figure is used only to give the maximum “point” temperature for a “hemisphere” at 93.5 million miles distance from the sun.
1366 times 1/4 A = Volume X

I suggest “we” are still “talking” with crossed wires.

Eli further states,
“Derek also does not believe that the oceans have a significant heat capacity.

Oh yes, Derek also doesn’t believe that the disc has two sides. . .”

Which means he has ignored, or does not understand my comments so far, or my daily water jacket plot I linked to,
and,
he has obviously not understood either the shape issue so far discussed, and my tentative first suggestion in post 60.
(Post 60 is about a model that IS the correct shape, AT the right distance from P, AND has two sides, one lit, one unlit.
Furthermore the proposed models resolution would allow for the whole shape / model to be rotated.)

Nor has Eli grasped that AGW’s 4 X times disc model DOES IGNORE the other side of the “disc”, so
it is AGW that models the disc as only having ONE SIDE.
Pot calling (incorrectly) the kettle black springs to my mind.

I prefer to sit back and wait for you both to catch up with what “we” are supposedly trying to discuss here.

102. ### Mark Fsaid

101 – Derek:
W/M^2 is INTENSITY. Energy is the product of intensity and time. Watts per second is meaningless.

103. ### Richard T. Fowlersaid

Derek,

I am not going to use your term “volume” for total incoming radiation. It is unnecessarily vague. For all I know, it may be the correct term. But I don’t see the need to be ambiguous about the matter.

I will use TIA, total incoming radiation.

You have now stated/shown at least twice that you agree that it is the same for the three models. The formulae you state apparently mean that you think it is 341.5*A watts. I say “apparently” because you have the 1366/m^2 figure in your workbook, though as I alluded before, you do not explicitly say that this is TIA. You call the number, “Power of beam”. This term, at least to my mind, does not necessarily imply “prior to any in-model adjustments”. But if my memory is serving me well, there are no adjustments between the initial TIA and the “power of beam” figure you quote.

Now in case you haven’t noticed, the problem with saying that “disc” TIA is 341.5*A watts is that, if there are no adjustments before getting to “power of beam” in W/m^2, then the “power of beam” would be, obviously (I hope), 341.5 W/m^2. Since this is not the figure you have used, and since there are no known intervening adjustments at this stage of the formula, your two numbers for TIA and “Power of beam” cannot both be correct, or else we are not talking about the same model.

How you get the “per second” is beyond me, as you have not shown how seconds could have anything to do with your answer, and indeed it is quite impossible for them to have anything to do with it. I want to know TIA for *any* given period of time. For this, I need watts. Not watt-seconds, not watt-hours, not watt-years, not watt-centuries. Just watts.

So, forget about the seconds, already! 🙂

All right, then. Next question. For this question, I will assume that you agree with the two points that I have just made, and I will assume that you are willing to accept a figure of 1366A watts for TIA in all three models. (Even if you are not comfortable with the number 1366, it’s, erm, okay, because the important thing is that you agree the number has to be the same for all three models.)

The question is:

Can you see how your article and some of your comments repeatedly suggest that in the hemispheric model, only half of the world is receiving any radiation at all?

I recommend you disregard Dr. Rabett’s comments for the time being, as they add complexity which we are not yet ready to consider. And if I were you, I wouldn’t hold my breath waiting for him to catch up with anything. I think he’s already past both of us. (Even if his knowledge is not perfect!)

Anyway, I’ll check back here to see if you’re still interested in continuing.

Regards.

RTF

104. ### Richard T. Fowlersaid

Ha! TIR, not TIA. That was sad.

Oh, that hurts. Sorry if I hurt anyone’s eyes.

RTF

105. ### Richard T. Fowlersaid

To all:

I think I have a correction to make to my comments. Anyone who sees this, feel free to chime in about whether my correction is correct.

I had said that the calculation for TIR, total incoming radiation, on each of Derek’s three models was as follows:

Let D be the radiation per m^2 for the “disc” model.

“Disc” incoming radiation = DA
Hemispheric incoming radiation = DA
Sphere incoming radiation = DA

The above also assumes that A is the surface area of our reference sphere.

I now see that by the disc being flat rather than curved, into a hemisphere, the use of the same formula for that model as for the other two results in the disc intensity (Watts per m^2 of A) being twice as high as the intensity for the other two. Thus, with the same value for A being used, the TIR is doubled versus the other two.

Thus, I submit that the following are the correct formulae for calculating TIR:

“Disc” incoming radiation = DA
Hemispheric incoming radiation = 2DA
Sphere incoming radiation = 2DA

In this way, all three models can be made to have the same total incoming radiation.

The reason why I believe the last two are the same, is because in the hemispheric model, the entire area of A is squashed onto a hemisphere — so even thougth the “sphere” surface area is twice the “hemisphere” surface model, the values for A are the same in either case. So then, with the same value D for intensity, you have the same coefficient of 2.

106. ### Richard T. Fowlersaid

Continuing my previous comment:

Just as important, of course, is what these three models DO with that radiation once they get it. So far, I am not seeing strong evidence that anyone other than me really cares to talk about this subject in any great detail on this page. Maybe some people think that what has already been discussed on this page IS great detail. But I want to say that, as regards the actual models that are in use (as opposed to Derek’s attempts to “fix” them) and what exactly they do with their radiation, I disagree.

I can see that Derek has offended some people, but I frankly don’t care because I see lives hanging in the balance, and I want some clear answers. The fact that Derek may not fully understand the models does not, by itself, validate them.

I understand that there appear to be many regulars here who believe they already know exactly what these three models do with the energy they are given. But, for what it’s worth, I don’t think this is something that one should just take for granted. I think Jeff Id did the right thing by posting this. To those who would say it has been a waste of time, I say: how do you know? Because you’ve proven that Derek doesn’t know some certain set of facts? So what? What does Derek’s knowledge have to do with the question of whether models are being manipulated? Not much, as far as I can see.

I think the real question here has to be: if models *were* going to be manipulated, in what way(s) could this possibly be done wherein it wouldn’t necessarily ring alarm bells from here to Timbuktu?

That is how a true investigator, who would be counted worthy of the history books, approaches such a situation. Even if the models aren’t being manipulated, he asks the question, and then he formulates, in his mind, an experiment — a test, something he could do or check such that, IF the models are being manipulated, this test would SHOW that fact, or at least would show an increased probability.

And then he formulates a plan for how to make that test happen, within a world where many people don’t *want* it to happen, and the rest basically don’t care and will not help.

It’d be nice if we had a few more investigators like *that* around in science, today.

RTF

107. ### DeWitt Paynesaid

Let the solar ‘constant’ for a plane surface perpendicular to the incoming radiation be 1366 W/m2. For a flat disk with the same mean radius as the Earth, 6,371 km, and an absorptivity of 1, the total incoming radiation is 1366*π*6.371E06^2 = 1.74187E+17 W. That will be the same total for any shape with the same effective radius. The area of the illuminated half of a non-rotating sphere is 2*π*6.371E06^2. That means the average incoming radiation is 1366/2 or 683 W/m2. For a rotating sphere, you have to divide by another factor of two because the surface is lit for only half as much time or 341.5 W/m2. It’s really simple.

108. ### Jeff Idsaid

DeWitt,

“For a rotating sphere, you have to divide by another factor of two because the surface is lit for only half as much time or 341.5 W/m2. It’s really simple.”

I agree with your calculation.

The average watts per meter^2 doesn’t get reduced for rotation it gets reduced for area only. One side of a disk is π r^2, one side of a sphere is 2 π R^2, the total area of a sphere is 4 π r^2. The total incoming power on the equal r disk or sphere is the same. The power/area of the sphere means that the power per area from a disk must be divided by 4 and power per area from half a sphere must be divided by two.

The power per area using the numbers above for a disk = 1366 W/m^2
for half a sphere = 683 W/m^2
for whole sphere 341.5 W/m^2

For all 3 calculations there is an incoming power of 1366 W/m^2. For all calculations, the result is the same.

109. ### Richard T. Fowlersaid

Mr. Payne,

That makes sense to me; it does appear to be reasonably simple, but trying to deduce all that using only Derek’s presentation is not exactly the easiest thing in the world. Hopefully you can see that.

Thanks for your help! And thank you, Jeff Id, for all your comments about this question.

RTF

110. ### Greenhouse Thought Experiment « Wott's Up With That?said

[…] the “back” half! Only they’re not of course…Science of Doom tries to turn the conversation back to reality on Jeff’s “No Consensus” website (sorry, I can’t write that without quotes […]

111. ### scienceofdoomsaid

Derek asked a trick question and should be congratulated for it.
I am probably too late to claim the prize.

As Derek asked in #44:

..No one appears to have read, or even skimmed the articles and papers to read first that I state at the start of my piece.
Hands up, did anyone read even one of them, particularly Alan Siddons, and Claes Johnson.

In “Computational Blackbody Radiation” by Claes Johnson – September 16, 2010:

..With Tin = 5778 K and k=0.005², this gives T ≈ 273K (including a factor 4 from the fact that the disc area of the Sun is πR² and the Earth surface area 4πr² with r the Earth diameter)..
(from p.12)

Derek was having us on all the time and we all missed the clue!