# Another Look at Climate Sensitivity

Guest Post by Willis Eschenbach

OK, a quick pop quiz. The average temperature of the planet is about 14°C (57°F). If the earth had no atmosphere, and if it were a blackbody at the same distance from the sun, how much cooler would it be than at present?

a) 33°C (59°F) cooler

b) 20°C (36°F) cooler

c) 8° C (15°F) cooler

The answer may come as a surprise. If the earth were a blackbody at its present distance from the sun, it would be only 8°C cooler than it is now. That is to say, the net gain from our entire complete system, including clouds, surface albedo, aerosols, evaporation losses, and all the rest, is only 8°C above blackbody no-atmosphere conditions.

Why is the temperature rise so small? Here’s a diagram of what is happening.

Figure 1. Global energy budget, adapted and expanded from Kiehl/Trenberth . Values are in Watts per square metre (W/m2). Note the top of atmosphere (TOA) emission of 147 W/m2. Tropopause is the altitude where temperature stops decreasing with altitude.

As you can see, the temperature doesn’t rise much because there are a variety of losses in the complete system. Some of the incoming solar radiation is absorbed by the atmosphere. Some is radiated into space through the “atmospheric window”. Some is lost through latent heat (evaporation/transpiration), and some is lost as sensible heat (conduction/convection). Finally, some of this loss is due to the surface albedo.

The surface reflects about 29 W/m2 back into space. This means that the surface albedo is about 0.15 (15% of the solar radiation hitting the ground is reflected by the surface back to space). So let’s take that into account. If the earth had no atmosphere and had an average albedo like the present earth of 0.15, it would be about 20°C cooler than it is at present.

This means that the warming due to the complete atmospheric system (greenhouse gases, clouds, aerosols, latent and sensible heat losses, and all the rest) is about 20°C over no-atmosphere earth albedo conditions.

Why is this important? Because it allows us to determine the overall net climate sensitivity of the entire system. Climate sensitivity is defined by the UN IPCC as “the climate system response to sustained radiative forcing.” It is measured as the change in temperature from a given change in TOA atmospheric forcing.

As is shown in the diagram above, the TOA radiation is about 150W/m2. This 150 W/m2 TOA radiation is responsible for the 20°C warming. So the net climate sensitivity is 20°C/150W-m2, or a temperature rise  0.13°C per W/m2. If we assume the UN IPCC canonical value of 3.7 W/m2 for a doubling of CO2, this would mean that a doubling of CO2 would lead to a temperature rise of about half a degree.

The UN IPCC Fourth Assessment Report gives a much higher value for climate sensitivity. They say it is from 2°C to 4.5°C for a CO2 doubling, or from four to nine times higher than what we see in the real climate system. Why is their number so much higher? Inter alia, the reasons are:

1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.

2. The climate models underestimate the increase in evaporation with temperature.

3. The climate models do not include the effect of thunderstorms, which act to cool the earth in a host of ways .

4. The climate models overestimate the effect of CO2. This is because they are tuned to a historical temperature record which contains a large UHI (urban heat island) component. Since the historical temperature rise is overestimated, the effect of CO2 is overestimated as well.

5. The sensitivity of the climate models depend on the assumed value of the aerosol forcing. This is not measured, but assumed. As in point 4 above, the assumed size depends on the historical record, which is contaminated by UHI. See Kiehl for a full discussion.

6. Wind increases with differential temperature. Increasing wind increases evaporation, ocean albedo, conductive/convective loss, ocean surface area, total evaporative area, and airborne dust and aerosols, all of which cool the system. But thunderstorm winds are not included in any of the models, and many models ignore one or more of the effects of wind.

Note that the climate sensitivity figure of half a degree per W/m2 is an average. It is not the equilibrium sensitivity. The equilibrium sensitivity has to be lower, since losses increase faster than TOA radiation. This is because both parasitic losses and albedo are temperature dependent, and rise faster than the increase in temperature:

a) Evaporation increases roughly exponentially with temperature, and linearly with wind speed.

b) Tropical cumulus clouds increase rapidly with increasing temperature, cutting down the incoming radiation.

c) Tropical thunderstorms also increase rapidly with increasing temperature, cooling the earth.

d) Sensible heat losses increase with the surface temperature.

e) Radiation losses increases proportional to the fourth power of temperature. This means that each additional degree of warming requires more and more input energy to achieve. To warm the earth from 13°C to 14°C requires 20% more energy than to warm it from minus 6°C (the current temperature less 20°C) to minus 5°C.

This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase. As a result, the equilibrium value of the climate sensitivity (as defined by the IPCC) is certain to be smaller, and likely to be much smaller, than the half a degree per CO2 doubling as calculated above.

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dp
March 16, 2010 8:46 pm

Hmmm – Isn’t the moon pretty much earth without air? The temperature there is entirely radiative to surface dwellers (the few that have dwelt), and the soil temperature is dependent upon it’s exposure to the sun. In shadows it is damned cold – I’d wager colder than 40 something degrees.
I think your science on this is not robust, or my understanding of what you mean by black body located in our celestial toroid is flawed.

enough
March 16, 2010 8:51 pm

For a real chart showing thermal balance go look up figure 2.3 Principles of Atmospheric Physics and Chemistry. Goody 1996
The back radiation term is pure bs. It is tied in with the nonsense argument of the atmospheric reradiation of 1/2 up /1/2 down. Absolute rubish

Bill Parsons
March 16, 2010 9:03 pm

Wiki Answer to: “What is the temperature on the moon?”
The average daytime temperature on the Moon is around 107°C (225°F), but can be as high as 123°C (253°F).
When an area rotates out of the sun, the “nighttime” temperature falls to an average of -153°C (-243°F).
253 F
-243 F
_______
10 F
Don’t know if that helps. 8 degrees is pretty close.

Jan
March 16, 2010 9:10 pm

I find this analysis be very important. Thanx

March 16, 2010 9:15 pm

Willis Eschenbach (21:07:38) :
enough (20:51:43) : edit
For a real chart showing thermal balance go look up figure 2.3 Principles of Atmospheric Physics and Chemistry. Goody 1996
The back radiation term is pure bs. It is tied in with the nonsense argument of the atmospheric reradiation of 1/2 up /1/2 down. Absolute rubish
Sorry, you have not given enough information for me to respond … plus you misspelled “rubbish”. If you post and link the figure you refer to, we can discuss it. In the interim, you might read the Kiehl/Trenberth paper referred to above.

Willis, I think he is talking about a figure numbered 2.3 in this book:

Principles of Atmospheric Physics and Chemistry (Hardcover)
~ Richard Goody (Author) “In the longest view, the earth is passing through a series of disequilibrium conditions between the origin of the universe, 15 billion years (By) ago,…”

http://www.amazon.com/Principles-Atmospheric-Physics-Chemistry-Richard/dp/0195093623
You can get it used for only \$50.00!

B. Smith
March 16, 2010 9:16 pm

Without an atmosphere there would be no oceans, ice, snow or plant life to affect albedo, so the albedo of the Earth should be more similar to that of the barren Moon. It would seem reasonable and logical to expect the high and low temperature variations on an atmosphere-less, barren, waterless Earth to more closely resemble those of the Moon. Am I missing something here?
“Lunar Surface Temperatures
Temperatures on the Lunar surface vary widely on location. Although beyond the first few centimeters of the regolith the temperature is a nearly constant -35 C (at a depth of 1 meter), the surface is influenced widely by the day-night cycle. The average temperature on the surface is about 40-45 C lower than it is just below the surface.
In the day, the temperature of the Moon averages 107 C, although it rises as high as 123 C. The night cools the surface to an average of -153 C, or -233 C in the permanently shaded south polar basin. A typical non-polar minimum temperature is -181 C (at the Apollo 15 site).
The Lunar temperature increases about 280 C from just before dawn to Lunar noon. Average temperature also changes about 6 C between aphelion and perihelion.
Marvin Ostrega
Reference: Heiken et al. Lunar Sourcebook: A User’s Guide to the Moon. Cambridge: University of Cambridge Press, 1991.”

March 16, 2010 9:19 pm

Willis Eschenbach (21:04:42) :
My calculations are based on a blackbody using the Stefan-Bolzmann equation. This equation allows us to convert between radiation intensity and temperature.
I would have said 20C, not 8C…

Matt
March 16, 2010 9:31 pm

The link about the manifold ways in which thunderstorms cool the earth appears not to be working. Pity – I really wanted to read it.

Jim Masterson
March 16, 2010 9:31 pm

>>
In the interim, you might read the Kiehl/Trenberth paper referred to above.
<<
The link isn’t working (for me). Which paper? KT1997? Or the update, TFK2009? I made an infinite layer model and the back radiation doesn’t match KT1997, but it comes close. That 40 W/m^2 for the atmospheric window may be bogus. KT1997 doesn’t calculate it correctly. It may be 80 W/m^2 or 87 W/m^2, depending on how you define “cloudy.”
Jim

March 16, 2010 9:35 pm

I keep reading a number of .05-.09C of CO2 warming and that may indeed be right.
less than .5C or a factor of TEN! decade vs century?
Thanks for the post, id bet is was a lot of work!
You will note that the closer we come to the TRUTH the more they attack, without sources, or logic.
Tim L

March 16, 2010 9:46 pm

How of the incoming energy from the Sun is converted into atmospheric and oceanic convection?
None, according to Kielh and Trenberth.
Which is amazing.
Climate physics is a fascinating subject.

jorgekafkazar
March 16, 2010 9:47 pm

dp (20:46:32) : “Hmmm – Isn’t the moon pretty much earth without air? The temperature there is entirely radiative to surface dwellers (the few that have dwelt), and the soil temperature is dependent upon it’s [sic] exposure to the sun. In shadows it is damned cold – I’d wager colder than 40 something degrees.
“I think your science on this is not robust, or my understanding of what you mean by black body located in our celestial toroid is flawed.”
Your understanding is flawed across the board, and your prose is impenetrable. What do you mean by “radiative to surface dwellers?” Where did you come up with “our celestial toroid?”

dp
March 16, 2010 9:52 pm

A thing you’ve raised to ponder is does the moon radiate solar heat as fast as it absorbs it? Well, obviously yes, at some temperature it does, and necessarily into that part of the sky not filled with sunlight. (And I’m perfectly happy to accept regional variances in temperature owing to maria vs disrupted lunar surfaces.) As would an airless earth. Actually, so too would an atmospheric earth as Venus demonstrates.
Since we’re talking about energy balance here – clearly the moon is now radiating back to space as much energy as it receives after all these billions of years, there being no mechanism for storage and transport as there is on earth, the current energy level, the black body temperature of the moon, allowing for the rotational gradient, is it really that close to earth’s temperature which is moderated by atmospheric and oceanic energy distribution? My problem is I don’t see how black body temperature translates to an embracing thermal experience as provided to living things by the atmosphere and oceans.
If what you say is true it is really quite profound as it suggests our global temperature is a matter of albedo, no?

March 16, 2010 10:09 pm

As is shown in the diagram above, the TOA radiation is about 150W/m2. This 150 W/m2 TOA radiation is responsible for the 20°C warming. So the net climate sensitivity is 20°C/150W-m2, or a temperature rise 0.13°C per W/m2. If we assume the UN IPCC canonical value of 3.7 W/m2 for a doubling of CO2, this would mean that a doubling of CO2 would lead to a temperature rise of about half a degree.

I think this first part is questionable. Nothing wrong with your division of 20/150 mathematically, just whether it means anything.
If you take a fixed albedo and increase the radiative forcing at the surface you can calculate the increase in temperature without feedbacks. You obviously know how to do this..
(For people who don’t, you can see more at The Earth’s Energy Budget – Part One and at the end of CO2 – An Insignificant Trace Gas – Part Seven )
So your calculation is basically saying that the response of the climate system from no atmosphere to current status is a linear model.
I know that of course, we can’t assume the same albedo. I see your comments on the various aspects of climate response, don’t know what to think about those yet.
Rightly or wrongly to assume no climate through to current climate is linear seems way way out there!
Perhaps you are aiming at a parody of GCMs producing a linear response?
I find their linear response hard to believe but perhaps you could say that if you look at one small part of a big curve, the slope is linear or can be approximated linear close to that area..

savethesharks
March 16, 2010 10:11 pm

Great post, Willis.
Maybe I missed it, but it should also be stated that, as a new point number 7, the general circulation models are also woefully lacking in accounting for the multi-decadal oscillations of the oceans, including, but not limited to the PDO and the AMO, among many others.
But that point on the importance of cooling with tropical thunderstorms, poorly understood, can not be overemphasized!
Dr. Spencer and Lindzen and others are cracking that block….and so little is known. But you could not be more right. Thunderstorms…are magnificent (albeit ephemeral) players in the earth’s climate.
They boil in atomic-bomb-scale local and mesoscale fury…to heights up to the tropopause…and we wonder how they can bring down airlines (Brazil to France, last year).
Then they fade away in cirrus wisps the next day….all of that heat escaping through an open “window” in the atmosphere…
Thanks again for this post.
Chris
Norfolk, VA, USA

Brian W
March 16, 2010 10:14 pm

enough (20:51:43)
Bingo! Spot On.
Willis Eschenbach
It’s pretty late here and I would like to debate your so called diagram. So a couple of things now and more tomorrow. Backradiation as well as feedbacks are pseudoscience. There is no real physical basis for back radiation. As enough(my feelings exactly) stated it’s nonsense and simply not true.
The holy grail of the black/grey body is inadequate to analyse the earth’s weather and climate. Does the earth look like a black/grey body? The warmers have concocted a heady brew from this as the diagram shows. Forget Kiehl/Trenberth they are climate quacks. Take clouds for example, according to the diagram 76w/m2 is not only reflected by the clouds, but also by aerosols and the atmosphere. Really? How is the number 76 arrived at? Clouds most certainly absorb the suns energy. Where is the figure for that? Another thing, with clouds constantly forming and dissipating across the planet, is this an averaged figure for the entire surface? Are all the numbers in the diagram static and unchanging or subject to WIDE variations?
It all looks like an add up the numbers exercise. My BS detector goes off the charts whenever I look at that diagram. Its a mess. So please explain to me your understanding of backradiation and how it occurs and I’ll debunk it.
By the way, in your post at 21:04:42 you misspelled Boltzman. Pretty easy to leave a letter out, eh.

JDN
March 16, 2010 10:15 pm

Let’s try it and find out. 🙂

stan stendera
March 16, 2010 10:21 pm

Sometimes my birdfeeder has hawks circling above it looking for easy pray. One is them is Eschenbach’s hawk. a subspecies of Cooper’s hawk; both of which are far too uncommen

JinOH
March 16, 2010 10:31 pm

‘Sometimes my birdfeeder has hawks circling above it looking for easy pray. One is them is Eschenbach’s hawk. a subspecies of Cooper’s hawk; both of which are far too uncommen’
LOL – I have hawks visiting my feeders too at times.
Nature is a bummer for science.

Ray Boorman
March 16, 2010 10:33 pm

The Moon is very different to Earth, which makes the simple comparisons of pundits above worthless.
1) It is 3,743 km in diameter versus earth’s 12,756 km;
2) Earth is over 1000 deg C at the top of the mantle, which is a mere 5km below the seabed in the deep ocean, & about 50km under the land surface.
3) Moon rotates once each 27 & a bit days, Earth rotates every 24 hours.
4) Earth has lots of volcanic activity, which brings heat to the surface. None has been detected on the Moon, yet.

Allan M R MacRae
March 16, 2010 10:41 pm

Good stuff Willis!
Posted on June 28, 2009
http://wattsupwiththat.com/2009/06/27/new-paper-global-dimming-and-brightening-a-review/#comments
The sensitivity of global temperature to increased atmospheric CO2 is so small as to be inconsequential – much less than 1 degree C for a doubling of atmospheric CO2. CO2 feedbacks are negative, not positive. Climate model hindcasting fails unless false aerosol data is used to “cook” the model.

March 16, 2010 11:02 pm

Brian W:

Backradiation as well as feedbacks are pseudoscience. There is no real physical basis for back radiation

How about measurement of it?
We can measure longwave radiation (>4um) downward at the earth’s surface. It’s not coming from the sun which radiates 99% of its radiation 4um coming from?
You can see one example of some specific measurements at CO2 – An Insignificant Gas? Part Six – Visualization
Strangely enough the peak radiation of this downwards radiation occurs in those wavelengths that match the absorption spectra of CO2, CH4, water vapor, ozone.
And at the top of the atmosphere the outgoing (upward) longwave radiation has a “notch” in those same wavelengths.

Brian Johnson uk
March 16, 2010 11:27 pm

dp….
“I think your science on this is not robust, or my understanding of what you mean by black body located in our celestial toroid is flawed.”
dp, Homer Simpson rules – we live on a celestial donut!
“Donuts. Is there anything they can’t do?”

stan stendera
March 16, 2010 11:32 pm

Way, way, way off topic!
for JinOH
I will never forget this: I put birdseed on my deckrail. And of course the squirrels get into it. I was playing on the compute , came to a stopping point,and glanced out the sliding door to the deck. The squirrel, stealing birdseed, suddently started running back and forth then leapt off the deck. Out of nowhere came a Cooper’s hawk and snatched it out of the air. That is why Altamonte is such a crime.
MODERATORS! I know this is off topic. You have my complete permission to forward it to jinOH including my E-mail!!!!

March 16, 2010 11:43 pm

(Something funny happened to my last post, I think because I used a less than sign..)
Brian W:

Backradiation as well as feedbacks are pseudoscience. There is no real physical basis for back radiation

How about measurement of it?
We can measure downward longwave radiation (greater than 4um) at the earth’s surface. It’s not coming from the sun which radiates 99% of its radiation less than 4um.
Where is this radiation coming from?
You can see one example of some specific measurements at CO2 – An Insignificant Gas? Part Six – Visualization
Strangely enough the peak radiation of this downwards radiation occurs in those wavelengths that match the absorption spectra of CO2, CH4, water vapor, ozone.
And at the top of the atmosphere the outgoing (upward) longwave radiation has a “notch” in those same wavelengths.

Steve Goddard
March 16, 2010 11:47 pm

Venus is the same size as earth and has a fairly uniform temperature of over 400C across the entire planet (day or night equator or pole) This is because of their atmosphere.
Sunlight on Mercury is four times stronger than Venus, yet temperatures average much cooler, because of the lack of an atmosphere.
My plane flight on Saturday descended 35,000 ft. and temperatures warmed by 150 degrees F as we descended – because of the blanket generated by the atmosphere.

anna v
March 16, 2010 11:48 pm

Willis,
All the temperatures we have are in the air , and there is a large difference between ground and air. I do not know of any measured ground temperatures. Where do you take the 14C from?
SSTs might be OK, so maybe you should take the SST average temperature as the 0 atmosphere temperature. Earth is 75% ocean after all. the AMSU plots give a scale from -2 to 30, that would give 14, but I would think a true average would come higher.
The albedo of the moon is around 0.1 so it is not a good analogue for your purposes.

anna v
March 16, 2010 11:49 pm
Spence_UK
March 16, 2010 11:59 pm

The figure of 0.13 K per W/m^2 is interesting.
Spencer and Christy found that the LT temperature varies by 2.3K over the year due to the eccentricity of the orbit of the earth. There is little or no perceptible phase lag in this variation wrt the orbital pattern, suggesting this is a full figure and there is no “warming in the pipeline”.
The orbital eccentricity results in a variation of solar radiance of 90 W/m^2. Of course, this arrives over a pi*r disc and is spread over 4*pi*r surface of the sphere, resulting in an equivalent of 17 W/m^2 averaged over the surface of the earth.
This yields a climate sensitivity of 2.3/17 = 0.135 K per W/m^2, entirely independently of Willis’ calculation.
Coincidence?

Sou
March 17, 2010 12:09 am

Is this post meant to be parody, or is the humour unintentional?

Spence_UK
March 17, 2010 12:09 am

re: my 23:59
Good grief, why don’t I check my posts before pressing submit 🙂
Firstly, the area of a circle is pi*r^2 and the surface of a sphere is 4*pi*r^2. Secondly, 90/4 is 22.5 not 17. I’m sure I got that number from somewhere, just can’t remember where…
This yields a climate sensitivity *lower* than Willis at 0.1 K per W/m^2. Not such a coincidence, but at least in the same ball park…

B. Smith
March 17, 2010 12:14 am

Thank you for clarifying, WE.

wayne
March 17, 2010 12:17 am

Willis :
enough (20:51:43) :
I follow you Willis. Good point and well done!
And as ‘enough’ pointed out about the 1/2 up and 1/2 down being rubbish, he is right. All of these energy balance charts are basically wrong, misleading, or incomplete but that doesn’t take one iota away from what you have just have shown us on the sensitivity. The one thing they all do agree on is output equal input.
You see, none of them take into account the dip of the horizon correction or just “the dip”. Any good sailor who knows celestial navigation is well aware of that factor and it applies whether radiation is inbound from stars or outbound from CO2 molecules. Half up and half down can only occur exactly at the sea’s surface and ALL radiation leaving randomly from atoms or molecules in the atmosphere must account for the dip factor.
If you even stand up and are 5 feet above sea level it is no longer half up and half down however tiny. At 20,000 feet it is sizeable. There is now much more going up to space and much less going down back to warm the surface and this factor increases with the height above sea level quickly at first and slower near satellite altitudes. At high tropic cloud top levels this imbalance can get rather large, shunting heat directly to space. If you want the equations, holler. I learned celestial navigation from a captain right out of college (had a dream of circumnavigating when young).
That reinforces your statements about the mention of wind speed on evaporation rates which feed the clouds which shunt the heat to space, but you can’t if “half is up and half is down” and that often used term IS rubbish.
This is only my opinion, wink.

kevoka
March 17, 2010 12:24 am

I am missing something. Everywhere I have read the calculations result in a blackbody temp of 255K.
No atmosphere , no blanket, the temp would be 255K. About 30K less than it is. 1C = 1K.

March 17, 2010 12:35 am

anna v:
You have a point when you say:

All the temperatures we have are in the air , and there is a large difference between ground and air. I do not know of any measured ground temperatures.

The air temperature is taken as the closest proxy to the land surface temperature. During the day it is a reasonably close value because the land heats up, heats the air directly over it and this air then expands and rises. The result is convective mixing which means that the air temperature six feet off the ground is a reasonable estimate.
At night it’s a different story. The temperature can vary significantly in the first few feet because the bottom layer is now colder and there is no convective mixing. “No wind” conditions give very different results from wind conditions. For this reason and UHI issues we should just give up on trying to measure land temperatures
But given what we do have our best estimate of annual average global surface temperature is around 15’C.

tallbloke
March 17, 2010 12:35 am

Hi Willis, great post for opening up discussion. The ‘atmospheric window’ is only 13W/M^2 according to Trenberth. You commented that thunderstorms would increase with increasing temperature. Presumably that would widen the window quite a bit. Would it also stick quite a lot more water vapour into the higher part of the troposphere? What effect would that have?

Nylo
March 17, 2010 12:35 am

You should not compare to a black body of an homogeneus temperature. Because of the emisivity dependence on t^4, two black bodies with the same average temperature but different distribution of it along their surface radiate different ammounts of energy. For a given average temperature, the more temperature differences you have along the surface, the more you radiate to space.
To continue with the moon example, if the average of the moon’s daily temperature is 107C and the average night temperature is -153C, you can be sure that the emisivity total is quite higher than that of a black body with a uniform average temperature of -23C, i.e. the same average temperature. That’s because the emisivity you gain with the +130C of the illuminated side is much bigger than the emisivity you lose with the -130C of the dark side. Actually, it would be equivalent to a black body with a uniform temperature of 320K (47C). Quite hot huh? But this is considering a moon with one half at some uniform 107C and the other half at some uniform -153C. In reality, there are also differences of temperature along both surfaces leading to those averages. And remember: any inhomogeneity in the temperature leads to more emisivity than that of a black body of the same average temperature. So the equivalence of the emisivity would be to a black body of even more than 47C average temperature.
Calculations:
Daily temperature = 107C = 380K
Night temperature = -153C = 120K
The black body with equivalent emisivity would have a temperature equal to the 4th root of ((380^4+120^4)/2) = 320K = 47C
(The day side, with 3.16 times the temperature of the night side, radiates more than 3.16^4= 100.5 times the energy radiated by the night side.

Brian W
March 17, 2010 12:44 am

Willis Eschenbach (22:57:51)
Well, lets see, you used the diagram to show us that “this is what’s happening”.
I and many others do not agree. If the diagram is wrong then using w/m2 and blackbody physics will not give you the right answer. If the numbers are wrong then how do you calculate anything? Yes indeed a very impressive link but since incoming sunlight is comprised of 10% ultraviolet, 44.8% visible (shortwave) and 45.2% INFRARED how is it these wonderful scientists can accurately separate upwelling from downwelling without proper hard experiments in the atmosphere? Sensitivity based on what, fancy algorithms with built in beliefs. Explaining part or all of the diagram shows me that you know what you are talking about, and you avoided it so I’m dubious.
As to the sensitivity. CO2 with a specific heat less than N2, O2 and even aluminum, poor absorption compared to air, fast emission and most importantly a concentration of .038% by volume contributes NO sensible (usable) heat to our atmosphere. A doubling of CO2 will not give even .1 degree increase. I wish people would stop fixating on CO2. So the models do overestimate sensitivity by a huge amount. The whole AGW/CO2 thing is a stinking scientific fraud. Period.

stan stendera
March 17, 2010 12:46 am

It was not a Cooper’s hawk. it was an Eschwnbach hawl!

stan stendera
March 17, 2010 12:47 am

It was not a Cooper’s hawk. it was an Eschenbach hawk!

March 17, 2010 12:58 am

Willis, you are only looking at the high frequency end of an amplifying process, there is definetely a frequency dependency in climate sensititivity noticable, although not as high on a century timespan as IPCC models make of it..
http://members.casema.nl/errenwijlens/co2/howmuch.htm
See also Pelletier for a temperature power spectrum
Jon D. Pelletier, 2002, Natural variability of atmospheric temperatures and geomagnetic intensity over a wide range of time scales, doi:10.1073/pnas.022582599 PNAS February 19, 2002 vol. 99 no. Suppl 1 2546-2553 Link

March 17, 2010 1:34 am

Brian W (00:44:44) :

but since incoming sunlight is comprised of 10% ultraviolet, 44.8% visible (shortwave) and 45.2% INFRARED how is it these wonderful scientists can accurately separate upwelling from downwelling without proper hard experiments in the atmosphere?

It’s easy.
Incoming sunlight is 99% less than 4um in wavelength. Climate science calls this conventionally “shortwave”.
Radiation from the surface of the earth is 99.9% greater than 4um in wavelength. Climate science calls this conventionally “longwave”.
Which means solar and terrestrial radiation can be easily distinguished and lots can be learnt even without “fancy algorithms with built in beliefs.
You can see some explanation about longwave and shortwave

March 17, 2010 1:34 am

Surprisingly, new NASA diagram shows no back-radiation..
http://eosweb.larc.nasa.gov/EDDOCS/images/Erb/components2.gif
If Kiehl-Trenberth diagram is correct and backradiation of 321W/m2 is responsible for the hypothetical +33K effect, it means additional 3.7W/m2 will cause 0.38°C net warming.

March 17, 2010 1:42 am

kevoka (00:24:44) :
Everywhere I have read the calculations result in a blackbody temp of 255K.

This calculation is wrong, since it is calculated with present albedo of 0.3. 2/3 of this albedo is made by clouds, which should actually not be there if there are no GH gases or atmosphere.
More, much colder Earth with (frozen) oceans would turn partially to white snowball and its albedo would be mightily increased.

Nylo
March 17, 2010 1:45 am

Re: Willis Eschenbach (00:44:35) :
“Regarding the moon, I’m not the one who brought it up”.
I know. I have used the moon as an example only because it is easy to see the implications as the temperaure swings are so huge. But the same idea could be applied to the Earth. The energy radiated by the Earth is not the same as the energy radiated by a black body of the same average but homogeneous temperature. It is quite bigger, because also the Earth has noticeable swings in the temperature between night and day (especially in the deserts), or the poles and the ecuator, or the summer and the winter hemispheres. The hottest parts radiate sooo much more.

Lindsay H
March 17, 2010 1:49 am

interesting:::
I have a problem with your black body figures for the earth
http://en.wikipedia.org/wiki/Black_body
 Temperature of Earth
If we substitute in the measured values for the Sun and Earth:
If we set the average emissivity to unity, we calculate the “effective temperature” of the Earth to be:
TE = 254.356 K or -18.8 C.
This is the temperature that the Earth would be at if it radiated as a perfect black body in the infrared, ignoring greenhouse effects, and assuming an unchanging albedo. The Earth in fact radiates almost as a perfect black body in the infrared which will raise the estimated temperature a few degrees above the effective temperature. If we wish to estimate what the temperature of the Earth would be if it had no atmosphere, then we could take the albedo and emissivity of the moon as a good estimate. The albedo and emissivity of the moon are about 0.1054[20] and 0.95[21] respectively, yielding an estimated temperature of about 1.36 C.
Estimates of the Earth’s average albedo vary in the range 0.3–0.4, resulting in different estimated effective temperatures. Estimates are often based on the solar constant (total insolation power density) rather than the temperature, size, and distance of the sun. For example, using 0.4 for albedo, and an insolation of 1400 W m−2), one obtains an effective temperature of about 245 K.[22] Similarly using albedo 0.3 and solar constant of 1372 W m−2), one obtains an effective temperature of 255 K.[23][24]
this is substantially different from your estimates is the author of the wiki misleading us?
the earth atmosphere albido is in the order of 30% based on satelliteMODIS values.
so the greengouse effect plus .3 albido gives an average temp of 14 deg c , a difference of 32deg c from the theoretical black body temp.
I’m scratching my ancient head a bit ! applying the albido of 30% should give the planet a temp of 279 deg k = 5.85 deg C meaning the greenhouse effect adds 8 deg c.
confusion abounds!

Brian W
March 17, 2010 1:59 am

Steve Goddard (23:47:55)
The shell of your airplane heated due to air friction! Blankets indeed!
Science of Doom (23:43:15)
“We can measure downward longwave radiation (greater than 4um) at the earth’s surface. It’s not coming from the sun which radiates 99% of its radiation less than 4um.”
The sun is a “broadband radiator” it radiates at ALL frequencies. Since everyone likes w/m2 here is some watts for you. Bright sunlight provides an irradiance of just over 1 kilowatt per square meter at sea level. Of this energy, 527 watts is INFRARED light, 445 watts is visible light, and 32 watts is ultraviolet light(wiki). And these are only three tiny bands of frequencies in a much larger spectrum. So good luck with your pseudoscience of doom. You are just another warmist site promulgating calculated nonsense.

Ronaldo
March 17, 2010 2:00 am

scienceofdoom (23:43:15) :
“Strangely enough the peak radiation of this downwards radiation occurs in those wavelengths that match the absorption spectra of CO2, CH4, water vapor, ozone.
And at the top of the atmosphere the outgoing (upward) longwave radiation has a “notch” in those same wavelengths.”
Isn’t this because the absorption of IR at specific wavelengths raises the internal (vibrational) energy of the molecule (of CO2, H2O), which then promptly re-emits energy at the same wavelength but in a random direction, hence the notches in the upwelling radiation? The absorption is an internal process, not a thermal process.

Frozen man
March 17, 2010 2:08 am

Hi Willis,
good post i think. I’ve been studying some time this kind of radiation balances, So some things can be added for further knowledge:
1. The graph is old in this authors (Trenberth et altea, 2008), the new one is clearly unbalanced to warm ( i couldn’t see clear what was based on)
2. I think they don’t have applied some basic laws of energy and thermodynamics like: internal energy of the system (U), all the kinds of energy that the earth also receipt, f. e., magnetic, gravitational, etc…. Also thety didnt take into account the effect of winds as cinetic energy, etc… this graph suffers from too much simplicity (only radiation).

Dermot O'Logical
March 17, 2010 2:29 am

I thought incoming solar radiation was about 1400W/m^2 i.e. the TSI figure.
Why is it 342 W/m^2 in the diagram? Or is it a different physical quantity?
Thanks
Dermot

Louis Hissink
March 17, 2010 2:39 am

Willis
You might have fun adding a graph showing the Earth’s temperature profile up to 500km above surface – it might surprise quite a few that the temperature does not continue dropping the further from Earth but actually rises significantly into the thermosphere.
So which temperature would be representative of the Earth’s temperature, and does the temperature at the thermosphere change diurnally etc etc etc.
The climate people are waxing thermally over the bottom part of the plot – but presumably remain ignorant of what causes the rest of the profile up to 500 km etc.
http://www.windows.ucar.edu/tour/link=/earth/images/profile_jpg_image.html
Have fun folks – I am bush till Easter drilling holes in the ground testing scientific hypotheses in situ.

John A
March 17, 2010 2:40 am

Steve Goddard:

My plane flight on Saturday descended 35,000 ft. and temperatures warmed by 150 degrees F as we descended – because of the blanket generated by the atmosphere.

But blankets don’t warm by suppressing radiation.

March 17, 2010 2:44 am

I think that in time the Kiehl/Trenberth diagram will be considered an even bigger joke than the Mann hockey stick.
Proof
The hockey stick had at least a possibility of being correct, if the predictions turned out to true.
Its a travesty for him that reality dealt a massive raspberry.
The Kiehl/Trenberth diagram is wrong from the start
Just two points
1. They seem at great pains to show that the radiation intensity is conserved.
There is no such conservation law in Physics.
2. The magnitude of so called back radiation from the trace gasses co2 and water vapour is shown as almost twice the total solar radiation must be physical nonsense.

Toho
March 17, 2010 2:55 am

Juraj V. (01:42:28) :
“More, much colder Earth with (frozen) oceans would turn partially to white snowball and its albedo would be mightily increased.”
There can’t be exposed ice if there is no atmosphere. Ice without an atmosphere vaporizes. That is why there is no exposed ice on Mars or the Moon.

Cement a friend
March 17, 2010 3:00 am

Willis,
Another consideration of the energy budget is in figure 2 of http://physicsworld.com/cws/article/print/17402 . You will need to log in to physicsworld.com (IOP website)but it is free. As far as I know has discredited this assessment and the work of Ahilleas Maurellis.
In my assessment (from texts books on heat transfer and personal experience and measurement) if there are two molecules of CO2 near each other in the atmosphere and they are at the same temperature there will be no heat transfer them because both molecules will be saturated at the wavelengths of absorption and emission. If the temperature is different, heat flux will be from the higher to the lower temperature and never the reverse. The same thing applies to applies to the earth’s surface which is about 70% water with an emissivity of about 0.95 (almost a black body). If the surface temperature is above that of the atmosphere, which is the case due to due the the lapse rate and radiation from H2O and CO2 heat will only from the surface.
In my assessment the water vapor (and water and ice in clouds) and to a minor extent CO2 can be compared to a perforated wide spaced sieve. A large quantity of heat flows straight through while some is absorbed and re-radiated upwards reducing the overall heat flux (or heat flow rate). The heat flux will vary considerable around the globe and at different times of the various cycles which affect the earth.
I have no respect for the Kiehl and Trenberth paper. However, I do not dispute (more correctly agree) some of the other points you make such as about evaporation and forced (wind) convective heat transfer.

Nylo
March 17, 2010 3:01 am

Dermot O’Logical (02:29:29) :
“I thought incoming solar radiation was about 1400W/m^2 i.e. the TSI figure.
Why is it 342 W/m^2 in the diagram? Or is it a different physical quantity?”
1400W/m2 is what you would get in space, in a surface perpendicular to the incident light. 342W/m2 represents an average in the surface of the Earth between night and day zones as well as the different angles of the incident light wrt the Earth’s surface. That is, if you multiply 342 times the area of the surface of the Earth you would get the total energy we receive from the Sun at any given time.

rbateman
March 17, 2010 3:20 am

So, Willis, if a station is located where it is naturally near maximum for it’s latitude (due to phsyical geography and circulation), then UHI from any C02 increase will be negligible.
Give me a day or so, I am almost done with just such a station.

Dermot O'Logical
March 17, 2010 3:21 am

Willis Eschenbach (02:51:43)
Much obliged. Kind of obvious when you…. think… about…. it. I must learn to do that.
But why divide by the _whole_ surface area of the Earth? We are only getting sunlight on half of the surface area i.e. the daylight side. It seems invalid to average the incoming insolation over the whole surface area, as that would imply complete absorption of the energy, then being evenly distributed across the whole surface, _before_ being re-radiated away. Stefans law would suggest the daylight side radiates away significantly more energy that the night side, and that’s non-linear, at T^4.
(Please forgive amateur analysis of a physics graduate 20 years out of University – the day job doesn’t require me to keep up to speed with thermodynamics and statistics)
Dermot.

anna v
March 17, 2010 3:24 am

Re: scienceofdoom (Mar 17 00:35),
It is not only in the night that the ground temperature is different than the air.
I live in Greece, in the summer the ground in sunlight is more than 50C, the air temperature may be anywhere from 38 to 42, not higher. One could fry eggs on the rocks.
Searching for ground temperatures, I found one study in Puerto Rico where they measured in the ground in the sun for planting studies, and the temperature 20cm in the ground was the same as the air temperature ( in that particular study 22C). This means that the surface, which is the radiator, would be much higher, as soil is not a good conductor.
That is why I think that average SST temperatures would be more realistic, since they are being measured.

March 17, 2010 3:24 am

Willis: Do the models take account of the endothermic photosynthesis process? The vegetation on the earth is cooling the surrounding air because conversion of CO2 and water through photosynthesis is an endothermic reaction. When biomass is burned, liberating CO2 and water again, this energy is released. As CO2 concentration increases the rate of photosynthesis of plants using the C3 photosynthetic pathway (trees, most grasses, most plants) increases much more rapidly than the increase in CO2 concentration.

March 17, 2010 3:30 am

Ronaldo asked:
On my comment about downward longwave radiation in the bands of various “greenhouse” gases and outgoing longwave radiation with an equivalent “notch” in it

Isn’t this because the absorption of IR at specific wavelengths raises the internal (vibrational) energy of the molecule (of CO2, H2O), which then promptly re-emits energy at the same wavelength but in a random direction, hence the notches in the upwelling radiation? The absorption is an internal process, not a thermal process.

It’s both. When a “greenhouse” gas absorbs energy there is the possibility that it will reradiate this same energy in a very short space of time in a random direction.
But especially in the troposphere and generally in the first 100km of the atmosphere, the gas in the excited state is rapidly deactivated by collisions and the energy absorbed from the original photon is distributed thermally.
This layer of the atmosphere therefore warms up and radiates energy both up and down. This process is the dominant one.

dr.bill
March 17, 2010 3:34 am

Willis is just “testing the class” by changing horses a little bit with the answers to his pop quiz.
Using 342 W/m&sup2; and 14°C, here’s the “point spread”:
The 8°C cooler is correct if comparing “now” to a blackbody (albedo = 0), but if comparing “now” to a denuded Earth with its current average surface albedo (≈ 0.16), the 20°C is correct. If we turn the Earth into a “big Moon” (albedo ≈ 0.12), we get 17°C. In any case, it’s nice to have our cosy blanket, and even nicer for it to be so stable. 🙂
/dr.bill

Recipy
March 17, 2010 3:39 am

Willis . You are confusing things. Is it deliberate?
1) You do not define which temperature you are talking about. From the 14degC average you’d think it was surface temperature, but then you stumble by looking at TOA fluxes. Try looking at the total non-reflected downwelling fluxes at the surface instead and apply stefan-boltzman to that. In that case you’d get an earth which is ~50 degC warmer than a blackbody. This is reduced to ~33C, if you account for non-radiative heat loss terms accounting for the mixing in the atmosphere (i.e. sensible and latent).
2) Wikipedia says: the surface temperature of Earth as a blackbody would be -19C. That is 33C colder than you 14C with an atmosphere. See T_E calculation here: http://en.wikipedia.org/wiki/Black_body.

March 17, 2010 3:58 am

Brian W said:

The sun is a “broadband radiator” it radiates at ALL frequencies. Since everyone likes w/m2 here is some watts for you. Bright sunlight provides an irradiance of just over 1 kilowatt per square meter at sea level. Of this energy, 527 watts is INFRARED light, 445 watts is visible light, and 32 watts is ultraviolet light(wiki). And these are only three tiny bands of frequencies in a much larger spectrum. So good luck with your pseudoscience of doom. You are just another warmist site promulgating calculated nonsense.

I don’t expect to convince Brian W.
Max Planck was the original promulgator of this “nonsense”. Born 1858, won the Nobel prize for physics in 1918. Many people know his name.
He found the equation that described radiation from a body according to its temperature, and the shape of the waveform. You can see the formula here:
http://en.wikipedia.org/wiki/Planck's_law – along with typical radiation vs wavelength curves.
The energy from the sun, as measured by satellite, amazingly follows this curve. So misguided, that sun..
It’s sad when so little is understood of the absolute basics and yet the passion and the belief is so strong.
Of course, Brian W. can win the Nobel prize by showing that Max Planck – and the sun – are wrong.

jmrSudbury
March 17, 2010 4:10 am

The http://www.worldscinet.com/cgi-bin/details.cgi?id=pii:S021797920904984X&type=html paper disagrees with the idea of back radiation too.
The net is not equivalent to 50/50. The atmosphere is not made up of a single atom re-emitting half of what it absorbed back to earth. For air particles that are 1km up, the earth would be about 170 degrees of the field of view. For those that are 10km up, the earth would be something like 80 degrees of the field of view. This is just an example. I made up the numbers, but the fact there is a big difference is what is important.
John M Reynolds

March 17, 2010 4:19 am

My quick 2 cents on this is I’m not sure Trenberth’s energy balance is accurate. It ignores readings we get from satellites and thermometers in favor of equations taken from Hansen’s climate model.
I’m not going to call Trenberth a quack, as someone else did, but I will say it’s not clear to me that we have the technology needed to build an accurate energy balance model at this time.
Which means it’s not clear to me that anything build upon a foundation of our current energy models is accurate.

Allen63
March 17, 2010 4:21 am

I seem to remember reading that Earth’s atmosphere causes 33C heat up vs no atmosphere. Sometimes writers say the “greenhouse effect” for earth is 33C.
Was 33C wrong?
Guess this echos a post just above.

wayne
March 17, 2010 4:27 am

Willis, I have a question for you. Want to see if this seems real to you. You say they measure the down dwelling LW and I assume you are speaking of 321 W/m2 but I can’t get the math to jive. Using their numbers I always come up with the surface being too hot.
If you take all radiation absorbed by the ground, 169 for SW Absorbed By Surface and 321 for LW Back Radiation Absorbed By Surface, you get 490 W/m2. That is the radiation from the sun the chart says you would need to warm the earth to 14 C as if there was no atmosphere at all, right?
Take (490/SB)^0.25 = 304.9 K or 89 F for surface temperature according to the chart. Does that seem correct to you? Am I viewing it wrong? To me it seems too hot and the back radiation is way too high.
To correct that would recalculate to SB*(14+273.15)^4 – 169 = 216.5 for the LW back radiation absorbed by the surface. That is where I don’t connect right to K&T math. Seems 216 not 321, or 105 Wm-2 too high.
Don’t think this would affect your sensitivity logic at all, just the internal numbers.

March 17, 2010 4:36 am

jmrSudbury:
Which one are you going with?
No “back-radiation” according to the entertaining Gerlich and Tscheuschner or back-radiation but modified by the volume integral?
Equally divided up and down is just a useful approximation to explain the basic principles to people who are starting out. Mathematic treatments take this effect into account.
The Planck formula for radiance is in units of W/m^2/sr. “sr” is steradians, or solid angle, i.e. the energy is radiated in all directions.

Basil
Editor
March 17, 2010 4:38 am

I’m confused. At the beginning, the answer to the Quiz is said to be (c), 8° C cooler. But latter on, I read
“If the earth had no atmosphere and had an average albedo like the present earth of 0.16, it would be about 20°C cooler than it is at present.”
And Leif chose (b). So which is it? (b) or (c)? While I don’t see eye to eye with Leif on some things, the “smart money” would bet (b).

1DandyTroll
March 17, 2010 4:39 am

Has anyone actually measured the incoming wattage per square meter?

March 17, 2010 4:43 am

[quote anna v (03:24:15) :]
Searching for ground temperatures, I found one study in Puerto Rico where they measured in the ground in the sun for planting studies, and the temperature 20cm in the ground was the same as the air temperature ( in that particular study 22C). This means that the surface, which is the radiator, would be much higher, as soil is not a good conductor.
That is why I think that average SST temperatures would be more realistic, since they are being measured.
[/quote]

I’m not sure if this is what you’re looking for, but the ISCCP does produce surface skin temperatures (as opposed to surface air temperatures).
Just follow the link below, select “Mean Surface Skin Temperatures” from the “Select a variable:” dropdown, select a time period, and click the “View” button to see a picture.
http://isccp.giss.nasa.gov/products/browsesurf1.html

March 17, 2010 4:48 am

magicjava:

I’m not sure Trenberth’s energy balance is accurate. It ignores readings we get from satellites and thermometers in favor of equations taken from Hansen’s climate model.

Which page on Kiehl and Trenberth’s 1997 paper are you referring to?

March 17, 2010 4:56 am

[quote Allen63 (04:21:29) :]
I seem to remember reading that Earth’s atmosphere causes 33C heat up vs no atmosphere. Sometimes writers say the “greenhouse effect” for earth is 33C.
Was 33C wrong?
[/quote]

The “greenhouse effect” isn’t going to be the same as “effect of greenhouse gasses”. The greenhouse effect is always a warming effect. The effect of greenhouse gasses can be warming or cooling.
An example of a cooling effect created by greenhouse gasses is water vapor becoming clouds. Overall, clouds cool the Earth by about 20 degrees, even though they’re made from what started out as a greenhouse gas.

March 17, 2010 5:03 am

[quote: scienceofdoom (04:48:54) :]
magicjava:
I’m not sure Trenberth’s energy balance is accurate. It ignores readings we get from satellites and thermometers in favor of equations taken from Hansen’s climate model.
Which page on Kiehl and Trenberth’s 1997 paper are you referring to?
[/quote]

It’s not the 1997 paper. It’s the updated paper (2005 I believe).
The exact quote is:
[quote: Trenberth :]
The TOA energy imbalance can probably be most accurately determined from climate models and is estimated to be 8.5 +/-0.15 Wm-2 by Hansen, et. al.
[/quote]

I’d consider any earlier paper to be even less accurate than this one.

dr.bill
March 17, 2010 5:09 am

Recipy (03:39:01) :
Allen63 (04:21:29) :
Regarding the 33°C issue:
That value comes from using an albedo of ≈ 0.31, which is roughly what you get with the atmosphere, clouds, etc, left intact. In that case, the 342 W/m² available for absorption by the surface becomes 236 W/m² (= 0.69×342), Willis’ values (and mine a little further up the page) are found by elimininating the atmosphere altogether. If you don’t have an atmosphere, you can’t have clouds, and you wouldn’t have such a high albedo, so the comparison is between “everything in place as now” versus “bare planet like the Moon”. You can’t logically use the albedo that one gets with the clouds and such if you don’t have an atmosphere to produce them.
/dr.bill

March 17, 2010 5:10 am

What about internal warming due to radioactivity and tidal forces (although, I guess those are mitigated by the fact that the oceans can move in response to the forces)? My understanding is that Lord Kelvin’s thermodynamic calculation of the age of the earth, which was based on a black-body irradiated by the Sun, failed to produce the correct age because he was unaware of the concept of radioactivity.

March 17, 2010 5:17 am

[quote 1DandyTroll (04:39:37) :]
Has anyone actually measured the incoming wattage per square meter?
[/quote]

Yes. The CERES satellite measures longwave and shortwave radiation. Their web site is here:
http://science.larc.nasa.gov/ceres/index.html
And an overview of the meaning of it all is here:
http://asd-www.larc.nasa.gov/ceres/brochure/intro.html

March 17, 2010 5:23 am

scienceofdoom:
Here’s the link to Trenberth’s paper. See page 3 for the quote:
http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/BAMSmarTrenberth.pdf
1DandyTroll:
On page 3 of that paper you’ll see Trenberth saying he ignores data from the CERES satellite when building his energy balance model.

Kay
March 17, 2010 5:30 am

@ Bill Parsons (21:03:27) :
Wiki Answer to: “What is the temperature on the moon?”
The average daytime temperature on the Moon is around 107°C (225°F), but can be as high as 123°C (253°F).
When an area rotates out of the sun, the “nighttime” temperature falls to an average of -153°C (-243°F).
253 F
-243 F
_______
10 F”
=============================
For a range of 496 degrees between the light and dark sides. That’s pretty extreme. What is the range on Earth? Thanks to the atmosphere, its nowhere near close to that.

March 17, 2010 5:42 am

Trenberth is far less certain than he used to be about the energy budget. Remember he said it was a ‘travesty’ that he couldn’t account for the lack of warming in the last decade. One of his recent papers talks about ‘missing energy’ that he simply can’t account for: ‘Tracking Earth’s energy: From El Niño to global warming’ (February 8, 2010, submitted to ‘Science’)
http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/Tracking%20Energyv5.pdf
“Increasing concentrations of carbon dioxide and other greenhouse gases have led to a post-2000 imbalance at the top-of–atmosphere (TOA) of 0.9±0.5 W m-2 that produces “global warming”. Tracking how much extra energy has gone back to space and where this energy has accumulated is possible, with reasonable agreement between model and observational results for 1993 to 2003…Because carbon dioxide concentrations have further increased since 2003 the amount of heat subsequently being accumulated should be even greater.”
“…although some heat has gone into the record breaking loss of Arctic sea ice, and some has undoubtedly contributed to unprecedented melting of Greenland and Antarctica, it does not add up to be anywhere near enough to account for the measured TOA energy. Closure of the energy budget over the past 5 years is elusive. Thus state-of-the-art observations are unable to fully account for recent energy variability implying error bars too large to make the products useful.”
[“not…anywhere near enough”, “elusive”, “unable to fully account” “too large to make the products useful” etc: read: we’re at a loss to know what’s going on – it’s a travesty.]
“After 2000, observations from TOA…referenced to the 2000 values, show an increasing discrepancy…relative to the total warming observed…The quiet sun changes in total solar irradiance reduce the net heating slightly …but a large energy component is missing…”
“A large energy component is missing” – you can say that again! The discrepancy is huge and increasing by the year – this is the ‘travesty’ because it is a mysterious component that Trenberth knows nothing about, so for five years he’s been unable to ‘close’ the energy budget. Big problem! The contributions from melting glaciers, ice caps, Greenland, Antarctica and Arctic sea ice plus contributions from land and atmosphere warming and changes in solar irradiance all taken together are completely swamped by the “missing energy” due to an energy flux that is not in any way understood.
Of course, it’s only ‘missing’ on the assumption that the greenhouse effect is working as it is alleged to, with positive feedbacks. Lindzen and Choi (though rubbished by Trenberth) showed that, far from positive feedback, there may even be negative feedback – the surface temperature increased AND the escaping IR was increasing just as fast (this is probably Trenberth’s TOA discrepancy, and where the ‘missing energy flux’ is going). This seems to suggest that surface warming through the 1990s was NOT largely due to greenhouse warming, and with increasing CO2 concentration year by year the effect is considerably less than assumed by IPCC. Wind out the positive feedbacks and this ‘missing energy’ component disappears and things start to look sensible again.

March 17, 2010 5:43 am

The label “back- radiation “Is causing needless confusion.
I take it to mean that all of the 321w/m2 has come “back” from somewhere.
The somewhere must be the atmosphere.
This label seems to separate it from all other IR and long wave from scattering and direct to surface by whatever means EM radiation.
“No “back-radiation” according to the entertaining Gerlich and Tscheuschner”
In fact what G&T say is no HEAT from back radiation which they further state is so small that it can be ignored.
Wayne
….If you take all radiation absorbed by the ground, 169 for SW Absorbed By Surface and 321 for LW Back Radiation Absorbed By Surface,…….
I agree with Wayne does all the 321 come back from somewhere?
This diagram was meant to educate but instead it seems full of obvious ambiguities

March 17, 2010 5:47 am

[quote Kay (05:30:20) :]
For a range of 496 degrees between the light and dark sides. That’s pretty extreme. What is the range on Earth? Thanks to the atmosphere, its nowhere near close to that.
[/quote]

Looking at the raw data for the aqua satellite from the last 13 months, it gives a high of 274.18 K and a low of 197.06 K, for a difference of 77.12 K.
Obviously, the Earth’s experienced greater extremes than what it’s seen in the last 13 month, but 77.12 K is a decent enough rough estimate.

David L
March 17, 2010 5:56 am

“What is the temperature on the moon?”
First, remember that heat is the thermal unit of measure that “flows”, not temperature. Temperature is a result of heat content, defined as the average translational kinetic energy of atoms and molecules, and is dependent on the mass and specific heat of the object (such as air). There is less specific heat (and less mass) for a planet with no atmosphere than one with atmosphere. So if the amount of energy were the same for both cases, the temperature would be higher for the planet with no atmosphere. But heat is lost radiatively (black body radiation) and the atmosphere will prevent some of this loss, but itself also radiates it’s heat back to the earth and out to space. Quite a complicated system to figure out exactly as Willis nicely shows in his graphic.
There is no atmosphere on the moon, so there’s no air temperature (there’s no air to absorb heat), only surface temperature (the surface absorbs heat). When the sun is hitting the surface, it’s very hot. When the sun isn’t hitting the surface, it’s very cold because, having lost whatever heat it had by radiation. One side is hot, the other cold, and the average works out to be about -10F.
This illustrates the problem I have with a single term like average global temperature. Statistically if I put one foot in boiling water and one foot in ice water, the average is luke warm and I should feel fine. Some stations show an increase in temperture, others show a decrease. What’s the average? If North America becomes an ice sheet but Antarctica melts and becomes a jungle, and the average global temperature is still zero, should we be concerned? Is this global climate change? It’s certainly local climate change!!!!
Scientists should be focussed on the thermal budget and not a meaningless average global temperature that is dependant on so many other factors. I guess thermometers are cheap and everyone can read them…it’s harder to track the energy. But that’s the core of the problem. Is the earth retaining too much of the solar energy it receives and not allowing the proper balance to radiate back into space? When a climate scientist shows me these fundamental calulations and experiments on the heat budget then I’ll believe we have a problem.
In closing, a guy who lived in my neighborhood in 1902 wrote in his diary that yesterday (in 1902) was 58F….Today (2010) the high is 47F……not warmer for you climate scientists keeping score at home.

March 17, 2010 5:59 am

Willis Eschenbach (01:08:38) :
On a blackbody earth the current distance from the sun, it would only be 8°C cooler.
You are misunderstanding this. There is a difference between a ‘blackbody’ [20C cooler] and a ‘black body’ [8C cooler]. The correct answer to your question as stated is 20C cooler.

March 17, 2010 6:02 am

[quote ScientistForTruth (05:42:08) :]
[Quoting Trenberth]
“Increasing concentrations of carbon dioxide and other greenhouse gases have led to a post-2000 imbalance at the top-of–atmosphere (TOA) of 0.9±0.5 W m-2 that produces “global warming”
[/quote]

I’ll just add that the 0.9 Wm-2 Trenberth uses _does not_ come from measurements. It comes from Hansen’s climate model. Actual measurements give a higher or lower number than 0.9, depending on what measurement you use.
So, Trenberth is basically saying “Global Warming is 0.9 Wm-2 because Hensen’s climate models say it is so.”

David L
March 17, 2010 6:05 am

This is an excellent website that explains the fundamental concepts of Physics for those of you new or “rusty” on these concepts.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Click on “Heat and Thermodynamics” for links to those concepts, under that link click on “Radiation” to read about Stefan-Boltzman.

Bill Illis
March 17, 2010 6:15 am

I think you are right Willis.
They have actually calculated the 33K greenhouse effect wrong.
The Albedo figure used in the 33K calculation is 0.3 which includes the effects of solar reflection by clouds. But then the calculation excludes the positive greenhouse contribution of those same clouds. You can’t include one without the other.
So everything should start over at the beginning and say an atmosphere without any greenhouse gases would be 20K cooler.
Put water vapour/clouds back in, and you have -13K from cloud Albedo and something around +17K to +25K from the water vapour/cloud greenhouse effect so you do get something like +16K to +8K greenhouse effect from CO2/non-water vapour GHGs.
After that, the Stefan Boltzmann equations are logarithmic so one needs to calculate the K/W/m2 impact at each individual differential rather than take the total temperature K / total forcing change [which is one of the problems with the greenhouse theory – they stop using the differential].
Once you get close to the 390 watts/m2 total forcing, the differential is 0.18K / W/m2.

Basil
Editor
March 17, 2010 6:33 am

“The surface reflects about 29 W/m2 back into space. This means that the surface albedo is about 0.16 (16% of the solar radiation hitting the ground is reflected by the surface back to space). ”
Another question. 29/.16 = 181.25. I don’t see ~181 in the diagram. I do see 29+169=198 (168 absorbed, and 29 reflected).
Is this CBO scoring, or .gov accounting? I.e. 181 is close enough to 198 for government work?

March 17, 2010 6:36 am

Bill Illis (06:15:41) :
I think you are right Willis.
I think he is not.

Anders L.
March 17, 2010 6:43 am

“If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?

Mike Haseler
March 17, 2010 6:57 am

“The Wikipedia article is using the amount of sun after albedo losses, which is about 235 W/m2. That gives the colder value you quote above.”
Just a simple question. The wikipedia equations regarding radiating in all directions and receiving only in one direction seem to suggest the calculation assumes the sun is a point source, and the earth a hard disk. In actual fact, the sun has a finite size and slightly more than half the earth’s surface is being irradiated at any time. In addition, the atmosphere itself tends to increase the size of the cross sectional area – which could both increase and decrease the temperature of the earth.
Finally, I really wish people would stop reproducing these rubbish static models of the atmosphere. The atmosphere is a dynamic system with convectional currents taking heat from the surface up above otherwise blocking layers. The hysteria about global warming, is essentially a static view of the atmosphere (both in terms of time and space), in contrast I would suggest the sceptic view is a dynamic view whereby the climate not only varies in time, but it doesn’t stick around the surface but in contrast there are dynamic convective currents taking hot air high into the atmosphere where gases with a high emissivity (CO2!) enable the heat to escape.
Think about it this way
Why does hot air rise to create thermals …. it is because cold air descends! (Ha ha got you!) OK, a bit flippant, but if cold air didn’t descend, then we’d be left with a vacuum at the surface much like the space between Al Gore’s ears.
Hot air rises because cold air descends and so we should stop focussing on what is heating the air, and start asking the question: What is cooling the air?
Now, the answer is pretty obvious: the heated air is loosing its heat by the emission of IR, and therefore anything that helps to increase the emission of IR … like e.g. a complex molecule with a high IR interaction … what could that be? CO2? helps to cool the planet! That is to say, any molecule able to absorb IR is also (at the same temperatures) able to emit CO2 and so is just as capable of being a global cooler as being a global warmer!
The warmists see the atmosphere as a big blanket keeping us warm. In contrast I think the sceptical view should be to see the atmosphere as a huge cooling system taking heat from the surface of the earth up into space. The more CO2, the better the cooling system, and if like the warmists, we were to ignore the other effects of CO2, we also could create a hysteria that adding CO2 will dramatically increase the effectiveness of the world’s cooling system and lead to runaway global cooling.
OK, that’s far too simplistic but no less simplistic than the rubbish about CO2 causing runaway warming!

David L
March 17, 2010 7:05 am

Anders L. (06:43:13) :
““If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?”
Understanding if the earth is warming due to greenhouse gases has everything to do with the thermodynamics of the system. If you can’t explain the heat gain and the heat loss of this system then you don’t understand the system and shouldn’t be scaring people into doom and gloom scenarios.
Consider this, satellites orbiting the earth are in “vacuum” and being irradiated by the sun. Plus they contain electronics that generate heat. Here on earth my Pentium stays cool because air circualtes around it. How do you cool a pentium in a vaccum??? I’ll give you a hint “Black Body Radiation”. You ever see those giant sails on the space station? Some are solar panels, some are cooling radiators.
Now if scientists and engineers can use the basic concepts of physics to allow satellites to work in orbit, the Rover drive all over Mars, and the hubble space telescope develop great images, and in every case without freezing or melting down in a vacuum, then those same concepts should be able to explain the temperature on earth and it’s response to solar flux, greenhouse gases, and radiative cooling to name just a few.
Comparing the system with an atmosphere and without leads to a basic understanding of the maximum and minimum of the system. In calculus it’s understanding the limits of the equation. This helps in framing what’s going on in the middle.

March 17, 2010 7:06 am

[quote Leif Svalgaard (06:36:25) :]
Bill Illis (06:15:41) :
I think you are right Willis.
I think he is not.
[/quote]

I think the skeptic community would be well served if we built out own energy balance model, rather than using one based on numbers James Hansen pulled out of his ass, which is all Trenberth’s model is.
Starting with data from the CERES satellite would be a good idea, IMHO. Building a model that changes with conditions rather than just giving a hard-coded answer would be a good idea too. And being honest about things we don’t know, like where extra energy is going, would also be a plus.

Claude Harvey
March 17, 2010 7:16 am

Willis:
According to the chart of the past 450,000 years of reconstructed global temperatures for planet earth (the one The Goracle likes to wave about), the last Ice Age saw temperatures of at least 9 to 10 degrees C below today’s. I believe we had an atmosphere back then.
I think your “black body” calculation is off by a country mile. I admire much of your work, but suspect you got wrapped around your own axle on this one.

Frank Schroeder
March 17, 2010 7:19 am

>Leif Svalgaard (05:59:34) :
>>Willis Eschenbach (01:08:38) :
>>On a blackbody earth the current distance from the sun, it would only be
>>8°C cooler.
>You are misunderstanding this. There is a difference between a ‘blackbody’
>[20C cooler] and a ‘black body’ [8C cooler]. The correct answer to your
>question as stated is 20C cooler.
Yes, this distinction is important – as I understand it, a “black” blackbody earth would be 8 degrees colder, whereas a blackbody earth would be 20 degrees colder. However, does this matter for the energy flux diagram and Willis’ conclusions?

harrywr2
March 17, 2010 7:21 am

Anders L. (06:43:13) :
““If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?”
Because it’s a discussion on how sensitive the earths climate is.
One set of scientists says that if you modify one little part of the atmosphere there are going to be gigantic changes the climate.
I believe Willis is attempting to show if you just completely got rid of the atmosphere altogether the climate wouldn’t change that much.
Of course it would be hard to breath and the polar bears would surely die in Willis’s example.

March 17, 2010 7:23 am

Basil (06:33:41) :
This means that the surface albedo is about 0.16 (16% of the solar radiation hitting the ground is reflected by the surface back to space). ”

Should we thank the chinese for so kindly lowering that albedo?..with Carbon soot they lower it but with SO2 they increase it. But all this is nonsense as just one humble volcano will surpass any amount of anthropogenic “forcings”.

Ryan Stephenson
March 17, 2010 7:28 am

Interestingly I don’t see any part of this diagram to account for geothermal heat from underground. Even under the UK (not known for its volcanic activity) the ground at 3000meters depth is about 90Celsius. That underground heat energy must be percolating up to the surface eventually and contributing to the background temperature at sea level.
Interestingly, the bottom of the ocean cannot be lower than 4Celsius. If it fell below this it would freeze, and then the ice, being less dense than water, would float to the surface, where it would then melt in the sun. So the deepest ocean trenches cannot be lower than 4Celsius, even though in principle they can be getting almost no energy directly from the sun. Complicated isn’t it?
The Moon doesn’t have a lot of geothermal energy or ocean’s that refuse to freeze. It is relatively easy to understand why it gets close to absolute zero at night and above boiling point during the day. Fortunately we are well insulated from such extremes.

mkelly
March 17, 2010 7:28 am

Why is it that PV=nRT is never considered when figuring a surface temperature of the earth. It is a valid scientific formula. It explains the normal use of STP when saying under what conditions you did your test etc. As such 0 deg C is what the earth with an atmosphere has as a temperature and the 33 deg C warm up from GHG is really only 15 deg. at max.
Also if IR goes from 4-60 micro and the band for CO2 is 14-15&16 accounting for any over absorbtion then all CO2 can only get 5.4% of total reradiation. Of that 5.4% human input to CO2 is roughly 3% ergo 3% of 5.4% is .0016. The best that human CO2 can do is .0016 of any longwave radiation. If you use TOA of 247 W/m2 then .4 W/m2 is max for human input. It is probably less than that.

March 17, 2010 7:43 am

Frank Schroeder (07:19:28) :
Yes, this distinction is important – as I understand it, a “black” blackbody earth would be 8 degrees colder, whereas a blackbody earth would be 20 degrees colder. However, does this matter for the energy flux diagram and Willis’ conclusions?
To be frank, I can’t tell what his conclusions are.

March 17, 2010 7:45 am

Your theory seems to be Swiss cheese on this one Willis.

David L
March 17, 2010 7:50 am

mkelly (07:28:59) :
“Why is it that PV=nRT is never considered when figuring a surface ”
I don’t believe it applies in any significant way. Gas temperature by this equation changes given a change in volume, moles of gas, or pressure. but it says nothing of surface temperature.
In the earth’s atmoshere there is no change of volume (it’s the boundless sky..bounded only by gravity), no change in pressure (except small barometric changes when fronts move around) and there’s really no change in moles of gas (unless you consider the tiny fractional increase in CO2). Plus the ideal gas law only applies in the ideal cases…in this marginal increases in moles of CO2 it wouldn’t be an accurate equation anyhow.

Jim
March 17, 2010 8:19 am

******************
David L (07:50:25) :
mkelly (07:28:59) :
“Why is it that PV=nRT is never considered when figuring a surface ”
I don’t believe it applies in any significant way. Gas temperature by this equation changes given a change in volume, moles of gas, or pressure. but it says nothing of surface temperature.
In the earth’s atmoshere there is no change of volume (it’s the boundless sky..bounded only by gravity), no change in pressure (except small barometric changes when fronts move around) and there’s really no change in moles of gas (unless you consider the tiny fractional increase in CO2). Plus the ideal gas law only applies in the ideal cases…in this marginal increases in moles of CO2 it wouldn’t be an accurate equation anyhow.
**************
Isn’t the ideal gas law used to calculate the adiabatic lapse rate?

mkelly
March 17, 2010 8:33 am

David L (07:50:25) :
Are you saying that the 5.3million giga tons of atmosphere has no effect on the earth’s temperature? If I leave T as an unknown in the formula and use standard pressure for P and figure for a volume up to say 100km I get a temperature of 0 deg C.
The term STP means standard temperture and pressure. If it is standard to have 0 deg C at one atm then that should apply to the surface of the earth also.
I realize air is not an ideal gas but then again we use radiation formula that were meant for cavities and black bodies with no depth.

March 17, 2010 8:46 am

Is anyone going to consider the second law of thermodynamics and “back radiation” supposedly “warming” the earth’s surface. ?
Quite simply anything based off the bunkum IR budgets,
is bunkum itself.
That is a shame Willis because your recent post(s) regarding thunderstorms (atmospheric heat pipes) is damned excellent, and a great leap forward in our understanding. I particularly have to mention the illuminating “sun’s eye view” you used, brilliant.
Excellant posts by Brian W, Ryan Stephenson, David Haseler and David L. to mention only a few.
The rest, this place is becoming a warmists site.

Richard M
March 17, 2010 8:47 am

As a corollary it would be interesting to look at historic situations. For example, when the Earth had no ice caps what was the sensitivity? At the height of glaciation what was the sensitivity? That might put some bounds on the problem.

Bill DiPuccio
March 17, 2010 8:48 am

“1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.”
I would agree with this statement as it applies to a global scale. But positive water vapor feedbacks do occur on smaller spatial and temporal scales (Dessler, Lindzen, Pielke).
The error occurs in trying to generalize these situations to the entire atmosphere. As you point out, all the negative feedbacks come into play canceling or even reversing the potential effect.

OceanTwo
March 17, 2010 8:53 am

harrywr2 (07:21:49) :
Anders L. (06:43:13) :
““If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?”
Because it’s a discussion on how sensitive the earths climate is.
One set of scientists says that if you modify one little part of the atmosphere there are going to be gigantic changes the climate.
I believe Willis is attempting to show if you just completely got rid of the atmosphere altogether the climate wouldn’t change that much.
Of course it would be hard to breath and the polar bears would surely die in Willis’s example.

That’s essentially what I got out of it, also.
The actual energy flows in the atmosphere in this case are (mostly) irrelevant. The linearity is irrelevant. It seems we are trying to determine what the (theoretical) temperature of the earth is without an atmosphere.
And honestly, even at 20 degrees, it seems like you’d need a radical atmospheric change to significantly alter the atmospheric contribution. (Of course some people claim that the current and predicted CO2 change is a ‘radical’ change).
Maybe Willis *does* hate polar bears, who knows? they aren’t particularly friendly creatures, after all. 😛

Charles Higley
March 17, 2010 8:56 am

Tempest in a teapot when you realize that there is not enough carbon out there for us to burn to double the CO2 in the atmosphere.
Why? Good question.
CO2 partitions 50 to 1 between water and air. It would take 50 times the CO2 that the atmosphere contains to double the CO2 in the atmosphere. One estimate is that, if we really tried, we might be able to raise the CO2 by 20%. We are fighting against a partition coefficient.

Alan D McIntire
March 17, 2010 9:00 am

The temperature for a blackbody earth should be
T = (1368 watts/square meter)/(4* 5.67*10^-8 watts/square meter K^4)
gives a temperature of 278.683 K- Willis Eschenbach’s figure. What you’re all arguing about is whether or not you should throw the factor (1 – albedo) into the numerator. If you throw
in the clouds, you throw a (1-.3) into the numerator, you decrease the prior figure of 278.683 to 278.683 * (0.7^.25) = 254.909 -blackbody earth without greenhouse gases but with clouds

mkelly
March 17, 2010 9:08 am

Derek (08:46:04) : No warmist I.
For anyone a question. If CO2 can cause an increase in temperature based on reradiation of IR then why are thermos bottles not filled with 100% CO2 instead of using a vacuum?
By the IPCC formula of 5.35ln(C/CO) you could get up to 75 W/m2 of extra “heat” going back into your coffee. OUCH!
Someone should patent this.

Allen63
March 17, 2010 9:09 am

dr.bill,
Thanks for the clear response. Choice of albedo it is, then.

March 17, 2010 9:18 am

Alan D McIntire (09:00:18) :
The temperature for a blackbody earth should be […] a temperature of 278.683 K- Willis Eschenbach’s figure.
No, this is for a black blackbody, but the Earth is not black. It is very nearly a blackbody, but the albedo of the Earth without atmosphere [and clouds] is not zero, it is [as Willis points out] more like 0.15, thus a real Earth with atmosphere removed would be 20C cooler, not 8C.

March 17, 2010 9:19 am

Alan D McIntire (09:00:18) :
The temperature for a blackbody earth should be […] a temperature of 278.683 K- Willis Eschenbach’s figure.
No, this is for a black blackbody, but the Earth is not black. It is very nearly a blackbody, but the albedo of the Earth without atmosphere [and clouds] is not zero, it is [as Willis points out] more like 0.15, thus a real Earth with atmosphere removed would be 20C cooler, not 8C.

Steve Goddard
March 17, 2010 9:19 am

Using this logic, Mt. Everest should be only four or five degrees cooler than Saudi Arabia. Another example of the misuse of mathematics.

March 17, 2010 9:23 am

David L (07:50:25) :
mkelly (07:28:59) :
“Why is it that PV=nRT is never considered when figuring a surface ”
I’m afraid mkelly is quite correct.
The atmosphere is subject to the Earths Gravitational Field
This causes gravitational compression of the gases.
High school physics student would be expected to answer question below.
Typical HSPQuestion is” using the kinetic theory of gases
a/calculate the rms speed of a N2 molecule at STP
b/If this molecule moved vertically upward and had no further interactions how high would it get
c/ What would happen to the temperature as it went higher”
(Answers 517m/s,13.6Km,drops constantly)

lgl
March 17, 2010 9:30 am

Close enough for the black-body sensitivity which is 0.18 °C per W/m2, but we want to know the ‘climate-body’ sensitivity, and then you have to include the feedbacks.
Trenberth’s diagram shows that the system amplifies the external input (sun) by 2.4 so the resulting sensitivity is somewhere around 0.43 °C per W/m2, or 1,6 C per CO2 doubling.
This assumes todays gain of 2.4 will not change if the forcing increases, but it probably will, probably not linear. But since a 3.7 W/m2 increase is <1% of the initial, setting a 2.2-2.6 boundary should be safe (+/- 8% of initial) and then the sensitivity range becomes 0.40-0.47 or 1.48-1.73 C per CO2 doubling.
calcs:
Solar input to surface: 169 + (58+10)/2=203 W/m2 (/2 because only 1/2 downward)
Total to surface: 321+169=490
Gain: 490/203=2.41
Black-body sensitivity: (289^4-288^4)*5.678*10^-8=5.45 W/C=0.18 C/W/m2
This holds if we do not reach a Tipping Point, but the good thing is we already reached the cold-to-warm TP 10000 years ago, so the next TP will be a warm-to-cold TP.

brent
March 17, 2010 9:34 am

Willis,
Thanks for your superb efforts and excellent way to get open discussion
I’m a little wary should we ourselves draw overly firm conclusions/convictions in the sense of taking the Kiehl/Trenberth graph figures as a given, and then drawing some overly firm conclusions therefrom
(All sorts of conjecture is absolutely warranted )
Isn’t this the same Kevin Trenberth referred to below?
I.E. Who admits in a private email we are not close to balancing the earths energy budget (and agreed by Tom Wigley) and therefore geoengineering would be hopeless, as we would not be in a position to evaluate it’s effect
If so what is the level of confidence in all the various fluxes shown in that diagram in the first place.?
It may well be that it’s just that I’ve not taken the time myself to study these particular aspects in depth required to get a personal comfort level
The Warmers are absolutely open to highly warranted criticism on multiple levels. I thought your robust 🙂 comments to Judith Curry and Ravetz were spot on. They reflected my views exactly.
Just a word of caution (very slight) that we not get ahead ourselves in term conviction/confidence
Warm Regards
brent
http://www.eastangliaemails.com/emails.php?eid=1056&filename=1255550975.txt
Kevin,
I didn’t mean to offend you. But what you said was “we can’t account
for the lack of warming at the moment”. Now you say “we are no where
close to knowing where energy is going”. In my eyes these are two
different things — the second relates to our level of understanding,
and I agree that this is still lacking.
Tom.
++++++++++++++++++
Kevin Trenberth wrote:
> Hi Tom
> How come you do not agree with a statement that says we are no where
> close to knowing where energy is going or whether clouds are changing to
> make the planet brighter. We are not close to balancing the energy
> budget. The fact that we can not account for what is happening in the
> climate system makes any consideration of geoengineering quite hopeless
> as we will never be able to tell if it is successful or not! It is a
> travesty!
> Kevin
>

March 17, 2010 9:44 am

An argument that is made on the basis of a Kiehl-Trenberth diagram makes me nervous for this type of diagram makes a muddle of thermodynamics. In the language of thermodynamics, there is no such thing as an “energy flow.” The only energy that “flows” is heat. In a Kiehl-Trenberth diagram, some of the “flows” are heat. Others are radiation intensities. It is clear that the “back radiation” is not heat for it “flows” from cold to hot matter; if it were heat, it could not flow in this manner, under the second law of thermodynamics. However, while we have a conservation principle for heat flows, we do not have one for radiation intensities. Thus, the proposition that the K-T diagram portrays some kind of “balance” is false.

KDK
March 17, 2010 9:48 am

Amazing to see so many using ‘wiki’… LOL
What is truly amazing is this: So many of you consider yourself scientists but, as seen in many, if not all posts, is that you cannot agree on actual numbers. I thought sciences, like mathematics, uses numbers that have been accepted by measurements that can be validated.
How is it that there are so many theories and only a few datasets to use? How can we almost NEVER agree on an answer? Why so many theories? Can’t someone just factually prove what is being presented… ever? Certainly, I am not saying we shouldn’t continue to push forward as new and enlightening evidence is brought forth via discovery, but, damn, I didn’t know science was so grey… shouldn’t it be either black or white, one or the other? I realize that some of these are just theories and can never be proven, but like AGW, the one’s claiming to KNOW MUST prove it via facts, or it should just be academia speaking for the sake of spurring on innovation/discovery and NOT be presented like it is the end all–unless it truly is.
Again, I am not saying that because Joe, Jim, Janet, and Jomojo all say it is fact, that one should stop looking, but let’s find out which measurements are the best, accurate and then all use the same info after analysis. How are there so many theories surrounding supposed fact-based research?

March 17, 2010 9:49 am

Steve Goddard (09:19:41) :
Another example of the misuse of mathematics.
Nonsense, it is of misuse of the physics.

renminbi
March 17, 2010 9:58 am

If the earth’s atmosphere were only N2 and O2 would either of these gases absorb or emit radiation from the sun or earth. Sorry to expose my ignorance,but it is better to ask than to remain ignorant. This is a great blog,with great comments.

Richard Sharpe
March 17, 2010 10:01 am

mkelly (09:08:39) said:

Derek (08:46:04) : No warmist I.
For anyone a question. If CO2 can cause an increase in temperature based on reradiation of IR then why are thermos bottles not filled with 100% CO2 instead of using a vacuum?
By the IPCC formula of 5.35ln(C/CO) you could get up to 75 W/m2 of extra “heat” going back into your coffee. OUCH!
Someone should patent this.

I don’t think this is a good argument. The devil is in the details.
1. Coffee is normally at a temperature that is outside the frequency range that CO2 efficiently absorbs.
2. Evacuating the thermos shell is to prevent conduction between the inner part of the shell and the outer part of the shell. CO2 will conduct much better than a vacuum.
So, it really depends on whether the losses due to conduction are higher or lower than the reduction in losses due to radiation.

March 17, 2010 10:20 am

Willis Eschenbach (10:09:49) :
8°C if the no-atmosphere earth is considered to be a blackbody.
You do not understand what a blackbody is. The 8C is for a black blackbody. The Earth is not black, but is nevertheless very close to a blackbody. Take a white sphere and heat it to 400C, then take a black sphere and heat it to 400C, they both radiate the same amount. If not, we would violate the thermodynamics, where we could get work out of the radiative difference [use it to power a steam engine] between two bodies at the same temperature.

March 17, 2010 10:25 am

Willis Eschenbach –
Do you think you’re up to creating an energy balance diagram from scratch using actual data rather than James Hansen’s climate model the way Trenberth did?
Somebody with motivation has to do it and my plate is full at the moment with the Aqua satellite.

Eric (skeptic)
March 17, 2010 10:26 am

Enneagram states ”just one humble volcano will surpass any amount of anthropogenic “forcings””
Pinatubo released 42 Mt of CO2 (see http://vulcan.wr.usgs.gov/Projects/Emissions/Reports/Pinatubo/pinatubo_abs.html). Mankind releases 6 or 7 Gt of carbon per year (multiply by 44/12 to get CO2)

March 17, 2010 10:41 am

[quote Willis Eschenbach (10:35:46) :]
Are there any substantive claims that I have not disputed?
[/quote]

Well, um, yes. Mine. Trenberth’s model uses made up numbers and ignores real life measurements. Using his work to check your results means you’re comparing your results to made up numbers.
That’s not a flaw in your line of reasoning, but it is a flaw with what you’re using for a sanity check on your results.
Which is why I was wondering if you feel up to the task of creating a more realistic energy balance. One that uses real-world numbers.

March 17, 2010 10:42 am

Willis Eschenbach (10:34:11) :
Is my writing really that opaque? My conclusion is that the climate sensitivity is on the order of half a degree C per doubling of CO2, and the equilibrium sensitivity is much less than that.
As far as I can see you concluded that based on the 20C figure. You never used the 8C figure for anything, which is good because it is wrong.

March 17, 2010 10:47 am

scienceofdoom (23:02:15)
Thanks for the response to brian W. I find your blog a wonderful resource for people get an understanding of things like back radiation.
There is enough detail there for those with technical backgrounds and your way of explaining things to those with “not so technical” backgrounds is very good. I often wonder what the blog wars would have been like if reasonable voices like yours had been in charge of real climate.
Scienceofdoom: WRT Willis contention. It seems a first good guess that
the first order effect of albedo is linear, it’s effectively a TSI knob, but one thing that bothers me about the diagram is what happens to the 29W/sq meter that get reflected from the surface. Shouldnt some portion of that be absorbed by trop and lower strat ( as is the case with that same radition when it is incoming)? And shouldnt some be relected back down by clouds and aerosals? is the 29 a net net net. Let me put it this way. Short wave reflected at the surface has to travel back through the same phyiscal medium it took to get to the earth.. probably should not read too much into the cartoon I suppose.

Steve Goddard
March 17, 2010 10:50 am

Willis,
There is no need for laboratory experiments. The earth itself is the laboratory, with some locations having much less atmosphere than others. Temperatures in the upper elevations of the Himalayas average 40-50C cooler than similar latitudes at lower elevations, because of the lack of atmosphere. Your claim below is simply incorrect.

This means that the warming due to the complete atmospheric system (greenhouse gases, clouds, aerosols, latent and sensible heat losses, and all the rest) is about 20°C over no-atmosphere earth albedo conditions.

Kevin
March 17, 2010 10:55 am

WTF? Environment Canada says that this winter was 4.0°C above average?
http://www.msc.ec.gc.ca/ccrm/bulletin/national_e.cfm
The national average temperature for the winter 2009/2010 was 4.0°C above normal, based on preliminary data, which makes this the warmest winter on record since nationwide records began in 1948. The previous record was 2005/2006 which was 3.9°C above normal. At 3.2°C below normal, the winter of 1971/1972 remains the coolest. As the mean temperature departures map shows all of the country, but for a small area over the southern Prairies, was above normal, with some areas of the arctic and northern Quebec more than 6°C above normal. Southern Saskatchewan did have a cooler than normal winter, with temperatures more than 1°C below normal. This past spring was the first season in the past 5 years with temperatures below normal, as shown in the consecutive seasons graph.

March 17, 2010 11:00 am

Willis Eschenbach (10:34:11) :
Is my writing really that opaque?
The initial quiz was a bit opaque. I would have agreed completely if the premises were ‘no atmosphere and totally black’. The ‘blackbody’ bit is somewhat off, because in the infrared, the Earth is an almost perfect blackbody and that is where it radiates.

March 17, 2010 11:08 am

Willis Eschenbach (10:59:56) :
Can’t have it both ways … the 8C is right for a black blackbody. Since I was not talking about a white blackbody, the 8C figure is not wrong.
But you were not explicitly talking about a ‘black’ blackbody either so the built-in assumption is that you were talking about a blackbody with an albedo [0.15] as that the surface actually has…
But perhaps we have generated enough heat by now without much new light. 🙂

David L
March 17, 2010 11:09 am

mkelly (08:33:01) :
David L (07:50:25) :
“Are you saying that the 5.3million giga tons of atmosphere has no effect on the earth’s temperature? If I leave T as an unknown in the formula and use standard pressure for P and figure for a volume up to say 100km I get a temperature of 0 deg C. ”
Not due to PV=nRT realistically. Are you saying that because there’s pressure and gas molecules and a volume (what’s the volume of the earth’s atmosphere by the way?) then we have a temperature? Turn off the sun, shut down the geothermal heat, and see how long it takes for the atmosphere to become pools of liquid N2 on a frozen ocean. Pressure is not driving temperature here. It’s the other way around. Classic example of mixing up Causation with Correlation. Temperature and Pressure are correlated…but which causes which? If your highschool student compresses air, the compressed air temp changes….but that temperature is soon absorbed by the surroundings. Ever touch a high pressure gas cylinder? How hot is it? It’s room temperature…But it’s under high pressure? how can that be? Now put that tank in the sunlight….guess what…temp goes up AND pressure goes up.
So why would the amount of air, the earths atmospheric volume, and pressure have anything to do with global temperatures according to PV=nRT? Look up R….what are the terms? J/K mol. It’s the amount of energy per temperture per mole….it’s all about the energy folks Know the energy flow and you’ll know all the temperatures!!!!!! If you’re not doing work on the system you aren’t moving energy around!! Just having an atmosphere is not doing work!!!! You have to do something to it…like heat it up with sunlight!!!!!!

Mike Haseler
March 17, 2010 11:15 am

“Take a white sphere and heat it to 400C, then take a black sphere and heat it to 400C, they both radiate the same amount. If not, we would violate the thermodynamics, where we could get work out of the radiative difference [use it to power a steam engine] between two bodies at the same temperature.”
This warmist logic! The body with the lower emissivity will radiate less, that is what emissivity means: it radiates less at the same temperature.
As for breaking fundamental laws of thermodynamics, if that were true, then thermos flasks and space blankets both break your “fundamental” laws as the low emissivity material prevents irradiation of heat compared to similar “blackbody” materials at the same temperature.

mkelly
March 17, 2010 11:18 am

Richard Sharpe (10:01:57) : I don’t have “smiley” faces to put at the end of sentences, but you it the nail on the head. It was a rhetorical question.
Most of the people on the earth live in areas that get above the temperature of where CO2 radiates. That’s part of the point.
If CO2 would not warm up coffee or keep gespacho luke warm how is it able to warm up an entire global. The atmosphere has conduction and convection. Gases dissipate heat. My frozen turkey will warm to room temperature but no more. No amount of CO2 in my house will heat the turkey higher than what the room temperature is.
Is there a ratio of convection/conduction by CO2 versus radiation by CO2 when dissipating heat?
If I remember my heat transfer class when two discs at different temperatures radiate at each other the lower will rise to the higher then it becomes a push/pull and nothing further happens. So the best that could happen with CO2, is it could rise to the temperature of the earth and then all stops. The pot does not warm the burner.

Micky C
March 17, 2010 11:21 am

Willis, I think inadvertently you have stumbled upon the key wrong assumption in this whole AGW debacle. The value for Trenberth back radiation is very wrong. In the paper they show the upward radiation from the surface and then from the top of the atmosphere. At the TOA we can see the absorption peaks and how there is a reduction in the spectra. The difference in W/m2 is the difference in the integrals of these curves i.e. what does not come out of the TOA is the difference in the integrals. If those curves are integrated an eyeball estimate is about 10 to 15% difference. If we have 339 going up how can we then have 321 coming back?
Secondly using the same curves, and using the measured CO2 15 µm band, which is heavily saturated for the current ppm of CO2. Any increase in ‘the wings’ is 4 orders of magnitude different than the main peak. This puts any decrease in transmissivity of that order of magnitude. Now considering that the difference in back radiation is around 30 to 50 W, how exactly is a doubling of CO2 going to produce 3W/m2 extra forcing. The simple answer is it can’t hence the assumption that water vapour increases; an assumption that as yet has not been tested.

Mike Haseler
March 17, 2010 11:22 am

Just realised, when I mentioned thermos flasks, I forgot to mention I was referring to the silver coating used to reduce heat loss.

anna v
March 17, 2010 11:34 am

groundRe: magicjava (Mar 17 04:43),
Thanks for the link. One can see the difference between air and skin surface. in the maps.

March 17, 2010 11:39 am

mkelly (09:08:39)
For anyone a question. If CO2 can cause an increase in temperature based on reradiation of IR then why are thermos bottles not filled with 100% CO2 instead of using a vacuum?
By the IPCC formula of 5.35ln(C/CO) you could get up to 75 W/m2 of extra “heat” going back into your coffee. OUCH!
Let’s see if I can explain. Heat is transferred by Convection, Conduction and radiation. See the sun? It’s in a vacuum. There is a lot of nothing around it.
So the heat of the sun get here by radiation. Take a metal rod in your hand hold it in a fire. That’s conduction.
In a thermos you build a vacuum chamber. You try to get as few as particles in there as possible to prevent conduction. Filling that chamber with anything would defeat the purpose of reducing the heat transfered by CONDUCTION. How do you prevent radiation through this vacuum? Easy, you put a SILVERED LINING on the metal.
That shiny metal “blocks” or reflects the IR. It does this much better than C02 does. C02 only “blocks” certain regions of Longwave.
http://www.howstuffworks.com/thermos.htm/printable
When you look at how radiation move through a material you have to look
at all the ‘windows’ in the material. .
This may help:
http://www.crisp.nus.edu.sg/~research/tutorial/atmoseff.htm
Then go google radiative transfer equations.

Bill Illis
March 17, 2010 11:45 am

Technically, if we have now reduced the greenhouse effect/atmospheric pressure impact to 20K, the TOA forcing relative to that figure is not 150 W/m2 anymore – it would only be about 95 W/m2.
The equilibrium emission temperature would now be 268K rather than 255K and 95 W/m2 gets you up to the surface temperature of 287K or 14C.
So the average over the entire 95 W/m2 would be 0.21K /W/m2. [As I said the first part of the 95 watts would have more impact than the last one which is just 0.18K /W/m2].
Technically, it also wouldn’t be the TOA forcing either but just the 3.1 km high forcing.

March 17, 2010 11:51 am

[quote Willis Eschenbach (10:57:20) :]
I greatly regret putting up the global energy budget diagram, as people are overlooking the subject of the thread, which is climate sensitivity.
[/quote.

My apologies for being one of those people. When I see Trenberth’s diagram, it just sets off a giant red flag for me.
There are so many topics covered in the post that it’s difficult for me to find the core substance of it all. I _think_ the point to the entire thing is:
[quote Willis Eschenbach:
This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase.
[/quote]
… but that was saved until the next-to-last sentence.
Also, you’re tossing around a lot of unsubstantiated claims. Examples include almost everything you say about the IPCC. For example, does the IPCC really calculate the effects of Evaporation incorrectly? And no one ever noticed this? With all the things coming out about the IPCC lately, this may very well be the case, but statements like this need some supporting evidence.
The use of 20°/8° is also very confusing to me. It’s not clear when one or the other applies to what you’re saying.
I’d suggest stripping out as many non-essential points as possible, so folks like me don’t get tripped up over Trenbreth, and folks like Lief don’t nit-pick you to death on black blackbody and white blackbody stuff.
If your main point is the increase in temperature due to increase in W/m-2 is logarithmic, start with that and make it clear.

March 17, 2010 12:02 pm

Willis Eschenbach (10:35:46) :
Are there any substantive claims that I have not disputed?
Derek says – Well, yes actually, for instance see the below.
AND my own post elsewhere.
http://www.globalwarmingskeptics.info/forums/thread-609.html
Terry Oldberg (09:44:22) :
In the language of thermodynamics, there is no such thing as an “energy flow.” The only energy that “flows” is heat. In a Kiehl-Trenberth diagram, some of the “flows” are heat. Others are radiation intensities. It is clear that the “back radiation” is not heat for it “flows” from cold to hot matter; if it were heat, it could not flow in this manner, under the second law of thermodynamics. However, while we have a conservation principle for heat flows, we do not have one for radiation intensities. Thus, the proposition that the K-T diagram portrays some kind of “balance” is false.

Frank
March 17, 2010 12:12 pm

John A wrote (21:46:51):
How of the incoming energy from the Sun is converted into atmospheric and oceanic convection? None, according to Kielh and Trenberth. Which is amazing. Climate physics is a fascinating subject.
What causes atmospheric and oceanic convection to slow down (the only reason solar energy would be needed to keep them going)? Friction. Where does the energy consumed in overcoming friction appear? As heat. Physics is a fascinating subject.

March 17, 2010 12:30 pm

magicjava

I’m not sure Trenberth’s energy balance is accurate. It ignores readings we get from satellites and thermometers in favor of equations taken from Hansen’s climate model.

And in response to my question..

It’s not the 1997 paper. It’s the updated paper (2005 I believe).
The exact quote is:
[quote: Trenberth :]
The TOA energy imbalance can probably be most accurately determined from climate models and is estimated to be 8.5 +/-0.15 Wm-2 by Hansen, et. al.
[/quote]
I’d consider any earlier paper to be even less accurate than this one.

The updated paper is 2008.
In fact, Kiehl and Trenberth don’t throw out any measurements in favor of climate models. In the 1997 paper they assume that incoming and outgoing energy balances. Why?
Because the instrument error means the measurement results are not accurate enough to be able to say whether energy in = energy out. In a non-warming and non-cooling world this would be true.
And there is a small challenge of whether the measurements of solar energy in is more accurate than the measurement of terrestrial energy leaving and reflected solar energy.
So there is an uncertainty around 5-10W/m^2.
In the updated 2008 paper, being good scientists they attempt to provide better numbers all around. In the case of the instrument error, there are still the same unknowns. So rather than fix energy in=energy out, they say, “well, the earth is warming up, how much by?”.
Anyhow, for those for whom the word “model” causes outrage.. look at the statement which was inaccurately quoted:

The TOA energy imbalance can probably be most accurately determined from climate models and is estimated to be 0.85±0.15 W m-2 by Hansen et al. (2005) and is supported by estimated recent changes in ocean heat content (Willis et al. 2004; Hansen et al. 2005).

As you can see -first the number is pretty small, and second, the calculation is supported by the measurement of the increase in ocean heat (as you can see in The Real Measure of Global Warming ).
1W/m^2 out of 240W/m^2 – and you want to throw out the results?
For anyone wanting to learn a little about climate basics, read the whole original 1997 paper.

Frank
March 17, 2010 12:32 pm

Willis: Technically speaking, climate sensitivity is the partial derivative of temperature with respect to energy (probably both incoming energy from the sun and energy radiated downwards by the atmosphere and clouds). Your post may be assuming that temperature is a linear function with respect to incoming energy. The relationship between T and W is certainly a very complicated one if conditions are dramatically different from today – a snowball earth or even an ice-age where the relative proportions of land and ocean change. For small changes in W and T, the relative is approximately linear and climate sensitivity can be treated as a constant.
The IPCC also worries about the time-scale associated with estimates of climate sensitivity. For a given dW, how long does it take for T to come to equilibrium? Does equilibrium include changes in the deep oceans and ice-caps? For models with the highest climate sensitivity, equilibrium of the atmosphere and upper ocean apparently requires decades.

Joel Shore
March 17, 2010 12:34 pm

Willis:
I am confused as to where you have gotten the 150 W/m^2 figure that you quote. I also think, in line with what scienceofdoom said, that what you have essentially calculated here is not the climate sensitivity but the sensitivity in the absence of feedbacks. In particular, what you have to recognize is that the distinction between forcings and feedbacks is somewhat arbitrary. For example, CO2 change is, in some sense, a feedback in the glacial – interglacial cycles but is a forcing in our current predicament. The thing is that you have to be consistent. What you have done is not consistent because you have presumably counted everything, e.g., including the water vapor, as a forcing in getting that 150 W/m^2 figure but then you have not counted any change in water vapor as a forcing in computing the resulting climate sensitivity.
Another way of putting it is this: If the warming due to the change in CO2 causes additional water vapor to go into the atmosphere, this water vapor will produce an additional radiative forcing…and if the additional warming causes ice melt that changes the earth’s surface albedo, this will also produce an additional radiative forcing. (And, the change in clouds would also produce an additional radiative forcing whose magnitude, and even admittingly, sign are uncertain.) Now, we usually call all of these things feedbacks instead of forcings and that is fine if we do so consistently. The problem is that I don’t think you have done so consistently…i.e., as near as I can tell, your “150 W/m^2” (which I don’t really understand anyway) counts everything as a forcing, not a feedback.
By the way, Jim Hansen has made a similar point recently about being consistent with what you call a feedback and a forcing…And, in fact, he has argued that what he calls the Charney sensitivity derived from the glacial – interglacial cycles considers changes in albedo due to changes in ice sheets as a forcing whereas any such changes in our current discussions are considered to be a feedback. Hence, he argues that the 3 C climate sensitivity for doubling CO2 is probably too small for our current “experiment”…and says it is more like 6 C when you properly consider the effects of the albedo changes due to changes in ice sheets. (Others question whether the ice sheet albedo effects would really be that large in our current climate and also how fast the ice sheet changes can occur…so, I am not saying his 6 C estimate is correct, but I think that his basic point about being careful what you consider to be forcings and feedbacks is.)

mkelly
March 17, 2010 12:51 pm

steven mosher (11:39:03) : Steve I guess you missed reading the part about patenting the idea. It was a rhetorical question. I cannot put smiley face at the end of the sentences.
However, it is possible to construct an idealized thermos floating in space, with an inside chamber transparent to IR but infinite R value, and an outside surface that blocks incoming radiation whose inside surface is a perfect reflector of IR. So no conduction no convection only IR from the soup, coffee whatever to the next chamber. Fill the space where the vacuum would normally be with CO2.
The IPCC formula makes no difference as to anything except the forcing of the log of (C/CO). So with that in mind using the IPCC 5.35ln(C/CO) we get
ln 1000000 is 13.81 times 5.35 for 73.91 W/m2. According to IPCC whatever you have in the thermos will heat by 74 W/m2 indefinitly to vaporization and explosion.
Please don’t miss the point Steve I don’t think CO2 can do what is claimed and none of the information I have read demonstrates via physics, math, chemisty etc or whatever, over turns my understanding of the laws of thermodymanics, heat transfer etc.
My ultimate point is that if CO2 could increase heat energy through radiation then someone somewhere would have found a way to make money off the idea. But as of yet nothing even though we are told this has been know for a 100 years. Hog wash.

OceanTwo
March 17, 2010 12:58 pm

Honestly, I can’t find anything about a ‘colored’ black body. A Black body is a black body (or blackbody if you are so inclined). Either way, anything which as an emissivity (not equal to 1) is not a black body.
Unless there’s something changed in the last 20 years. Maybe: they are teaching all sorts of funky stuff these days (perhaps an object with a specific emissivity is classified as a specific ‘black body’ color? ). Even so, it’s an irrelevant argument: as I understand it, based on what had been demonstrated, if earth was a black body it [surface] would be 8 degrees cooler; as it has an emissivity, it is theorized to be 20 degrees cooler than what we measure today.
It seems like the prosecution is stating that the perpetrator accelerated from a stand still to 100mph when they breezed through the red light. The defense [Willis] is arguing that it’s only 50 yards from where the perpetrator started to the red light, so how can they be going 100 mph? Weather it’s 50 yards, or 60, or 20, or 70.235 is quite irrelevant; or what the shade of the red light is, or even the decimal accuracy of the radar gun used for measurement.
Does the theory hold? Are the results of Willis calculation so far off ? Is the principle employed sound? If not, why not? Can it be demonstrated that his theory is false? Are there any significant factors not accounted for?
In other words…

Mattias, Sweden
March 17, 2010 1:09 pm

8 degrees cooler? That contradicts what can be found using Stefan-Boltzmann law, 255K or -18C. Or have I missed something?
http://answers.yahoo.com/question/index?qid=20090408183916AASBK1D
http://en.wikipedia.org/wiki/Black_body#Temperature_of_Earth

March 17, 2010 1:23 pm

OceanTwo (12:58:04) :
if earth was a black body it [surface] would be 8 degrees cooler; as it has an emissivity, it is theorized to be 20 degrees cooler than what we measure today.
albedo, perhaps, rather than emissivity.
The 8C was not really used by Willis for anything, so it is perhaps not worth harping too much on it. There is also a possible confusion because you can talk of blackbody radiation as that which has a blackbody spectrum without the body be required to be black [the Sun is not black, for example]. More confusion comes from the statement: “That is to say, the net gain from our entire complete system, including clouds, surface albedo, etc..” where the clouds and surface albedo are not gains but deficits. Anyway, I may just have been too pedantic about this, but I was disappointed that the lead paragraphs were on something not actually used further on. The important things was the 20C difference.

wayne
March 17, 2010 1:32 pm

Willis:
Re: wayne (04:27:21) :
I may have answered my own question to you.
Seems the 105 W/m2 might very well come from the 22 for Sensible Heat and 76 for Latent Heat. Those total 98, close but not exact to the 105.
Seems the K/T chart is organized in a confusing manner. Their fault, not yours. It should have been broken in three sections. The Input, the internal flows, and the output. That would tie the back radiation to the back radiation mathematically.
On the left, the SW radiation from the sun, the input, and it is pretty clear, then a dashed light gray vertical division.
Next is the internal energy flows. That is the atmosphere, clouds, the two upper pointing heats, sensible and latent, and the down dwelling back radiation on the far right. Then another vertical dashed light gray line.
What remains on the far right is the output section. That is all of the up dwelling radiation that gathers from the ground and from the center internal section.
That would de-mystify it a bit, mathematically anyway.
Willis, somewhat OT:
Some of your graphs are coming out vary fuzzy on this end. If you are getting these via screen captures, this might help. It works on xp anyway. Before capturing, turn the screen font edge smoothing off. On xp it’s standard or ClearType. Capture the picture then turn the smoothing back on. It does wonders on my machine, no more fuzziness! You get am exact pixel for pixel copy.

Anders L.
March 17, 2010 1:37 pm

Willis Eschenbach (10:25:22) :
Anders L. (06:43:13)
“If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?
“Thought experiments” have a long and proud history in science. They are widely used to examine conditions that we cannot replicate in a laboratory.
I agree wholeheartedly, but I am not convinced that anything meaningful
about the detailed properties of our climate system can be inferred by
thinking about a planet completely devoid of a climate system.

March 17, 2010 1:55 pm

[quote scienceofdoom (12:30:53) :]
In fact, Kiehl and Trenberth don’t throw out any measurements in favor of climate models.
[/quote]

They do, readings that are more than 6 times higher than their estimate of 0.9. The CERES satellite shows an imbalance of 6.4 W/m-2. It’s right there in the paper. The sentence before the one I quoted. So is the reason why they threw it out: because it’s not expected by their models.
And yes, I think Trenberth’s model needs to be redone. I’ve given reasons why on this thread. Willis Eschenbach has given reasons why he believes the model can’t even work. In a nutshell, we need to start looking at reality rather than models.
Anyway, Willis has stated that Trenberth’s energy budget is not central to what he was trying to say, so I don’t want to clog up this thread with debates over it. If you’d like to respond to what I’ve posted here, I’ll let you have the last word on the matter.

DavidB
March 17, 2010 2:03 pm

Just curious (really) about one point. If the average temperature of the moon in sunlight is about 225F, how come the average temp of the earth in sunlight is so much lower? As far as I can see from the energy budget diagram, about half of the incoming radiation is either reflected by the atmosphere or the earth’s surface itself, or absorbed by the atmosphere on the way in (and some of the latter would be radiated down to the surface). So it seems that we should expect the average surface temp of the earth in sunlight to be not less than about 110F, i.e. about half that of the moon. True, this is reached or exceeded in the middle of the Sahara or Death Valley, but it seems very high for a global average. Am I wrong about this, or is there something wrong in my reasoning?

DavidB
March 17, 2010 2:05 pm

….maybe cooling by evaporation of water accounts for the apparent shortfall?

Richard Sharpe
March 17, 2010 2:08 pm

steven mosher (11:39:03) said:
In a thermos you build a vacuum chamber. You try to get as few as particles in there as possible to prevent conduction. Filling that chamber with anything would defeat the purpose of reducing the heat transfered by CONDUCTION. How do you prevent radiation through this vacuum? Easy, you put a SILVERED LINING on the metal.
That shiny metal “blocks” or reflects the IR. It does this much better than C02 does. C02 only “blocks” certain regions of Longwave.
Ahhh, that’s a much better explanation than I managed to produce. I forgot about the silver coating on the inside side of the outer sleeve … and yes, almost complete reflection is much better than 50% reflection.
Looks like I didn’t get my invite from CTM. Was hoping to meet you.

Joel Shore
March 17, 2010 2:16 pm

mkelly says:

However, it is possible to construct an idealized thermos floating in space, with an inside chamber transparent to IR but infinite R value, and an outside surface that blocks incoming radiation whose inside surface is a perfect reflector of IR. So no conduction no convection only IR from the soup, coffee whatever to the next chamber. Fill the space where the vacuum would normally be with CO2.
The IPCC formula makes no difference as to anything except the forcing of the log of (C/CO). So with that in mind using the IPCC 5.35ln(C/CO) we get
ln 1000000 is 13.81 times 5.35 for 73.91 W/m2. According to IPCC whatever you have in the thermos will heat by 74 W/m2 indefinitly to vaporization and explosion.

There’s an awful lot of confusion in these couple of paragraphs but I’ll try to point out some of the major problems:
(1) In fact, if you postulate a system where incoming radiation gets in and (at least some of it) gets absorbed but any outgoing radiation cannot escape then indeed that system will keep heating up indefinitely. That’s what energy balance (essentially the 1st Law of Thermodynamics) says it has to do!
(2) In reality, in the case that you described, the system could never get hotter than the surface of the sun because, as the temperature rises, its emission of radiation will shift toward the shorter wavelengths and eventually it won’t be emitting mainly in the IR but rather in the visible where, by the assumptions of the problem (e.g, such radiation is able to get into the system), the radiation is able to escape.
(3) The IPCC formula is not a magical formula that holds for any container with a concentration of CO2 in it. It is a formula obtained from detailed line-by-line radiative (or even radiative-convective) transfer calculations in the actual earth’s atmosphere. Among other things, the actual temperature profile of the atmosphere with height and the other IR-active constituents like water vapor play an important role. See here for further discussion: http://www.aip.org/history/climate/simple.htm#L_0623

daniel
March 17, 2010 2:22 pm

I have one question regarding the Kiehl/Trenbeth radiation budget scheme : I have the understanding that since some quite extraordinary papers
Gerlich & Treuschchner (1989), Robitaille, etc. this kind of ‘budget’ was heavily falsified for basic physics reasons ?
I’m wondering whether these papers had eventually to face rebuttals in peer reviewed papers…
Ar this stage I’m not aware of any, and would just conclude that they represent the current stage of physics, which mean that next IPCC report’s science basis would be interesting to draft

Joel Shore
March 17, 2010 2:25 pm

DavidB says:

Just curious (really) about one point. If the average temperature of the moon in sunlight is about 225F, how come the average temp of the earth in sunlight is so much lower?

It is the global radiation in and out that half to balance, not the local one. The heat transport in the earth’s atmosphere and the thermal inertia relative to the diurnal variation means that the earth’s temperature goes through less extreme ranges. If the earth rotated much slower and/or if there was less thermal inertia (e.g., not so much ocean) and/or if there was less convective processes in the atmosphere to mix things then there would be more extreme temperature ranges.

It's always Marcia, Marcia
March 17, 2010 2:38 pm

Roy Spencer on clouds and negative feedback. May be evidence for Lindzens’ Ifrared Iris
Part 1

It's always Marcia, Marcia
March 17, 2010 2:38 pm

Part 2

Brian G Valentine
March 17, 2010 3:03 pm

Thank you, Willis, this is interesting.
It’s a common calculation in the senior-level course of radiation heat transfer to show the Earth’s radiation exchange temperature with the Sun is (about) 280K, then the extra 8K are observed from satellites
In other words, isn’t this pretty common knowledge to people with a BS in mechanical engineering etc?

pochas
March 17, 2010 3:07 pm

Also recommended:
Nir Shaviv: “On Climate Sensitivity and Why it is Probably Small”
http://www.sciencebits.com/OnClimateSensitivity

Brian P
March 17, 2010 3:29 pm

Willis
A bit old but I really liked your thunderstorm post from some months ago

DavidB
March 17, 2010 3:30 pm

To Joel Shore: Thanks. Those are good points. The influence of the oceans as a heat store is probably the key factor. In the interior of the continents there is a much larger diurnal range than over or near the oceans, and even there convection currents must carry the extremes of daytime heat away from the surface. I’m just a curious newbie to these subjects, so please forgive my errors and oversights.

Alan D McIntire
March 17, 2010 3:31 pm

You’ve also got to adjust for emissivity. The earth’s albedo would be maybe 0.15 without clouds, but it would also would not emit as a black body- what would the emissivity be? I’ve read estimates that the oceans’ emissivity is about 0.94, what would it be for land only, about 0.9?
in that case, you’d figure in a correction of ( 1 – 085/.9)^0.25

DocMartyn
March 17, 2010 3:35 pm

‘David L (07:50:25) :
mkelly (07:28:59) :
“Why is it that PV=nRT is never considered when figuring a surface ”
I don’t believe it applies in any significant way. Gas temperature by this equation changes given a change in volume, moles of gas, or pressure. but it says nothing of surface temperature.’
One liter of liquid water contains 55.56 moles of water molecules and occupies 1000 mls or one litre. At STP, as a gas, this would generate a volume of 1244.544 liters; giving a ratio of 1,244.544:1.
The globally-averaged annual precipitation is 990 millimetres (39 in), or 0.27 mls/cm2 per day; so that during the day/night cycle, on average, each cm2 of the planet generates and then collapses one third of a liter of volume, or an ingot 336 cm or 3.36 meters.

Colin Davidson
March 17, 2010 3:36 pm

I think Willis’s estimate makes sense on other grounds.
Using the diagram the atmosphere receives its energy primarily from the Surface.
If you look at the Surface energy flux balance, you get a sensitivity between 0.095 and 0.15 DegC/W/m^2, depending on what assumption is made on evaporation.
(This number makes no allowance for “feedbacks” eg greater back-radiation due to increased CO2 and water vapour concentrations, or lesser solar radiation absorbed into the surface due to increased cloud cover. However it is interesting that if the surface temperature rises by the median IPCC estimate of 3 DegC for a doubling of CO2, the 3.7W/m^2 “radiative forcing” at the top of the atmosphere would necessarily need to be a huge “surface forcing” of between 22 and 32 W/m^2 at the surface. )
The usual IPCC calculations, without feedbacks, use a TOA sensitivity of around 0.3. This is then assumed to apply throughout the troposphere, right down to ground level. But that sensitivity is between twice and thrice the surface sensitivity, and I reckon that the assumption that the temperature change at TOA translates to the same temperature change at the surface cannot be sustained without invoking magic, or unless the sensitivities are in fact similar.
Willis’s calculation brings the TOA sensitivity in line with the surface.

Alex Heyworth
March 17, 2010 4:13 pm

magicjava (07:06:00) :
[quote]
I think the skeptic community would be well served if we built out own energy balance model, rather than using one based on numbers James Hansen pulled out of his ass, which is all Trenberth’s model is.
Starting with data from the CERES satellite would be a good idea, IMHO. Building a model that changes with conditions rather than just giving a hard-coded answer would be a good idea too. And being honest about things we don’t know, like where extra energy is going, would also be a plus.
[/quote]
You mean something like this? http://www.palisad.com/co2/eb/eb.html

Alan D McIntire
March 17, 2010 4:21 pm

Sorry, that should be a correction factor of roughly
(albedo/emissivity) ^0.25 = (.85/,9)^.25 times the original temperature.
Another point, Using Trenbeth ballpark figures, if we get 240 watts from the sun, and the greenhouse effect makes that an effective 390 watts and a 33 C increase, isn’t the increase in temperature per delta watt increase in the ballpark of
33 C/(390-240) = about 0.22C per watt increase?
A thrid point using Trenbeth ballpark figures. The surface flux is about 490 rather than 390 watts, with 100 watts going into conduction and convection rather than sensible heat. Figuring that half of that is radiated to space, half back to earth, about half of 100 watts or 50 watts is removed from sensible heat by convection and conduction. 50 watts latent heat + 150 watt sensible heat implies that 25% of any past wattage increase has gone into latent heat. Shouldn’t that apply to the 3.8 watt increase for a doubling of CO2? Shouldn’t at least 0.95 watts go into latent heat, only 2.85 watts or less go into sensible heat?
A fourth point, since emissivity is less than 1, assuming it’s about 0.9, isn’t the actual greenhouse effect something like 100 latent+ 150*.9 = 135 watts sensible heat?

Brian G Valentine
March 17, 2010 5:07 pm

On a “small” climate sensitivity:
Estimates from Lindzen, Shaviv, Monckton, Schwarz, others, put the value BELOW what is known to be “natural” variability – partly stochastic, partly determinate.
My question is: if there is no way to observe a physical quantity, does it have meaning at all?
I can construct a thousand quantities that I can calculate in some way (e.g, change in Earth’s rotational angular momentum from buildings, transportation, etc) but the construct is entirely meaningless because there is no way to physically measure it

cba
March 17, 2010 5:25 pm

Willis,
there are some significant problems with the albedo numbers. If you go by K&T, it would seem the 0.3 total albedo is 0.22 for clouds and 0.08 for surface with about 0.62 cloud cover fraction. In essence, the 0.08 is the average albedo where there are no clouds present.
Your number of 0.15 would be fairly reasonable based upon the Moon and Mars. It is not reasonable for Earth as over 70% of the surface is covered by oceans and the oceans have less than 0.04 albedo for incoming solar radiation, at least for high angles of incidence (wrt horizon where the incoming flux is most significant).
Try these numbers:
surface (ocean – use 0.7 ocean fraction) 0.04
surface( land) 0.19 (includes snow cover, forests, sand etc)
average surface albedo 0.085 (corresponds to 29/342 on chart )
the land albedo is calculated from the 0.085 total surface and the 0.04 ocean.
clouds (62% cover, averaged value) 0.36
this corresponds to the assumption of 0.22 for cloud albedo and fractional contribution.
NOTE that these numbers do not add up.
The weighted average for 0.62 cloud cover leaves 0.38 surface contribution.
Using the 0.085 overall surface albedo and fraction visible, one gets
0.38*0.085 = 0.032 contribution to the overall albedo which conflicts with the chart.
Since the overall albedo listed is 0.307, the cloud contribution would actually be 0.307 – 0.032 = 0.275 at a 0.62 fractional cover.
That puts the cloud albedo at 0.275/0.62 = 0.44
Note that this leaves the clouds with what would seem a more realistic albedo value than the original 0.38 albedo value.
Your question of temperature drop really depends upon the conditions.
if the surface is stripped of ocean and ice and vegetation, then it’s going to have an albedo of around 0.17 or 0.15. That’s around 265K or -8 deg C or 22 C less (20 C less if 0.15). If the surface stayed the same, the drop would be 15 deg C. If the cloud albedo somehow stayed along with the surface albedo so that the total was 0.307 the temperature drop would be 34 deg. lower.

Joel Shore
March 17, 2010 5:26 pm

daniel says:

I have one question regarding the Kiehl/Trenbeth radiation budget scheme : I have the understanding that since some quite extraordinary papers
Gerlich & Treuschchner (1989), Robitaille, etc. this kind of ‘budget’ was heavily falsified for basic physics reasons ?
I’m wondering whether these papers had eventually to face rebuttals in peer reviewed papers…

That Gerlich & Treuschchner saw the light of day in even a second-rate physics journal is extremely embarrassing, as its fundamental error regarding the 2nd Law of Thermodynamics is basic enough that one could lead a group of bright first-year physics students through the logic. (And, some of its other arguments are just plain bizarre.) A comment on G&T that I am a party to and that will essentially trash it completely has been accepted by the journal and should appear within the next few months.

Brian G Valentine
March 17, 2010 5:36 pm

Thank you, Mister Troll Shore, for your erudite analysis of Gerlich and Tscheuschner’s paper, whether you believe it or not, the analysis is correct, and if you think Phys B is second rate, don’t look at it, read some first rate stuff like “Scientific American” instead, the favored bedtime reading material of the first-rate watermelon

Joel Shore
March 17, 2010 5:38 pm

Alan DMcIntyre says:

Shouldn’t that apply to the 3.8 watt increase for a doubling of CO2? Shouldn’t at least 0.95 watts go into latent heat, only 2.85 watts or less go into sensible heat?

Well, the issue of latent heat is really one of distribution of the temperature change with height in the atmosphere since what evaporates at the surface has to condense higher in the atmosphere, and this in turn can affect the radiative balance since most of the emission back out into space escapes from the upper troposphere. But, now the punchline: What you are describing is, I believe, just another way of talking about the (negative) lapse rate feedback, i.e., the notion that because of the condensation of water vapor, the upper troposphere tends to warm faster than the surface. This is a feedback that is already included in all the climate models…and, in fact, because much of the physics governing this feedback also governs the (positive) water vapor feedback, models that have a higher magnitude for the water vapor feedback tend to have a higher magnitude for the lapse rate feedback…and hence the sum of these two feedbacks tends to vary less from model-to-model than each of them individually.

March 17, 2010 5:40 pm

Willis Eschenbach (17:36:52) :
G&T is bad, bad, bad.
I’ll agree with that.

Joel Shore
March 17, 2010 5:51 pm

Willis says:

Dang, here I am agreeing with Joel again … will wonders never cease?

Watch out…It could be addicting! 😉

Francisco
March 17, 2010 5:52 pm

yan Stephenson (07:28:41) :
Interestingly I don’t see any part of this diagram to account for geothermal heat from underground. Even under the UK (not known for its volcanic activity) the ground at 3000meters depth is about 90Celsius.
—————–
Geothermal is supposed to play its biggest role in warming the bottom of the oceans, as the crust under the ocean floor is much thinner than under most land. The amounts of geothermal energy and its variations and distributions are largely unknown. Considering how close it is from us, it’s amazing how little we really know about what is going on at the earth’s core. Some interesting new theories are emerging:
http://tinyurl.com/y8pguwr
Beginning in 1969, astronomers discovered that three of the giant planets, Jupiter, Saturn, and Neptune, each radiate about twice as much energy as they receive from the Sun. Those planets each contain a powerful energy source which was inexplicable until J. Marvin Herndon, pictured at left, demonstrated in 1992 the feasibility of natural, nuclear fission reactors as the energy source for those planets
http://tinyurl.com/y8pguwr

Harry Lu
March 17, 2010 5:59 pm

Leif Svalgaard (13:23:50) :
A questions perhaps you could answer please?
Albedo refers to reflectivity.
The normally referred albedo is to the visible light reflectance. But surely the reflectance in the LW IR will be different. Is this known?
Emissivity of a surface is not he same as albedo. Having used an IR camera a few times I know that the IR temperature of an object changes with emissivity. and corrections can be incorporated in the IR to temp conversion (this is why Mr Watts IR photos in his surface station project are not very valid – he need to correct for surface emissivity.) When using IR cameras one usually coats the objects with a thin layer of mat paint to equalise (approx) emissivity.
Surely albedo controls the absorbed solar radiation, but emissivity controls the radiated LW energy.
If so then one needs to know the emissivity of the earth in the 4u range not its albedo.
/harry

March 17, 2010 6:00 pm

Leif Svalgaard (10:20:53) :
You do not understand what a blackbody is. … Take a white sphere and heat it to 400C, then take a black sphere and heat it to 400C, they both radiate …

“Surface emissivity” must be considered; both may not have equal emissivity figures … which will affect radiation.
An Emissivity Primer
Emissivity Coefficients of some common Materials
Examples, emissivity:
Aluminum Foil – 0.04
Beryllium —- 0.18
Granite —– 0.45
Sand —- 0.76
Wrought Iron – 0.94
Plaster —– 0.98
.
.

March 17, 2010 6:02 pm

Francisco (17:52:28) :
J. Marvin Herndon, pictured at left, demonstrated in 1992 the feasibility of natural, nuclear fission reactors as the energy source
Recent measurements of geo-neutrinos rule out the hypothesis that most of the planet’s internal heat is generated by a uranium-fuelled nuclear geo-reactor in the Earth’s core.:
http://www.physorg.com/news187946006.html

Francisco
March 17, 2010 6:08 pm

yan Stephenson (07:28:41) :
Interestingly I don’t see any part of this diagram to account for geothermal heat from underground. Even under the UK (not known for its volcanic activity) the ground at 3000meters depth is about 90Celsius.
———————-
See also letter by Herndon in Current Science, titled “Variables unaccounted for…”
http://tinyurl.com/yfx4dax

sky
March 17, 2010 6:15 pm

Nothing impedes clear physical reasoning than a wrong conceptual model. A blackbody Earth without an atmosphere might be a good starting point for teaching students, but no realistic analysis of actual temperatures can be made without fully mastering the implications of enthalpy. That atmospheric pressure is a vital variable seems to have been forgotten.

Harry Lu
March 17, 2010 6:15 pm

Thermal imaging and emissivity here
http://en.wikipedia.org/wiki/Thermal_imaging
“Since there is no such thing as a perfect black body, the infrared radiation of normal objects will appear to be less than the contact temperature. The rate (percentage) of emission of infrared radiation will thus be a fraction of the true contact temperature. This fraction is called emissivity.
Some objects have different emissivities in long wave as compared to mid wave emissions. ”
/harry

March 17, 2010 6:53 pm

Update, pls:

Joel Shore (17:38:05) :

Well, the issue of latent heat is really one of distribution of the temperature change with height in the atmosphere since what evaporates at the surface has to condense higher in the atmosphere, and this in turn can affect the radiative balance since most of the emission back out into space escapes from the upper troposphere.

Hmmm … which space-based sensor (and what wavelength) can allow measurement, can allow this to be actively seen during convective T-storm activity?
I’m going to posit the theory that a largish amount of thermal energy is _not_ radiated directly from the troposphere (from the gases constituting the air e.g. CO2 which are *capable* of EM radiation) into space, but rather becomes part of Hadley Cell circulation and the bulk of that sensible heat is released back into space is at latitudes greater than ~45 degrees due to *surface cooling* …
Also note that thermal budgets show a ‘deficit’ of incoming solar energy at those latitudes (for the average temps seen), so it is circulation from the lower latitudes that keeps those temperatures elevated relative to what might otherwise be experienced over ~50 latitude.
.
.

HankHenry
March 17, 2010 7:12 pm

Willis tells us “The average temperature of the planet is about 14°C.” Of course that’s disregarding the massive cold of the ocean which is more like 4°C – averaged out. What is it that makes the oceans so cold? Two miles down on the continents the earth is quite warm, but two miles down in the oceans it’s quite cold. It suggests to me there is a refrigeration process of some sort extracting heat from the depths. Also, do I remember my grade school science correctly when I say the total mass of the atmosphere is represented by the mass of just 33 feet of water – the theoretical lifting limit of a suction pump?

March 17, 2010 7:24 pm

Willis Eschenbach (19:16:37) :
a warming from a no-atmosphere situation of ~ 18°C rather than the ~ 20°C I used above.
I like your number now.

Brian G Valentine
March 17, 2010 7:25 pm

Willis and Leif think Gerlich & Tscheuschner’s analysis is bad.
Sorry folks, it’s good. Two legitimate criticisms leveled at it: it does not account for convection heat transfer in the atmosphere (considers conduction only), and it does not consider clouds.
But “convection” heat transfer is nothing more than conduction heat transfer (through a thermal boundary layer), and all radiation in the atmosphere is eventually lost via degradation to 20+ micrometer wavelengths, for which everything in the lower atmosphere is transparent
– or it increases the internal energy of the atmosphere, the contribution (via increasing the average k.e. of molecules) is negligible.
Ladies and Gentlemen: Either Gerlich and Tschneuschner are correct, or the atmosphere is a perpetual motion machine.
If it is, then copy the principle with a scale model and make yourself wealthy. Power your invention up a little bit with cold fusion.

Joel Shore
March 17, 2010 7:59 pm

Brian G Valentine says:

Ladies and Gentlemen: Either Gerlich and Tschneuschner are correct, or the atmosphere is a perpetual motion machine.

That’s a “head-I-win, tails-you-lose” style of argumentation. Actually, either Gerlich and Tschneuschner are correct and the greenhouse effect describes a perpetual motion machine or they are wrong and it doesn’t. It is very easy to show that the second of these possibilities is in fact the case…And, I think Brian that you are intelligent enough to comprehend this.

Joel Shore
March 17, 2010 8:05 pm

Willis Eschenbach says:

But none of these change my point. Use whichever TOA and albedo figures you want, you won’t get anything near the UN IPCC canonical sensitivity of 3°C per doubling.

But, I’d like to see you address the more fundamental objections that I raised in my post of (12:34:08), namely that I don’t see how your calculation includes feedback effects. I think you are just calculating the sensitivity before feedbacks and somehow getting the wrong number (as I don’t understand where the 150 W/m^2 value you use comes from).

March 17, 2010 8:08 pm

Brian G Valentine (19:25:34) :
But “convection” heat transfer is nothing more than conduction heat transfer (through a thermal boundary layer)
It is a lot more. Conduction is mediated by micro-scale vibration without any bulk transfer of matter, while convection is macro-scale movements of matter.

Tsk Tsk
March 17, 2010 8:11 pm

Leif,
I don’t buy the white 400C sphere and the black 400C sphere radiating the same. The fact that the white sphere is white has altered its emissivity and its radiation. It’s no longer a black body but a grey body. You can’t extract energy from the two spheres initially as the white sphere will also have lower absorption to go with the lower emissivity and so you won’t get a net flow of heat from the black sphere to the white. Maxwell’s demon still naps even though the total radiation from the two spheres initially is not the same.
Willis,
Yes, your post was a bit confusing because you didn’t show the math for the two cases until your follow up comments. I originally agreed with Leif until I re-read your post a couple of times to get your point. And, as others have pointed out, why didn’t you use the derivative with respect to power of S-B at mean global temp to get your sensitivity? The error is relatively small (0.18 vs. 0.13 K(C)/W) but it’s still a bug, or were you claiming that the sensitivity is something other than black body?

cba
March 17, 2010 8:14 pm

Willis,
to measure the actual sensitivity of the atmospheric absorption you’re going to have to have similar albedo, despite its unreal conditions. That means before and after albedo of 0.307 despite a lack of atmosphere or of atmospheric absorption. The numbers are an average temperature of 288K with the associated surface radiation of 390 w/m^2. Balance occurs with about 239w/m^2 making it out of the atmosphere – leaving 150 w/m^2 to be absorbed in the atmosphere or blocked in the atmosphere. Note that some of this is due to ghgs and some due to cloud blocking.
doing the comparisons for various warming values:
33 deg C = 34/ 150 = 0.23
18 deg C = 18/150 = 0.12
8 deg C = 8/150 = 0.05 deg C rise per W/m^2 power increase
keeping albedo constant means we have 342 W/m^2 – 107 = 235 w/m^2 which corresponds to a bb temperature of 254 or about 34 deg C.
Note that even the 34 deg C rise due to the current atmosphere corresponds to only 0.8 deg C rise for a 3.7 w/m^2 forcing increase due to a co2 doubling.
Applying an increase in absolute humidity based on constant RH and a 2 deg C increase in temperature, one finds that h2o vapor (ignoring cloud formation) results in only about 1/3 of the power increase due to the co2 doubling (it’s about 3.1 w/m^2 for an increase of 30% based on a 5 deg C rise)which means that most of such a 2 deg C rise doesn’t have a mechanism to cause the increase. That is even assuming the rise of 5 deg. C, the co2 + h2o vapor increase doesn’t have enough forcing to generate even a 2 deg. C rise. Less warming results in even lower forcing from h2o vapor.

Glenn Tamblyn
March 17, 2010 8:44 pm

Willis
A few points
Your TOA figure of 147 seems wrong, from the diagram. You seem to be using the radiation that ORIGINATES at the TOA and not including the radiation making it THROUGH FROM LOWER LEVELS – 237 w/M^2 on the diagram
“1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.”
Well I would say rather estimated from calculations. Just because it may not have been adequately demonstrated does not automatically mean it isn’t valid
“2. The climate models underestimate the increase in evaporation with temperature.”
I can’t comment on whether they do but the question is what impact does this have on energy balance. What altitude does this additional heat get transported to before it condenses out?
“3. The climate models do not include the effect of thunderstorms, which act to cool the earth in a host of ways . “.
Yes, conceivably they do, but what is the quantitative impact of that? What percentage of the energy transport in the atmosphere could be attributed to Thunderstorms. Is it a significant percentage or small enough that it can be ignored as a 2nd or 3rd order effect. Or simply subsumed into the general circulation within a cell in the model. A qualitative idea about a possible factor is only useful if you can get a handle on how important that factor is quantitatively.
“4. The climate models overestimate the effect of CO2. This is because they are tuned to a historical temperature record which contains a large UHI (urban heat island) component. Since the historical temperature rise is overestimated, the effect of CO2 is overestimated as well.”.
How can the effect of CO2 be ‘tuned to the temperature record’? Surely the effect of CO2 is used to estimate the radiative balance consequences. Any tuning to the temperature record is then about relating radiative balance to temperature change. And the UHI effect won’t be that big. 70% of the surface temperature is oceans – no UHI. Most of the worlds land area isn’t in cities, no UHI. If it is the historical record being used, even in cities, UHI was much lower in the past.
“5. The sensitivity of the climate models depend on the assumed value of the aerosol forcing. This is not measured, but assumed. As in point 4 above, the assumed size depends on the historical record, which is contaminated by UHI. See Kiehl for a full discussion.”
See my comments at 4 about UHI. Also isn’t aerosol effects also analysed by modelling and measuring the optical properties aerosols?
“6. Wind increases with differential temperature. Increasing wind increases evaporation, ocean albedo, conductive/convective loss, ocean surface area, total evaporative area, and airborne dust and aerosols, all of which cool the system. But thunderstorm winds are not included in any of the models, and many models ignore one or more of the effects of wind”
See my comments at 3 about quantifying such statements. How much do winds speeds increase relative to the current speed of the Jet Stream for example. Also, what if any impact does any of this have on vertical mixing and energy transport to higher altitudes which could increase outgoing radiation. Greater mixing at one altitude does not necesarily result in changing the vertical heat balance for that level, just the horizontal distribution of that heat.
You don’t mention increasing contributions from Methane, Ozone, Nox. You don’t discuss long term albedo change due to decline of ice coverage or vegetation change. You don’t discuss the consequences if cooling due to aerosols declines due to humanity trying to clean up our emissions. You don’t mention that the capacity of the air to hold water vapour does not go up exponentially with temperature, so increased evaporation will be balanced by increased precipitation.
There is a lot you don’t mention Willis

March 17, 2010 9:00 pm

HankHenry:

What is it that makes the oceans so cold? Two miles down on the continents the earth is quite warm, but two miles down in the oceans it’s quite cold. It suggests to me there is a refrigeration process of some sort extracting heat from the depths.

The sun warms the bottom of the atmosphere and the top of the oceans. When a gas or liquid warms it expands and so rises because it is less dense.
So the oceans are very stratified in temperature – because the top is warmer and gets further warmed by the sun.
But the atmosphere experiences a lot of convective overturning as the warmed air at the bottom rises up. So the atmosphere is more mixed.
You can see more about this at Why Global Mean Surface Temperature Should be Relegated, Or Mostly Ignored
PS More technical note, yes the top of the stratosphere gets warmed as well due to the O2-O3 process. But most of the sun’s energy then travels through the “transparent” atmosphere and so warms the surface of the earth and the surface of the oceans.

March 17, 2010 9:20 pm

Tsk Tsk (20:11:34) :
I don’t buy the white 400C sphere and the black 400C sphere radiating the same.
My argument goes like this:
Consider a solid white sphere with an opaque interior in empty space far from everything. In its interior it has a nuclear reactor that heats the sphere [by conduction] to a temperature of 400C. This means that the electrons just inside the surface due to thermal jiggling will be accelerated constantly [in random directions]. Accelerated charges radiate determined by their acceleration. Since we stipulate that the sphere stays at the same temperature it must radiate corresponding to the thermal jiggling determined by that temperature.
The same must hold for a black sphere, or a green sphere, etc. The radiating, accelerated electrons don’t know what color the sphere has. The energy emitted must equal that generated by the reactor.

gbaikie
March 17, 2010 9:53 pm

“Temperature of a body at the distance of the earth from the sun, with an albedo equal to that of the earth: about 20°C cooler than the present earth.
TOA radiation from the present earth: about 150 W/m2
Since 150 W/m2 TOA radiation gives a warming of about 20°C, this gives a climate sensitivity of about half a degree for a doubling of CO2. This is only one-sixth of the UN IPCC canonical value of 3°C.
I don’t care if I’m off by 10%, this is a first-order analysis. My answer is way below the IPCC answer, that’s the issue, not the exact details.”
Couple questions.
First, it’s well known that earth’s average temperature is about 15 C. Is that suppose to be air temperature or temperature of the ground or surface of oceans. I assume it’s air temperature in the shade and on sea level elevation. Or simply the average air temperature of the ocean [since a large part of the surface of the Earth is covered by oceans- and therefore one can basically ignore all the land with it’s all of varying elevation.
Now you talking about an imagined airless world with the same albedo as earth. It seems then that you talking about the surface temperature- as you have removed the air.
Take Mars [a nearly airless world] it’s “air temperature” varies widely depending upon whether it is a meter or two above the surface:
“The daytime SURFACE temperature is about 80 F during rare summer days, to -200 F at the poles in winter. The AIR temperature, however, rarely gets much above 32 F. ”
http://www.astronomycafe.net/qadir/q2681.html
“The problem with measuring the temperatures on Mars is that they change in time and space in radical ways,” he said. “On Mars Pathfinder, for example, over a distance of only 40 centimeters (about 16 inches) up and down a little mast, we had three temperature sensors. They often measured temperatures that differed by about 15 degrees Celsius — that’s about 25 degrees Fahrenheit.”
http://www.space.com/news/mplmet_991201.html

anna v
March 17, 2010 10:14 pm

Re: DavidB (Mar 17 15:30),
A contributor posted the following link
http://isccp.giss.nasa.gov/products/browsesurf1.html
where one can find the “surface skin temperature”.
It is higher than the air at 2meters temperature that is used in all the calculations.
Mind you, it is the surface skin temperature that should be used in black body formulae, because it is the solids/liquids that radiate as T^4.( I think the atmosphere is something like a T^6 and in any case has small heat capacity with respect to the surface). Nobody uses the “surface skin temperature” in all these radiation budgets.
In any case, my physicist trained brain gives up with all the hand waving and double counting that goes on in “climate science”.

LightRain
March 17, 2010 11:16 pm

Why are we worried about a black body earth when there is, and never will be such a thing. I see the point, saying a perfect black body would only be 8°C colder doesn’t allow the tiny CO2 increase much latitude to increase temperatures. But in reality we have a 30°C difference caused by our atmospheric blanket, therefore a 3-4°C increase is less impossible, and at the same time quite significant.

anna v
March 17, 2010 11:29 pm

Re: Willis Eschenbach (Mar 17 23:19),
They are repulsive in any case, and totally unmerited in Joel’s case. I disagree with him a lot, but he is absolutely not a troll.
Truly said. I have had interactions with Joel and I confirm this.