While we are marveling at the recent revelation out of Serbia that shows a connection between cosmic rays, clouds and temperature, our own volunteer moderator, Roger (Tallbloke) noticed and collated some comments from Bill Illis which are well worth repeating here. Thanks Rog for catching this while I was otherwise engaged. I repeat his post here, which consists of a WUWT comment, but be sure to bookmark Tallbloke’s Talkshop
Over on the Spencer Good, Bad and Ugly response to Dessler 2011 thread on WUWT, Bill Illis quietly drops this little bombshell:
While we are having no luck finding a good correlation between clouds and temperatures in a feedback sense (the scatters are providing r^2 of 0.02) which indicates there is probably NO cloud feedback either way (and the IPCC calculates that positive cloud feedback might be half of the total feedbacks so that is very clearly in question now) …
There is a very interesting relationship between the Net Cloud Radiation levels and the Total Global Net Radiation as measured by the CERES satellite (which I don’t think anyone has looked yet being busy trying to find the temperature feedbacks).
I’m getting Cloud variability being a very large part of the variability in the total Global Net Radiation Budget – anywhere from 65% to 100% (with R^2 between 0.29 and 0.77).
First the (not really convincing but better) scatter using the CERES data (that Steve McIntyre and Roy Spencer made available).
And then the (much, much better) relationship over time.
And then the versions of the data that Dessler provided (where adjustments where made according to the ERA reanalysis dataset which some think is actually a little more accurate). 100% of Net Radiation governed by Clouds with R^2 at 0.77 .
And then over time, a really tight relationship.
So, do Cloud Variations affect the Earth’s Energy Budget? – the title of Dessler’s new paper – His own data says: holy moley!
ECEGeorgia;
Thanks again for the compliment. Just for the record, I’m not a scientist, I’m just a salesman (I sh*t you not). Mind you, I’ve spent a considerable number of years selling very high tech products to solve bleeding edge problems in both industry and academic work. I picked up a lot of science from a lot of different disciplines over the years. At days end though, the science I’ve been expounding upon in this thread isn’t exactly bleeding edge. You need some understanding of calculus to understand how the formulas are derived, but applying them is just plain math coupled with some 1st and 2nd year university physics. Every number I quote, every formula, can be found on wikipedia or in the IPCC reports themselves, it just takes some wading through the minutia to get a grip on it. But that’s what makes the debate so frustrating. Try and get an alarmist into a discussion about the physics, and they (for the most part) run and hide. You’ll get responses like “the polar bears are going extinct” and “the blue toed sloth is running out of habitat” or “the glaciers are shrinking” and on and on…but the actual DIRECTLY MEASURABLE PHYSICS? They don’t want to discuss that, and it takes very little self education to figure out why.
Re your question above, yes you’ve got the right idea. Energy can get moved around the system as a whole as lot of different ways. Convection for example. Rain. Snow. Wind. Energy gets transferred by conduction (molecules actually hitting each other) and so on. But let’s just focus on photons for the moment. Photons just being uhm… light. Not particles, not waves, photons share the characteristics of both. Wavicles, each carrying a bit of energy and travelling at…the speed of light.
So, at earth surface, if you think about if for a moment, photons can only be released in one direction. Up. Every photon has (in theory anyway) a small percent chance that it will hit nothing, and just zip on out to space. In practice, it will probably hit something, and be absorbed. That something then gains the energy of the photon, and becomes warmer. The warmer something is, the more photons it emmits. But it doesn’t emmit the SAME photons, it emmits “new” photons based on the temperature. The sun being very hot, emitts short wave, high energy photons, while the earth surface being cool by comparison, emitts long wave low energy photons. So, SW goes in, LW comes out. The total amount of energy equals out over time though.
Back to your question, imagine a photon (for the sake of simplicity) having a 50% chance of getting to space, and 50% being absorbed. The ones that are absorbed, get re-emitted, but since they are now in the atmosphere, in a water droplet or a CO2 molecule, or what ever, the “new” photon gets emitted in ANY direction. again, for saked of simplicity, let’s pretend they can only go “up” or “down”. Any given photon could be aborbed instantly upon leaving earth surface, or at the very last molecule at TOA, but “on average”, half way up. So, upon being absorbed and a “new” photon being emitted, half go up, half go down. Of the half that go up, they would have a 75% chance of zipping straight out to space. 25% would be absorbed again, but their average absorption altitude would be 75% of the way to TOA. Of the ones that went down, each would have a 75% chance of being absorbed by earth surface, 25% would be absorbed by something else. Keep in mind that this all happens, literaly, at the speed of light, and the number of photons is big. How big? I don’t know. Ten to the Big.
So, per your question, insert cloud at night. Being made of water droplets, it is much more likely to block the path of a photon than plain old air. Since there is no significant downward bound photons coming from space at night, the only possible effect the cloud can have is to absorb photons from below, temporarily store them before emitting new phtoons to space, and emitting some back downward…where they contribute to the temperature of the earth surface. Not warming per se, I call it “less cooling”.
Day time rolls around, and the sun starts beating down on them thar clouds with massive amounts of shortwave. Without the clouds there, the SW would have a MUCH higher chance of getting to earth surface. With the cloud there…some gets absorbed, and re-emitted (as LW) back to space, and some gets to keep going down toward eaerth. Not cooling per se… more like “less warming”.
Hope that helps!
RE: davidmhoffer: (September 13, 2011 at 11:23 am)
“Wrong. Wrong. OH SO WRONG!
“How much IR the earth gets from the sun means diddly squat.”
Quite true, but to keep cool, the Earth must be able to radiate *all* energy it receives from the sun back out in the IR band, more particularly in that band characteristic of current ambient temperatures. If we were in close orbit around a cool star, it would get a little more complicated as we might have a significant overlap of stellar and terrestrial radiation. The hypothetical fly might have a problem telling the difference between back radiation and stellar radiation–perhaps he would not care.
On Earth, I believe the -55 deg C temperature of the tropopause indicates that the narrow absorption/emission frequency range at this altitude intercepts such a small slice of the total solar energy spectrum, that extreme cooling is allowed in the otherwise transparent atmosphere. This extreme cooling may be critical to life on Earth.
RE: Clouds
If a static cloud were being cooled by radiation from above and below, then it is reasonable to assume that this cloud would gradually lose altitude unless that heat loss were made good by radiation from below. If the cloud were a marker for a rising column of condensing air, the lost energy would be replaced by fresh convection and heat from the condensation process. Ionizing radiation may increase the likelihood of the initiation of such processes
Fred H. Haynie says:
September 11, 2011 at 12:13 pm
Well Fred, that’s an incomplete analogy. The top of the cloud acts like a diode in series with the resister.
Let’s call the ocean a battery. It charges during the day and discharges at night. A cloud acts like a resister limiting current flow at night but during the day the cloud top acts like diode blocking the charging current. The crux of the matter is which effect is greater – limiting discharge current or blocking the charging current. My bet is on the latter.
Fred H. Haynie says:
September 13, 2011 at 1:14 pm
“The “blocking actions” you suggested are high resistances: but remember, clouds do not blanket the earth and when there are no clouds, there is very little resistance. Think about parallel resistances. Click on my name.”
Clouds cover is about 50-60% of the earth’s surface on any given day. Think about a diode in series with each of the resisters.
Dave Springer says:
September 14, 2011 at 4:29 am
I agree. I think that such a model would be useful in quantifying the energy flow rates both ways.
Data from the Arctic where days and nights are six months long and clouds are fewer and less dense than at the equator can reveal the relative resistance.
Just for reference, MODTRAN, a program generated by the Air Force, intended to predict IR levels that might be observed by aircraft flying at various levels in the atmosphere, (as measured in energy flow, W/m2 over the wavenumber range of 100 to 1500 cycles per centimeter) seems to be indicating that most of the clear-air outgoing IR energy originates in the atmosphere. This is based on the assumption that the net outgoing radiation is radiation seen coming up from below minus energy seen coming down from above.
This data was obtained using the web tool provided by Dr. Archer at the University of Chicago. It somewhat complicates the determination of the effect of clouds as they can only block or ‘resist’ the escape of energy emitted by the Earth from below.
Spector;
You are putting your trust in a calculator without understanding what it is calculating.
You are of course correct that as the altitude of the cloud increases, the less resistance is presents to upward bound LW. I said as much in my explanations above. Now, here’s some problems to think about:
CO2 is at 398.ot the 375 you used. You also used Clear Air Tropical which has massive amounts of water vapour, a completely diffwerent story over land at high altitude in the winter (as one example). But you are also ignoring time of day. The resistance the cloud represents to incoming solar is at a MINIMUM at high noon and at maximums at dawn and sunset. The sharper angle of the sun means more solar gets reflected instead of absorbed and more gets absorbed ALSO, the cloud casts a longer shadow so to speak.
So yes, any given cloud, at any given altitude, at any given TIME, might be positive as a feedback…or negative. Sorry, but running some numbers through MODTRAN only gives you what a pilot needs to know about his plane, and that is it.
RE: davidmhoffer: (September 15, 2011 at 9:00 am)
“Spector;
“You are putting your trust in a calculator without understanding what it is calculating.”
If someone has a better description for what this is indicating, I would be glad to hear it. The 375 PPM CO2 setting is the normal MODTRAN web tool default. I limited my changes to sensor altitude and viewing directions. This data was only intended to illustrate an apparent principle of the atmosphere.
The results, as returned, seem to be indicating that water vapor is a leaky greenhouse gas. While it does trap heat very effectively near the surface, it would also seem that water vapor does allow an important fraction of that trapped heat to be radiated out at all altitudes where water vapor is a significant component of the atmosphere. I am restricting my comments to water vapor because MODTRAN shows a deep hole in the outgoing radiation spectrum around the CO2 absorption wavenumber of 667 cycles/cm.
If there were a perfect mirror just over the surface, I assume that one would also see 417 W/m2 looking up as well as looking down and thus there would be no heat gain or loss. As there are no clouds in this model, I assume the reading of 348 W/m2 must come from atmospheric IR radiation emitted above. Thus it seems that net *radiative* ground level cooling can only be about 69 W/m2. But the program also says that looking down at 70 km we should expect to see about 288 W/m2 actually escaping to outer space. If most of this radiation does not come from the surface, it must come from the atmosphere. Looking level by level seems to confirm that this is the case–at least in the MODTRAN universe.
As the upper air cannot be a continuous source of heat, the outgoing radiation would cause continual cooling, eventually forcing dry air parcels to the ground where they could cool the surface by absorbing moisture and heat and perhaps once again rise to form clouds.
For this type of static calculation, MODTRAN should be assuming a typical atmosphere as established by normal convective activity.