Which way to the feedback?

Guest Post by Willis Eschenbach

There is an interesting new study by Lauer et al. entitled “The Impact of Global Warming on Marine Boundary Layer Clouds over the Eastern Pacific—A Regional Model Study” [hereinafter Lauer10]. Anthony Watts has discussed some early issues with the paper here. The Lauer10 study has been controversial because it found that some marine stratocumulus clouds decrease with increasing warming. This is seen as an indication that (other things being equal) clouds are a net positive feedback, that they will amplify any warming and make it even warmer. This finding has engendered much discussion.

I want to do a different analysis. I want to provide a theoretical understanding of the Lauer10 findings. Figure 1 shows the larger picture, within which Lauer’s results make sense. This is the picture of part of the Earth as a solar-driven heat engine.

Figure 1. Very simplified picture of the main driving loop of the tropospheric circulation. A large counter-rotating cell (a “Hadley Cell”) of air exists on each side of the equator. Energy enters the system mostly around the equator. Thunderstorms (shown with rain) drive deep convection currents from the surface to the upper troposphere. Some of the energy is transferred horizontally by the Hadley Cells to the area at 30N/S. There, some the energy is radiated out to space. A large amount of the radiation occurs in the clear dry desert regions. Other parts of the atmospheric circulation not shown.

Lauer10 is discussing the low cloud decks found off the western edges of the continents at around 30°N/S, as illustrated in Fig. 1.

Considering the earth’s climate as a heat engine can lead us to interesting insights. First, we can see how the heat engine works. The thunderstorms in the wet tropics convert some of the incoming solar energy to work. The work consists in part of moving huge amounts of warm air vertically. In the process, most of the moisture is stripped out of the air, producing the rain shown in Fig. 1. After rising, some of this now-drier air travels polewards. It descends (subsides) in the region around 30° north and south of the Equator. This dry descending air forms the great desert belts of the planet. The air then returns equator-wards to repeat the cycle.

A closed system heat engine (like the climate) needs some form of radiator to cool the working fluid before it returns to be recycled through the engine. In the climate, the areas around 30°N/S serve as the main radiators for this loop of the atmospheric circulation. There, excess energy is radiated to space.

Now, here’s the theoretical question:

What would we expect to happen to this flow system if there is an increase in the temperature?

The Constructal Law says that in such a case, a flow system like the climate will rearrange itself to “speed up the wheel”. That is to say, it will change to increase the throughput of the system. The system reorganizes itself to increase the total of work plus turbulence.

How can the circulation shown in Fig. 1 become more efficient and increase its throughput? There are not a whole lot of control points in the system. The main control points are the clouds at both the hot and the cool ends of the heat engine.

The Constructal Law suggests that as the system warms, two things would happen. First, there would be an increase of cumulonimbus (thunderstorm) clouds at the equatorial end of the system. This would increase the speed and volume of the Hadley circulation. Next, there would be a decrease of clouds in the area around 30° latitude. This would increase the amount of radiation leaving the system. These changes would combine to increase the total throughput of the system.

In that light, let us re-consider the results of Lauer10. What they show is that as more heat passes through the system, as expected, the clouds at the radiator end of the system decrease. This increases the amount of energy that can pass through the system in a given time. In other words, they are an expected result of the system warming.

Lauer10 appears to discount this possibility when they say:

The radiative effect of low marine clouds is dominated by their contribution to the planetary albedo as their impact on outgoing longwave radiation is limited because of the small temperature difference between cloud tops and the underlying surface.

I found this doubtful for a number of reasons. First, the cloud top for marine stratiform clouds is typically at an altitude of ~600-700 metres, and the cloud bottom is at around 400-500 metres. The dry adiabatic lapse rate (cooling with increasing altitude in dry air) is about 1°C per hundred metres. This puts the cloud base at around five degrees C cooler than the surface. Then we have 200 metres at the wet adiabatic lapse rate, that’s about another degree. Total of six degrees cooler at the cloud tops.

The annual average surface temperature at 30°N is about 20°C, which puts the cloud tops at about 14°C. While this doesn’t seem like a lot, it gives a blackbody radiation difference of about 30 W/m2 … hardly a “limited” difference. Even if it is “only” half of that, 15 W/m2, that is the equivalent of four doublings of CO2.

Next, the strength of the solar contribution at 30° latitude is only about 60% of equatorial sunshine. This is due to the greater angle to the sun, plus the greater distance through the atmosphere, plus the inherent increase in albedo with decreasing solar angle.

Next, there is a fundamental difference between equatorial clouds (cumulus and cumulonimbus) and the stratocumulus decks of the area at 30° latitude. This difference is ignored by the averaging, with which climate science is unfortunately rife.

The problem is that the timing of clouds is often more important than the amount. Consider someplace in the tropics that has say eight hours of clouds per day. If those clouds are in the afternoon, the reflection of the sunlight will dominate the effect of the clouds on radiation. The clouds will cool the afternoon, as we all know from our common experience.

If that same eight hours of clouds occurs at night, however, the situation is reversed. Clouds are basically an impervious black body to outgoing longwave radiation. Because of this, they increase the downwelling LW when they are overhead. During the day this is usually more than offset by the reduction in solar radiation.

But at night there is no sun, so the effect of night-time clouds is almost always a warming. Again this is our common experience, as clear winter nights are almost always colder than winter nights with clouds.

However, all of this is obscured by the averaging. In both the day and night cases above, we have the exact same amount of clouds, eight hours per day. At night the cloud warms the earth, during the day the same cloud cools the earth, and averages can’t tell the difference.

The relevant difference between stratocumulus at 30° latitude and the equatorial clouds is that the equatorial clouds die out and vanish at night. This allows for free radiation from the surface. The stratocumulus deck, on the other hand, persists day and night. This means that it has much more effect on radiation than equatorial cloud.

Finally, I think that there is a fundamental misunderstanding in their claim that the maritime stratocumulus cloud “impact on outgoing longwave radiation is limited” because of the small temperature difference.

It is true that between the upwelling longwave from the surface and from the low clouds is about 10% (30W/m). The temperatures are not hugely dissimilar. But the internal energy flows are very different under the two conditions (clear and cloudy).

Consider a night-time hour with cloud. The cloud is radiating through clear dry air above to space at something like 370 W/m2. In addition, the cloud is radiating roughly the same amount back to the surface, something like 370 W/m2. Meanwhile, the ocean surface is radiating (losing) around 400 W/m2.

So the ocean loses 400 and gains 370 W/m2, so it is losing 30 W/m2 in this part of the transaction.

Now take away the cloud for an hour. The surface is still radiating something like 400 W/m2, this time out to space. So the authors of Lauer10 are correct, there’s not much change in outgoing LW, “only” 15 to 30 W/m2. But what they are neglecting is that the ocean is no longer receiving 370 W/m2 of LW from the cloud. Instead, above the ocean is mostly dry air, which provides little downwelling radiation to the surface. In this case the surface itself is losing about 400 W/m2.

So despite having identical energy flows to space, these two conditions have two very different net internal energy flows. When the sky is clear, the ocean is losing energy rapidly. When it is overcast with marine stratocumulus, the ocean loses energy much more slowly. The difference in ocean loss is 370 W/m2, which is a large difference. That is why I don’t agree that the clouds make little difference to the radiation balance. They make a big difference to net energy flows (into and out) of the ocean.

And why are oceanic net energy flows important to the outgoing radiation? It is the long-term balance of these flows across the ocean surface that determines the oceanic (and therefore the atmospheric) temperature. As a result, small sustained imbalances can cause gradual temperature shifts of the entire system.

I think I notice the problem because of my training as an accountant. A small difference in the amount of payments can mask a huge difference in the source of those funds. And a small amount of income or expense adds up over time.

My conclusions?

1. I think it quite possible that Lauer’s findings are correct, that increased warming in the area of the persistent marine stratiform layers at 30°N/S leads to decreased clouds in those areas.

2. I think that Lauer’s finding are an expected effect when we consider the Earth as a heat engine operating under the Constructal Law. With increasing heat, the Constructal Law says the system will adapt by increasing throughput. Reduced cloudiness at the cold end of the heat engine is an expected change in this regard, just as we expect (and find) increased cloudiness at the hot end of the heat engine with increasing heat.

3. Of course, for this study to truly be science I need to insert the obligatory boilerplate. So let me note that mine is a preliminary study, that “further investigation is warranted”, that I could use a big stack of funds to do just that, that I will require a personal assistant to undertake the onerous task of archiving a few datasets per year, and that Exxon has been most dilatory in their payment schedule …

FURTHER INFORMATION

Constructal Theory Web Portal

Constructal Law and Climate (Adrian Bejan, PDF)

The constructal law of design and evolution in nature (Adrian Bejan, PDF)

A previous post of mine on Constructal Law and Flow Systems

The constructal law and the thermodynamics of flow systems with configuration (Adrian Bejan, PDF)

Addendum before posting. After writing the above, I noted today a new paper published in Science (behind a paywall) entitled Dynamical Response of the Tropical Pacific Ocean to Solar Forcing During the Early Holocene, Thomas M. Marchitto et al. It is discussing one of the geographical areas that Lauer10 analyzed, the eastern Pacific off of Mexico. The abstract says:

We present a high-resolution magnesium/calcium proxy record of Holocene sea surface temperature (SST) from off the west coast of Baja California Sur, Mexico, a region where interannual SST variability is dominated today by the influence of the El Niño–Southern Oscillation (ENSO). Temperatures were lowest during the early to middle Holocene, consistent with documented eastern equatorial Pacific cooling and numerical model simulations of orbital forcing into a La Niña–like state at that time. The early Holocene SSTs were also characterized by millennial-scale fluctuations that correlate with cosmogenic nuclide proxies of solar variability, with inferred solar minima corresponding to El Niño–like (warm) conditions, in apparent agreement with the theoretical “ocean dynamical thermostat” response of ENSO to exogenous radiative forcing.

In short, their study reports that when the ocean gets warmer at the equator, it gets cooler at 30°N, and vice versa. They also find that this effect is visible on annual through millennial timescales. Unsurprisingly, this is not found in the GCMs.

Intrigued by the idea of a “ocean dynamical thermostat”, I read on:

Values in the middle of this range are sufficient to force the intermediate- complexity Zebiak-Cane model of El Niño–Southern Oscillation (ENSO) dynamics into a more El Niño–like state during the Little Ice Age (A.D. ~1400 to 1850) (3), a response dubbed the “ocean dynamical thermostat” because negative (or positive) radiative forcing results in dynamical ocean warming (or cooling, respectively) of the eastern tropical Pacific (ETP) (4). This model prediction is supported by paleoclimatic proxy reconstructions over the past millennium (3, 5, 6). In contrast, fully coupled general circulation models (GCMs) lack a robust thermostat response because of an opposing tendency for the atmospheric circulation itself to strengthen under reduced radiative forcing (7).

Now, consider this finding in light of Figure 1. Yes, it is a simple “thermostat” in the sense that as the equator heats up, the area around 30°N/S cools.

But in the light of the climate heat engine it is much more than that. The Constructal Law says in response to increased forcing the climate system will respond by increasing throughput. One way to increase the throughput of a closed cycle heat engine is to cool the radiator.

And that is exactly what their “ocean dynamical thermostat” is doing. By cooling the radiator of the climate heat engine, the engine runs faster, and moves more heat from the tropics. Conversely, when the earth is cooler than usual, the engine runs slower, and less heat is transported from the tropics. This warms the tropics.

I started this by saying that I would provide a theoretical framework within which the Lauer10 findings would make sense. I believe I have done so. My theoretical results were strengthened by my subsequent finding that Marchitto et al. fits the same framework. However, this is only my understanding. Additions, subtractions, questions, falsifications, confusions, expansions, and just about anything but conflagrations gratefully accepted.

Finally, testable predictions lie at the heart of science, and they are scarce in climate science. If I am correct, the kind of study done by Lauer et al. of the persistent stratocumulus decks in e.g. the Eastern Pacific should reveal that in the observations, changes in night-time cloud cover are greater than changes in day-time cloud cover. My check from the Koch brothers must have gotten lost in the mail, so I don’t have the resources for such a study, but that is a testable prediction. It would certainly be a good and very easy direction for Lauer et al. to investigate, they have the records in hand. Here’s their chance to prove me wrong …

My regards to all,

w.

References and Notes for the above quotations from Marchitto et al.

3. M. E. Mann, M. A. Cane, S. E. Zebiak, A. Clement, J. Clim. 18, 447 (2005).

4. A. C. Clement, R. Seager, M. A. Cane, S. E. Zebiak, J. Clim. 9, 2190 (1996).

5. K. M. Cobb, C. D. Charles, H. Cheng, R. L. Edwards, Nature 424, 271 (2003).

6. M. E. Mann et al., Science 326, 1256 (2009).

7. G. A. Vecchi, A. Clement, B. J. Soden, Eos 89, 81 (2008).

PS – Both papers, one discussing the atmosphere and the other the ocean, explicitly note that this thermostatic effect is not correctly simulated by the climate models (GCMs). The Marchitto paper is very clear about exactly why. It is because of one of the most glaring and under-reported shortcomings of the models. Here’s Marchitto again, in case you didn’t catch it the first time through (emphasis mine):

In contrast, fully coupled general circulation models (GCMs) lack a robust thermostat response because of an opposing tendency for the atmospheric circulation itself to strengthen under reduced radiative forcing (7).

Say what? Model circulation strengthens under reduced forcing?

In a natural heat engine, when you add more heat, the heat engine speeds up. We can see this daily in the tropics. As the radiative forcing increases, more and more thunderstorms form, and the atmospheric circulation speeds up. It’s basic meteorology.

In the models, amazingly, as the radiative forcing increases, the atmospheric circulation actually slows down. I might have missed it, but I’ve never seen a modeller address this issue, and I’ve been looking for an explanation since the EOS paper came out. Although to be fair the modellers might have overlooked the problem, it’s far from the only elephant in the model room. But dang, it’s a big one, even among elephants.

So yeah, I can see why the models are missing the proper thermostatic feedback. If your model is so bad that modelled atmospheric circulation slows down when the forcing increases, anything’s possible.

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Jim D
December 12, 2010 8:57 am

Willis asked for the matchbook version.
Basically it stems from how you dismissed the solar effect of low clouds, but there are two parts.
1. Removing low clouds means there may be 300 W/m2 of solar radiation, daily-averaged, going into the ocean instead of being reflected from the tops of these clouds, which is net warming in the earth system.
2. When there are no clouds the downwelling longwave at the surface doesn’t go to zero (from 370 W/m2 with clouds), it goes to about 250-300 W/m2 from clear sky, even if it is quite dry. The loss of 100 W/m2 is more than balanced by the 300 W/m2 gain from solar.

Brian H
December 12, 2010 9:28 am

This has become an incredibly important thread.
OHD’s observation is shattering. It breaks every GCM into smithereens, IMO.

MaxL
December 12, 2010 10:58 am

Willis Eschenbach says:
December 12, 2010 at 12:33 am
“There’s a very interesting journal article here that discusses the formation of stratus clouds. On p. 2153 they talk about what factors determine which kind of clouds form.
They find that if the sea is cooler than the air, the solid horizon to horizon bank of stratus forms. But if the sea is warmer than the air, individual cumulus form, with areas of clear descending air between them.”
You have to be careful which type of stratus clouds you are talking about as you have found out. Plain stratus generally forms under an inversion in a stable lapse rate in the low levels. The tops of the clouds are about the same temperature as below or even a bit warmer. These clouds are often difficult to pick out on infrared satellite photos because they are close to the temperature of the land/sea. This becomes a big problem in Canada when it is often hard to determine which areas are cloudy. When the land becomes colder is get easier because the stratus clouds tops are then actually warmer than the land. Forecasting stratus clouds is always a big headache because you are never sure how long an inversion will persist and when the low level moisture may finally get used up. It plays havoc with airlines and their landing limits.
Stratocumulus clouds form due to instability in the lower levels, as you noted happens when the water is warmer than the air. This is what produces lake effect snow squalls off the Great Lakes at this time of year, as the very cold Arctic air moves over the relatively very warm open water. This has been in great effect the past week or so as anyone living around the London Ontario region can tell you. They have been getting buried in snow squalls off the lake.

Joel Shore
December 12, 2010 2:00 pm

O H Dahlsveen says:

The Watts per square meter (W/m²) of incoming solar irradiation is a given value averaging1368W/m² pr. year. It has nothing to do with the size of the Earth because as for irradiation pr m² there can be no, or very little, difference between the Earth and its moon (because as averaged over one year, the difference in the two heavenly bodies’ distance from the Sun is close to Zero)

The solar constant tells you how much radiation each m^2 on an imaginary spherical surface centered around the sun and having a radius of the sun-earth distance. To get the total amount of radiation in Watts that hits the earth, you take this and multiply it by area of a disc having the radius of the earth (pi*r^2) because that is how much radiation from the sun the earth will intercept. However, to figure out how much radiation that corresponds to for each m^2 of the earth’s surface, we have to account for the fact that the earth’s surface area is 4*pi*r^2.
The end result is that there is a factor of 4 conversion between the solar constant and the average intensity of the radiation per unit area of the earth’s surface. (The reason that it is not simply a factor of 2 is because even most of the radiation that hits the half of the earth that is in daylight hits it at an oblique angle.)
Brian H:

This has become an incredibly important thread.
OHD’s observation is shattering. It breaks every GCM into smithereens, IMO.

I really hope that you are kidding and are not actually that easily led astray!

Editor
December 12, 2010 2:21 pm

Willis – you say “I have most assuredly considered the idea that clouds drive the warming, and I find it lacking a cause. If the clouds are driving the warming, what is driving the clouds?
Henrik Svensmark’s well-known theory, supported by actual observation and experiment, says that cosmic rays (GCRs) do in fact play a role in the formation of clouds, both over long periods and in very short periods (eg. Forbush decreases). But maybe that is not the only driver of clouds. Roy Spencer says (http://www.drroyspencer.com/research-articles/global-warming-as-a-natural-response/) “What has caused the warming seen over the last 100 years or so?
Here I present new evidence that most of the warming could be the result of a natural cycle in cloud cover forced by a well-known mode of natural climate variability: the Pacific Decadal Oscillation (PDO). […] Since this timing between the phase of the PDO and periods of warming and associated climate change seems like more than mere coincidence, I asked the rather obvious question: What if this known mode of natural climate variability (the PDO) caused a small fluctuation in global-average cloud cover?
Such a cloud change would cause the climate system to go through natural fluctuations in average temperature for extended periods of time. The IPCC simply assumes that this kind of natural cloud variability does not exist […] This is an assumption that many of us meteorologists find simplistic and dubious, at best. Spencer and Braswell (2008) showed theoretically that daily random variations in cloudiness can actually cause substantial decadal time-scale variability on ocean temperatures.
” [and more…].
Unfortunately, when Roy Spencer tried to get a paper published in GRL along these lines, it was violently rejected.
I would add that the albedo measures from the Earthshine project tally very nicely with observed global temperature changes. ie, further evidence that clouds do drive temperature. Graph is here: http://members.westnet.com.au/jonas1/PalleInterAnnualAlbedo.JPG – my link to the paper is broken but I think the paper was Palle, E., P. R. Goode, and P. Montanes-Rodriguez (2009), Inter-annual trends in Earth’s reflectance 1999-2007, J. Geophys. Res., doi:10.1029/2008JD010734. [NB. albedo relates to rate of change of temperature (1st derivative)].
Coming back to your post. Your post states clearly “What would we expect to happen to this flow system if there is an increase in the temperature?“. As I read it, you are looking exclusively at the reaction of clouds to temperature – in other words you are interpreting the observed behaviour of clouds only as a reaction to temperature – the exact same mistake that is made by Lauer and the IPCC. Now there may indeed be a reaction of clouds to temperature along the lines that you suggest, but I would argue that you are putting the cart before the horse : surely by far the more important effect is the reaction of temperature to clouds.

E.M.Smith
Editor
December 12, 2010 3:11 pm

I find it fascinating that the majority of the “heating” found by GISS and the Warmers is in the polar part of your diagram where “nothing happens” from a heat engine point of view…
I’m going to have to ponder this posting a bit more. Thanks for that!

E.M.Smith
Editor
December 12, 2010 3:57 pm

Ok, Willis, I’ve thought about this for a while. There is only one clear response I can make. So I’ve got to set this on your table. Here’s my reply:
http://chiefio.files.wordpress.com/2010/12/dscf0540.jpg

Stephen Wilde
December 12, 2010 4:07 pm

Mike Jonas says:
December 12, 2010 at 2:21 pm
Mike,
My preferred mechanism for those cloudiness and albedo changes is a shift in the latitudinal position of the jetstreams as I have explained elsewhere.
The change in trend towards increases for both cloudiness and albedo as noted by the Earthshine project was in the late 90s and by 2000 I first noted that the jets were no longer so persistently poleward as they had become by the mid 90s.
That change in jetstream behaviour has greatly intensified since then and from 2007 to date has become obvious to all but some are still in denial and others lamely try to say it is down to AGW despite the correlation with the quiet sun and the failure of the present situation to have developed steadily during the entire course of the late 20th century warming spell from about 1975 onwards.
I prefer my explanation to the Svensmark hypothesis because his proposition involves cloudiness and albedo changes without any need for changed behaviour and/or positioning on the part of the jets.
Given that the jets do shift in correlation with the cloudiness and albedo changes we do not need the Svensmark idea. Unless of course Svensmark also claims that more cosmic rays can shift the jets but I don’t think he has ever suggested that.
Instead it is more likely to be a matter of a solar effect on the size and intensity of the AO and AAO operating via ozone chemistry changes within the polar vortices.

December 12, 2010 4:32 pm

Joel Shore: December 12, 2010 at 2:00 pm
Joel Shore you tell me how to work out the total average solar irradiation the Earth receives from the Sun in much the same way as Richard Sharpe did in a post on December 11, 2010 at 6:07 pm which was in response to my original post on December 11, 2010 at 3:59 pm. A posting which you seem not to have read.
To be absolutely correct you say;
“The solar constant tells you how much radiation each m^2 on an imaginary spherical surface centered around the sun and having a radius of the sun-earth distance. To get the total amount of radiation in Watts that hits the earth, you take this and multiply it by area of a disc having the radius of the earth (pi*r^2) because that is how much radiation from the sun the earth will intercept. However, to figure out how much radiation that corresponds to for each m^2 of the earth’s surface, we have to account for the fact that the earth’s surface area is 4*pi*r^2.”
I do NOT want to know how to “work out” the total amount of radiation in Watts that hits the earth.
I know that already!
Nor do I wish to know whatever an imaginary spherical surface centered around the sun and having a radius of the sun-earth distance may be – or what it may have to do with what I wrote.
My “Thread” was briefly:
1.) According to NASA: “The solar constant is the amount of energy received at the top of the Earth’s atmosphere on a surface oriented perpendicular to the Sun’s rays (at the mean distance of the Earth from the Sun). The generally accepted solar constant of 1368 W/m2 is a satellite measured yearly average.
2.) How come the “Incoming Solar Radiation” is 341.3 W/m² as in a “Global Energy Plan W/m²” posted earlier? (By the way all other plans of that ilk that I have ever seen show very similar values.)
3.) As the incoming “solar constant” is accepted to be 1368 W/m² it should only be necessary to divide that figure by 2 (to account for day and night) and arrive at a figure of 684 W/m².
Which may only mean that all these plans can be binned.

December 12, 2010 4:50 pm

PS
Brian H seems to have grasped the thread ok.

Jim D
December 12, 2010 5:17 pm

OHD, your factor of 0.5 does not account for the reduction in solar radiation when it hits the surface obliquely. If the sun was overhead everywhere for half the day, and nowhere the other half, 0.5 would apply. Clearly this is not a good approximation to the oblique angles over most of the earth and most of the day, and that is where the other factor of 0.5 comes from to make it 0.25.

Editor
December 12, 2010 5:33 pm

Stephen Wilde – do you have a link where I can see more about the ‘shift in the latitudinal position of the jetstreams’?
I don’t see why all the possible influences have to be mutually exclusive. Isn’t it possible/probable that two or more of them genuinely exist and have an influence? Of course, there might also be interaction between them.

Joel Shore
December 12, 2010 5:47 pm

O H Daahlsveen:

I do NOT want to know how to “work out” the total amount of radiation in Watts that hits the earth.
I know that already!

Apparently not.
The quote that you give from NASA contains the answer to your query. Let me point out the important part to you that you have clearly failed to grasp the meaning of – “oriented perpendicular to the Sun’s rays”. Hint: The whole daylight side of the earth is not oriented perpendicular to the sun’s rays.
And, to quote what I told you above, which part of this do you not understand?

The end result is that there is a factor of 4 conversion between the solar constant and the average intensity of the radiation per unit area of the earth’s surface. (The reason that it is not simply a factor of 2 is because even most of the radiation that hits the half of the earth that is in daylight hits it at an oblique angle.)

Stephen Wilde
December 13, 2010 5:14 am

“Mike Jonas says:
December 12, 2010 at 5:33 pm
Stephen Wilde – do you have a link where I can see more about the ‘shift in the latitudinal position of the jetstreams’?
I don’t see why all the possible influences have to be mutually exclusive. Isn’t it possible/probable that two or more of them genuinely exist and have an influence? Of course, there might also be interaction between them.”
This is the most specific item I have found relating to the pre 2000 poleward drift:
http://www.msnbc.msn.com/id/24228037/
I have not yet found a paper that admits that the trend has now reversed but there is lots of comment on the issue and of course we can see for ourselves on a day by day basic that the jets are now looping much more equatorward than they did during the late 20th century.
As for Svensmark’s idea I don’t dismiss it completely. I just think it is more likely to be a lower order modulating efect rather than a climate driver.
Since the jets appear to shift in response to solar forcing the separate cosmic ray effect could be just a coincidental consequence of the same solar changes.

Editor
December 13, 2010 11:41 am

Stephen Wilde – interesting. Since they don’t know the cause, “look south of where you are and that’s probably a good guess of what your weather may be like in a few decades” should have been “look south of where you are and that’s probably a good guess of what your weather may be like if the jetstream continues to shift north“. Quite a difference.
And, as always : “proving it is a rigorous process, using complex computer models“.
I love the question “what are oak trees going to do?“. The answer is so simple (thx to Bob Carter & polar bears) : They can’t do anything, because if you look at climate history – the onset of ice ages for example – they have already gone extinct several times.
Your comment “the jets appear to shift in response to solar forcing” leaves us with not very many primary drivers of climate. It seems likely that virtually all of the climate mechanisms are non-linear / chaotic to some extent, so sorting them all out will be a challenge. Yes, computer models will be needed, but will they be the primary driver of the solutions?

December 13, 2010 12:21 pm

Jim D and Joel Shore – I know what you are on about. However think outside “The Accepted Box” for a minute; if, in your mind, you replace the Earth with a giant ring (hula hoop) with its opening facing the Sun, then how many W/m² would pass through the opening? Would it be1368 or would it be 342 W/m²? If the answer is 342 then I give in. – However if the answer is 1368 then my original question:”—- how come the “Incoming Solar Radiation” is 341.3 W/m² as in a “Global Energy Plan W/m²” posted earlier?” is a valid one which has not yet been answered. You two have been at pains to tell me about “An accepted way” of working out the total solar irradiance of the planet. You have also told me about the approximation to the oblique angles over most of the earth and most of the day. Joel – you even say: “The reason that it is not simply a factor of 2 is because even most of the radiation that hits the half of the earth that is in daylight hits it at an oblique angle.”
It is quite correct that the oblique angle has a big influence. I will never dispute that, but it does not influence the incoming irradiation pr. m². I still believe that what I learnt in secondary (middle) school more than 55 years ago still is correct. -That went something like this: “Sunlight can be refracted away from the surface (sent back to space) by the atmosphere. It can also be reflected back to space(before it is absorbed) by the surface. The whole surface of the Earth, for ex. mountain sides, hillocks and rocks on dry land and even the crests of waves on the oceans as well as the Earth’s curvature face the incoming sunlight at an oblique angle. How oblique that angle is varies and is impossible to work out ”(The angle of the Earth’s curvature varies as well. Hence I Suppose scientists have found it acceptable to divide the incoming radiation by 4 instead of 2.
But that only proves that the figures on the various “plans global energy flows W/m² are invalid. That includes the one on feed back

Joel Shore
December 13, 2010 3:14 pm

O H Dahlsveen says:

I know what you are on about. However think outside “The Accepted Box” for a minute; if, in your mind, you replace the Earth with a giant ring (hula hoop) with its opening facing the Sun, then how many W/m² would pass through the opening? Would it be1368 or would it be 342 W/m²

That question doesn’t make sense. W/m^2 is an intensity…It is not an amount of stuff passing through. What you want to ask is how much power (in watts) passes through the ring. The answer to that question is (1368 W/m^2)*pi*r^2 where r^2 is the radius of the earth [which works out to about 1.748 *10^17 W].
Then the next question you want to ask is that if you consider that power spread out over the earth’s surface, what is the (average) intensity on that surface. Since the earth has a surface area of 4*pi*r^2, it is 1.748 *10^17 W / (4*pi*r^2), which works out to be 342 W/m^2. Of course, if you look at the calculation we have done, you can see that the pi*r^2 parts have canceled and the net effect of what you did was to divide the solar constant by 4. That 4 is simply the geometrical conversion factor between m^2 of the imaginary surface across the ring you talked about and m^2 of the earth’s surface.
This isn’t a matter of what scientists find “acceptable” to do or not. It is a matter of what is scientifically and mathematically correct.

Richard Sharpe
December 13, 2010 3:32 pm

Jim D says on December 12, 2010 at 5:17 pm

OHD, your factor of 0.5 does not account for the reduction in solar radiation when it hits the surface obliquely. If the sun was overhead everywhere for half the day, and nowhere the other half, 0.5 would apply. Clearly this is not a good approximation to the oblique angles over most of the earth and most of the day, and that is where the other factor of 0.5 comes from to make it 0.25.

Hmmm, I don’t like your explanation very much. Since we already account for albedo separately, it should not matter whether or not the solar radiation strikes the surface obliquely or not.
What is important, as Joel Shore says, and I tried somewhat clumsily to point out above, is that the incoming radiation is averaged across the whole surface of the earth, which has an area four times that of its cross-section.

Stephen Wilde
December 13, 2010 5:41 pm

Mike Jonas said:
“Your comment “the jets appear to shift in response to solar forcing” leaves us with not very many primary drivers of climate.”
Exactly. The secret is in the word ‘primary’.
The only primary driver is the sun. The oceans are a secondary driver that modulates the effects of the primary driver over time.
There is then a plethora of lower order drivers but sun and oceans remain way out ahead of all the others combined.
As often as not the lower order drivers act against one another so that the net effect of all those drivers combined is further reduced over time.
The equilibrium that the system always seeks to retain is that between sea surface temperature and surface air temperature. The mechanism is the speed or intensity of the hydrological cycle operating via the phase changes of water. The equilibrium temperature is set by the pressure and density differentials between sea and air at the point of contact.
It is the pressure and density of the atmosphere as a whole that matters more than the composition. More GHGs other than water (unless they significantly affect total atmospheric density and pressure) will have little or no effect because the water cycle just ramps up to accommodate the extra non water GHGs and thereby maintain the sea surface / surface air temperature equilibrium. The shifting of the jets is the visible sign of that process but the shift required to deal with more CO2 would be too small to measure. The shifting that we actually do see is virtually all sun and ocean induced.
That explains the findings of Ferenc Miskolczi who appears to have ascertained that the optical depth of the atmosphere has not changed despite more CO2 content.

Joel Shore
December 13, 2010 5:44 pm

Richard:
The oblique angle doesn’t have anything to do with albedo. What Jim D is saying is that if the sun doesn’t hit perpendicularly to a surface, the intensity is not the 1368 W/m^2 on that surface. In particular, the intensity on a surface will go as (1368 W/m^2) * cos(theta) where theta is the angle between the sun’s rays and the normal to the surface.
Jim D’s statement and my statement are really saying the same thing in different ways: The average value of cos(theta) over a hemisphere turns out to be 1/2, which is why the factor of 1/2 that O H Dahlsveen thought should be applied to the solar constant (because half the earth is facing the sun) becomes instead a factor of 4.
The mathematics is all nice and consistent whichever way you choose to look at it.

Jim D
December 13, 2010 5:51 pm

Richard Sharpe, the first explanation given by Joel was the traditional one, the amount intercepted depends on the cross section area, but it is averaged over the surface which is four times larger. Mine was a second way, since it was clear the first one wasn’t understood. I was trying to show why it is less than 1368 W/m2 qualitatively.
Let’s try a third way. 1368 W/m2 is intercepted over pi*r^2. This is distributed over half a sphere which is 2*pi*r^2 in surface area, so a factor of 2 comes from that. The other factor of two comes from averaging over the dark side too. So 342 W/m2 is the average instantaneous flux on the earth’s surface.

Editor
December 13, 2010 8:22 pm

Joel Shore – “The mathematics is all nice and consistent whichever way you choose to look at it.“. Well, be a little bit careful. You have looked at the Earth as a smooth sphere. But if I choose to look at it as having an undulating surface …..

December 14, 2010 4:11 pm

Ok guys let me try once more “a fourth way” –
1) “Irradiance – The amount of electromagnetic energy incident on a surface per unit time per unit area. In the past this quantity has often been referred to as “flux”.
The Book says: “When measuring solar irradiance (via satellite), scientists are measuring the amount of electromagnetic energy incident on a surface perpendicular to the incoming radiation at the top of the Earth’s atmosphere, not the output at the solar surface.”
I say: if the intensity of the radiation that hits the top of the atmosphere above the Earth’s Equator at an angle of 90° – this spot is to be taken as a data point or “Earth’s Zenith” – is measured to be 1368 Watts per square meter (m²) then that means, as far as I understand it, that at all points in space with a clear view of the sun and at that distance from the Sun (about 149600000 km,up down or sideways) the intensity of solar radiation is 1368 W/m². It has absolutely nothing to do with the curvature of the Earth! The same intensity hits at 20 000 km further away north, south, west, or east. You may work out the surface area of a sphere in order to find out the total irradiation of that sphere (in this case the Earth.) But why do you want to do that? You just get a much larger number to work with! Why not just work with the base number of radiation intensity in Watts per square meter? That is after all what you want to end up with.
And furthermore, by accepting that the measured radiation is correct we must also accept that; all incoming solar radiation has got the potential to be equal in intensity of 1368 W/m² wherever it strikes the surface.
How much radiation (intensity) is lost because the route of radiation through the atmosphere gets longer as it flattens out, and because more and more is reflected (glanced off) from the surface as the Earth’s curvature changes is a completely different matter. But to say that the known incoming radiation must be devided by 4 because the earth is spherical does NOT COMPUTE!
Why anybody should deem it appropriate to divide incoming radiation by any factor at all because of the shape of the object it hits is a mystery to me. To divide the Earth’s annual radiation by 2 to account for 6 months of full radiation and 6 months of none however seems quite acceptable. But only for this particular kind of calculation.

December 14, 2010 4:31 pm

Once again I have looked at your posts. And once again I must ask: How can anything that happens on earth i.e reflection –albedo – rising sea levels, finger in the dike or anything else you like to mention have any influence on the Sun?