Cloud Radiation Forcing in the TAO Dataset

Guest Post by Willis Eschenbach

This is the third in a series ( Part 1Part 2 ) of occasional posts regarding my somewhat peripatetic analysis of the data from the TAO moored buoys in the Western Pacific. I’m doing construction work these days, and so in between pounding nails into the frame of a building I continue to pound on the TAO dataset. I noticed that a few of the buoys collect data on both shortwave (solar) radiation and longwave (infrared or greenhouse) radiation at two-minute intervals. For a data junkie like myself, two-minute intervals is heaven. I decided to look at the data from one of those buoys, one located on the Equator. at 165° East.

Figure 1. Location of the buoy (red square) which recorded the data used in this study. Solid blue squares show which of all the buoys have the two-minute data. DATA SOURCE

It was a fascinating wander through the data, and I found that it strongly supports my contention, which is that the net effect of clouds in the tropics is one of strong cooling (negative feedback).

To start with, I looked (as always) at a number of the individual records. I began with the shortwave records. Here is a typical day’s record of the sun hitting the buoy, taken at two-minute intervals:

Figure 2. A typical day showing the effect of clouds on the incoming solar (shortwave) radiation.

In Figure 2 we can see that when clouds come over the sun, there is an immediate and large reduction in the incoming solar energy. On the other hand, Figure 3 shows that clouds have the opposite effect on the downwelling longwave radiation (DLR, also called downwelling infrared or “greenhouse” radiation). Clouds increase the DLR. Clouds are black-body absorbers for longwave radiation. After they absorb the radiation coming up from the ground, they radiate about half of it back towards the ground, while the other half is radiated upwards  The effect is very perceptible on a cold winter night. Clear nights are the coldest, the radiation from the ground is freer to escape to space. With clouds the nights are warmer, because clouds increase the DLR. Figure 3 shows a typical 24 hour record, showing periods of increased DLR when clouds pass over the buoy sensors.

Figure 3. A typical day showing the effect of clouds on the downwelling longwave radiation (DLR).

Once again we see the sudden changes in the radiation when the clouds pass overhead. In the longwave case, however, the changes are in the other direction. Clouds cause an increase in the DLR.

So, here was my plan of attack. Consider the solar (shortwave) data, a typical day of which is shown in Figure 2. I averaged the data for every 2-minute interval over the 24 hours, to give me the average changes in solar radiation on a typical day, clouds and all. This is shown in gray in Figure 4.

Then, in addition to averaging the data for each time of day, I also took the highest value for that time of day. This maximum value gives me the strength of the solar radiation when the sky is as clear as it gets. Figure 4 shows those two curves, one for the maximum solar clear-sky conditions, and the second one the all-sky values.

Figure 4. The clear-sky (blue line) and all-sky (gray line) solar radiation for all days of the record (2214 days). 

As expected, the clouds cut down the amount of solar radiation by a large amount. On a 24-hour basis, the reduction in solar radiation is about 210 watts per square metre.

However, that’s just the shortwave radiation. Figure 5 shows the comparable figures for the longwave radiation at the same scale, with the difference discussed above that the clear-sky numbers are the minimum rather than the maximum values.

Figure 5. The clear-sky (blue line) and all-sky (gray line) downwelling longwave radiation (DLR) for all days of the record.

As you can see, the longwave doesn’t vary much from clouds. Looking at Figure 3, there’s only about a 40 W/m2 difference between cloud and no cloud conditions, and we find the same in the averages, a difference of 36 W/m2 on a 24-hour basis between the clear-sky and all-sky conditions.

DISCUSSION

At this location, clouds strongly cool the surface via reflection of solar radiation (- 210 W/m2) and only weakly warm the surface through increased downwelling longwave radiation (+ 36 W/m2). The net effect of clouds on radiation at this location, therefore, is a strong cooling of – 174 W/m2.

This likely slightly overstates the radiation contribution of the clouds. This is because, although unraveling the effect on shortwave is simple, the effect on longwave is more complex. In addition to the clouds, the water vapor itself affects the downwelling longwave radiation. However, we can get an idea of the size of this effect by looking at the daily variation of longwave with and without clouds in more detail. Figure 6 shows the same data as in Figure 5, except the scale is different.

Figure 6. As in Figure 5 but with a different scale, the clear-sky (blue line) and all-sky (gray line) solar radiation for all days of the record.

Note that the minimum (clear-sky) DLR varies by about 10 W/m2 during the 24 hours of the day. Presumably, this variation is from changes in water vapor. (The data is there in the TAO dataset to confirm or falsify that presumption, another challenge for the endless list. So many musicians … so little time …). Curiously, the effect of the clouds is to reduce the underlying variations in the DLR.

This warming due to water vapor, of course, reduces the warming effect of the clouds by about half the swings, or 5 W/m2, to something on the order of 30 W/m2.

Finally, to the perplexing question of the so-called “cloud feedback”. Here’s the problem, a long-time issue of mine, the question of averages. Averages conceal as much as they reveal. For example, suppose we know that the average cloud cover for one 24 hour period was forty percent, and for the next 24 hours it was fifty percent. Since there were more clouds, would we expect less net radiation?

The difficulty is, the value and even the sign of the change in radiation is determined by the time of day when the clouds are present. At night, increasing clouds warm the planet, while during the day, increasing clouds have the opposite effect. Unfortunately, when we take a daily average of cloud cover, that information is lost. This means that averages, even daily averages, must be treated with great caution. For example, the average cloud cover could stay exactly the same, say 40%, but if the timing of the clouds shifts, the net radiation can vary greatly. How greatly? Figure 7 show the change in net radiation caused by clouds.

Figure 7. Net cloud forcing (all-sky minus clear-sky). Net night-time forcing is positive (average 36 W/m2), showing the warming effect.

In this location, the clouds are most common at the time they reduce the net radiation the most (mid-day to evening). At night, when they have a warming effect, the clouds die away. This temporal dependence is lost if we use a daily average.

So I’m not sure that some kind of 24-hour average feedback value is going to tell us a lot. I need to think about this question some more. I’ll likely look next at splitting the dataset in two, warm dawns versus cool dawns, as I did before. This should reveal something about the cloud feedback question … although I’m not sure what.

In any case, the net cloud radiative forcing in this area is strongly negative, and we know that increasing cloud coverage and earlier time of cloud onset are functions of temperature. So my expectation is that I’ll find that the average cloud feedback (whatever that means) to be strongly negative as well … but in the meantime, my day job is calling.

A final note. This is a calculation of the variation in incoming radiation. As such, we are looking at the throttle of the huge heat engine which is the climate. This throttle controls the incoming energy that enters the system. As shown in Figure 7, in the tropics it routinely varies the incoming energy by up to half a kilowatt … but it’s just the throttle. It cools the surface by cutting down incoming fuel.

The other parts of the system are the tropical thunderstorms, which further cool the surface in a host of other ways detailed elsewhere. So the analysis above, which is strictly about radiation, actually underestimates the cooling effect of tropical clouds on surface temperature.

All the best, please don’t bother questioning my motives, I sometimes bite back when bitten, or I’ll simply ignore your post. I’m just a fool like you, trying to figure this all out. I don’t have time to respond to every question and statement. Your odds of getting a reply go way up if you are supportive, on topic, provide citations, and stick to the science. And yes, I know I don’t always practice that, I’m learning too …

w.

PS — Here’s a final bonus chart and digression. Figure 8 shows the average of the actual, observed, measured variation in total downwelling radiation of both types, solar (also called shortwave) radiation and longwave (also called infrared or “greenhouse”) radiation.

Figure 8. Changes in average total forcing (solar plus longwave) over the 24 hours of the day.

Here’s the digression. I find it useful to divide forcings into three kinds, “first order”, “second order”, and “third order”. Variations in first order forcings have an effect greater than 10% of the average forcing of the system. For the system above, this would be something with an effect greater than about seventy W/m2. Figure 7 shows that the cooling from clouds is a first order forcing during the daytime.

Variations in second order forcings have an effect between 1% and 10% of the average. For Figure 8 that would be between say seven and seventy W/m2. They are smaller, but too big to be ignored in a serious analysis. With an average value of 36 W/m2, the warming from night-time clouds is an example of a second order forcing.

Finally, variations from third order forcings are less than 1%, or less than about seven W/m2 for this system. These can often be ignored. As an example of why a third order forcing can be ignored in an overall analysis, I have overlaid the Total Radiation (red line in Figure 8) with what total radiation would look like with an additional 7 W/m2 of radiation from some hypothetical CO2 increase (black line in Figure 8). This seven watts is about 1% of the 670 W/m2 average energy flowing through the system. The lines are one pixel wide, and you can scarcely see the difference.

Which is why I say that the natural governing mechanisms that have controlled the tropical temperatures for millions of years will have no problem adjusting for a change in CO2 forcing. Compared to the temperature-controlled cloud forcing, which averages more than one hundred and fifty W/m2, the CO2 change is trivial.

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183 Responses to Cloud Radiation Forcing in the TAO Dataset

  1. Stephen Wilde says:

    “Which is why I say that the natural governing mechanisms that have controlled the tropical temperatures for millions of years will have no problem adjusting for a change in CO2 forcing. Compared to the temperature-controlled cloud forcing, which averages more than one hundred and fifty W/m2, the CO2 change is trivial.”

    Exactly and a nice post overall. Just extend the principle to the entire globe rather than limiting it to the tropics. And acknowledge that a miniscule latitudinal shift of the surface air pressure distribution would adequately deal with the thermal effect of more CO2. We would not be able to measure such a shift compared to the shifts that occur naturally from solar and oceanic variability.

    One of the fundamental issues as regards the net cloud effect is the balance between shortwave denied to the system and downward longwave enhanced by energy retained in the system for longer.. AGW theory implies that the latter exceeds the former as regards the energy budget for an overall system warming.

    The logical flaw in the AGW position is that shortwave denied to the system is lost completely whereas it needs to get into the system in the first place to provide the energy for downward longwave.

    Thus, if solar shortwave never gets into the system it cannot provide the energy that would otherwise have been available to fuel the downward longwave and warm the surface.

    If solar shortwave into the oceans declines for whatever reason then the fuel for the downward longwave declines too. The additional delay in transmission of energy to space would need to be large enough to more than offset the energy value of the shortwave denied to the system. On an intuitive basis I cannot see that to be possible and this post gives some idea as to why it cannot be possible. We would need a far denser atmosphere to make any difference and extra CO2 barely affects atmospheric density at all.

  2. dcfl51 says:

    Willis, I am not a scientist so forgive me if this is a silly question. I thought that just under 50% of the radiation emitted by the sun was long wave. So why do the instruments not detect more DLR during the day than during the night ? Is the long wave radiation from the sun all absorbed by the atmosphere before it reaches the surface of the earth ? If so, how do we know how much of the DLR is back radiation originating from the earth’s surface and how much comes straight from the sun having stopped for a coffee in the atmosphere on the way down ?

  3. Steve Keohane says:

    Thanks Willis for putting some numbers on this. I like the way you lay it out.

  4. Ken Methven says:

    Willis,
    Thank you so much for elucidating so much from a focused, objective and scrupulous treatment of the data. Even I got it!

    Well done and keep it up when the motivation takes you.

    Ken

  5. Dr T G Watkins says:

    Science at its best. Empirical data explained in a sensible, logical and ‘simple’ way.
    Thanks Willis.

  6. Derek Farmer says:

    I’ve found all of your posts on cloud feeback in the tropics both facinating and very informative. The best compliment I can give is that your ideas seem simple and obvious! (with 20-20 hindsight!!!) It does beggar the question as to why ‘professional’ climatologists (being paid large amounts of our money) don’t appear tomake similar levels of progress. These numbers you are coming out with seem very significant.

    Needless to say – none are in current GCMs.

  7. Dolphinhead says:

    Willis

    I like your work. I am not a scientist but in looking at the question of weather and climate I came to the conclusion that if I were a climate scientist and I wanted to understand the climate I would start by looking at the area where where the big players in the drama strut their stuff, or as you put it ‘…we are looking at the throttle of the huge heat engine which is the climate. This throttle controls the incoming energy that enters the system.’

    Without knowing a great deal about the nuances of the climate system it seems obvious that any modulation of the forces that drive the climate that take place within the ‘heat engine’ are liklely to have a proportionally greater impact there than they would elsewhere in the system. In my view understanding what happens in the tropics – how much heat energy enters the system through the front door and how the opening of the door is modulated by low clouds – is the necessary first step to glimpsing the wonders of what must be the finest physical system of checks and balances in the universe.

    keep up the good work

  8. gnomish says:

    wow- willils, you did it again – another brilliant piece.
    it won’t be long before you’ve got enough data and analysis for a serious publication about the ‘thunderstorm thermostat’.
    plz consider it – even if it takes until gore dies to get published.

  9. Peter Pond says:

    Thanks Willis – an easy to understand explanation.

    However, your use of clear, readable English, rather than pretentious and obfuscatory polysylabbic terminology, means that “academe” will unfortunately fail to take you seriously.

    PS “This warming due to water vapor, of course, reduces the warming effect of the clouds by about half the swings, or 5 W/m2, to something on the order of W/m2.” There appears to be a number missing at the end of this sentence (between Figs 6 & 7).

    [Thanks, fixed. w.]

  10. cba says:

    Willis,
    This is an excellent post. It has some information I’ve not found anywhere that goes right to the heart of the matter. It seems that this is a dagger in the heart for CAWG modellers who believe that clouds must be almost neutral and that combined with THE ASSUMPTION that cloud cover decreases with rising temperature creates the ‘positive’ feedback that permits their claims that there is great sensitivity to CAWG. Most of what I’ve seen out of Lindzen seems tied with the concept of cloud cover reflectivity rather than cloud cover fraction. I noticed too that your resulting reflectivity is around 61% which is definitely in the realm of realistic for clouds while the reflectivity that is associated with the modellers is half that and lower than the average of most types of clouds.

    BTW, any conversion of SW incoming to LW incoming by clouds is removing incoming power capable of penetrating beyond the skin of the ocean’s surface. That means it must either heat up the surface skin or it must evaporate more h2o and increase the water vapor cycle activity.

    I look forward to seeing more on this track, especially dessler’s backpeddling response.

  11. Septic Matthew says:

    I think that you are on to something, but have you incorporated independently acquired information on cloud cover? It looks like you infer the presence of clouds from the change in radiation, but it would be nice to related the presence of clouds to the changes in radiation.

    Your comment about the small size of the CO2 effect compared to the natural radiation is only relevant to the speed at which the steady-state induced change occurs, not whether it occurs.

    What you want to show is that transport of heat from the surface of the ocean to the upper atmosphere is supralinear in temperature, that a change from 31C to 32C increases heat transport more than a change from 30C to C1C. With radiation, that’s true because heat radiates proportionally to T^4. For net heat transport above the ocean, that may not be relevant to much.

    Overall, I think you have found a gold mine, and I look forward to reading more.

  12. if youi find the time, Willis, perhaps you might look at this post from John Eggerts
    http://johneggert.wordpress.com/2010/09/26/the-path-length-approximation/
    which might maybe give another clue to your climate thermostat idea.

    And also pay attention to what commentator Bart has been posting on climate audit about how established system control theory could inform the Spencers and Dessler’s of our world. My worry is that using these techniques on monthly averaged cludged data does not mean a lot. But it might set something going for you.

  13. JamesD says:

    Bump of dcfl51 question. How does you analysis take into account LR from the sun?

  14. Bernie McCune says:

    Willis

    An interesting value that we found when monitoring SW radiation at a couple of solar furnaces over several years in the southern deserts of NM has to do with effects of water vapor. During our monsoon season in late summer atmospheric moisture content is often “high” while in the fall moisture content is low and days are very clear and “crisp”. Our Normal Incident Pyrheliometer (NIP) on clear summer days often read 200 w/m^2 lower (850 w/m^2 at noon) than on dry fall days where readings would often peak at 1050 w/m^2 at noon. There were no clouds during the time of measurement. Water vapor has a significant affect on incoming SW radiation. In the early ’80s El Chicon volcanic gases and ash over a period of months dropped our year round noon NIP readings by up to 100 w/m^2 for about a year. I have always agreed with your ideas about H2O (clouds and vapor) being a very significant moderator of solar radiation be it in the tropics (I spent a year in the Seychelles) or here in the deserts of NM. The data just keeps on confirming it. Thanks for your efforts.

    Bernie

  15. Steve in SC says:

    Willis I agree with your contention.
    A small personal anecdote if I may, I spent a week after the 4th of July down in St. Croix of the US Virgin Islands. When I left South Carolina it was 99+ deg F with about 80+% RH. At St Croix it was about 85 and 75 decidedly more comfortable. You could tell that the sun was a bit stronger but clouds would pop up and it would rain for 20 minutes 3 times a day. Very comfy.
    You must try the rum down there it is most highly excellent!

  16. Willis Eschenbach says:

    Septic Matthew says:
    September 15, 2011 at 3:57 pm

    I think that you are on to something, but have you incorporated independently acquired information on cloud cover? It looks like you infer the presence of clouds from the change in radiation, but it would be nice to related the presence of clouds to the changes in radiation.

    Thanks, Matthew, and the other posters. The buoys do not collect any information about clouds. However, I know (from the maxima and minima) what the clear-sky values are, and from the means (averages) I know what the all-sky values are. The difference has to be the clouds …

    w.

  17. Rhoda Ramirez says:

    Willis, I have to agree with most of the commentators above, outstanding work. Your stuff is easy to understand and follow.

  18. Willis Eschenbach says:

    Dolphinhead says:
    September 15, 2011 at 3:03 pm

    … Without knowing a great deal about the nuances of the climate system it seems obvious that any modulation of the forces that drive the climate that take place within the ‘heat engine’ are liklely to have a proportionally greater impact there than they would elsewhere in the system. In my view understanding what happens in the tropics – how much heat energy enters the system through the front door and how the opening of the door is modulated by low clouds – is the necessary first step to glimpsing the wonders of what must be the finest physical system of checks and balances in the universe.

    I’m glad that someone gets the importance of the cloud throttle and the tropics. And indeed, what a magnificent system, it has kept the temperature of the Earth within a half a degree for a hundred years, and within a few degrees for thousands of years. That’s good regulation, particularly for a system governed by something as ephemeral as clouds …

    w.

  19. dlb says:

    A very interesting post Willis, you were able to do what I couldn’t, that is finding a bouy measuring DLWR. I’d like to think you are reinventing the wheel, but you never know :) anyway it is the best way to learn to figure it out for ones self. Just something else to consider, NASA’s Terra satellite flies over of a morning when there is less cloud over the land while Aqua flies in the afternoon when there is supposed to be less cloud over the ocean.

  20. I think if you did a similar analysis at the poles where water vapor and clouds are a minimum and days are a year long, you would find that the 50 year rise in atmospheric CO2 hasn’t altered the flow of energy measurably. Finding the data to analyze is the problem.

  21. Willis Eschenbach says:

    man in a barrel says:
    September 15, 2011 at 4:01 pm

    if youi find the time, Willis, perhaps you might look at this post from John Eggerts
    http://johneggert.wordpress.com/2010/09/26/the-path-length-approximation/
    which might maybe give another clue to your climate thermostat idea.

    That citation argues that the atmosphere is saturated for CO2, because you can’t get “blacker than black”. Which kinda makes sense … until you think about it.

    What it ignores is the effect of altitude. The “blackness” of any given layer varies linearly with pressure altitude. Half the pressure, half the molecules to absorb, half the “blackness”. In those upper altitude “gray” regions is where additional CO2 has an effect.

    And also pay attention to what commentator Bart has been posting on climate audit about how established system control theory could inform the Spencers and Dessler’s of our world. My worry is that using these techniques on monthly averaged cludged data does not mean a lot. But it might set something going for you.

    I’m struggling to keep up and to understand the discussion between Bart and the other gentlemen. I do think he’s on to something, and I wish I knew more about it. Always more to learn …

    Part of the problem is the issue of causality. Suppose the day warms, and at some point a thunderstorm forms. That thunderstorm wanders over the ocean, and it leaves the ocean cooler, often cooler than it was when the thunderstorm formed.

    What would you say is the cause of that cooling? The thunderstorm? The high temperature that precipitated the thunderstorm?

    This is why I’m not real comfortable with the idea of “cloud feedback”. I’m still wrestling with that concept, along with the question of whether clouds are a forcing or a feedback. I’d have to answer “both”, but I’m not sure of that …

    Brings to mind the story my mom told, of how the son of a Unitarian minister said “Dad, is there a God?” “Yes, there is”, his father said. “Are you sure?” asked the son. His father replied, “If I was sure of anything, I wouldn’t be a Unitarian …”

    So as to thunderstorms being feedback or forcing … count me as a Unitarian.

    w.

  22. Willis Eschenbach says:

    JamesD says:
    September 15, 2011 at 4:16 pm

    Bump of dcfl51 question. How does you analysis take into account LR from the sun?

    The answer is, there’s longwave, and there’s longwave. The longwave absorbed by the atmosphere might be though of as “long longwave”. It is radiated by things at a temperature from say 30°C down to -40°C or so.

    The longwave from the sun, on the other hand, is “short longwave”, also called the “near infrared” because it is near to the visual spectrum. That infrared is included in the value listed as “shortwave”. I looked in the sensor specs, but found nothing about the bandwidth. However, it says they are using a pyranometer, which is designed to be as flat as possible across the entire solar spectrum.

    In any case, by the time you get out to the frequencies of the “long longwave”, there is very little power from the sun in those “long longwave” frequencies. Which is why little of the solar radiation is absorbed by the pure atmosphere.

    w.

  23. Allan Tereba says:

    Very nice post as usual and easy to follow. I am a firm believer that any system that has lasted billions of years needs to have strong buffering mechanisms. This is one good example of a strong temperature buffering mechanism. For your analysis you used an equatorial location over the ocean, which is appropriate to examine energy balance and the effect of clouds. It would be interesting to compare these results with a location at considerably higher latitudes, especially in the winter. Short wave radiation would be considerably less due to the angle of the sun and shorter daylight period. Long wave radiation may be somewhat lower due to lower temperatures but relatively more due to the longer night. In general though long wave radiation and cloud cover may have a bigger relative warming (less cooling) effect at night than it does in the tropics due to the lower total radiation. I have not seen cloud formation patterns in these latitudes but expect them to be different than equatorial cloud formation.

  24. Bill Illis says:

    Now if you check the temperature levels (in a Stefan Boltzmann sense), you’ll find they make no sense at all compared to the net radiation levels at any given time.

    The temperature increases by a tiny fraction of a single W/m2 during the sunshine day and decreases by a tiny fraction of a single W/m2 during the no sunshine night. (Noting that a Watt equals 1 joule/second and one heck of website host as well).

  25. paulhan says:

    Excellent post, Willis and very apt in the context of Spencer – Dessler. I’m surprised that the cloud forcing (cooling) is so much – thanks for putting it in plain old watts. It beggars belief that this has not been factored in to the IPCC equations, no wonder Trenberth can’t find the heat.

    Sure, it gets more complicated as you go to higher latitudes, and when you factor in time of day and land, but with a 6:1 forcing/feedback ratio in the throttle of the climate engine (tropics), there’s no way you could come away from that analysis with anything other than the assertion that clouds are a negative feedback, and a large one at that.

    I liked the way you introduced first, second and third order effects into the mix. I’ve never had it satisfactorily explained to me how it is that a third order effect can have such an influence on a first order effect that it multiplies the original effect by three. Surely if that were the case we would have cracked free energy.

  26. Mac the Knife says:

    Willis,

    You are “the gift that keeps on giving”! Excellent analysis, presented in an easily digestible style!
    One comment and one question:

    Comment: You may have a typo at
    “Consider the solar (shortwave) data, a typical day of which is shown in Figure 1.”
    Did you mean Figure 2?

    Question: In your ‘bonus Figure 8, you applied a 7 watts/meter2 of hypothetical radiation addition from CO2, to illustrate it may be a tertiary and negligible contributor. What source (or sources) were the basis for this hypothetical 7w/m2 CO2 contribution?

    Thank You for an excellent contribution! This is really interesting and fundamental work!!!

  27. Bob in Castlemaine says:

    Great post Willis, very clearly explained.
    Your day job no doubt allows time for your unstructured creative thought processes.

  28. KingOchaos says:

    Interesting stuff Willi’s.

    Just one point in you analysis of back radiation from clouds, from the surface you will only see what is coming back through the atmospheric window directly from the clouds/in wavelengths that the atmosphere is transparent to. But it would also be blocking and re radiating in more opaque wavelengths, but these will be absorbed higher up in the atmosphere above the surface.

    So i think it would be a difficult task to infer the insulating effects of clouds purely from surface measurements… cloud altitude, humidity/ the variable path length would effect it.

    But interesting stuff, the diurnal variation in clouds/ or trend in them, is certainly an interesting area of inquiry.

  29. James Sexton says:

    Willis,
    Nice post! Your averaging problem……yes, it is very much dependent upon the time of day. But, you have a wonderful bell curve right in front of you in which you can use to solve half the problem. I think, but don’t know, that the curve would depend very much on latitude and time of year, but that, too, may be averaged. The question then would be, do some places typically have more clouds during different times of the day and when would it occur? It may be difficult to answer, but, it may not. Meteorologists often inundate us with seemingly obscure data. Perhaps they track that sort of stuff in a manner that can be interpreted. I’d ask Anthony.

    To the forcing/feedback…… whatever. It is a feedback that alters forcing and changes the dynamics of our climate, which, causes it to be a forcing. It is simply a matter of believing H2O’s natural state. Was it a cloud first or part of a body of water? I’m not big on labels. But, what if all water started as a solid? Is then the oceans a feedback? What if it was all clouds and then snowed?

    As to your response about CO2 and pressure, and just because its fun, does the ideal gas law come in to play there anywhere? ;-)

    Thanks again,

    James

  30. Mark says:

    Willis, in your reply to “Man in a barrel” about the John Eggerts post, you state:

    “That citation argues that the atmosphere is saturated for CO2, because you can’t get blacker than black”. Which kinda makes sense … until you think about it.

    “What it ignores is the effect of altitude. The “blackness” of any given layer varies linearly with pressure altitude. Half the pressure, half the molecules to absorb, half the “blackness”. In those upper altitude “gray” regions is where additional CO2 has an effect.”

    If nearly all of the LW-IR is absorbed within 100 m of the surface, how does it matter what happens at high altitude or the top of the atmosphere? I fail to see how the “re-radiation” argument works. Wouldn’t collisions with other molecules near the ground affect the re-radiation? If the difference is that increased CO2 raises the altitude where the radiation finally “escapes to space”, and therefore the temperature at high altitude increases, how does that heat get back to the surface? If re-radiation is really occurring, how come UV that is absorbed by ozone not get re-radiated and passed through the ozone layer and reach the surface? I would appreciate any direction on how this works.

    The rest of your series is wonderfully written. I can’t do the math but it looks like an interesting application of chaos theory. The concept of meta-stable strange attractors for the climate seems to match the historical record of rapid transitions between ice ages and inter-glacial temperatures with fairly stable temperatures between the transitions. Have you considered engaging a chaos theory specialist to assist? I look forward to additional posts!

  31. gnomish says:

    Bill Illis says:
    September 15, 2011 at 5:42 pm
    “Now if you check the temperature levels (in a Stefan Boltzmann sense), you’ll find they make no sense at all compared to the net radiation levels at any given time.
    The temperature increases by a tiny fraction of a single W/m2 during the sunshine day and decreases by a tiny fraction of a single W/m2 during the no sunshine night. (Noting that a Watt equals 1 joule/second and one heck of website host as well).”

    that is because temperature and heat are not the same thing – you can not convert degrees to watts.
    also, a phase change releases releases or sequesters lots of heat with no temperature change.

    Watts, of course, is inimitable. Long may he reign.

  32. Mac the Knife says:

    Willis,
    One more question:

    Are the data plots in Figures 2 and 3 derived from data recorded by the same buoy on the same day?

    Thanks Again!

  33. Willis Eschenbach says:

    Mac the Knife says:
    September 15, 2011 at 5:54 pm

    Willis,

    You are “the gift that keeps on giving”! Excellent analysis, presented in an easily digestible style!
    One comment and one question:

    Comment: You may have a typo at
    “Consider the solar (shortwave) data, a typical day of which is shown in Figure 1.”
    Did you mean Figure 2?

    Thanks, fixed.

    Question: In your ‘bonus Figure 8, you applied a 7 watts/meter2 of hypothetical radiation addition from CO2, to illustrate it may be a tertiary and negligible contributor. What source (or sources) were the basis for this hypothetical 7w/m2 CO2 contribution?

    I just picked a number. It happened to be 1% of the dataset average, and it’s larger (almost twice) the 3.7 W/m2 change in TOA forcing from a doubling of CO2, so it made my point well.

    Thank You for an excellent contribution! This is really interesting and fundamental work!!!

    You’re welcome.

    w.

  34. Willis Eschenbach says:

    Bob in Castlemaine says:
    September 15, 2011 at 6:00 pm

    Great post Willis, very clearly explained.
    Your day job no doubt allows time for your unstructured creative thought processes.

    Actually, indulging in unstructured creative thought processes when one is running a table saw is not necessarily the best plan …

    w.

  35. George E. Smith says:

    “”””” Once again we see the sudden changes in the radiation when the clouds pass overhead. In the longwave case, however, the changes are in the other direction. Clouds cause an increase in the DLR. “””””
    And by inference those same clouds would also be increasing the LWIR to space, since as you say, half of the cloud absorbed radiation is re-emitted to space, so the cloud which you say is an efficient black body absorber (and emitter) whereas the atmosphere is not.

  36. Willis Eschenbach says:

    Mark says:
    September 15, 2011 at 6:53 pm

    Willis, in your reply to “Man in a barrel” about the John Eggerts post, you state:

    “That citation argues that the atmosphere is saturated for CO2, because you can’t get blacker than black”. Which kinda makes sense … until you think about it.

    “What it ignores is the effect of altitude. The “blackness” of any given layer varies linearly with pressure altitude. Half the pressure, half the molecules to absorb, half the “blackness”. In those upper altitude “gray” regions is where additional CO2 has an effect.”

    If nearly all of the LW-IR is absorbed within 100 m of the surface, how does it matter what happens at high altitude or the top of the atmosphere? I fail to see how the “re-radiation” argument works. …

    Thanks, Mark. The issue is not where the radiation is intercepted on its way to space. It is how many times, on average, it is intercepted. This is because every time the radiation is absorbed, half of it goes back towards the surface. This is true even if 100% of the radiation is intercepted just off the deck.

    The “greenhouse effect” works because a shell has two sides. See my discussion here and here. The heating is proportional to how many times (on average) the radiation is absorbed on its way to space.

    w.

  37. Willis Eschenbach says:

    Mac the Knife says:
    September 15, 2011 at 7:16 pm

    One more question:

    Are the data plots in Figures 2 and 3 derived from data recorded by the same buoy on the same day?

    Thanks Again!

    No, I just picked a couple of different days that each illustrated the points I wanted to make.

    w.

  38. dp says:

    Willis – what would the effect of sea surface wave action be on the sensors and what would a churning sea vs a glassy sea look like in the data?

  39. Willis Eschenbach says:

    dp says:
    September 15, 2011 at 8:18 pm

    Willis – what would the effect of sea surface wave action be on the sensors and what would a churning sea vs a glassy sea look like in the data?

    Good question, dp. I don’t have a clue … but I figure the scientists who designed, built, and installed the sucker thought about it, and they chose and designed the instrument packages accordingly.

    w.

  40. T.A. says:

    Willis,

    You said, “I know (from the maxima and minima) what the clear-sky values are, and from the means (averages) I know what the all-sky values are. The difference has to be the clouds …”

    Would you kindly explain more specifically how you go from the maxima and minima to determining when there is cloud cover? Are you making assumptions, or is your conclusion based on measurement or observation of actual cloud cover?

    I am not disputing your work, just want to be clear on what the evidence is. Thanks.

  41. R. Gates says:

    Interesting as usual Willis. This comment however, may be a bit of an oversimplification:

    “After they absorb the radiation coming up from the ground, they radiate about half of it back towards the ground, while the other half is radiated upwards.”

    How did you compute this percentage? And even if correct (50% seems way too high), it doesn’t of course mean that 50% is lost to space. LW radiation, striking a greenhouse gas molecule, which of course is not static, but moving and tumbling rapidly, would tend to be re-radiated on average in a spherical pattern, such that 50% would be too high of a percentage to be transmitted directly upward, in fact, a very small percentage would actually be transmitted directly upward, as there would be lots more angles of transmission that are possible short of directly upward . And of course, even any LW radiation going directly upward could then be absorbed and re-transimtted by another greenhouse molecule, etc, such that the LW radiation that does actually go upward from the tops of clouds is not equal to the top of the atmosphere (TOA) LW actually transmitted into space, and is in fact, far less.

    The other observation I would have is that the area along the equator, while receiving the most direct insolation, is not of course indicative of the way clouds and radiation interact in other areas further from the equator. As everyone is aware, there are those large regions of descending air on the north or south sides of the Hadley cells where convection is obviously not the rule, but rather a different kind of cloud dynamics exist. See:

    http://en.wikipedia.org/wiki/File:Omega-500-july-era40-1979.png

    These regions, while not receiving as the same intensity of SW, actually cover a larger area of the earth both north and south of the equator, so some analysis of the balances of SW and LW in relation to clouds would be necessary in those larger areas. Finally of course, while not receiving but a minor portion of the total SW radiation striking earth, the role of clouds in the Arctic environment also needs to be considered, such as found in this study:

    http://www.esrl.noaa.gov/search/publications/5724/

    But your posts are always enjoyable!

  42. anna v says:

    Willis Eschenbach says:
    September 15, 2011 at 8:33 pm
    ” Willis – what would the effect of sea surface wave action be on the sensors and what would a churning sea vs a glassy sea look like in the data?”

    Good question, dp. I don’t have a clue … but I figure the scientists who designed, built, and installed the sucker thought about it, and they chose and designed the instrument packages accordingly.

    It would be interesting seeing wavy days versus quiet days. The fractal effect, waves can increase a lot the surface available for absorbing and radiating and evaporating surface.

  43. Septic Matthew says:

    Willis, you wrote: This is why I’m not real comfortable with the idea of “cloud feedback”. I’m still wrestling with that concept, along with the question of whether clouds are a forcing or a feedback. I’d have to answer “both”, but I’m not sure of that …

    To me, that is the only sensible evaluation of the evidence right now. My earlier question was sort of just wishing … .

  44. James Sexton says:

    R. Gates says:
    September 15, 2011 at 8:58 pm

    Interesting as usual Willis. This comment however, may be a bit of an oversimplification:

    “After they absorb the radiation coming up from the ground, they radiate about half of it back towards the ground, while the other half is radiated upwards.”

    How did you compute this percentage? And even if correct (50% seems way too high), it doesn’t of course mean that 50% is lost to space.
    ===================================================
    Gates, of course, I can’t speak for Willis, but yes, the 50% is too high. Given the multi-directional release, and the spherical nature of the earth, I’m guessing only 40% or so heads back towards the earth. Obviously, it depends upon the height of the molecule. As to the other part about not being lost to space……. I think it is, at least the majority of it is……… Given the multi-directional absorption and consequent emission, and the point that Willis earlier made about pressure and density, there should be less molecules to absorb going upwards, and more to absorb going down. As established earlier, then, more would be emitted upwards than downwards. So, more than 50% gets lost to space.

  45. Willis Eschenbach says:

    T.A. says:
    September 15, 2011 at 8:56 pm

    <Willis,

    You said,

    “I know (from the maxima and minima) what the clear-sky values are, and from the means (averages) I know what the all-sky values are. The difference has to be the clouds …”

    Would you kindly explain more specifically how you go from the maxima and minima to determining when there is cloud cover? Are you making assumptions, or is your conclusion based on measurement or observation of actual cloud cover?

    I am not disputing your work, just want to be clear on what the evidence is. Thanks.

    I’ll give it another shot. Let’s consider shortwave. If we take an average of all the shortwave data for each two minute interval, T.A., we are looking at what happens during “all-sky” conditions. This average includes both when the sky is clear, and when it is cloudy. It is shown as the gray line on Figure 4.

    The maximum values for each two minute time span, on the other hand, have to be for times when there are no clouds at all. If there were clouds, the total insolation would be less. So that is what happens during “clear sky” conditions. This is the blue line in Figure 4.

    The difference between the two is the amount that the clear-sky shortwave radiation is decreased by the addition of the clouds. If for some time around noon when there are no clouds the shortwave is 1400 W/m2, and on an average day at the same time the shortwave is 900 W/m2, then it is clear that the effect of the addition of clouds is a reflection of half a kilowatt per square metre back to space.

    If you still have questions, ask’em …

    w.

  46. Richard111 says:

    Willis Eschenbach says:
    September 15, 2011 at 10:59 pm
    ” If for some time around noon when there are no clouds the shortwave is 1400 W/m2, and on an average day at the same time the shortwave is 900 W/m2, then it is clear that the effect of the addition of clouds is a reflection of half a kilowatt per square metre back to space.”

    Thanks for for this. It makes the data on this site easier to understand.
    http://www.milfordweather.org.uk/solar.php
    Since it NEVER reaches 1400 W/m2 I assume is because the location is at 51N. Sun simply too low in the midday sky. We’ve been getting a lot of cloud this year it seems.

  47. AusieDan says:

    Willis – congratulations on this and your earlier posts in ths series.
    I have a question which I fear discloses my ignorance, but here goes:

    If the average 24 hour temperature does not vary very much, I would expect that the difference between SW and LW radiation would be a sine wave, more or less, over 24 hours, centered on zero.

    Can you comment or straighten me out when you have a chance?
    Regards.

  48. AusieDan says:

    Willis,
    Having now re-read your two earlier posts and thought more about the charts in this one,
    I’m feeling quite foolish.
    I see now that the incoming heat is greater than the outgoing and that the balance is transferred via wind and ocean to higher latitudes.

    Now two (probably even more stupid) questions arise:
    1. Why is just enough heat moved up latitude (north and south) to keep equitorial temperate constant?
    2. Why do the poles hold some net heat and not pass them into space, during years when the global temperature is rising (the reverse during ice ages)?

    You are saying in effect that the equator acts as a very good airconditioner but that the poles can’t control their temperatures very well.
    Have I grasped it and if so, why is it thus?
    What drives the climate?
    (Easy question).

  49. Stephen Wilde says:

    In order to deal with the ‘problem’ that global cloudiness decreased during the warming spell/period of more active sun it is necessary to recognise three things:

    i) That a reduction in cloudiness during a period of more active sun introduces a positive system feedback by allowing more solar shortwave into the oceans.

    ii) Something else has to happen to negate that positive feedback from cloud reduction.

    ii) We have observed a consequent poleward shift of the climate zones with more poleward/zonal jets. That is what alters the rate of energy transfer from surface to space and that is the negative feedback. Its power comes from the water cycle and the energy transfer capabilities of the phase changes of water.

    Shifting the air circulation systems latitudinally has two contradictory effects:

    i) Altering energy flow into the oceans by moving the cloud bands latitudinally

    ii) Altering the speed with which energy is transferred from surface to space by moving the climate zones latitudinally.

    The first is a positive feedback and the second is a negative feedback and the second always limits the first with the climate price to be paid for the necessary adjustment being a change in the relative sizes and positions of all the permanent climate zones. The sign of i) and ii) can each be reversed, thus:

    i) A reduction in cloudiness for a warming effect will be countered by a poleward shift increasing energy flow to space

    ii) An increase in cloudiness for a cooling effect will be countered by an equatorward shift decreasing energy flow to space.

    Note that it is NOT a zero sum process because the natural flow of energy from warm Earth to cold space is dominant so whereas scenario i) will limit warming of the system very effectively scenario ii) will only restrain cooling for a short while because if energy doesn’t get into the system in the first place slowing down the energy flow to space is only a short term solution.

    That is all that climate change amounts to. Anything that tries to warm or cool the system is countered by a change in the rate of energy transfer to space which alters the relative sizes and positions of the permanent climate zones. All observed climate changes fit that description.

    The underlying equilibrium is set by atmospheric pressure which in turn sets the energy cost of evaporation by fixing the amount of energy required to achieve the phase change of water from liquid to vapour.

    Much more detail here:

    http://www.irishweatheronline.com/news/environment/climate-news/wilde-weather/setting-and-maintaining-of-earth%E2%80%99s-equilibrium-temperature/18931.html

    as regards the setting of the basic equilibrium temperature and here:

    http://www.irishweatheronline.com/news/environment/climate-news/wilde-weather/feature-how-the-sun-could-control-earths-temperature/290.html

    as to how the solar changes seek to alter the equilibrium temperature (but largerly fail as per the so called ‘faint sun’ paradox).

  50. Willis Eschenbach says:

    AusieDan says:
    September 16, 2011 at 12:04 am

    Willis – congratulations on this and your earlier posts in ths series.
    I have a question which I fear discloses my ignorance, but here goes:

    If the average 24 hour temperature does not vary very much, I would expect that the difference between SW and LW radiation would be a sine wave, more or less, over 24 hours, centered on zero.

    Can you comment or straighten me out when you have a chance?
    Regards.

    You’d think it would be a sine wave but nature is rarely that simple. The day goes through various circulation regimes that mean it ends up far from a sine wave, just as the temperature is far from a sine wave. See my post “It’s Not About Feedback” for some discussion of this.

    w.

  51. Bart says:

    Willis Eschenbach says:
    September 15, 2011 at 5:12 pm

    “Part of the problem is the issue of causality. Suppose the day warms, and at some point a thunderstorm forms. That thunderstorm wanders over the ocean, and it leaves the ocean cooler, often cooler than it was when the thunderstorm formed.

    What would you say is the cause of that cooling? The thunderstorm? The high temperature that precipitated the thunderstorm?”

    The cooling from the thunderstorm is a feedback response.

  52. Viv Evans says:

    Thank you, Willis, for this excellent ‘Third Instalment’.

    Using observational data, thinking about them, and finding this leads to further questions and debates: well, that is how science used to be done.

    I am very glad to learn from you and from the debates following your essays.
    For me, it’s like being back in the research lab, brainstorming at the end of a productive day.

  53. TomVonk says:

    Re: Averaging

    Let us focus on figures 2 and 4.
    In Figure 2 you show a “typical day”. You don’t say what a “typical” day is so I will assume that you just randomly choose a day which is neither more nor less typical than any other day.
    However this figure clearly shows when there were clouds.
    Between sunrise and about 11:00 the curve follows approximately the clear sky curve with very short jumps downwards.
    The distance between the downward jumps decreases as one approaches 11:00 what means that the morning began cloudless and then more and more clouds passed over the detector.
    At 11:00 the sky is almost wholly covered and the clear sky curve joins the cloudy sky curve and stays there untill 13:00.
    Afterwards it jumps again up and joins the clear sky curve where it stays without interruption untill sunset.
    So clearly during this “typical day” the clouds gathered in the morning, reached a full cover around 11:00 and then moved away at 13:00.
    From the cooling “efficiency” point of view the clouds acted “maximally” – they removed SW at a time of the day when it is maximal.

    The Figure 4 tells another story. The grey all sky curve which is an average of every 2 minute interval over 2214 days is perfectly smooth and symmetrical. At least it looks like that.
    If there is any assymmetry, it could only be detected by computing the differences F(12+t) – F(12-t) for all t between 2 min and 12 hours where F is the SW all sky flux.
    But if the curve is perfectly symmetrical then it means that the clouds distribution in day time is perfectly random.
    In other words if Fclear(t) = k.Fallsky(t) with k some constant for all t as Figure 4 seems to say, then Fallsky(t)/Fclear (t) which represents the fraction of SW removed by the clouds is independent of time.
    It is easy to check with the data – just compute Fallsky(t)/Fclear (t) for all 12×30 2 minute day time intervals. If this ratio is approximately constant then clouds are random and make the same “cooling” work on any time of the day. From that follows that averages can be used because there is no privileged time where clouds would preferentially work.

    Of course if this ratio was not constant or if you amused yourself by computing the grey curves for shorter time intervals (f.ex 3 months averages) and observed that they are not identical, then the time would matter and averages couldn’t be used.

    P.S
    Like you deduced the clear sky curve by taking the maximum value for every 2 min interval, you could also deduce the fully cloud covered sky by taking the minimum value for every 2 min interval.
    On an updated Figure 4, you would then see that the allsky grey curve would be between the clear sky and the fully clouded sky curves (as expected :)) and from there you could deduce the cloudiness for every interval if you wished to do so.

  54. Will says:

    Willis

    “After they absorb the radiation coming up from the ground, they radiate about half of it back towards the ground, while the other half is radiated upwards “

    What ground are we talking about out there over the Pacific ocean?

    Over the ocean, any so called DLR would be that from incoming Solar (at least 50% LW). This appears to be supported by figure 3 where DLR seems to pick up most around the hours of daylight.

    Clouds can only block the shortwave portion of the incoming.

  55. LazyTeenager says:

    I thought that was a pretty good article Willis. Very illuminating.

  56. richard verney says:

    Willis

    An interesting post and nice to see an analysis performed on empirical observational data. For telling you what is going on in the real world, it trumps model projections anytime.

    No surprise that cloud formation around midday reduces incoming solar radiation by a substancial extent whereas cloud cover at night increases DWLWIR by only a relatively small extent. Most people have experienced the effect of clouds passing over on a bright sunny summer day and know how this can cause temperature on your skin to drop by 20deg or so. Most people have witnessed the difference between clear night and cloudy night with the latter being a couple or so degrees warmer. .

    I consider that it would be useful to see the data plots from data taken from the SAME buoy at (or around) the turn of each season so that we can see the differing strength of the input radiation.

    On a related point, would an ocean receiving between 1000 to 1400 w per sqm for 12 hours a day freeze even without the benefit of DWLWIR? As you are aware 1000 to 1400 w per sqm is a lot more than the Trenberth aveage figure he uses for solar irradiance in his cartoon.

    I look forward to hearing from you on that question.

  57. John W says:

    That bonus chart (figure 8) is the pièce de résistance!

  58. cba says:

    “George E. Smith says:

    September 15, 2011 at 7:26 pm

    “”””” Once again we see the sudden changes in the radiation when the clouds pass overhead. In the longwave case, however, the changes are in the other direction. Clouds cause an increase in the DLR. “””””
    And by inference those same clouds would also be increasing the LWIR to space, since as you say, half of the cloud absorbed radiation is re-emitted to space, so the cloud which you say is an efficient black body absorber (and emitter) whereas the atmosphere is not.

    Such things should not be a suprise. However, cloud radiation downward is at the bottom of the cloud and radiation upward is at the top and there are temperature differences. At the top, typically there is little to no h2o vapor above to absorb the new continuum but it’s still less than what radiates downward. Note also that it must be less than what escapes to space from the surface radiation. Otherwise Earth would have to be somewhat colder on average.

  59. Alan D McIntire says:

    Great post, Mr Eschenbach.

    In reply to

    dcfl51
    September 15, 2011 at 2:38 pm

    “Willis, I am not a scientist so forgive me if this is a silly question. I thought that just under 50% of the radiation emitted by the sun was long wave. So why do the instruments not detect more DLR during the day than during the night ?”

    I suppose in theory there IS more DLR during the day, by about 1/2%.

    I did a rough calculation of daily cooling of
    the atmosphere.

    mass atmosphere = 5* 10^18 kg=5*10^21gm
    temp atmosphere 255K (effective radiating temp to space- underestimates heat content)
    specific heat 1.01 joules/gm C
    5* 10^21*1.01*255= 1.288 * 10^24 joules

    radius earth = 6400km= 6.4*10^6 meters.
    area earth = 4 pi r^2 =514,718,540,364,021.76
    240 watts/sq meter = 240 joules/sec per square meter
    60 sec/min*60 min/hr*24hr/day=86,400 secs per day

    5.147* 10^14 sq meters*240 joules/sec/sq meter *8.64*10^4 secs/day= 1.067*10^22 joules per day radiated away
    1.067*10^22/1.288*10^24 = 0.83%

    So the atmosphere as a whole cools by less than 1% over the course of a day. That figure makes sense when you figure that the earth’s surface temperature my change by 10 C or more overnight far more than average changes over a week, but weather patterns persist for several days, and that’s why meteorologists can predict daily highs out a week or so. That cooling is obviously mostly from the
    earth’s surface and air near the surface ,leaving most of the atmosphere unchanged.

  60. TimTheToolMan says:

    Willis writes “The issue is not where the radiation is intercepted on its way to space. It is how many times, on average, it is intercepted. This is because every time the radiation is absorbed, half of it goes back towards the surface.”

    My understanding is that the average time between collisions of molecules (near sea level) is much less than the average time to re-radiate. Consequently at sea level the energy isn’t re-radiated “up or down” but rather its conducted to the N2 and O2.

    Of course collisions can go the other way and give the CO2 molecule sufficient energy to radiate and this does happen sometimes…its just that when thinking about CO2′s role at sea level its more of a transference of energy to the atmosphere effect.

    Further up in the atmosphere where there are fewer molecules and lower rates of collision, the radiation “up and down” becomes more prominent.

  61. daniel kaplan says:

    Mark says:
    September 15, 2011 at 6:53 pm
    “If nearly all of the LW-IR is absorbed within 100 m of the surface, how does it matter what happens at high altitude or the top of the atmosphere?”
    You have to work this out wavelength by wavelength in the IR. For wavelengths for which CO2 is highly absorbing (e.g. the main C02 absorption line), the TOA radiation comes from the stratosphere, yielding a negative greenhouse effect change with increasing CO2. For most of the central part of the spectrum, the TOA radiation comes from the tropopause, i.e. zero greenhouse effect change. For the tails of the band and some wavelengths between lines, the TOA comes from the troposphere and produces the greenhouse effect change through a rise of the effective radiation altitude with increasing CO2 concentration. These wavelengths dominate the concentration dependence of TOA radiation, not the TOA radiation itself. The overall concentration dependence corresponds to a positive greenhouse effect change. Downward radiation variations, on the other hand can come from a variety of wavelengths regions outside the main CO2 band where the absorption is not saturated, and be due to multiple causes.

  62. jim says:

    Willis – there is no way you have added value to the conversation. You haven’t used a computer model to verify and amplify your result :)

  63. Willis Eschenbach says:

    R. Gates says:
    September 15, 2011 at 8:58 pm

    Interesting as usual Willis. This comment however, may be a bit of an oversimplification:

    “After they absorb the radiation coming up from the ground, they radiate about half of it back towards the ground, while the other half is radiated upwards.”

    How did you compute this percentage? And even if correct (50% seems way too high), it doesn’t of course mean that 50% is lost to space.

    Thanks, R. When there is emission from the atmosphere, it is emitted spherically. Half of it is going up, and half going down.

    As you note, the horizon is not horizontal, and this increases as you go upwards. But the majority of the absorption and emission is taking place at lower elevations. In addition, even from a height of say 15 kilometres, the depression of the visual horizon is only about four degrees.

    So if the horizon depression is zero at the ground, and it is 4° at 15 kilometres, the average is about 2°.

    So we could say that rather than 50% going up and 50% going down, that 92/180 goes toward space and 88/180 goes towards the earth … which is 51% going up and 49% going down.

    In other words, R., you’ve busted me about a 1% error, in a system where it is very hard to measure just about any variable to better than 5% …

    Yes, I knew going in it wasn’t exactly 50/50 … but that is the usual assumption made, to simplify calculation, since the error is so small.

    w.

    PS – You are correct that 50% (or 51%) going upwards doesn’t mean that 51% goes directly to space.

    PPS – This hightlights an important issue, that of size. There are many things that look like they should be variables in the climate mix, until one looks at them closely and realizes that they are too small to make a difference. Run your numbers first, folks, I’m often ignoring these third order effects because I’ve run the numbers and I know they are small.

  64. Willis Eschenbach says:

    Bart says:
    September 16, 2011 at 1:41 am

    Willis Eschenbach says:
    September 15, 2011 at 5:12 pm

    “Part of the problem is the issue of causality. Suppose the day warms, and at some point a thunderstorm forms. That thunderstorm wanders over the ocean, and it leaves the ocean cooler, often cooler than it was when the thunderstorm formed.

    What would you say is the cause of that cooling? The thunderstorm? The high temperature that precipitated the thunderstorm?”

    The cooling from the thunderstorm is a feedback response.

    Thanks, Bart, but that is a naming without explanation. If I’m mean to somebody and they punch me in the nose, you could describe that as a “feedback response” as well … but like with your claim, it leaves a whole lot out of the explanation.

    In any case, since the thunderstorm can leave the local ocean much cooler than when it started, are you saying that that part of the ocean is cooling because it is warming? This is a recurring problem with self-organized criticality …

    w.

  65. R. Gates says:

    Willis Eschenbach says:
    September 16, 2011 at 9:05 am
    R. Gates says:
    September 15, 2011 at 8:58 pm

    Interesting as usual Willis. This comment however, may be a bit of an oversimplification:

    “After they absorb the radiation coming up from the ground, they radiate about half of it back towards the ground, while the other half is radiated upwards.”

    How did you compute this percentage? And even if correct (50% seems way too high), it doesn’t of course mean that 50% is lost to space.

    Thanks, R. When there is emission from the atmosphere, it is emitted spherically. Half of it is going up, and half going down.

    ______

    If you agree that it is essentially a spherical transimission pattern for LW being re-emitted from a greenhouse gas molecule, then I think we may differ in our definition of what “up” means. If by “up” you mean 180 degrees from the ground, then of course far less than 50% goes “up”, and this is important as the majority of the other angles, would either be back to the ground or in some other angle tangential to the ground (but not 180 degrees from the ground). These tangential angles and ground directed angles give much more opportunity for the the continually absorption and re-emission by either the ground or other greenhouse gas molecules. Thus, of course, the greater amount of greenhouse gas molecules present, the greater the opportunity for the absorption and re-emission of LW.

    In short, maybe only something less than 5% actually goes “up” and this could make a difference in the long-run in how you calculate the effects of both water vapor and other greenhouse gases.

  66. Joe Born says:

    Mark:
    “If the difference is that increased CO2 raises the altitude where the radiation finally “escapes to space”, and therefore the temperature at high altitude increases, how does that heat get back to the surface? . . . If the difference is that increased CO2 raises the altitude where the radiation finally “escapes to space”, and therefore the temperature at high altitude increases, how does that heat get back to the surface?”

    I don’t profess to be knowledgeable in the field, but one comment of Richard Lindzen’s here: http://www-eaps.mit.edu/faculty/lindzen/230_TakingGr.pdf, p.940, regarding what he feels is misnamed the “greenhouse” effect may supplement what those who answered above said. My no doubt inaccurate paraphrase is that CO2 enrichment raises the altitude at which, seen from space, a given optical depth is reached. Given the lapse rate, that means that the earth’s effective radiation temperature for a given surface temperature decreases: there is less radiation into space.

    Perhaps you will find that helpful, at least when taken together with the comments above regarding differing opacities at different wavelengths.

  67. richard verney says:

    @ Willis Eschenbach says:September 16, 2011 at 9:05 am
    in the course of his comment in response to R. Gates says:September 15, 2011 at 8:58 pm

    “…PS – You are correct that 50% (or 51%) going upwards doesn’t mean that 51% goes directly to space….”
    /////////////////////////////////////////////////////////////

    Much may depend upon the meaning attributed to the word “directly”, but, Willis, to the extnt that your observation is correct, it equally follows that of the 50% (or 49%) DWLWIR going downwards this does not go directly to the ground.

    Do you not consider that your assessment of 51% up/49% down might be a little bit of an under-assessment of the odds of the LWR radiating upwards? I say this, since, if you consider the path of a downward radiated photon from high in the atmosphere, each time it is involved in a collision on its downwrd path it either gives up its energy in heat (which may not necessarily result in another photon being radiated), or if colliding with a GHG molecule the resultant photon then radiated has a 51% prospect of radiating upwards and only a 49% prospect of being radiated downwards. Consider the effect of this in a scenario where there a re multiple collisions on the downward path. Do you not consider that the odds are more adversely stacked against the DWLWIR photon reaching the ground (or ocean) than you are actually suggesting.

    Some consider that the role that CO2 may play, in the overall workings of the atmosphere when convection, conduction, energy transfer consequent upon molecular collsions etc are taken into account, is predominantly in delaying energy finding its way out to space.

  68. richard verney says:

    Further to my last post, I accept that nearer to the ground, the curvature/horizon point becomes less of an issue such that the odds of a re-radiated photon going downwards tends towards becoming more equally balanced.

  69. Pete says:

    Mr. Eschenbach …

    Thanks for the informative article and the discussion it generated.

    As a non-scientist, I have an observation and query regarding your comments at 7:32pm on 9/15/11:

    “It is how many times, on average, it is intercepted. This is because every time the radiation is absorbed, half of it goes back towards the surface. This is true even if 100% of the radiation is intercepted just off the deck.”

    Observation: As a person with many hours flying aircraft, I know that oftentimes weather system clouds are layered, perhaps 3 or 4 layers (with clear air in between) in 20 – 25 thousand feet of altitude.

    Query: What, if any, effect do the several layers have on the absorptiion and return properties of cloud layers as opposed to your apparently simple illustration?

    Keep up the great work … the truth is out there, but it takes open minds like yours and some honest initiative to find it.

    Pete

  70. Bart says:

    Willis Eschenbach says:
    September 16, 2011 at 9:21 am

    “…but like with your claim, it leaves a whole lot out of the explanation.”

    In a feedback loop, such as depicted here, there is perfect symmetry if you ignore the input. Resolution of the ambiguity is found by where the outside forcing comes in. I explained this to Nick Stokes here.

    “In any case, since the thunderstorm can leave the local ocean much cooler than when it started, are you saying that that part of the ocean is cooling because it is warming?”

    You have to look at the system in its entirety to see where all the energy ends up. But, yes, it is entirely possible to have a sensitivity function which attenuates an input in one particular variable in the system.

  71. richard verney says:

    Mr Gates

    I would appreciate you answering a question arising out of your comment:
    R. Gates says:
    September 16, 2011 at 10:15 am

    “…If you agree that it is essentially a spherical transimission pattern for LW being re-emitted from a greenhouse gas molecule, then I think we may differ in our definition of what “up” means. If by “up” you mean 180 degrees from the ground, then of course far less than 50% goes “up”, and this is important as the majority of the other angles, would either be back to the ground or in some other angle tangential to the ground (but not 180 degrees from the ground)…”
    ///////////////////////////////////////////////////////
    Do you not accept that this comment applies equally to the ‘down’ scenario? If not, why not?

    It appears to me that on your reasoning,something less than 5% would go down. Should you disagree please explain why you disagree and what percentage goes ‘down’ and why that is so.

    As I see matters, the logical conclusion of your comment is that when one also takes into account the effect of the horizon/curvature point (which slightly favours the upward direction), the less than 5% going ‘down’ would be less than the less than 5% going ‘up’. Should you disagrre, I would appreciation your full explanation/reasoning. .

  72. coturnix says:

    After they absorb the radiation coming up from the ground, they radiate about half of it back towards the ground, while the other half is radiated upwards The effect is very perceptible on a cold winter night. Clear nights are the coldest, the radiation from the ground is freer to escape to space.
    ——–
    Half and half? Thats simply wrong. The amount radiated down through the window depends only and only on cloud’s own temperature. If we to disregard changes of emissivity with temperature, got this: for a stratus cloud 1 km aloft, with 6.5 *C/km lapse rate and +20C ground surface temperature, the percentage of down welling radiation from the cloud would be ((273+20-6.5*1)/(273+20))**4 = 91%, for cloud at 2 km it would be 83%; for altostratus at 5 km it would be 63%, and for high cirrus at 12 rm it would be a mere 29%. See, high clouds cool earth more than low clouds.

    To me, all this shit about ‘clouds warm earth at night’ is preposterous. They don’t. The only thing low clouds do is prevent to some degree formation nighttime inversions – but that can only cool planet, as warmer surface of cloud emits more efficiently than cold ground. High clouds however don’t really warm the earth underneath them. They do warm the planet as a whole, though.

  73. coturnix says:

    To add, Even more preposterous is the popular fable of ‘water vapor at night keeps earth warmer’. What a bullshit. Locally, more wv would decrease depth of inversion, and therefore make night colder but make earth warm up quicker in the morning, thus warming the earth slightly – but not locally. The common perception of warm nights being moist is from confusion of cause and effect. The cause is advection of warm, moist air. The effect is warmer nights – because the air is already warm.

  74. Willis Eschenbach says:

    Pete says:
    September 16, 2011 at 11:17 am

    Mr. Eschenbach …

    Thanks for the informative article and the discussion it generated.

    As a non-scientist, I have an observation and query regarding your comments at 7:32pm on 9/15/11:

    “It is how many times, on average, it is intercepted. This is because every time the radiation is absorbed, half of it goes back towards the surface. This is true even if 100% of the radiation is intercepted just off the deck.”

    Observation: As a person with many hours flying aircraft, I know that oftentimes weather system clouds are layered, perhaps 3 or 4 layers (with clear air in between) in 20 – 25 thousand feet of altitude.

    Query: What, if any, effect do the several layers have on the absorptiion and return properties of cloud layers as opposed to your apparently simple illustration?

    Keep up the great work … the truth is out there, but it takes open minds like yours and some honest initiative to find it.

    Pete

    Multiple cloud layers mean that photons will perforce be absorbed and emitted by each cloud on the way up, increasing the number of “shells” in the greenhouse.

    w.

  75. kwinterkorn says:

    Re R Gates’ issue:
    Try the following as a thought experiment:

    Draw a circle and draw a line tangent to the circle. This models the Earth and an IR and CO2 molecule interaction. An IR photon meeting a CO2 molecule near the surface of the Earth will be absorbed and then, assuming spherical pattern of re-radiaton, has a 50% chance of being radiated outside of that tangential line—ie away from the Earth.

    Then draw a second line, parallel to the first, but further from the circle. This represents the next interaction of that original photon’s energy with a second CO2 molecule, which 50% of the time will be further out from the Earth. Again, 50% of the time the photon will be re-radiated outside the line, ie away from the Earth and 50% of the time toward the Earth.

    Those photons (initially 50%) that are re-radiated back toward the Earth have as their next encounter either the Earth’s surface or another CO2 molecule nearer to the Earth. The 50% rule continues to apply for all CO2 encounters near the surface. The photon’s that end back up at the surface are also re-radiated ultimately, though they may spend some time as kinetic energy (ie heat) at the surface.

    For the photon’s that do get reradiated away from the Earth, two changes in their chances of escaping the Earth occur:
    1. With altitude, the CO2 concentration rapidly diminshes, and the chances of getting to space without another encounter increases.
    2. The further the Photon/CO2 encounter occurs from the surface, the lower the chance of hitting the Earth even if reradiated below the line parallel to the tangent line.

    Thus as the photon “bounces around” from CO2 encounter to CO2 encounter, there is an inevitable bias for the photon to escape to space. Yet compared to a “no greenhouse gas” atmosphere, that escape takes longer when greenhouse gases are present and therefore the energy content of that atmosphere—-for the same temperature at the surface—-will be higher—-ie the temperature will be higher—-ie warmer.

    Any “saturation” effect from risng CO2 concentration (ie black cannot get any blacker than black) does not address the issue that rising atmospheric CO2 results in a taller column of CO2 above the gound, and therefore CO2 heating effects continue to increase, though not linearly, because to photons trying to escape have further to do before getting free of CO2 interactions.

    Thus the CO2 greenhouse effect is more like running through a minefield than just shining through a single window of variable opacity. The longer the minefield, the harder to get through free of getting blown up.

    So the CO2 Greenhouse effect is real. Fortunately it seems to be countered by negative feedback related to clouds, storms, etc. and certainly is not an apocalyptic concern.

  76. tallbloke says:

    Willis, this is great stuff. Now we just need to replicate this for each latitude over a year and get a real idea of the overall cloud feedback to temperature.

    My money is on -20W/m^2 or thereabouts.

    Thing is, it looks like the cloud feedback to temperature is only a minor part of the story. What we really want is the feedback to solar variation, via whichever mechanisms affect it. GCR’s, solar UV/plankton aerosols, stratospheric chemistry – the complexity is bewildering. Thanks for making a good start on it. 2 minute data points, now we’re talking!

  77. TimTheToolMan says:

    Hi Willis, I just thought I’d post the reference to the fact that its too simple (and not particularly useful IMO) to think of the IR as going up and down with more CO2 involving more jumps. This is from Pierrehumbert’s article “Infrared Radiation and Planetary Temperature”

    http://climateclash.com/2011/01/15/g6-infrared-radiation-and-planetary-temperature/

    “Coupled vibrational and rotational states are the key players in IR absorption. An IR photon absorbed by a molecule knocks the molecule into a higher-energy quantum state. Those states have very long lifetimes, characterized by the spectroscopically measurable Einstein A coefficient. For example, for the CO2 transitions that are most significant in the thermal IR, the lifetimes tend to range from a few milliseconds to a few tenths of a second. In contrast, the typical time between collisions for, say, a nitrogen-dominated atmosphere at a pressure of 104 Pa and temperature of 250 K is well under 10^−7 s. Therefore, the energy of the photon will almost always be assimilated by collisions into the general energy pool of the matter and establish a new Maxwell–Boltzmann distribution at a slightly higher temperature. That is how radiation heats matter in the LTE limit.”

  78. timetochooseagain says:

    Willis: it might help to illustrate the complexity of the relationship between the cloud radiative forcing and sea surface temperatures to do short period regressions. A thermostat is probably going to show up as the relationship between the variables changing sign at certain “thresholds”, I would think, which would illustrate the problems with the traditional approach of simply regressing the full sets of anomalies of temperature on radiation flux, as most feedback analyses have done.

  79. Alan D McIntire says:

    In response to

    Willis Eschenbach says:
    September 16, 2011 at 3:39 pm

    Pete says:
    September 16, 2011 at 11:17 am

    Mr. Eschenbach …

    ” Observation: As a person with many hours flying aircraft, I know that oftentimes weather system clouds are layered, perhaps 3 or 4 layers (with clear air in between) in 20 – 25 thousand feet of altitude.

    Query: What, if any, effect do the several layers have on the absorptiion and return properties of cloud layers as opposed to your apparently simple illustration?

    “Multiple cloud layers mean that photons will perforce be absorbed and emitted by each cloud on the way up, increasing the number of “shells” in the greenhouse.”

    Browsing online, I read about a NEGATIVE greenhouse effect on Saturn’s moon Titan.

    Using the greenhouse model here

    http://www.geo.utexas.edu/courses/387H/Lectures/chap2.pdf

    I did a few simple calculations.

    When the atmosphere is transparent to incoming radiation from the sun and opaque to outgoing radiation from the planet, you get a positive greenhouse effect.

    When the atmosphere is opaque to incoming radiation from the sun and transparent to outgoing radiation to the planet you get a negative greenhouse effect, as happens in some frequencies on Titan.

    When the atmosphere is opaque to both incoming radiation from the sun and outgoing radiation from the earth you get a zero greenhouse effect.

    Roughly 60% of earth’s surface is covered by clouds, but earth only has a 30% albedo.
    That means roughly 30% of incoming radiation is absorbed by clouds- the fraction of the atmosphere below the point of absorption will have a ZERO greenhouse effect on that 30% of photons radiated from that could to the earth and back to a cloud at a similar height.

  80. gnomish says:

    “Thus the CO2 greenhouse effect is more like running through a minefield than just shining through a single window of variable opacity.” ?

    maybe more like tossing ping.pong balls in the river to speed up the flow…lol

  81. Willis Eschenbach says:

    Bart says:
    September 16, 2011 at 12:16 pm

    Willis Eschenbach says:
    September 16, 2011 at 9:21 am

    “…but like with your claim, it leaves a whole lot out of the explanation.”

    In a feedback loop, such as depicted here, there is perfect symmetry if you ignore the input. Resolution of the ambiguity is found by where the outside forcing comes in. I explained this to Nick Stokes here.

    “In any case, since the thunderstorm can leave the local ocean much cooler than when it started, are you saying that that part of the ocean is cooling because it is warming?”

    You have to look at the system in its entirety to see where all the energy ends up. But, yes, it is entirely possible to have a sensitivity function which attenuates an input in one particular variable in the system.

    Thanks, Bart. So if everything can be seen as being caused by the initial forcing and the rest is seen as feedback, can we then say that sunshine is the forcing of Shakespeare’s sonnets and that everything else is just feedback?

    You are using what I might call a classic feedback paradigm, which you understand and explain very well, although I can’t follow some parts, that’s me, not you.

    I made the point in “It’s Not About Feedback“, or tried to make the point, that your classical feedback paradigm is inadequate to explain the evolution of the tropical day. The evolution of the typical tropical day involves the sequential replacement of the basic atmospheric circulation pattern. The circulation starts out weak and random. At a certain point a temperature threshold is passed, and by a process described as “self-organized criticality”, that circulation pattern is rapidly replaced by a brand new pattern involving cumulus cloud and local shallow atmospheric circulation cells around those clouds.

    If the day continues to warm, the circulation pattern changes abruptly for a second time. A new self-organized pattern, which involves deep tropospheric circulation driven by spontaneously generated, mobile, warmth-quenching heat engines called thunderstorms. These function as giant independently moving air conditioning machines that cool the surface.

    Now, there’s lots of ways we can seek to understand and analyze that hugely complex sequence of events.

    But thinking that the whole intricate story can be reduced to a Lissajous pattern and profitably analyzed in terms of classic feedback caused by the initial sunlight striking the system?

    Well, maybe … but only if we can also see Shakespeare’s odes as being caused by sunlight …

    w.

  82. Willis Eschenbach says:

    coturnix says:
    September 16, 2011 at 3:23 pm

    To me, all this shit about ‘clouds warm earth at night’ is preposterous.

    Then you should spend more time outdoors at night in the winter. Anyone who does so knows that the coldest nights are the clear nights. And during a clear night, when a cloud comes over, you can feel the difference immediately.

    As you should, since instead of having the cold of outer space at a temperature of 3 kelvins as a backdrop when you look up, you have a cloud at around 270 K as a backdrop when you look up.

    w.

  83. Willis Eschenbach says:

    TimTheToolMan says:
    September 16, 2011 at 5:12 pm

    Hi Willis, I just thought I’d post the reference to the fact that its too simple (and not particularly useful IMO) to think of the IR as going up and down with more CO2 involving more jumps. …

    Tim, it seems you are objecting to what people (not me) call “re-radiation”. This is the incorrect idea that the CO2 molecule captures and then re-radiates a photon. I agree with you that what happens is the photon is absorbed by the GHG, and then released as thermal energy through collisions with other molecules .

    In terms of whether it is useful, however, the average number of times thermal radiation is absorbed on its way to space is a factor in the math of calculating how much thermal advantage is provided by the system (and thus the surface temperature).

    Let me add in passing that a whole lot of the thermal energy goes to space without ever interacting with the troposphere. It is piped their by thunderstorms. Consider a molecule of water evaporated from the surface. It rises, carrying the latent thermal energy upwards. It releases that energy in the dark heart of the cloud, where the warm air drives the chimney circulation up the middle of the thunderstorm tower to the upper troposphere.

    During that whole time the thermal energy travels from the surface to the upper troposphere, it does not interact with any part of the atmosphere—not with the surface mixed layer, the lower troposphere, or the middle troposphere. It does not mixed mechanically with those layers, nor does it radiate even the slightest amount to those layers of the lower atmosphere. It took the secret passage from the surface to the upper troposphere and indeed, at times into the stratosphere itself. It did not take the radiative path.

    w.

  84. TimTheToolMan says:

    Willis writes “In terms of whether it is useful, however, the average number of times thermal radiation is absorbed on its way to space is a factor in the math of calculating how much thermal advantage is provided by the system (and thus the surface temperature). ”

    Again as I understand it, once the GHG’s have initially absorbed the IR from the ground (or ocean) within the first 100m or so, only about 2% of that energy is radiated again from the pointy end of the Maxwell–Boltzmann distribution. This means the rest is transmitted through the atmosphere via “other means”

    This is where I wholeheartedly agree with your emphasis on the roles of thunderstorms, the importance on the latent heat of vaporisation and so on.

    Meh. This is the second time I’ve insisted on better explanations of physical processes from your excellent articles. I guess I’m even more of a stickler for understanding and conveying the detail of the process than you ;-)

  85. richard verney says:

    Willis

    Willis Eschenbach says:
    September 16, 2011 at 7:25 pm
    coturnix says:
    September 16, 2011 at 3:23 pm

    To me, all this shit about ‘clouds warm earth at night’ is preposterous
    /////////////////////////////////

    In mid lattitudes. the difference between a clear night and a cloudy night is usually just a few degrees notwithstanding your point regarding space at 3K and clouds at 270K. The reality is that for the main part, irrespective of clouds, the atmosphere radiates at about 270K.

    Clouds do not warm. They may slow down the rate of cooling. Slowing down the rate of cooling is not warming. Being less cold, is not warming.

  86. Tim Folkerts says:

    Willis Eschenbach says: September 15, 2011 at 10:59 pm
    ” If for some time around noon when there are no clouds the shortwave is 1400 W/m2 …:”

    This number seems just plain wrong. First of all, the radiation at the top of the atmosphere is typically ~ 1365 W/m^2. UV accounts for ~ 10% of that, and much of the UV is blocked long before reaching the ground, so that is perhaps -100 W/m^2. Even with no clouds, there must be a fair amount of humidity, and H2O vapor absorbs a noticeable fraction of the solar IR – perhaps another -100 W/m^2. I can’t see how solar energy could possibly be more than ~ 1200 W/^2 at the surface. I CERTAINLY can’t see how the solar energy could be more more than 1365 W/m^2.

  87. Bart says:

    Willis Eschenbach says:
    September 16, 2011 at 7:17 pm

    “…can we then say that sunshine is the forcing of Shakespeare’s sonnets and that everything else is just feedback?”

    You can say sunshine is a forcing, but you’ve expanded the “system” now to include an awful lot of stuff. There are other forcings, too, and not all signal paths are feedback – some are feed-forward.

    “But thinking that the whole intricate story can be reduced to a Lissajous pattern and profitably analyzed in terms of classic feedback caused by the initial sunlight striking the system?”

    Feedback means nothing more nor less than a reaction to an input which influences how the overall system responds to that input. It is a very general concept. Feedback is not necessarily, or even usually, linear and smooth (even nonlinear systems can be linearized if they are “smooth” enough). In fact, engineers make use of bang bang controls and hard limiters and hysteresis loops and pulse width modulators and other gross nonlinearities in common mechanical and electrical feedback systems all the time. Lyapunov analysis can often be used to prove local or sometimes global stability of systems employing such elements.

    But, feedback often is linear, or nearly so, in natural systems, or can be profitably modeled as such, so it’s always something to watch out for in terms of common signatures it leaves behind. Even highly discontinuous feedback can often be approximated as smooth and continuous under appropriate assumptions. For example, pulse width modulators are devices which allows us to produce a train of pulses with frequency and/or width proportional to the feedback we want to provide. If the bandwidth of the system is low compared to the pulse frequency, then we can approximate the feedback as a linear , and analyze the loop using linear control analysis techniques.

    It is also possible to predict limit cycles of systems with highly nonlinear and discontinuous elements using linear systems theory by considering the first harmonic response of the element, on the assumption that higher harmonics will be attenuated in the overall feedback loop. This is called describing function analysis.

    In summary re feedback: You keep using that word. I do not think it means what you think it means ;-)

  88. Richard Sharpe says:

    Tim Folkerts says on September 16, 2011 at 8:59 pm

    Willis Eschenbach says: September 15, 2011 at 10:59 pm
    ” If for some time around noon when there are no clouds the shortwave is 1400 W/m2 …:”

    This number seems just plain wrong. First of all, the radiation at the top of the atmosphere is typically ~ 1365 W/m^2. UV accounts for ~ 10% of that, and much of the UV is blocked long before reaching the ground, so that is perhaps -100 W/m^2. Even with no clouds, there must be a fair amount of humidity, and H2O vapor absorbs a noticeable fraction of the solar IR – perhaps another -100 W/m^2. I can’t see how solar energy could possibly be more than ~ 1200 W/^2 at the surface. I CERTAINLY can’t see how the solar energy could be more more than 1365 W/m^2.

    In addition, something around 40-45% is infrared, which is, we are told, absorbed in the atmosphere and likely does not make it to the ground.

  89. Willis Eschenbach says:

    richard verney says:
    September 16, 2011 at 8:57 pm

    … Clouds do not warm. They may slow down the rate of cooling. Slowing down the rate of cooling is not warming. Being less cold, is not warming.

    I object strongly to this semantical nonsense. It doesn’t matter what you call it. The surface at night is warmer with clouds than without. Thus we can say that the clouds make the night warmer than it would be if they were not there. In common parlance, we call that warming.

    For example, we say “The room is cold, let’s close the windows to warm it up.” We don’t say “Let’s close the windows to reduce the heat loss”, that part is understood. We say “I’m cold, I’m going to put on a jacket to warm up”. We don’t say “Gosh, gotta put on a jacket to cut down my heat loss.”

    People keep flogging this issue over and over, as if what we called it made the slightest difference. Here’s the bottom line.

    It’s warmer with clouds than without, call it what you will. The mechanics are the same. The physics are the same. The transfer of thermal radiation energy from the cloud to the surface is the same. Call it what you will.

    For me, the clouds radiate energy to the surface, the surface absorbs that energy, and the surface ends up warmer as a result. I call that “warming the surface”. It is not just slowing the heat loss. It is actively adding energy to the surface … if you have another name than “warming” to describe “actively adding energy to the surface, as a result of which the surface ends up warmer than if was without the added energy”, let me know. “Slowing the rate of cooling” doesn’t work, because it is actively adding energy, not just slowing the loss.

    w.

  90. Willis Eschenbach says:

    Richard Sharpe says:
    September 16, 2011 at 9:17 pm (Edit)

    Tim Folkerts says on September 16, 2011 at 8:59 pm

    Willis Eschenbach says: September 15, 2011 at 10:59 pm

    ” If for some time around noon when there are no clouds the shortwave is 1400 W/m2 …:”

    This number seems just plain wrong. First of all, the radiation at the top of the atmosphere is typically ~ 1365 W/m^2. UV accounts for ~ 10% of that, and much of the UV is blocked long before reaching the ground, so that is perhaps -100 W/m^2. Even with no clouds, there must be a fair amount of humidity, and H2O vapor absorbs a noticeable fraction of the solar IR – perhaps another -100 W/m^2. I can’t see how solar energy could possibly be more than ~ 1200 W/^2 at the surface. I CERTAINLY can’t see how the solar energy could be more more than 1365 W/m^2.

    In addition, something around 40-45% is infrared, which is, we are told, absorbed in the atmosphere and likely does not make it to the ground.

    Thanks, Richard. You’re correct, except you’ve neglected the eccentricity of the orbit. This varies the “solar constant” from a low of about 1345 to a high of 1435 W/m2 over the course of the year. In addition, the infrared from the sun is definitely not absorbed in the atmosphere, it’s the wrong frequency (as I discussed upthread), so it is measured by the pyranometers.

    It is quite possible that the sensors are occasionally reading high due to unusual meteorological conditions, and in hindsight it likely would have been better to use say the median of the top five readings as the peak, rather than a single reading. Let me see what that does … OK, that reduces the peak values down to about 1350 W/m2, which seems more reasonable. It also reduces the cloud shortwave effect slightly, bringing it down from 210 W/m2 to 185 W/m2 …

    Finally, you say “… H2O vapor absorbs a noticeable fraction of the solar IR – perhaps another -100 W/m^2.” This is way high. The total atmospheric absorption of shortwave is about 70 W/m2. However, much of this is from black carbon, brown carbon, aerosols, smog, and other things in the air.

    So over the ocean, I would say (and the sensors agree) that 1350 W/m2 is certainly possible. As you point out, the 1400 W/m2 I quoted before, while theoretically doable (max solar is 1435), is not that likely.

    All the best,

    w.

  91. coturnix says:

    Clouds do not warm. They may slow down the rate of cooling. Slowing down the rate of cooling is not warming. Being less cold, is not warming.
    ——-
    Same could be said about greenhouse gases – they do not warm, they only slow down the rate of cooling, but actual warming is done by the sun. it is with presence of ghg that sun can warm earth more than without. of course, real world is more complex, like during polar night poles do not freeze to -200.

    However, what i want to say is that may be reported ‘warming effect’ of clouds may not be radiative after all, simply because when there are clouds, there is some sort of convection, that mixes air at the ground all the way though out boundary layer, or even with the free troposphere. Free troposphere only warms and cools very little if at all during diurnal cycle, and if there is something that prevents formation of inversion during night, clouds are most likely too a result of this mixing, and not the direct cause.

  92. R. Gates says:

    richard verney says:
    September 16, 2011 at 3:14 pm
    Mr Gates

    I would appreciate you answering a question arising out of your comment:
    R. Gates says:
    September 16, 2011 at 10:15 am

    “…If you agree that it is essentially a spherical transimission pattern for LW being re-emitted from a greenhouse gas molecule, then I think we may differ in our definition of what “up” means. If by “up” you mean 180 degrees from the ground, then of course far less than 50% goes “up”, and this is important as the majority of the other angles, would either be back to the ground or in some other angle tangential to the ground (but not 180 degrees from the ground)…”
    ///////////////////////////////////////////////////////
    Do you not accept that this comment applies equally to the ‘down’ scenario? If not, why not?

    It appears to me that on your reasoning,something less than 5% would go down. Should you disagree please explain why you disagree and what percentage goes ‘down’ and why that is so.

    As I see matters, the logical conclusion of your comment is that when one also takes into account the effect of the horizon/curvature point (which slightly favours the upward direction), the less than 5% going ‘down’ would be less than the less than 5% going ‘up’. Should you disagrre, I would appreciation your full explanation/reasoning.

    ————–
    I actually don’t disagree with this general reasoning if geometric concerns were all that mattered. My original question to Wills was simply to find out how he arrived at his 50% up/ 50% down figure for his cloud model. The actual circumstances of the way LW radiation interacts with both clouds and greenhouse gases is far more complicated than the simple geometric and billard ball type models we’ve been discussing. I actually think that, as far as the narrow band of the tropical oceans are concerned, Willis’ model is probably roughly correct. The problem of course is that it deals with a narrow region of the earth, and only considers one very specific and simple cloud interaction. if this was all there was on this planet in terms of cloud feedbacks I would think that we would have long ago figured out the exact sensitivity of the climate to the additional greenhouse forcing.

  93. Willis Eschenbach says:

    Bart says:
    September 16, 2011 at 9:11 pm

    Willis Eschenbach says:
    September 16, 2011 at 7:17 pm

    “…can we then say that sunshine is the forcing of Shakespeare’s sonnets and that everything else is just feedback?”

    You can say sunshine is a forcing, but you’ve expanded the “system” now to include an awful lot of stuff. There are other forcings, too, and not all signal paths are feedback – some are feed-forward.

    Thanks for your thoughts as always, Bart. You miss my point, which is likely from my lack of clarity. My point is that when a natural system (humans on the earth, say, or the global climate system) reaches a certain level of complexity (Shakespear’s sonnets, say, or the unexpected emergence of a thunderstorm spitting lightning), it is no longer profitable to analyze it using classical feedback theory. We are not dealing with the kinds of systems you are used to. In climate we are looking at intricately entangled emergent systems which are continuously adapting, have thresholds for self-organized criticality which when passed lead to total reorganization of circulation, spawn independently mobile heat engines that have a lifespan and then die, systems which are often running at the edge of turbulence, and are ruled by the Constructal Law.

    As a result, we need to use other analysis methods which are appropriate to that situation.

    “But thinking that the whole intricate story can be reduced to a Lissajous pattern and profitably analyzed in terms of classic feedback caused by the initial sunlight striking the system?”

    Feedback means nothing more nor less than a reaction to an input which influences how the overall system responds to that input. It is a very general concept. Feedback is not necessarily, or even usually, linear and smooth (even nonlinear systems can be linearized if they are “smooth” enough). In fact, engineers make use of bang bang controls and hard limiters and hysteresis loops and pulse width modulators and other gross nonlinearities in common mechanical and electrical feedback systems all the time. Lyapunov analysis can often be used to prove local or sometimes global stability of systems employing such elements.

    Perhaps because your background is signal processing, you are using “feedback” in a much wider, more inclusive sense than is common in the climate science field. You say, for example,

    “… engineers make use of bang bang controls and hard limiters and hysteresis loops and pulse width modulators and other gross nonlinearities in common mechanical and electrical feedback systems all the time.”

    And you are correct. But when people in climate science talk about “feedback”, they are talking about plain old, garden variety feedback, what I have described as “classical” feedback, not any kind of “bang bang” feedback or pulse width modulators. This concept of feedback involves some mystical percentage by which clouds are supposed to increase the warming. It is seen as a constant, and it has nothing to do with the kinds of feedback you are (correctly) referring to.

    I still say, however, that the whole concept of “feedback” is a very minor part of understanding the self-organized criticality which produces a thunderstorm. Of much more importance is the existence of a critical threshold, and the fact that when it is passed, a whole reorganization of the entire circulation regime takes place … that’s the kind of thing that “feedback”, no matter how broadly defined, can’t much help us with. Sure, feedback is in there, at every level actions have consequences, feedback is assuredly taking place … but the change in the circulation pattern or the creation of say a tornado is what rules the effects, not the feedback.

    My best to you,

    w.

  94. TimTheToolMan says:

    Willis writes “For me, the clouds radiate energy to the surface, the surface absorbs that energy, and the surface ends up warmer as a result. I call that “warming the surface”. It is not just slowing the heat loss. It is actively adding energy to the surface …”

    Not for 71% of the “surface” Willis. As discussed at some length, the oceans dont absorb that energy to become warmer as a direct result of that energy. There is no energy actively added in the sense it enters the ocean bulk. There are no mechanisms for it to pass further than the 10um skin. Thermodynamics doesn’t allow it. Clouds DO cool less and warm the ocean as a result though.

    Your argument is more valid for the ground although its still a slowing cooling effect simply because the energy you say “warms the surface” originally came from the ground.

  95. coturnix says:

    There are no mechanisms for it to pass further than the 10um skin
    —-

    Sure there is, it is mixing due to waves. Have no idea how important it may be away from hurricanes, but seas are rarely calm. Conduction may not be taken into account – water conducts heat much worse than even rock.

  96. TimTheToolMan says:

    coturnix writes “Sure there is”

    There has been a lot more discussion of this here
    http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/

    But essentially the top of the ocean is cold and mixing it down for the next few mm cools the ocean. Conduction wont happen because where the IR is absorbed at the very top of the ocean is cooler than just below. For it to conduct, it would have to break the laws of thermodynamics…

    There is a reason the skin of the ocean got this temperature gradient though and thats because the net flow of energy is upwards, not downwards. It all hinges on this “slower cooling” rather than “warming” thing that Willis and many others dont like to use.

  97. tallbloke says:

    TimTheToolMan says:
    September 16, 2011 at 11:15 pm

    coturnix writes “Sure there is”

    There has been a lot more discussion of this here
    http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/

    But essentially the top of the ocean is cold and mixing it down for the next few mm cools the ocean. Conduction wont happen because where the IR is absorbed at the very top of the ocean is cooler than just below. For it to conduct, it would have to break the laws of thermodynamics…

    :-)

    Willis chose to abandon that thread without responding to these arguments, and others we provided wich show that very little energy from DLR ‘back radiation’ is getting mixed into the ocean bulk.

  98. TimTheToolMan says:

    Tallbloke writes “Willis chose to abandon that thread ”

    Yeah but this thread isn’t really the forum to explore that further. This should be more about the clouds and I somewhat regret bringing it up here.

  99. coturnix says:

    The second law of thermodynamics states, clearly – heat only flows from (relatively) hot to (relatively) cold, if it flew the other way perpetuum mobile of second type would be possible. This also includes thermal radiation. The NET heat transfer between a warm ocean at 20 C and a cold cloud at 5 C is ALWAYS from ocean to cloud. From which it is instantly obvious that the so called DLR can’t increase ocean’s temperature. Unless air is already warmer than water – it could probably happen at western subtropical shores, where frigid water upwells. In arctic ocean, sometimes water is so much warmer than the air that once exposed it creates thick fog through evaporating.

    For some ideal ocean to warm up would mean to pump up some more energy into it. Because sun shines the same all the time, it must be DLR that heats it during warm-up phase. And yet, the net thermal emission is still from the ocean to whatever is above it. One can say that backradiation started warming the ocean, but unless you can catch and count photons with your bare hands, one can just as correctly say that suddenly ocean’s emissivity dropped, and now instead of emitting all the sun’s energy into space it retains bit of it to itself. Note, that NEVER in the process of ocean warming did net thermal radiation flew towards ocean from atmosphere. Net radiation always flows from ocean to atmosphere.

  100. Willis Eschenbach says:

    tallbloke says:
    September 16, 2011 at 11:51 pm

    TimTheToolMan says:
    September 16, 2011 at 11:15 pm

    coturnix writes “Sure there is”

    There has been a lot more discussion of this here
    http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/
    But essentially the top of the ocean is cold and mixing it down for the next few mm cools the ocean. Conduction wont happen because where the IR is absorbed at the very top of the ocean is cooler than just below. For it to conduct, it would have to break the laws of thermodynamics…

    :-)

    Willis chose to abandon that thread without responding to these arguments, and others we provided wich show that very little energy from DLR ‘back radiation’ is getting mixed into the ocean bulk.

    Nonsense. I got tired of asking the following question: if the DLR isn’t absorbed by the ocean, then what is keeping the ocean from freezing? The ocean gets about 170W/m2 from solar, and it’s losing about 390 W/m2 by radiation alone, not counting sensible and latent heat loss … if DLR isn’t being absorbed by the ocean, then why isn’t it frozen?

    After about the fourth time I’d heard every possible tap-dance around the subject, without anyone being able to answer my question, I said “Feh”. Tallbloke’s explanation was particularly hilarious, involving some kind of invisible mist just above the water that absorbed the DLR in some dance above the surface … but even that couldn’t explain the missing energy needed to keep the ocean from freezing.

    w.

  101. Myrrh says:

    dcfl51 says:
    September 15, 2011 at 2:38 pm
    Willis, I am not a scientist so forgive me if this is a silly question. I thought that just under 50% of the radiation emitted by the sun was long wave. So why do the instruments not detect more DLR during the day than during the night ? Is the long wave radiation from the sun all absorbed by the atmosphere before it reaches the surface of the earth ? If so, how do we know how much of the DLR is back radiation originating from the earth’s surface and how much comes straight from the sun having stopped for a coffee in the atmosphere on the way down ?

    JamesD says:
    September 15, 2011 at 4:16 pm
    Bump of dcfl51 question. How does you analysis take into account LR from the sun?

    Willis Eschenbach says:
    September 15, 2011 at 5:33 pm
    JamesD says:
    September 15, 2011 at 4:16 pm

    Bump of dcfl51 question. How does you analysis take into account LR from the sun?

    The answer is, there’s longwave, and there’s longwave. The longwave absorbed by the atmosphere might be though of as “long longwave”. It is radiated by things at a temperature from say 30°C down to -40°C or so.

    The longwave from the sun, on the other hand, is “short longwave”, also called the “near infrared” because it is near to the visual spectrum. That infrared is included in the value listed as “shortwave”. I looked in the sensor specs, but found nothing about the bandwidth. However, it says they are using a pyranometer, which is designed to be as flat as possible across the entire solar spectrum.

    In any case, by the time you get out to the frequencies of the “long longwave”, there is very little power from the sun in those “long longwave” frequencies. Which is why little of the solar radiation is absorbed by the pure atmosphere.

    #####################

    Shrug. AGWScience Fiction Inc has written out of the picture (see Kiehl/Trenberth ’97 cartoon) longwave infrared, aka thermal infrared aka heat, direct from the Sun to the Earth, which is what heats the land and oceans and us.

    It has been replaced by the fictional science meme that shortwave radiation from the Sun, aka solar, sunlight, which is visible and the two shortwaves either side of uv and near infrared, are what actually heat the land and oceans of Earth. Of course, those not modern day scientists might recall that these are reflective energies, not thermal, they are not hot, and they do not heat molecules. They don’t have the power to move molecules of water into rotational resonance which is what heats up water, thermal infrared does have that power, and so does heat up water.

    Blue light for example is in the set of reflective energies, it is absorbed briefly by the electrons of nitrogen and oxygen in our atmosphere, note electrons, and sent out the way it came, this is called reflection/scattering in the electonic transition category of effects. It is not moving the molecules, note whole molecules, of nitrogen and oxygen into rotational resonance which is what would actually heat them. Ditto, blue light does not heat the molecules of water in the ocean, not by the same process, but because blue light doesn’t even get in to dance with the electrons of water, it isn’t let in, a slight delay as it tries, and then it’s passed on, this is called transmission.

    That’s the basic mechanism from real world traditional science. AGWScience Fiction Inc has given the properties of the real heat we feel from the Sun, thermal infrared, to the light we see from the Sun.

    I suggest that none believing this AGWSF meme look at the blue sky, they’ll bake their eyeballs.

    Willis, I too enjoy your writing, you are exceptionally clear in telling a story and enjoyable in that and in the things you find. It’s such a great pity that your basic premise is based on the junk science fictional meme now widespread through mis-education. If you could get that sorted, you’d be truly brilliant.

    There are other stories around, for example:

    http://www.school-for-champions.com/science/infrared.htm

    “Although nitrogen and oxygen gases make up a large portion of the atmosphere, they do not absorb infrared. However, water vapor carbon dioxide methane and ozone molecules in the atmosphere absorb much of the infrared radiation coming from the Sun.

    There is a band of wavelengths between 8 and 12 microns where little infrared radiation is absorbed in the atmosphere. Radiation in this band of wavelengths is what reaches the ground to heat things up.”

    What do you make of that?

    Or this from the beginnings of the search for knowledge about Heat energy from the Sun?: http://docs.lib.noaa.gov/rescue/mwr/056/mwr-056-08-0322.pdf

    From which: ”
    Emden found that the stratosphere sends no radiation clownwards,
    and of, course the same result came out of my
    previous work. The new investigation shows that the
    stratosphere sends on the average a downward flux of
    longwave radiation of more than. 120 cal./cm.2/min., which
    is more than 43 per cent of the effective solar radiation.
    This agrees with the observations made by Angstrom on
    mountain peaks and in balloons, which revealed a downward
    radiation of between .13 and .16 cal./cm.2/min.
    at heights between 4,000 and 5,000 metres, where, according
    to Emden,’ there should have been less than .05
    cal./cm.2/ min.

    There’s some recent work here in this paper, for which I can’t get access, they don’t allow pay per view: http://www.agu.org/pubs/crossref/2011/2010JD015343.shtml Perhaps someone here can enlighten us as to contents.

    NASA used to teach traditional well tried and tested and understood physics about this: From this NASA page:

    “Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.

    Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.

    Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.

    Infrared light is even used to heat food sometimes – special lamps that emit thermal infrared waves are often used in fast food restaurants!

    Now NASA teaches that thermal infrared doesn’t even get to us here on Earth’s surface..

    What do you make of that?

    How does a non-thermal energy, near infrared, manage to send whole molecules into rotational resonance?

    So yes, there’s longwave and longwave, and these are heat on the move from the extremely hot Sun to us on Earth eight minutes later.

    The heat we feel every day from the Sun, a fire, is thermal infrared.

    So, dcfl51′s question is best answered, imho, by saying, ‘we who promote the AGWScience Fiction Inc’s meme which has reversed the properties of Heat and Light from the Sun, have no idea how to answer you, we actually don’t even understand the question.’

    I on the other hand would answer, ‘You’ll have to look elsewhere for a real physics answer to you question because, as interesting as Willis is, he bases his work on a fictional through the looking glass AGWSF premise; on a different world where impossible things happen, where Light energies from the Sun heat matter and direct Heat from the Sun plays no part in heating its imaginary organic matter.’

    At the very least, note there is a disjunct; both stories can’t be real physics because they each give contrary explanations of properties and processes. I would have thought this would be of interest to scientists in this subject..

  102. Myrrh says:

    Sorry, missed out the URL link to the NASA page I quote: http://science.hq.nasa.gov/kids/imagers/ems/infrared.html

    While I’m here, let me fetch what I’ve found on the manipulation of mass education on this point: http://wattsupwiththat.com/2011/07/28/spencer-and-braswell-on-slashdot/#comment-711886

  103. dlb says:

    T the T at 10.42pm “Clouds DO cool less and warm the ocean as a result though”

    Not sure on your logic? Clouds are not just some sort of valve that slows the release of heat from the ocean, they are warm entities and will direct radiation back to the surface thus reducing any net loss of energy from the ocean. If the ocean is warm it will emit radiation at a known rate for this temperature, regardless of whether there is cloud or a dry CO2 free atmosphere above it.

  104. R. Gates says:

    Just one thought for those who are insistent that downwelling LW radiation can’t enter the oceans– if you accept the notion that SW radiation stays more or less constant at the ocean surface when averaged over longer periods then how would you explain the increases in global ocean heat content when looked at over a longer period?

  105. R. Gates says:

    And some of you might benefit from looking over this excellent chart that shows the spectrum of sunlight at the top of the atmosphere compared to what is at sea level:

    http://en.wikipedia.org/wiki/File:Solar_Spectrum.png

  106. R. Gates says:

    And finally, this chart:

    http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png

    Which shows that Willis’ estimate of 50% of the LW going up into space is way too high as the actual amount is somewhere around 15 to 30%.

  107. Tom in Florida says:

    coturnix says:
    September 17, 2011 at 1:22 am
    “For some ideal ocean to warm up would mean to pump up some more energy into it. Because sun shines the same all the time, it must be DLR that heats it during warm-up phase. ”

    You may want to rephrase that. Although the “sun shines the same all the time”, the amount of energy arriving at the surface is dependent on lattitude and time of year. Hence, Gulf of Mexico waters are colder in winter than in summer.

  108. cba says:

    Joe Born says:

    September 16, 2011 at 10:46 am


    Mark:
    “If the difference is that increased CO2 raises the altitude where the radiation finall
    My no doubt inaccurate paraphrase is that CO2 enrichment raises the altitude at which, seen from space, a given optical depth is reached. Given the lapse rate, that means that the earth’s effective radiation temperature for a given surface temperature decreases: there is less radiation into space.

    There’s two factors going on that will affect the lapse rate. First off, the lapse rate is merely related to the energy flow and conservation of energy. A ‘chunk’ of atmosphere will be at a temperature where the energy coming in equals the energy going out. A change in ghg concentration will cause a bit of shifting around of temperature in order to balance again. A change upward in ghg concentration will also increase the radiative ability of the chunk of atmosphere so that it can radiate more energy at a given temperature which means the temperature change doesn’t have to be as much as a very simplistic calculation might suggest.

    Perhaps Willis should embrace a little more of the averages approach to hone his current (excellent) idea to a finer point and sharper cutting edge.

    Earth’s avg. T is about 288K which means that 390 w/m^2 escapes the surface. The incoming solar is about 342w/m^2 average with about 30% of that amount reflected back into space, which leaves about 239 W/m^2 to be absorbed by Earth and atmosphere. For balance, the Earth must radiate back into space about 239W/m^2 on average. Something like modtran calculator will show that about 259w/m^2 escapes to at least 70km above with clear skies and there’s not much up there. Note that we’re 20W/m^2 too much radiation over our needed average. Also note that with 390 w/m^2 leaving the surface and 239 w/m^2 being our necessary average, that there is 151 W/m^2 that must be blocked on average and ghgs only block about 130 w/m^2. Clouds, etc. are responsible for the rest. Since cloud cover is close to 50/50 (or 62/38 and variable) cloud tops can only radiate around 220 w/m^2 into space for our 239w/m^2 balance. At the ground level, there must be an average of over 151 w/m^2 LW radiation reaching the ground along with what reaches the ground from the solar.

    Note that solar may be almost half IR, but it is shortwave, mostly close to visible light, not LW such as the peak of emissions for Earth temperature objects.

    The nice thing about Willis’ thread here is that one can now use all those weapons from the enemy camp against them.

  109. TimTheToolMan says:

    Willis writes… “The ocean gets about 170W/m2 from solar, and it’s losing about 390 W/m2 … if DLR isn’t being absorbed by the ocean, then why isn’t it frozen?”

    …but doesn’t accept that the energy from the DLR forms part of the stefan-boltzmann energy requirement for the ocean to radiate. The upshot of this is that the DLR doesn’t warm the ocean. At leaast not in the way Willis and many others believe can happen.

    OK Willis, from what we know about the skin, where does the DLR energy go? And how does it do it? You need to be quite precise because waving your hands around and saying “it goes into the ocean” simply wont cut it as an answer.

  110. Willis Eschenbach says:

    R. Gates says:
    September 17, 2011 at 5:35 am

    And finally, this chart:

    http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png

    Which shows that Willis’ estimate of 50% of the LW going up into space is way too high as the actual amount is somewhere around 15 to 30%.

    R. Gates, please learn to read. I said at any given radiation event, the radiation is emitted spherically, with just under half headed back TOWARDS the earth, and the rest headed TOWARDS space … sheesh, I get into enough trouble with what I do say, no need for you to falsify things I’m supposed to have said …

    w.

  111. JKB says:

    Your figure 6. reminded me of an observation I made long ago on the hundreds of mornings watched sunrise from a ship in the tropics. It became sport to try to catch the morning green flash. Problem was, as sunrise approached the clouds (broken cumulus) formed to the East, essentially leading the sun. They would continue to build until the sun was almost at zenith then dissipate until the setting sun caused them to reform due to cooling. This is purely anecdotal but I’d expect the weather observations reported by ships working the TAO array would confirm with their cloud reports. I wouldn’t be surprised if those ships haven’t carried a cloud cover sensor for some period either.

  112. Willis Eschenbach says:

    TimTheToolMan says:
    September 17, 2011 at 8:47 am

    Willis writes… “The ocean gets about 170W/m2 from solar, and it’s losing about 390 W/m2 … if DLR isn’t being absorbed by the ocean, then why isn’t it frozen?”

    …but doesn’t accept that the energy from the DLR forms part of the stefan-boltzmann energy requirement for the ocean to radiate. The upshot of this is that the DLR doesn’t warm the ocean. At leaast not in the way Willis and many others believe can happen.

    OK Willis, from what we know about the skin, where does the DLR energy go? And how does it do it? You need to be quite precise because waving your hands around and saying “it goes into the ocean” simply wont cut it as an answer.

    First answer my question and we can avoid talking past each other. Does the DLR get absorbed by the ocean, and if not, what is warming the ocean? Once we can see what we agree on, we can move forwards.

    w.

  113. Tim Folkerts says:

    Willis,

    “You’re correct, except you’ve neglected the eccentricity of the orbit. This varies the “solar constant” from a low of about 1345 to a high of 1435 W/m2 over the course of the year. “

    Both my own calculations and wikipedia disagree with these numbers. Wikipedia & I get 1412 and 1320 for the two values. Earth moves about 1.67% closer Jan and 1.67% farther in July. This makes the sun about 2 * 1.67% = 3.35% brighter and dimmer than average thru the year, which gives the numbers above.

    “In addition, the infrared from the sun is definitely not absorbed in the atmosphere, it’s the wrong frequency (as I discussed upthread), so it is measured by the pyranometers.”

    These familiar images show that water vapor does indeed absorb noticeable parts of the incoming solar spectrum.
    http://www.howtopowertheworld.com/image-files/solar-spectrum.png
    http://www.sunwindsolar.com/a_images/co2_water_vapour.gif

    The one mechanism I can imagine that might boost the surface irradiance is reflection from clouds. If the sun is shining on the surface, but nearby white clouds are also reflecting some light, that might boost the numbers a bit.

  114. tallbloke says:

    Willis Eschenbach says:
    September 17, 2011 at 2:48 am
    Tallbloke’s explanation was particularly hilarious, involving some kind of invisible mist just above the water that absorbed the DLR in some dance above the surface … but even that couldn’t explain the missing energy needed to keep the ocean from freezing.

    What I was pointing out was that our instrumentation can’t differentiate between upwelling longwave emitted from the contiguous ocean surface, and downwelling longwave absorbed and re-emitted up and down by the highly humid air just above the ocean surface. This is a fact you didn’t try to refute.

    But even if we ignore the reality of the high humidity just above the ocean surface to simplify the argument, there is the issue of what happens to the downwelling longwave after it has been absorbed in the first few um of the ocean surface, which as Tim the Toolman correctly states, is cooler than the rest of the ‘skin layer’ below it. It can’t conduct downwards without defying the second law of thermodynamics, and if the ocean is disturbed by wind the additional downward mixing is more than offset by the extra evaporation wind causes.

    You’re ocean freezing argument fails because we’re not saying the downwelling longwave radiation isn’t absorbed. What we are saying is that it is re-emitted without affecting the bulk temperature of the ocean in any significant way.

  115. Lonnie E. Schubert says:

    Great post Willis.

    dcfl51: One doesn’t have to be a scientist to Google or look it up on wikipedia. http://en.wikipedia.org/wiki/Sunlight It only makes sense that we can see in the spectrum available from the sun. Most of the Sun’s energy is in the visible range, especially when including the near-IR. I find this graphic informative: http://en.wikipedia.org/wiki/File:Solar_Spectrum.png

  116. gnomish says:

    You’re ocean freezing argument fails because
    it ain’t frozen.
    period,

  117. gnomish says:

    it’s surely not up to anybody to explain why the ocean is not frozen because it isn’t.
    but perhaps you can make a model and tweak up some data and maybe you can convince yourself it should be?
    i think you are buying into travesties.

  118. Tim Folkerts says:

    gnomish says: September 17, 2011 at 12:32 pm

    it’s surely not up to anybody to explain why the ocean is not frozen because it isn’t.
    but perhaps you can make a model and tweak up some data and maybe you can convince yourself it should be?

    The point is that Willis believes (as do I) that several models that people are presenting would indeed lead to massive and continued heat loss from the ocean. Therefore their models must be incorrect (or our understanding of their models must be incorrect) since the ocean is indeed observed not be continuously cooling.

  119. Bart says:

    Willis Eschenbach says:
    September 16, 2011 at 10:40 pm

    Without a doubt, on a “micro” level, climate mechanisms are devastatingly complex. So is quantum mechanics. Yet, on a macro level, atomic and molecular systems regress to a mean behavior described by Ehrenfest’s theorem.

    Many (really all, depending on how far down you go) systems are like this – very complicated in the details, yet manifesting a simple order when you pull back from the trees to view the forest. I gave the example of a pulse modulator as a very nonlinear element whose overall actions in a feedback loop can nevertheless be described using linear systems theory.

    So, I think my answer to you is, yes, things may be very complicated. They may not be amenable to such simple modeling. But, then again… they may.

  120. Tim Folkerts says:

    Tallbloke says:

    But even if we ignore the reality of the high humidity just above the ocean surface to simplify the argument, there is the issue of what happens to the downwelling longwave after it has been absorbed in the first few um of the ocean surface, which as Tim the Toolman correctly states, is cooler than the rest of the ‘skin layer’ below it. It can’t conduct downwards without defying the second law of thermodynamics, and if the ocean is disturbed by wind the additional downward mixing is more than offset by the extra evaporation wind causes.

    The top layer doesn’t NEED to conduct any energy downward for the depper layers to warm up.

    Solar radiation already penetrates thru the surface and into the bulk of the ocean (up to ~ 100 m) quite well. To prevent massive warming of the bulk of the ocean, the surface layer needs to conduct energy upwards, which it does because of the aforementioned temperature gradient.

    My point that I was arguing in the previous thread is that if you reduce that temperature gradient of the skin layer by adding energy to the top of that layer — perhaps by adding more downward IR — then the conduction upward will be decreased. The energy from the sun will then not be able to escape from the bulk of the ocean, and the bulk of the ocean will warm.

    To repeat – no conduction downward of energy from IR hitting the surface is needed to warm the bulk! It is sufficient for the downward IR to throttle the escape of energy.

  121. TimTheToolMan says:

    Willis askes “Does the DLR get absorbed by the ocean”

    We’ve already covered this. The DLR is absorbed into the top 10um of the ocean. This much we agree on. The ocean “warms” as a result. We also agree on that. They’re the major points in terms of a macro *effect* but the problem comes when taking a macro *view* of it. Specifically you’ve written…

    “It is actively adding energy to the surface … if you have another name than “warming” to describe “actively adding energy to the surface, as a result of which the surface ends up warmer than if was without the added energy”, let me know.”

    Because no. The surface doesn’t end up warmer. If there were suddenly no GHG’s then instantaneously the surface would be exactly the same temperature despite an instantaneous “drop” of all that energy you believe “warms” it. The reason the surface temperature will drop is because the energy being lost from the ocean is now greater (the DLR used to account for some of this) and it will cool more quickly with an eventual corresponding drop in the SST.

    This as far as I can see is the fundamental difference in our understandings on the ocean and how its effected by GHGs.

  122. Septic Matthew says:

    Willis, I have read all of the posts and I think that what you wrote above is indeed a good next step: I’ll likely look next at splitting the dataset in two, warm dawns versus cool dawns, as I did before.

  123. daniel kaplan says:

    cba says: s
    “”””
    Mark:
    “If the difference is that increased CO2 raises the altitude where the radiation finall
    My no doubt inaccurate paraphrase is that CO2 enrichment raises the altitude at which, seen from space, a given optical depth is reached. Given the lapse rate, that means that the earth’s effective radiation temperature for a given surface temperature decreases: there is less radiation into space.

    There’s two factors going on that will affect the lapse rate. First off, the lapse rate is merely related to the energy flow and conservation of energy. A ‘chunk’ of atmosphere will be at a temperature where the energy coming in equals the energy going out. A change in ghg concentration will cause a bit of shifting around of temperature in order to balance again. A change upward in ghg concentration will also increase the radiative ability of the chunk of atmosphere so that it can radiate more energy at a given temperature which means the temperature change doesn’t have to be as much as a very simplistic calculation might suggest.

    “”””
    Nice comment.
    This is how it goes : In the normal troposphere, the lapse rate is close to constant (temperature decreasing linearly with altitude). This is because thermal exchanges are mainly by molecular motion and a simple thermodynamic transfer process(adiabatic transfer) yields this result. This means that the radiation exchanges are negligible and the difference of temperature between a given altitude and ground is fixed independently of ghg concentration, i.e no corrections of the type you suggest. However with increasing altitude the transition from troposphere to stratosphere is in fact the transition from temperature being controlled by molecular motion to controlled by radiation. In the stratosphere the corrections should be made.

    This is why the IPCC AR4 defines ghg forcing as the change of IR radiation emitted for a given change of ghg concentration, AFTER the stratosphere has been allowed to reequilibrate.

    Note : If the effective altitude of emission (altitude at which the radiation will escape to space without being reabsorbed) is indeed in the stratosphere (for a given IR wavelength), then the lapse rate is of the opposite sign and will lead to a negative change of the greenhouse effect. Corrections are only needed for this negative abnormal response.

    Two remarks :
    1:This business sems to me a major cause of uncertainty in evaluating the CO2 forcings: namely how do you treat the intermediate altitudes of the tropopause?
    2: You can get a derivation of the logarithmic law for the troposphere forcing from a simple heuristic argument based on the lapse rate. I can post it if anyone is interested.

  124. gnomish says:

    Tim Folkerts says:
    September 17, 2011 at 1:13 pm

    so true!
    the facts are right, therefore the models are incorrect.
    maybe radiation physics doesn’t cover phase change work at all…
    maybe the models don’t have a bazillion tons of vapor condensing at sea level every time the sun sets and releasing jiggawatts of latent heat.
    maybe the satellite infrared sensors can’t see thru that and aren’t really reading the ocean at all.
    maybe there’s lots of unknown vulcanism.
    but the models are wrong.

  125. cba says:

    “daniel kaplan says:

    September 17, 2011 at 5:42 pm

    cba says: …

    Daniel,
    There is NO altitude which allows radiation to escape. This is a function of wavelength and on strong line centers, it never happens. Elsewhere in the spectrum, emissions from the surface go straight through. As one goes up in the atmosphere, line widths narrow, increasing the liklihood of capture, but over narrower and narrower bandwidths – hence allowing more energy through. WHen dealing with smaller areas and lower temperature differences, the liklihood of absorption for strong lines tends to approach the liklihood of emission and so there is very little net radiative transfer between these areas. Given two areas of the same temperature, there is no net thermal transfer of any kind. This doesn’t and can’t happen in the atmosphere because of the geometry of the system and we can use the plane radiative transfer approximation for this.

    The lapse rate is strictly a conservation of energy situation. Conduction, convection, and radiation heat transfer in and out define what the lapse rate can be. When one hits the tropopause, that’s where convection is dwindling to a very low contribution. Conduction is not a great factor because air is a superb insulator. Without convection, all you have going is radiation.
    BTW, I tend to use a line-by-line one dimenisional atmospheric model rather than a heuristic approximation. However, the true problem is that even a 1-d model of radiative transfer works well for clear skies but clouds change everything dramatically.

  126. Tim Folkerts says:

    Tim the Toolman,

    Couldn’t I just as well say the following?

    Because no. The surface doesn’t end up warmer. If there were suddenly no GHG’s sunlight then instantaneously the surface would be exactly the same temperature despite an instantaneous “drop” of all that energy you believe “warms” it. The reason the surface temperature will drop is because the energy being lost from the ocean is now greater (the DLR sunlight used to account for some of this) and it will cool more quickly with an eventual corresponding drop in the SST.

    So it would seem that you must also conclude that sunlight does not “warm the ocean”.

  127. Willis Eschenbach says:

    tallbloke says:
    September 17, 2011 at 11:12 am

    Willis Eschenbach says:
    September 17, 2011 at 2:48 am

    Tallbloke’s explanation was particularly hilarious, involving some kind of invisible mist just above the water that absorbed the DLR in some dance above the surface … but even that couldn’t explain the missing energy needed to keep the ocean from freezing.

    What I was pointing out was that our instrumentation can’t differentiate between upwelling longwave emitted from the contiguous ocean surface, and downwelling longwave absorbed and re-emitted up and down by the highly humid air just above the ocean surface. This is a fact you didn’t try to refute.

    Tallbloke, I didn’t try to refute it because it made no sense, you didn’t think it all the way to the end. You know very well that radiation goes (approximately) half upwards and half downwards.

    If we are actually measuring the upwelling longwave “re-emitted up … by the highly humid air just above the ocean surface”, then you are just kicking the can down the road.

    Because if what you say is true, you then have to explain what is happening to the equal but opposite downwelling longwave “re-emitted … down by the highly humid air just above the ocean surface”.

    Either way, you have the full amount impinging on the surface, TB, either directly from higher up or as DLR from a layer just above the surface …

    In any case, we don’t need to measure the upwelling longwave. We can calculate it from the temperature of the ocean water … and you STILL HAVE NOT EXPLAINED WHY IT IS NOT FREEZING TEN FEET BELOW THE SURFACE.

    Again, it’s not a difficult couple of questions:

    1. Is the DLR absorbed by the ocean? Y/N?

    2. If no, what provides the ~ 320 W/m2 of energy that is keeping the ocean from freezing ten feet below the surface? Enter answer here: _________

    w.

  128. Tim Folkerts says:

    Dang, the strikethroughs didn’t work above.
    Lets try again.

    Because no. The surface doesn’t end up warmer. If there were suddenly no GHG’s sunlight then instantaneously the surface would be exactly the same temperature despite an instantaneous “drop” of all that energy you believe “warms” it. The reason the surface temperature will drop is because the energy being lost from the ocean is now greater (the DLR sunlight used to account for some of this) and it will cool more quickly with an eventual corresponding drop in the SST.

  129. TimTheToolMan says:

    Tim Folkerts rewrites “Because no.”…

    Correct. Turn off the sunlight and what happens? The ocean cools because it radiates away its energy. Its a cooling effect. Everything is a cooling effect except for the sunlight itself and that has implication on how clouds behave.

    From what I’ve seen (eg Dessler) there is a general acceptance in the AGW community that more clouds are probably going to be a positive feedback but I think this is nonsense and people who dont call reduced cooling for what it is…reduced cooling…are helping perpetuate these myths particularly with “casual climate observers”.

    On the larger question of whether GHGs play a role in heating the ocean, Willis and I agree. Its the mechanism that we (appear to) disagree on. I think its important to understand the process because without an understanding of the details of the process, how are you expected to understand implication of change that impacts on that process?

    Willis understands this concept because he analyses thunderstorms and their effects in more detail than the AGWers who simply average them, parameterise them and stick them in the models. How are those models expected to get it right when the environment within which those thunderstorms are formed changes…particularly when its projected to change outside of conditions we’ve seen?

  130. Willis Eschenbach says:

    Tim Folkerts says:
    September 17, 2011 at 10:51 am

    Willis,

    “You’re correct, except you’ve neglected the eccentricity of the orbit. This varies the “solar constant” from a low of about 1345 to a high of 1435 W/m2 over the course of the year. “

    Both my own calculations and wikipedia disagree with these numbers. Wikipedia & I get 1412 and 1320 for the two values.

    Thanks, Tim. Actually, upon further investigation I now think that both of us were wrong. We were both looking at the global average, rather than the actual by-latitude figures. From the NASA Monthly Latitude Insolation Calculator, here are the monthly average figures for the insolation at the equator:

    Jan, 1680 W/m2
    Feb, 1736 W/m2
    Mar, 1756 W/m2
    Apr, 1700 W/m2
    May, 1612 W/m2
    Jun, 1552 W/m2
    Jul, 1572 W/m2
    Aug, 1648 W/m2
    Sep, 1716 W/m2
    Oct, 1728 W/m2
    Nov, 1688 W/m2
    Dec, 1652 W/m2
    Ann, 1669.6 W/m2

    The corresponding figures for the global average are:

    Jan, 1412 W/m2
    Feb, 1400 W/m2
    Mar, 1380 W/m2
    Apr, 1356 W/m2
    May, 1336 W/m2
    Jun, 1324 W/m2
    Jul, 1324 W/m2
    Aug, 1332 W/m2
    Sep, 1352 W/m2
    Oct, 1376 W/m2
    Nov, 1396 W/m2
    Dec, 1412 W/m2
    Ann, 1367.2 W/m2

    In summary, you were right and I was wrong about the global average, and we both were wrong about the equator. The figures from the TAO buoys at the equator are well below the theoretical monthly averages at the equator, giving room for the known absorption of incoming sunlight you point out above. Both the size and the interannual range of the equatorial theoretical insolation were larger than I had thought, at about 1550 to 1750 W/m2.

    Always more to learn, I’ll have to think about the implications of all this for my analysis.

    Many thanks,

    w.

  131. Werner Brozek says:

    Let me try to tackle

    “R. Gates says:
    September 17, 2011 at 5:35 am

    Which shows that Willis’ estimate of 50% of the LW going up into space is way too high as the actual amount is somewhere around 15 to 30%.”

    Let us assume we have 360 photons of light that goes at all angles from 0 degrees to 359 degrees. By using sin or cos, we can figure out what fraction of each photon goes in the upward or downward direction and what goes in the horizontal direction. Neglecting the curvature of Earth for a moment, and working in two dimensions, we would find that the total of all components going left would be the same as the total components going right. In addition, the total of all components going up would equal the total of all components going down. And this is where the 50% comes from. Of course, 2 of the 360 rays would have neither an upward nor a downward component.
    Also, a 5% figure was mentioned earlier. If we measure to the nearest degree, then 1/360 of all photons go straight up and also 1/360 of all photons go straight down. On the other hand, if we measure to the nearest 1/10 of a degree and talk about 3600 photons being emitted equally in all directions, then 1/3600 go straight up and 1/3600 go straight down.

  132. Werner Brozek says:

    Alan D McIntire says:
    September 16, 2011 at 6:25 am

    Alan, are you attempting a Q = mcT here with “5* 10^21*1.01*255= 1.288 * 10^24 joules”? If so, you need to realize that the “T” is the change in temperature and NOT the temperature itself. So the 255 is only correct if it goes from 0 to 255 or from 510 to 255 or some other difference between two temperatures of 255.

  133. Spector says:

    Radiation from the Atmosphere
    I have noted elsewhere that the MODTRAN web tool hosted by the University of Chicago, seems to be indicating that most of the Earth’s heat energy radiated to outer space appears to have been emitted from the atmosphere.

    With its default Tropical Atmosphere, No Clouds or Rain settings, MODTRAN seems to show a surface emission power level of 417 W/m2 (altitude = 0 km) as seen looking down and integrated over the wavenumber range of 100 to 1500 cycles per cm. (I understand the official designation for this is 100 to 1500 kayzers.) At the surface, MODTRAN also shows an incoming radiation level of 348 W/m2 arriving from the atmosphere above. Based on this it would seem that the surface can only be losing *radiant* power at a rate of 69 W/m2. Yet out at an altitude of 70 km, I see a net energy flow of 288 W/m2 actually leaving the Earth.

    On a level by level basis, I see progressively more energy escaping to higher altitudes, based on the differences between looking down and looking up. As the high altitude spectrum has a deep hole around the CO2 absorption line at 667 kayzers, I can only assume that MODTRAN believes that H2O is a leaky greenhouse gas and it allows energy to escape to outer space at all altitudes where it is a significant component of the atmosphere.

    I assume atmospheric radiation would force air to drop ever lower as it cooled until it reaches the surface, where it would have an opportunity to cool the surface by contact and evaporation.

  134. TimTheToolMan says:

    @ The other Tim

    One more thinkg that I feel needs to be pointed out, though. Where I say…

    “If there were suddenly no GHG’s then instantaneously the surface would be exactly the same temperature despite an instantaneous “drop” of all that energy you believe “warms” it.”

    The point is that DLR is absorbed in the top 10um of the ocean. The top 10um has a certain heat capacity and the temperature gradient in the region is such that any heat cannot be conducted downwards into the ocean.

    So there is a fundamental difference between stopping DLR and DSR instantaneously. DSR effects the bulk of the ocean where there is a lot of heat capacity whereas DLR only effects the top 10um and that has implications on what the process is in that region based on the time taken to heat the top 10um vs observed changes there.

  135. Tim Folkerts says:

    Willis,

    I think you must be mis-interpreting their calculations.

    The total energy from the sun (ie luminosity) is 3.84E+026 W. If you divide this evenly over a sphere the radius of the earth’s orbit, you get 1365 W/m^2, which is the solar constant.

    Every square meter directly facing the sun will get this energy. Since there is no more than 1365 W/m^2 available, I cannot imagine any way that some areas could get nearly 30% more power than the output of the sun. Your idea to simply multiply the numbers on the chart by 4 are not correct.

  136. Willis Eschenbach says:

    Tim Folkerts says:
    September 17, 2011 at 10:30 pm

    Willis,

    I think you must be mis-interpreting their calculations.

    The total energy from the sun (ie luminosity) is 3.84E+026 W. If you divide this evenly over a sphere the radius of the earth’s orbit, you get 1365 W/m^2, which is the solar constant.

    That’s what I thought too, Tim … but when I use their figures for the global average it works out to 1365 W/m2 (actually 1367.2 W/m2). So that seems to indicate it’s real … but I’m always willing to learn, and it still seems strange to me. Maybe the units are not W/m2 … but then why would the global average be 1365?

    Always more mysteries,

    w.

  137. Willis Eschenbach says:

    Upon further thought, Tim, I think you’re right, I can’t just multiply them by 4. The reason that some monthly averages are high is the amount of time under the sun, not the overall intensity of the sun.

    Hmmm …

    w.

  138. tallbloke says:

    Tim Folkerts says:
    September 17, 2011 at 1:13 pm
    our understanding of their models must be incorrect

    Yes. Your understanding of our understanding of the observations (not a model) is incorrect. I take no pleasure, because it means we’re not explaining it clearly enough. Although I suspect there’s an element on your side of it which is also preventing your understanding of what we are saying.

    Tim Folkerts says:
    September 17, 2011 at 4:00 pm

    My point that I was arguing in the previous thread is that if you reduce that temperature gradient of the skin layer by adding energy to the top of that layer — perhaps by adding more downward IR — then the conduction upward will be decreased.

    And yet you didn’t, despite several requests, shown us any empirical data which demonstrates that:
    a) There has been any such additional DLR (might have been offset by something else changing). or
    b) That the ocean skin gradient would be affected by a tiny change in DLR (the surface pressure and other lower toposphere and boundary layer factors completely overwhelm the effects of small changes in the IR flux).

    and c) all people on your side of the debate are unable/unwilling to discuss the empirical data presented here:

    http://tallbloke.wordpress.com/2011/09/17/cloud-albedo-what-does-it-respond-to/

  139. tallbloke says:

    TimTheToolMan says:
    September 17, 2011 at 9:51 pm
    Everything is a cooling effect except for the sunlight itself and that has implication on how clouds behave.

    Agreed, see my new post linked below.

    people who dont call reduced cooling for what it is…reduced cooling…are helping perpetuate these myths particularly with “casual climate observers”.

    Agreed, it causes endless confusion and unnecessary argument.

    On the larger question of whether GHGs play a role in heating the ocean, Willis and I agree. Its the mechanism that we (appear to) disagree on. I think its important to understand the process because without an understanding of the details of the process, how are you expected to understand implication of change that impacts on that process?

    Agreed, except once again, it’s about the extent to which the GHG’s allow the ocean to cool from the solar input. GHG’s cannot significantly warm the ocean directly, only by changing the air temperature, and this mechanism, while effective over aeons, is a very, very slow way to change the bulk temperature of the ocean and cannot explain the warming rate from 1980-1998. It’s the reduction in low tropical cloud empirically measured by ISCCP and the Earthshine project which explains that.
    http://tallbloke.wordpress.com/2011/09/17/cloud-albedo-what-does-it-respond-to/

  140. tallbloke says:

    Willis Eschenbach says:
    September 17, 2011 at 9:11 pm

    2. If no, what provides the ~ 320 W/m2 of energy that is keeping the ocean from freezing ten feet below the surface.

    Since the DLR doesn’t get down ten feet, the ocean isn’t trying to lose 320W/m^2 from 10 feet down. It’s losing the 170W/m^2 put there by the Sun which builds up the ocean temperature. Nearly all the DLR absorbed in the first few nm (answer to Q1) is re-emitted or goes into latent heat of evaporation from the first few nm.

    (empirical facts highlighted in bold)

    I agree with Tim the Tool Man and you that the whole of the ocean is warmer than it would be with no GHG’s. But as Tim says, you need to understand the mechanism by which the GHG’s bring that about so you can correctly see what changes will do. DLR doesn’t cause warming by direct radiation and the mixing of energy downwards, that doesn’t happen to any significant extent. It makes the air warmer, and this change in air temperature will only change the bulk ocean temperature very, very slowly.

    Too slowly to account for the warming 1980-1998. The reduction in tropical low cloud did that, by allowing more Solar shortwave radiation into the ocean.
    http://tallbloke.wordpress.com/2011/09/17/cloud-albedo-what-does-it-respond-to/

  141. Willis Eschenbach says:

    OK, the useful figure is the actual average insolation. For the buoy above, it was 245 W/m2. The NASA data puts the theoretical average at 417 W/m2.

    The maximum data I used (blue line in Figure 4) gives an average of 420. However, that needs to be corrected to give us the mean, rather than the maximum, value. This is a double correction, for both the eccentricity and for the north/south movement of the sun (declination). The net of those is about 0.92. This gives 420 * .92 = 386 W/m2 for the clear-sky value.

    So clear-sky is 386 and all-sky is 245, which gives a cloud shortwave forcing of – 141 W/m2.

    Add in the + 36 W/m2 of LW forcing, and we get a net cloud forcing of – 105 W/m2, still strong cooling.

    w.

  142. tallbloke says:

    Willis Eschenbach says:
    September 18, 2011 at 3:05 am

    OK, the useful figure is the actual average insolation. For the buoy above, it was 245 W/m2. The NASA data puts the theoretical average at 417 W/m2.

    So clear-sky is 386 and all-sky is 245, which gives a cloud shortwave forcing of – 141 W/m2.

    Add in the + 36 W/m2 of LW forcing, and we get a net cloud forcing of – 105 W/m2, still strong cooling.

    Is your buoy under the ITCZ? Maybe you’ll get a less seasonally biased result a bit (but not much) further from the equator.

    If you are interested in the maximum effect, maybe consider averaging the data from two buoys, one either side of the ITCZ?

  143. TimTheToolMan says:

    “Add in the + 36 W/m2 of LW forcing, and we get a net cloud forcing of – 105 W/m2, still strong cooling.”

    A simple and effective way of looking at it…in the tropics at least. You’re ignoring aerosols and any other factors but as a first approximation I suspect its a good one and certainly seems useful.

    I wonder if thats a better way of calculating the effect of clouds. Doing it in 5×5″ grids in a similar way temperature is calculated. Has anyone done that?

  144. cba says:


    Willis Eschenbach says:

    September 18, 2011 at 3:05 am

    OK, the useful figure is the actual average insolation. For the buoy above, it was 245 W/m2. The NASA data puts the theoretical average at 417 W/m2. …”

    You might want to check out my second post on this thread as it has some of the rough global averages.

    You must take care fooling around with the variation to the solar insolation due to Earth’s orbit. It has a peak to peak change of around 100w/m^2 but it is a sine function not some straight line function.

    The cloud factor is actually the vast majority of the albedo for visible light. Using your original numbers, it showed that the cloud albedo for what you were dealing with had to be around 60% plus for its reflectivity. In LW clouds will block a lot of radiation but the tops will emit radiation based upon their temperature just as cloud bottoms do.

  145. Martin Lewitt says:

    Willis,

    “The ocean gets about 170W/m2 from solar, and it’s losing about 390 W/m2 by radiation alone, not counting sensible and latent heat loss … if DLR isn’t being absorbed by the ocean, then why isn’t it frozen?”

    If the ocean were a perfect reflector of the DLR accounting for most of the 390W/m^2 that it is “losing by radiation alone” would it still be a mystery to you why it isn’t frozen? It has been warmed by the sun. Radiation, latent heat flux and conductive coupling to the atmosphere is how the solar energy is lost.

    Very little of the upward longwave radiation that the ocean is emitting is due to reflection, because the albedo of the ocean is low at IR wavelengths, but since IR penetrates mere microns into the ocean, the argument is that as a skin effect, it is more directly and efficiently converted to upward LR and latent heat, You should be able to conceptualize how efficient upward re-emission would similar to reflection.

    Solar short wave radiation penetrates 10s of meters into the ocean and the energy deposited in a much larger thermal mass may be transported thousands of miles before being radiated or being converted to latent heat.

  146. Septic Matthew says:

    Willis, you wrote: In any case, we don’t need to measure the upwelling longwave. We can calculate it from the temperature of the ocean water … and you STILL HAVE NOT EXPLAINED WHY IT IS NOT FREEZING TEN FEET BELOW THE SURFACE.

    I think you have gotten distracted. I think that you want to show, from your data set, that something intervenes between morning and evening to turn above average am temperatures into below average evening temperatures. You have hypothesized that above average early am temps produce above average mid-day cloudiness which results in below average afternoon insolation, thus producing below average evening temperatures. You have shown that you can infer cloudiness from the temperature profile, which is good, but you can’t test from your data set whether other mechanisms are at work or can be excluded. What you can do, that is more direct and less bound to any particular mechanism, is test whether above average am temperatures are correlated with below average afternoon downward total radiation.

    To do this, sort the data by buoy (so each buoy is its own control) and month (to control for seasons, though there is a better way) and (maybe) by year (to remove another source of variation, though you could model it), and then correlate 9 am temp ( or a mean of 8 – 11, or some other composite) with 4 pm downward radiation (or a mean of 3 – 5 pm). If your hypothesis is correct, you ought to get negative correlations mostly, and a strong negative mean correlation across the sample of buoys. If these correlations are mostly clustered around 0, with a near 0 mean, then your hypothesis is not supported.

    Of course, you can modify this procedure, but you get the idea.

    Radiation is not the only mechanism for differential rate of heat removal/addition am vs. pm, but it is one that you can test from your data. You could also try computing the total energy drained by the evaporation of water, above average am vs below average am (again, sorted by buoy and season.) I think that introduces more problems than it solves, but it is probably worth a try eventually.

  147. Tim Folkerts says:

    “people who dont call reduced cooling for what it is…reduced cooling…are helping perpetuate these myths particularly with “casual climate observers”. “

    Actually, I would go a step further and say ““people who don’t call reduced cooling for what it is…reduced cooling “less energy loss”…are helping perpetuate these myths particularly with “casual climate observers”. “

    The distinction is even more important for the reverse — is the opposite of “cooling” the same as “warming” or the same as “heating”? “Warming” has the very clear meaning of “raising the temperature”, whereas in thermodynamics, “heating” mean “the net transfer of energy due to a temperature difference.” People who don’t appreciate the difference are doomed to argue semantics rather than science. It is quite possible to heat something without warming it. It is quite possible to warm something without heating it.

    Discussing from a vantage point of energy considerations is more fundamental that than discussing from a vantage point of temperature. (For one thing, conservation of energy is such a basic principle, that if there was an experiment that claimed to violate conservation of energy, my first (and second) reaction would be to question the data, not question the theory.)

    So “the energy of the ocean is increased by incoming thermal IR photons (which happen to come from GHGs and clouds)” just like “the energy of the ocean is increased by incoming UV/visible/NearIR photons (which happen to come from the sun)”. There is no need for “reduced cooling” or “increased warming”. Either source of energy adds energy to the ocean (which almost sounds redundant, but I think is an important point) and once the energy has been added and converted to thermal energy, it matters not one iota what the original source was. (The only distinction would be if you then want to look at specific parts of the ocean — skin vs bulk; tropical vs polar; …)

    Similarly, we can look at the energy loss mechanisms (eg upward IR, convection, evaporation). These all remove energy. If any one of these increases call it what it is “an increase in the rate of energy transfer from the ocean.” Calling it either “increased cooling” or “decreased warming” distracts from what is fundamentally happening!

  148. Tim Folkerts says:

    Willis posits: “The reason that some monthly averages are high is the amount of time under the sun, not the overall intensity of the sun.”

    You hit the nail on the head. For instance, the value for the North Pole in June is very close to
    ___ 1365 W/m^2 sin(23)
    since the sun is ~ 23 degrees above the horizon 24/7

    The value at the equator for March or Sep is close to
    ___ 1365 W/m^2 * pi
    where I am convinced that pi=3.14 is the appropriate geometric factor, not 4

  149. Tim Folkerts says:

    @ tallbloke

    And yet you didn’t, despite several requests, shown us any empirical data
    which demonstrates that:
    a) There has been any such additional DLR (might have been offset by something else changing). or
    b) That the ocean skin gradient would be affected by a tiny change in DLR (the surface pressure and other lower toposphere and boundary layer factors completely overwhelm the effects of small changes in the IR flux).
    and c) all people on your side of the debate are unable/unwilling to discuss the empirical data presented here:

    http://tallbloke.wordpress.com/2011/09/17/cloud-albedo-what-does-it-respond-to/

    a) I wish I had data handy. Remember, I am just a guy interested in science commenting on a blog — I don’t have access to data anymore than you do.

    Measuring DLR on a global scale would be very challenging, since there are only limited pyrgeometers around the world collecting. And a quick look around the web suggests they are a bit challenging to keep in calibration and that much of the work has only been done in the last decade. http://www.pmodwrc.ch/pmod.php?topic=irc

    b) The experiment you describe would also be interesting. I suspect that it has been done in various forms over the years (since evaporation is a rather basic concept).

    c) Ideally, there are not “people on your side of the debate”, but only scientists interested in deeper understanding. The cloud/abledo effect you describe is clearly as part of the situation and bears further study to see how much affect it has. What papers on the topic have you found and read?

    Three notes:
    1) Your second graph should be titled “Specific Humidity and Sunspot Number vs Time”
    2) The sunspot numbers are odd. SSN drops very close to zero at the end of every cycle, yet your plot shows large variations for the minima.
    3. I disagree with the line “Something had to be responsible, because as we know from previous lengthy debates here and on WUWT, back radiationdoesn’t heat the ocean”. If science is not settle by consensus of experts, it is most certainly NOT settled by consensus on blogs.

  150. TimTheToolMan says:

    The other Tim writes (amongst other things) ““Warming” has the very clear meaning of “raising the temperature””

    The problem is that warming doesn’t have a very specific meaning in common parlance. You say “raising the temperature” but warming can equally mean warming by putting on a jumper or warming by standing in front of the fire.

    Warming by standing in front of the fire is precisely not what GHGs do even though a simplistic energy flow consideration, mathematically, amounts to the same thing. Its only when you properly consider conservation of energy or in a general sense that the energy that comes from the clouds came from the ocean in the first place that an appreciation for the effect can be made.

    As an example, what would happen to the earth if somehow it was completely covered in cloud? Some people might be tempted to think that with all that increased DLR and a believed positive feedback associated with more cloud that the earth would actually warm up.

  151. Tim Folkerts says:

    TimTheToolMan says:

    The problem is that warming doesn’t have a very specific meaning in common parlance. You say “raising the temperature” but warming can equally mean warming by putting on a jumper or warming by standing in front of the fire.

    I’ll give you that — which is yet one more reason to avoid “warming”, “reducing the cooling”, etc and stick to statements about energy.

    Warming by standing in front of the fire is precisely not what GHGs do even though a simplistic energy flow consideration, mathematically, amounts to the same thing. Its only when you properly consider conservation of energy or in a general sense that the energy that comes from the clouds came from the ocean in the first place that an appreciation for the effect can be made.

    This confuses me. I could say that ALL the energy came from the sun — including the energy from the burning logs. So now we have the energy coming from the same source AND being the “same mathematically”. What other reason can there be to call the energy different? Does a 3 um photon from a cloud add energy to you in any different way than a photon from the fire would? How does a “proper consideration of conservation of energy” differentiate a photon from a GHG molecule vs a photon from a fire vs a photon from the sun?

    At a more fundamental level, I would say that the source of the energy is irrelevant. Some of the energy in the cloud came from the ocean; some came directly from the sun; some came from radioactive decay; some from kinetic energy of in-falling asteroids stored as thermal energy since the earth was formed. This doesn’t make an iota of difference. Once the energy is in the cloud, it is the cloud’s energy, pure and simple. There is no little tag attached to a bit of energy saying “I came from convection from the ocean” or “I came from a photon produced by a GHG molecule”. The cloud doesn’t “re-radiate” IR energy it received, it simply radiates ITS OWN energy.

  152. daniel kaplan says:

    cba says

    There is NO altitude which allows radiation to escape. This is a function of wavelength and on strong line centers, it never happens. Elsewhere in the spectrum, emissions from the surface go straight through. As one goes up in the atmosphere, line widths narrow, increasing the liklihood of capture, but over narrower and narrower bandwidths – hence allowing more energy through. WHen dealing with smaller areas and lower temperature differences, the liklihood of absorption for strong lines tends to approach the liklihood of emission and so there is very little net radiative transfer between these areas. Given two areas of the same temperature, there is no net thermal transfer of any kind. This doesn’t and can’t happen in the atmosphere because of the geometry of the system and we can use the plane radiative transfer approximation for this.
    —————-
    It does escape even at the central C02 line near 670 cm-1 , at a sufficiently high altitude. If the absorption length is 10 meter, then at altitudes where the pressure is 1/1000 of ground atmospheric pressure, the absorption length becomes comparable to the atmospheric thickness constant. z0 (pressure dependence p =p0 exp(-z/z0)), even allowing for line narrowing.This is why the central CO2 line appears as a strong peak in satellite top of the atmosphere radiation spectra, corresponding to the high temperatures in the high stratosphere.

    Your consideration of pressure dependence of the spectral width is the reason I like to use the following heuristic.

    Compute the absorption length of a cylinder of atmosphere of length z0, under the conditions (pressure and temperature) of the tropopause. Any wavelength that is strongly absorbed, e.g. 30% transmission or less, has to correspond to an atmospheric radiation effective altitude in the stratosphere. Any wavelength that is weakly absorbed e.g. 90% transmission or more , corresponds to radiation from below, mostly close to the ground. All the others yield the “normal” greenhouse effect dependent of concentration through the troposphere lapse rate.

    This exchange being somewhat out of topic in this post on cloud effects, with multiple comments, it is probably advisable to continue in another setting.

  153. tallbloke says:

    TimTheToolMan says:
    September 18, 2011 at 7:01 pm
    Its only when you properly consider conservation of energy or in a general sense that the energy that comes from the clouds came from the ocean in the first place that an appreciation for the effect can be made.

    This is another source of confusion. Most of the DLR came from the ocean before it came back down from the clouds. But according to the Trenberth-Keihl cartoon, around 66W/m^2 is ‘new’ LR derived from the solar energy absorbed in the atmosphere. Coincidentally this is around the same as the ~66W/m^2 net radiative loss from the ocean. 33 of this will come downwards, but it seems the amount of solar energy absorbed by clouds has been underestimated by around 30W/m^2 according to measurements taken by planes simultaneously flying above and below cloud decks. This is likely due to a models being based on imperfect physics around the issue of Mie scattering and forward propogation.
    http://tallbloke.wordpress.com/2010/11/14/alistairmcd-aerosols-cause-warming/#comment-7645

    Tim F says:
    The sunspot numbers are odd. SSN drops very close to zero at the end of every cycle, yet your plot shows large variations for the minima.

    The sunspot numbers are averaged over 96 months.

    Measuring DLR on a global scale would be very challenging, since there are only limited pyrgeometers around the world collecting. And a quick look around the web suggests they are a bit challenging to keep in calibration

    Douglas Hoyt (the worlds foremost expert in pyrgeometry) shows they don’t need to be calibrated against each other, and that optical depth hasn’t changed in seventy years at various locations.

  154. TomVonk says:

    Willis

    I hope that you have read my first post even if you did not comment on what seems a very important result.
    Indeed do you realize that the Figure 4 falsifies your favorite “thermostat” theory based on the daily asymmetrical tropical cloud cycle (e.g clear mornings and cloudy afternoons) ?
    Figure 4 (all skies) shows exactly the contrary – there are (in average !) as many clouds in the morning as in the afternoon, at sunrise as at sunset.
    Did you understand my suggestion to show the curve of the minimal SW radiation which corresponds to fully clouded skies?
    You can then count how often the skies were fully cloudy at every hour and see if it is still symmetrical.

    Based on the Figure 4 you must now either abandon idea of asymmetrical daily cycles of clouds because they are contradicted by the data or if you want to maintain the “thermostat” theory and the existence of daily cloud cycles then you must explain why the all sky curve of Figure 4 is symmetrical and shows no trace of a daily cloud assymmetry.

    Otherwise the Figures show there is less SW during cloudy skies than during clear skies but everybody will agree that this is a very trivial observation. Producing from this data saisonal averages of cloudiness f.ex 3 months (cloudiness definition ; SW all skies/SW clear skies) and looking if there are either global or temporally localised trends would be obviously less trivial and certainly more interesting.

  155. Tim Folkerts says:

    Tallbloke says:

    “Douglas Hoyt (the worlds foremost expert in pyrgeometry) shows they don’t need to be calibrated against each other, and that optical depth hasn’t changed in seventy years at various locations.”

    I find it fascinating how people (on both sides) accept information that agrees with their views, and are skeptical only of information that contradicts their views.

    * Douglas Hoyt is THE foremost expert? Even his own bio says nothing about prygeometry (although he has a couple papers on phyheliometry). http://www.warwickhughes.com/hoyt/bio.htm
    * Instruments don’t need to be calibrated??? Honestly, would you be so generous if Mann claimed such a thing? In many cases, only relative change is important and uncalibrated instruments work pretty well. But for long-term studies, involving small changes in difficult to measure numbers, I would think calibration would be quite important. Do you have a reference for this claim?
    * Typically optical depth for the atmosphere refers specifically at visible light, which would be immaterial for measurements of thermal IR from the atmosphere. Do you have a reference for your claim about optical depth, specifically as it relates to thermal IR?

  156. tallbloke says:

    Tim F
    Sorry, my mistake. I was getting mixed up with pyrheliometry. See the paper Hoyt co-wrote with Frohlich on pyrheliometry measurements between 1920-90 at Davos. Changes in DLR due to co2 change should be smaller than those due to aerosols absorbing/ reflecting sunlight. Not that the Team ever seems to be able to make up it’s mind which it is or what the overall effect is…

  157. Willis Eschenbach says:

    TomVonk says:
    September 19, 2011 at 4:06 am

    Willis

    I hope that you have read my first post even if you did not comment on what seems a very important result.
    Indeed do you realize that the Figure 4 falsifies your favorite “thermostat” theory based on the daily asymmetrical tropical cloud cycle (e.g clear mornings and cloudy afternoons) ?
    Figure 4 (all skies) shows exactly the contrary – there are (in average !) as many clouds in the morning as in the afternoon, at sunrise as at sunset.

    I saw it, and it got lost in the flood. Thanks for your thoughts.

    The morning/afternoon cloud shortwave forcing is indeed asymmetrical. It is lost in the graph, because the maximum is not smoothed. In fact, the average SW forcing from 6AM to Noon is -339 W/m2, while the Noon to 6 PM average forcing is -361 W/m2.

    It is important to note that this underestimates the actual forcing, because this is an average. Remember that on a hot day, the clouds form earlier, and on a cold day they form later. When you average them, it temporally smears out (and thus diminishes the amplitude of) the actual on-the-ground effect because the timing and amount of clouds is temperature dependent.

    Which is another example of why averages need to be interpreted with great caution.

    w.

  158. Tim Folkerts says:

    Just a couple other comments/questions, Tallbloke

    “The sunspot numbers are averaged over 96 months.”
    If that is the case and you are averaging last 48 months and the next 48 months (as it appears you are doing), then you are correlating today’s humidity to the solar cycle over the next 4 years. It seems very odd to try to predict current humidity using data you will not have for 4 more years. Or perhaps you aer saying you can reverse the process and use humidity to predict the upcoming solar cycle?

    “Most of the DLR came from the ocean before it came back down from the clouds.”
    OK, but I could just as well say “Most of the ULR came from the atmosphere before it came back up from the oceans.” (After all, the ocean received 325 W/m^2 from the atmosphere, but only 168 W/m^2 from the sun) What relevance does either have on the physics? The ocean radiates according to the energy it CONTAINS (as expressed in its temperature); the atmosphere radiates according to the energy it CONTAINS (as expressed in its temperature). Neither the atmosphere nor the surface knows nor cares where that energy may have been earlier.

    “Coincidentally this is around the same as the ~66W/m^2 net radiative loss from the ocean. 33 of this will come downwards…”
    In the trenberth diagram, the atmosphere radiates about ~ 325 W/m^downward (to the surface), and ~ 200 W/m^2 upward (to space), so ~ 62% of the energy goes down and 38% goes upward. If you really want to break it down by “sources” you could claim:
    62% of ~ 350 W/m^2 (IR energy received from the ocean) goes down
    62% of ~ 80 W/m^2 (evaporative energy from the ocean) goes down
    62% of ~ 25 W/m^2 (convective energy from the ocean) goes down
    62% of ~ 65 W/m^2 (SW photon energy from the sun) goes down

    NET: ~ 325 W/m^2 goes down. If you re going to treat the atmosphere as one object (a la Trenberths diagram), then you have to treat all the energy as going to/from the atmosphere as a whole.

    I really don’t see how you can talk about where “net loss” goes.

  159. Willis Eschenbach says:

    TomVonk says:
    September 19, 2011 at 4:06 am

    … Producing from this data saisonal averages of cloudiness f.ex 3 months (cloudiness definition ; SW all skies/SW clear skies) and looking if there are either global or temporally localised trends would be obviously less trivial and certainly more interesting.

    There’s lots and lots of things to do with the TAO dataset that would be interesting, Tom … but I’m one guy with no associates and no graduate assistants and no funding and no supercomputer. It’s just me with a day job, my trusty Mac, and my imagination … and all of that can only do so much in a given time.

    I’m sorry if my ongoing series of reports of my preliminary results aren’t up to your standards, and you find them trivial and less than interesting … so how about you produce some non-trivial results from the TAO dataset that you think are more interesting and report back to us?

    What’s that you say? You’re busy and that kind of analysis would take a while? Yes, I understand … and I hope you will as well.

    I’m doing my best with the time I have available, following the avenues I see and reporting on what I find. As I mentioned above, my next move will likely be to split the data into the warm dawns and the cool dawns and re-analyze it to see what happens in the ensuing days. As far as I know, I’m the only person doing a “warm dawn vs. cool dawn” type of analysis, and it provided fascinating results when I last used it on the TAO dataset, so I hope you will find it more interesting …

    w.

  160. Willis Eschenbach says:

    TimTheToolMan says:
    September 18, 2011 at 7:01 pm

    Warming by standing in front of the fire is precisely not what GHGs do even though a simplistic energy flow consideration, mathematically, amounts to the same thing. Its only when you properly consider conservation of energy or in a general sense that the energy that comes from the clouds came from the ocean in the first place that an appreciation for the effect can be made.

    Mathematically, it’s the same thing as warming, but it’s not warming because of where the energy came from?

    That is a most curious claim. I’ve never heard that whether something warmed something else depended on where the energy came from … particularly since in this case, all of the energy came from the sun.

    w.

  161. Septic Matthew says:

    Willis wrote: There’s lots and lots of things to do with the TAO dataset that would be interesting

    I have created a directory called “Eschenbach”, and I have started downloading the TAO data into it. I hope to expand upon your work in the coming weeks. I think, as I wrote, that you had a good idea. There’s wind speed data, rainfall data, and so forth. It’s a magnificent data set, apparently neglected before you.

  162. gnomish says:

    http://modis.gsfc.nasa.gov/data/atbd/atbd_mod05.pdf details galore

    well, if the sea’s skin is cooler than what’s above and below – obviously it’s not because of conduction and obviously not by radiation or obviously it can’t be convection.
    that means it’s by phase change – evaporation.
    that means there is a saturated layer of water vapor over the skin.
    the layer of dense water vapor has optical thickness, doesn’t it?
    it’s not ‘net releasing’ heat – it’s evaporating – it’s is sequestering heat – and the thermometers don’t measure the quantity of heat.

  163. TimTheToolMan says:

    Willis writes “Mathematically, it’s the same thing as warming, but it’s not warming because of where the energy came from? That is a most curious claim. I’ve never heard that whether something warmed something else depended on where the energy came from … particularly since in this case, all of the energy came from the sun.”

    If the sun increased its average output by 3.7W then there aren’t too many people in the world who would believe the earth wouldn’t heat up. Thats the fire analogy. However adding a thicker jumper isn’t necessarily going to heat you up if say you went from a thinner white jumper to a thicker black jumper which radiates better.

    This is essentially your argument (and my belief too) that the climatic processes will tend to arrange themselves to maximise energy loss. Its much harder to escape heating from an actual energy increase from the sun.

    So why does it matter that its cooling not heating? Well because its the jumper analogy that we’re looking at and not the fire analogy. A slower rate of cooling due to GHGs can be trumped (or at least offset) by a higher rate of cooling from other processes.

  164. Tim Folkerts says:

    gnomish says:
    September 19, 2011 at 4:22 pm

    “well, if the sea’s skin is cooler than what’s above and below …
    The clouds above and much of the atmosphere above and outer space above are colder than the surface. So this statement is wrong.

    – obviously it’s not because of conduction and obviously not by radiation or obviously it can’t be convection.
    Since the original hypothesis was false, this conclusion is false. Radiation can and does contribute to the cooling of the surface of the ocean. As does evaporation.

    that means it’s by phase change – evaporation.
    that means there is a saturated layer of water vapor over the skin.

    There is only a saturated layer if the relative humidity is 100%. (And at that point there is no more net evaporation, since the saturated vapor is returning to the liquid as fast as the liquid is evaporating into the gas phase. That is pretty much the definition of “saturated”.)

    the layer of dense water vapor has optical thickness, doesn’t it?
    Sure, any layer of gas has an optical thickness. What does this relate to?

    it’s not ‘net releasing’ heat – it’s evaporating – it’s is sequestering heat – and the thermometers don’t measure the quantity of heat.
    Evaporation only sequesters (isolates and/or stores) energy to the extent that there is a NET evaporation of water. Since water cycles from evaporation (sequestering energy) thru condensation (releasing the stored energy), the net change is basically zero. (When there is a net warming of the atmosphere & oceans, then could be a slight increase in global humidity, resulting in a slight sequestration of energy, but I am sure this is a tiny number and not important on a global level. )

    All that evaporation really accomplishes is a net transfer of energy from the oceans to the atmosphere (~ 80 W/m^2).

    It is true that thermometers do not measure this energy, but since this energy it is pretty much constant on a global, long-term basis (since humidity is pretty much constant on a global, long-term basis), it doesn’t really matter.

  165. gnomish says:

    i meant that ‘that which is in contact and immediate proximity’ with that surface layer, of course – not outer space or clouds a mile above it- you know- the saturated air near in contact – the interface.
    experiment i just conducted:
    heat cup of water in microwave. leave cup 1 inch underfilled.
    take reading with IR thermometer: 149-159F (don’t fog the thermometer!)
    cross ventilate to remove saturated vapor layer
    take reading with IR thermometer: 135-141F
    do it several times with the same cup to make sure it’s not a fluke.
    so – what material object is actually being detected?
    NOTE: ir thermometer can not ‘see’ water below the surface.
    NOTE: ir thermometer can not ‘see’ water beneath vapor.

    and of course there is a net evaporation of the water from that surface – that’s how we get rain, right? it doesn’t stay on the surface, either, as it rises, being the least dense gas in the atmosphere by a significant margin – that’s how we get rain clouds, right?
    and when it rises, it’s replace by other, dryer air that moves in along the lowest surface, sweeping along those water molecules that randomly acquired enough energy to evaporate- thus cooling the ‘skin’ which provided the calories and becoming saturated again.

    care to try again with these constraints clarified? i tried to reduce the quibbling quotient.

  166. gnomish says:

    heh- but you knew that there was net evaporation, for you say:
    “All that evaporation really accomplishes is a net transfer of energy from the oceans to the atmosphere (~ 80 W/m^2).”
    and so your remark about
    “Evaporation only sequesters (isolates and/or stores) energy to the extent that there is a NET evaporation of water. Since water cycles from evaporation (sequestering energy) thru condensation (releasing the stored energy), the net change is basically zero.”
    is not a refutation of anything.

    no rhetorical roulette, plz, if that’s what it was.

  167. gnomish says:

    “humidity is pretty much constant on a global, long-term basis”
    of course. but every single day/night cyle it varies hugely – definitely not constant.
    this is an example of how ‘averaging’ simply destroys information and an illustration of the faulty conclusions that proceed from relying on data homogenized beyond recognition.

  168. Martin Lewitt says:

    @Tim Folkerts,

    “All that evaporation really accomplishes is a net transfer of energy from the oceans to the atmosphere (~ 80 W/m^2).

    It is true that thermometers do not measure this energy, but since this energy it is pretty much constant on a global, long-term basis (since humidity is pretty much constant on a global, long-term basis), it doesn’t really matter.”

    You might be thinking of relative humidity rather than absolute humidity. Negative feedback from a couple extra turns of the water cycle might well matter for both the actual climate sensitivity and the model projections. Models which couple CO2 forcing to the whole mixing layer of the ocean rather than properly representing the shallow skin coupling and resulting differences in climate feedback vis’a’vis solar and other forcings are arguably not yet skillful in projection of either warming or precipitation. Consider the issues Wentz raised in the journal Science.

    “Climate models and satellite observations both indicate that the total amount of water in the atmosphere will increase at a rate of 7% per kelvin of surface warming. However, the climate models predict that global precipitation will increase at a much slower rate of 1 to 3% per kelvin. A recent analysis of satellite observations does not support this prediction of a muted response of precipitation to global warming. Rather, the observations suggest that precipitation and total atmospheric water have increased at about the same rate over the past two decades.”

    “There is a pronounced difference between the precipitation time series from the climate models and that from the satellite observations. The amplitude of the interannual variability, the response to the El Niños, and the decadal trends are all smaller by a factor of 2 to 3 in the climate model results, as compared with the observations.”

    “The difference between a subdued increase in rainfall and a C-C increase has enormous impact,
    with respect to the consequences of global warming. Can the total water in the atmosphere
    increase by 15% with CO2 doubling but precipitation only increase by 4% (1)?Will warming
    really bring a decrease in global winds? The observations reported here suggest otherwise, but
    clearly these questions are far from being settled.”

    How Much More Rain Will Global Warming Bring? Frank J. Wentz,* Lucrezia Ricciardulli, Kyle Hilburn, Carl Mears
    Published online 31 May 2007; 10.1126/science.1140746

    I heard a talk by Lindzen in which he claimed that results for latent heat flux are consistent with the observed increase in precipitation rather than with the models, but I didn’t get the reference. Hopefully, someone will supply that.

  169. Tim Folkerts says:

    Gnomish,

    Certainly variations in water evaporation from day to night or from summer to winter are important. More water evaporates during the day and less water evaporated during the night. This is one of the key reasons oceans moderate temperatures, which is well know to amateurs and professionals alike.

    I have not personally worked with climate models, but I have to assume that one way or anther they build in this moderating effect of oceans.

    “i meant that ‘that which is in contact and immediate proximity’ with that surface layer, of course – not outer space or clouds a mile above it- you know- the saturated air near in contact – the interface.”
    I also meant ‘in contact and immediate proximity’ with that surface layer. Radiation easily passes back and forth from the surface layer to the clouds, upper atmosphere, and space. This puts the surface in thermal contact with these regions. The large amounts of thermal energy involved suggest rather good thermal contact.

  170. TomVonk says:

    Willis

    The morning/afternoon cloud shortwave forcing is indeed asymmetrical. It is lost in the graph, because the maximum is not smoothed. In fact, the average SW forcing from 6AM to Noon is -339 W/m2, while the Noon to 6 PM average forcing is -361 W/m2.

    It is important to note that this underestimates the actual forcing, because this is an average. Remember that on a hot day, the clouds form earlier, and on a cold day they form later. When you average them, it temporally smears out (and thus diminishes the amplitude of) the actual on-the-ground effect because the timing and amount of clouds is temperature dependent.

    This is a great result and exactly what I was going after in my post !
    Of course as I had not the actual data, by eyeballing only the Figure 4 it seemed that your asymmetry hypothesis was in trouble.
    So, actually this analysis supports your hypothesis instead of falsifying it.
    Your remark about a different answer of the system at “cold” and “hot” days is also interesting.

    To avoid misunderstandings.
    There was no criticism intended when I was speaking about “trivial” and “non trivial” results.
    If I don’t do the work you did, it is not that I lack time. I lack the skills and energy.
    In science there have always been theorists and experimenters. I am a theorist and my skills are elsewhere. I would certainly be slower and less efficient than you in manipulating numerical data and writing computer programs.
    That’s why my comments regarding your results were always guided by a genuine interest and what I tried was to provide you with possible theoretical interpretations of your data as well as showing possible research directions which could show interesting results.

    I find that what I call “Willis’ asymmetrical system answer” makes sense and seems to be supported by the data you showed. I find it also highly interesting because you explore here the least understood aspects of the dynamics – the clouds. Admittedly only tropical clouds but it is already something.
    Now obviously if the asymmetrical mechanism was established beyond any reasonable doubt, the second and more ambitious stage would be to analyse the temperature dependence of the mechanism.
    Indeed if the mechanism was stronger with increasing temperatures, then you would have discovered a real negative feedback.
    If it was only slightly dependent on temperature or even independent then the mechanism would play no important role in the energy management of the system.

    Btw you should not call the negative number you find “forcing”. The – 339 W/m² number is just saying that there are sometimes clouds instead of clear sky 100% of the time. It tells us how many clouds in average there are but this is no “forcing”. You could call it “cloud albedo effect” if you want.

  171. Tim Folkerts says:

    Martin Lewitt says:
    September 20, 2011 at 2:46 am

    You might be thinking of relative humidity rather than absolute humidity. Negative feedback from a couple extra turns of the water cycle might well matter for both the actual climate sensitivity and the model projections. …
    “Climate models and satellite observations both indicate that the total amount of water in the atmosphere will increase at a rate of 7% per kelvin of surface warming. “

    That is a good point. The total humidity would almost certainly increase as temperatures increase. While that will almost certainly have a major impact on energy balances by affecting radiation balances and cloud cover, I don’t think it will specifically have a significant “energy sequestration” affect, as Gnomish seemed to suggest. A quick calculations supports my contention:

    The following are estimates from the internet:
    1.30E+16 kg = mass of water vapor
    18 = molar mass water
    7.22E+14 = moles of water
    0.01 % = increase of water
    7.22E+12 = Extra moles of water evaporated
    44000 J to evaporate 1 mole of water @ 290 K
    3.18E+17J to evaporate 1% more water vapor
    5.10E+14 m^2 = area of earth
    6.2E+02 J/m^2
    31557600 sec/year
    0.00002 W/m^2

    In other words, to raise the total humidity in the air by 1% over the course of a year would require about 0.00002 W/m^2 extra input. Using your numbers, even if a warming of 1 K occurred in only 7 years, the system would only require would require about 0.00002 W/m^2 each year to account for the evaporation. This could easily be off a little bit, but it is pretty clear to me that it is a drop in the bucket (pun intended) compared to other energies and other changes.

  172. gnomish says:

    let’s examine of the surface of the sea is ‘in contact and immediate proximity’ with the skin of the sea surface where evaporation takes place…
    the experiment i reported demonstrates that:
    ir thermometer can not ‘see’ water below the surface. ir is blocked by absorption in the first microns.
    ir thermometer can not ‘see’ water beneath vapor. ir is blocked by absorption, again.

    That means that a layer of vapor or water blocks LW transmission because it absorbs it, right?
    That means that the surface of the sea is NOT in direct ‘radiative contact’ with the clouds or outer space because it has an IR absorbing layer between, right?
    My experiment can’t really be interpreted any other way- it proves the existence of a layer of water gas blocks the reading of the surface.
    It also means that DLR, in such a situation, can NOT reach the surface but is absorved by a vapor layer.
    That’s how come DLR doesn’t directly heat the ocean.
    It’s been stated that the skin is cooler than the air above or the water below because of evaporation – that’s empirical, right?
    if that is so, then there is no net conduction of heat from cold to hot, right? conduction would convey heat from the warmer bodies and raise the temperature, right?
    if there were convection, it would have cooler gas rising thru warmer gas – that’s not the definition of convection.
    and my experiment demonstrated that a layer of vapor BLOCKS ir radiation, so there’s no direct radiation from atmosphere to skin- DLR doesn’t warm it – it’s not warm; it’s the coolest layer at that interface.

  173. gnomish says:

    oh, i need a do.over-
    let the first sentence of my previous post read:
    let’s examine if the skin of the sea surface, where evaporation takes place, is ‘in contact and immediate proximity’ with clouds or outer space …
    ima dock the pay of my proof reader…

  174. Tim Folkerts says:

    Gnomish,

    The experiment you described is interesting, but you will have to forgive me if “blowing on a cup of coffee” doesn’t 100% convince me. There are too many details missing. For one thing, water vapor only emits at certain frequencies, so you will always be seeing partly thru the water vapor to the liquid water behind. There will always be some wavelengths that can transmit from the “cup of coffee” far up thru the atmosphere, connecting it thermally with these colder areas. What wavelengths does your IR thermometer use?

    “cross ventilate to remove saturated vapor layer”
    This could be doing several things. For instance, it could be inducing evaporation/conduction at the surface, cooling the top layer of water.

    I can think of lots of other variations to try and variables to try.

    Let me leave you with one final thought. If less than 1″ of “saturated water vapor” is the main material being measured (before you blow on the hot water), then partly saturated water vapor a few times thicker should have the same effect. That means that your thermometer would not work on a humid day if it was more than a few inches from the item being measured, since it would read the temperature of the water vapor in the air instead. Is that indeed the case?

  175. gnomish says:

    well, it’s just about useless for measuring water temperature because of that, unless the water is at ambient anyhow- readings can be all over the place-
    the temps read from the coffee cup are consistent with that – for the vapor may not be hotter than the water, but you can only read hot vapor or evaporatively cooled skin (lower temp than the vapor)
    in normal use it’s not a noticeable problem. usually there isn’t a fog of dense vapor in the way. often the object to be measured is at ambient anyway- same as the air. usually one puts it very close to the object, too

    i don’t know for certain what is the band of ir the device is reading but it’s probably the same as other ir ‘thermometers’, IR spectrum sensor w/ sensitivity 4μm – 8μm with a log circuit to mimic the boltzman formula and an a/d to make it numeric, all on one chip.

  176. TimTheToolMan says:

    The other Tim writes “A quick calculations supports my contention:”

    I believe your moles of water vapour is out be a factor of 1000 because its 18 grams per mole not 18kg per mole which appears to be what you used.

    Also I’ve seen the stat of 4% more water vapour in the atmosphere used rather than 0.01%. Now I believe you actually meant 1% rather than 0.01% but that makes for a further factor of 4 out (if you use the 4% figure) and so makes your figure 4,000 times larger …or about 0.08W/m2.

    Not saying thats a correct figure, just how I believe the maths should have come out for you. And…that doesn’t support your contention as strongly as before.

  177. TimTheToolMan says:

    TTTM wrote “and so makes your figure 4,000 times larger …or about 0.08W/m2.”

    …but of course forgot that the calculated energy must be over the ocean because whilst there is evaporation happening over the land, its going to be much less than the evaporation over the ocean becasue..well..land dries out.

    so when further divided by 0.7 for the ocean proportion of the earth and rounded down for some evaporation from the land that makes the result around 0.1W/m2

  178. TimTheToolMan says:

    And finally I’ll finish off with an alternative way of looking at the change in energy from increased water vapour and that is the trenberth-kiehl diagram shows energy of latent heat as being 80W. A 4% increase in water vapour ought to directly increase this number proportionally to 83.2W or an increase of 3.2W

    Then the question becomes what happens to the energy released when the clouds form. That is another problem but working with 3.2W rather than your initial guess of 0.00002W makes quite a difference to potential outcomes.

  179. Tim Folkerts says:

    TTTM,

    Thanks for checking my back-of-the envelope calculations.

    1) Yes, it should have read “0.01 = relative change” or “1% increase”. That is what I calculated, even if I wrote it incorrectly.

    2) Yes, i messed up kg g. And you are also correct that even this 1000x increase,still puts the energy for evaporation way smaller than other typical energy flows.

    3) You say “Also I’ve seen the stat of 4% more water vapour in the atmosphere used rather than 1%”. I was using 1 % as a starting point. The source quoted to me earlier in the thread said 7% for a 1K increase. If a 1 K increase takes 70 years, this is only 0.1 % per year extra water into the atmosphere. Giving this estimate, a 1% net increase in water vapor over the course of a year seems like a very liberal estimate.

    3) I disagree with “A 4% increase in water vapour ought to directly increase this number proportionally to 83.2W or an increase of 3.2W”. The estimate given was for change in water content in the atmosphere, not change in rates into/out of the atmosphere (ie evaporation rate and precipitation rate). While more water in the atmosphere would logically lead to more rainfall, they do not need to be the same percentage. For example, a 4% increase in the water into a lake will eventually lead to a 4% increase in the water leaving the lake, but this does not mean the depth of the lake will rise 4% (it could be more or it could be less).

  180. TimTheToolMan says:

    The other Tim writes “I disagree with “A 4% increase in water vapour ought to directly increase this number proportionally to 83.2W or an increase of 3.2W”. ”

    The reason I went down the trenberth-kiehl path is because your original (corrected) 0.1W/m2 fails my sensible sniff test. The implication is that to double the amount of water vapour cycling in the atmosphere will requires only 25*0.1 = 2.5W/m2 which seems entirely unreasonable to me.

  181. Tim Folkerts says:

    TTTM,

    The original statement (that I took at face value) was:
    “Climate models and satellite observations both indicate that the total amount of water in the atmosphere will increase at a rate of 7% per kelvin of surface warming. However, the climate models predict that global precipitation will increase at a much slower rate of 1 to 3% per kelvin. “

    I think we are discussion the two different aspects of this claim. I was addressing the increase in total atmospheric H20; you are addressing, I think, the second number, which is how much would a 7% increase in total atmospheric H2O this speed up the water cycle. The relationship between these two rates seems to be the point of the original post.

  182. Spector says:

    RE: Tim Folkerts: (September 20, 2011 at 4:38 pm)

    “For one thing, water vapor only emits at certain frequencies, …”

    Is this really true? Unlike CO2, the water molecule is has a polar electrical field and they are strongly self attractive. It would seem that water molecules would be more likely to emit or absorb an odd frequency IR photon as they collide with each other and perhaps form temporary aggregates, which would have different sets of eigenfunction frequencies.

  183. TimTheToolMan says:

    The other Tim writes “I think, the second number, which is how much would a 7% increase in total atmospheric H2O this speed up the water cycle. The relationship between these two rates seems to be the point of the original post.”

    It is generally believed that water vapour has a relatively short residence time in the atmosphere. Assuming this is correct, we can conclude that a X% increase in atmospheric water vapour does mean an X% increase in the amount cycled and hence the amount of energy transported higher into the troposphere.

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