Estimating Cloud Feedback Using CERES Data

.Guest Post by Willis Eschenbach

As usual, Dr. Judith Curry’s Week In Review – Science Edition contains interesting studies. I took a look at one entitled “Cloud feedback mechanisms and their representation in global climate models“, by Ceppi et al., hereinafter Ceppi2017. The paper looks at the changes in the radiative effects of clouds. From the paper:

The radiative impact of clouds is measured as the cloud-radiative effect (CRE), the difference between clear-sky and all-sky radiative flux at the top of atmosphere. Clouds reflect solar radiation (negative SW CRE, global-mean effect of −45 W m−2) and reduce outgoing terrestrial radiation (positive LW CRE, 27 W m−2), with an overall cooling effect estimated at −18 W m−2 (numbers from Henderson et al.[36]).

The Ceppi2017 Figure 1 shows that almost all of the models report that as the modeled surface warms, the modeled clouds change in such a way as to increase the modeled warming. On average, Figure 1 shows that for every degree C that the modeled surface warms, the modeled clouds add on another ~ 0.5 W/m2 of additional modeled forcing.

cloud feedback cepp1 fig 1

Figure 1. First figure in Cepp1017, as detailed in their caption.

Let me say that I find such a large positive cloud feedback to be very doubtful. Setting that aside for the moment, the Ceppi2017 authors have included a graphic showing the average change in the modeled cloud radiative effect from a number of models.

cloud feedback ceppi fig 3

Figure 2. Average modeled net cloud feedback, from Ceppi2017.

I thought, hmmm … I wondered how that compared to the CERES data. Here’s a look at the same thing, net cloud feedback … except theirs is modeled and the CERES satellite data below is observational.

net cloud feedback CERES

Figure 3. Average of 180 months of CERES data showing the relationship between changes in temperature and corresponding changes net cloud feedback. The calculations are done on a gridcell by gridcell basis, with the monthly gridcell climatology removed before the calculations.

Now, while the models kind of get it right, there are several problems with them. In the CERES data above, you can clearly see the Inter-Tropical Convergence Zone (ITCZ) as the yellow/green area above/below the equator in the Atlantic, Pacific, and Indian Oceans. In the models, it is only weakly visible in the Atlantic and is missing in the Pacific and Indian Oceans.

Next, the CERES data shows that much more of the planet has negative net feedback than the models claim. The entire southern extra-tropics shows negative cloud feedback, some of it quite strong.

Next, because theirs is an average of various models, it doesn’t capture the full variation in the net cloud feedback. In the real world, there are areas of both strong positive and strong negative feedback.

Finally, on average the CERES data shows that the net cloud feedback is negative. Now, we have to take the accuracy of that number with a grain of salt, in that we are looking at trends. Trends are a ratio, and ratios tend to distort averages. For example, the area-weighted average of the trends as shown in Figure 3 is -1.4 W/m2 per °C. A better measure is likely the area-weighted median of the trends, which is -0.5 W/m2 per °C.

Alternatively, we can look at the relationship on a global basis. Here’s a scatterplot of the monthly residual changes in CRE versus the monthly residual changes in temperature (after removing the global monthly climatology).

scatterplot cre vs temperature

Figure 4. Scatterplot of the monthly global CRE and temperature data.

This gives us a third estimate of the relationship between CRE and temperature. This one is between the other two; we have estimates of the cloud feedback factor of -1.4, -1.0 ± 0.3, and -0.5 W/m2 per °C

Whichever way we estimate it, however, the CERES data shows that the net effect of clouds is negative, not positive as the models claim. The average of the models’ estimates of cloud feedback is about plus one-half of a W/m2 per °C. The CERES data, on the other hand, gives a value of about minus one W/m2 per °C.

This is a net swing on the order of ~ 1.5 W/m2 per degree C between the model estimates and the CERES data … and thus a 1.5 W/m2 reduction in the estimated climate sensitivity.


Let me say in closing that I don’t think that “climate sensitivity” is a real thing. I say this because of ample evidence that the climate is a governed system, with a variety of thermoregulatory climate phenomena that work together to constrain the global temperature to a very narrow range (e.g. ± 0.3°C variation over the entire 20th Century). When such a system is in a steady state like that of the earth, the temperature is essentially decoupled from the “forcing” (the changes in downwelling radiation). And because it is decoupled, there is no such thing as “climate sensitivity”.

My best to all,


My Usual Note: When you comment, please QUOTE THE EXACT WORDS YOU ARE DISCUSSING, so we can all understand the exact subject of your response.

Data: The CERES data is here.

Further Reading: Here are a few of my previous posts on the subject of the regulation of global temperature by emergent phenomena:

The Thermostat Hypothesis 2009-06-14

This was my original post on the subject, in 2009. Abstract: The Thermostat Hypothesis is that tropical clouds and thunderstorms actively regulate the temperature of the earth. This keeps the earth at an equilibrium temperature.

The Details Are In The Devil 2010-12-13

I love thought experiments. They allow us to understand complex systems that don’t fit into the laboratory. They have been an invaluable tool in the scientific inventory for centuries. Here’s my thought experiment for today. Imagine a room. In a room dirt collects, as you might imagine. In my household…

It’s Not About Feedback 2011-08-14

The current climate paradigm believed by most scientists in the field can be likened to the movement of balls on a pool table. Figure 1. Pool balls on a level table. Response is directly proportional to applied force (double the force, double the distance). There are no “preferred” positions—every position…

Emergent Climate Phenomena 2013-02-07

In a recent post, I described how the El Nino/La Nina alteration operates as a giant pump. Whenever the Pacific Ocean gets too warm across its surface, the Nino/Nina pump kicks in and removes the warm water from the Pacific, pumping it first west and thence poleward. I also wrote…

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Stephen Richards
May 25, 2017 10:57 am

When such a system is in a steady state like that of the earth, the temperature is essentially decoupled from the “forcing” (the changes in downwelling radiation). And because it is decoupled, there is no such thing as “climate sensitivity”.
Very profound

Reply to  Stephen Richards
May 25, 2017 2:40 pm

Rather than “profound” the word you are looking for is “wrong”. Suggesting that the earth’s temperature is decoupled from the changes in downwelling radiation would violate conservation of energy. If the downwelling radiation changes then the amount of energy received by the earth changes and so the temperature must change so that the amount of energy radiated equals the amount received.

Thin Air
Reply to  Germinio
May 25, 2017 3:24 pm

Not necessarily true. Whenever there is more “downwelling radiation” could be a nearly equal amount energy moving up and out of atmosphere (as latent heat, convection, reflection, etc). due to changes in cloud cover and other factors. All depends on how your prefer to build your model –still being compliant with the laws of physics.

Reply to  Germinio
May 25, 2017 4:25 pm

Germinio- you are absolutely incorrect as you have ( deliberately, I’m sure) left out any and all heat rejection routes from your ,-let’s call it an equation- to be kind. In fact, ALL downwelling radiation is radiated away from the Earth. The only question is how long that takes. The fact that the planet is cooling as we speak means that heat is radiating away faster than it is coming in. This is the victory of the Earth’s natural heat rejection mechanisms over your pathetic CO2 influence. You could make the case that this is temporary ( nearing 20 years now), but you cannot make the case that it isn’t happening without exposing yourself as a persistent purveyor of pessimism. I make the case that if the Earth can easily reject 50 years of warming in a year or so then CO2 is a meaningless canard!

Reply to  Germinio
May 25, 2017 4:29 pm

Thin Air, John,
Both of your points are valid and show that the earth is not “decoupled” from downwelling radiation. You are describing feedback mechanisms that respond to a change in downwelling radiation and thus the earth is coupled (and probably strongly coupled) to the amount of radiation received from the sun.

Reply to  Germinio
May 25, 2017 5:52 pm

I don’t agree with this logic. Willis has made the point that temperature is decoupled from forcings because there are built in compensating mechanism that constrain temperature excursion regardless of forcings. Your statements that “feedback mechanisms that respond to change in downwelling radiation and thus the earth is coupled” is not about temperature but rather describing the very feedbacks mentioned by Willis. You’ve supported rather than refuted his point.

Reply to  Germinio
May 25, 2017 6:46 pm

would violate conservation of energy
nope. for example: adding 1x energy could warm the upper atmosphere by 2x and cool the lower atmosphere by 1x without violating conservation of energy (2x-1x=1x). however, the net effect off adding energy could be to cool the surface.

Reply to  Germinio
May 26, 2017 7:42 am

This analysis was conducted during a period mostly during “The Pause”, when global temperature rate of change was below its longer term trend due to downswing of the AMO or some set of multidecadal oscillations. Clouds were likely changing for reasons including ones other than global temperature.

Mark - Helsinki
Reply to  Germinio
May 26, 2017 8:05 am

The 1880s mid to late temps show that there is no climate sensitivity. most of the 20th century warming happened before 1940. The majority of worst extremes in drought and heatwaves happened before 1940.
US 100f+ days have plummeted since the 1940s.
All when forcing was lower, and our emissions were FAR FAR lower.
The fact is the 1800s for stability were far worse than the 1900s, and the 2000s are not seeing anything like early 1900 and 1800 extremes

Reply to  Germinio
May 26, 2017 8:12 am

However, I think the Ceppi2017 Figure 1 model feedback averages are higher than actuality. Were the models CMIP5 ones, selected and/or tuned for hindcasting what happened before 2006, especially the 1975-2005 or 1970-2005 period? That was mostly during an upswing of the AMO, or some set of multidecadal oscillations.
For one thing, I think positive cloud albedo feedback, especially positive SW cloud feedback, requires the water vapor feedback to be less than its value with constant relative humidity.
The water vapor + lapse rate feedback ensemble median is shown as +1.1 W/m^2 per degree, and I think actuality is around .6, and will decrease as the world warms due to the lapse rate feedback (negative) getting stronger. The albedo (surface albedo?) is shown as .4, and I think the actual surface albedo feedback is around .3, and will decrease as the world warms due to decreasing snow and ice cover. The cloud albedo feedback is shown as +.5, and I think actuality is around +.1 with the LW and SW components being not much different from each other due to clouds not changing much in the longer term (a full cycle of the AMO). This means I think the net of the listed feedbacks is about +1 W/m^2 per degree K and decreasing as the world warms, rather than the alarming +2 W/m^2/degree.
Also there are other feedbacks, all minor but probably net negative. I wonder if anyone took into account increased CO2 masking some of the absorption bands of other greenhouse gases including water vapor, which would reduce the water vapor feedback. I think a more realistic net feedback is around +.5 W/m^2/degree K.

May 25, 2017 11:04 am

Maybe it is wrong to have a forcing with feedback model. Why not consider CO2, water vapor, and clouds as parallel resistances in the transfer of energy to space by radiation. Clouds provide the greatest resistance.

Reply to  fhhaynie
May 25, 2017 1:33 pm

Like the parallel thermal resistances of the insulated envelope of a building that cause the inside to be warmer than the outside – or cooler depending on the direction of heat flow – from hot to cold.
Except the governing formula is Q = U * A * dT and not S-B BB. LWIR just carries a share of the heat transfer as seen with the low emissivity. 1/U = Conduction p+ convection + latent + radiation.—We-don-t-need-no-stinkin-greenhouse-Warning-science-ahead-

Reply to  Nicholas Schroeder
May 27, 2017 6:17 pm

I am interested in views, science, to explain how ozone layer warms. Most say its from space via UV. A claim suggests its from earths emitted IR (60W/m2 or suchlike) because ozone can absorb at 9.6um. I am troubled by diiference in intensity as wavelengths get longer. Just because there is radiation may not lead to warming. … Anyone?

Joe Born
May 25, 2017 11:15 am

I welcomed another of Mr. Eschenbach’s thermostat posts, but, when I went to bookmark it, I was surprised to see that my “thermostat” folder had only four entries.
However, I then discovered that one of the entries itself had a more-extensive list:

Lance Wallace
May 25, 2017 11:20 am

Is there a reference for the Ceres data?

May 25, 2017 11:27 am

Let me say in closing that I don’t think that “climate sensitivity” is a real thing. I say this because of ample evidence that the climate is a governed system, with a variety of thermoregulatory climate phenomena that work together to constrain the global temperature to a very narrow range (e.g. ± 0.3°C variation over the entire 20th Century).

Even with such thermoregularory mechanisms in action it should be possible to find a sensitivity. It will be very small, ie insenstivie.
You could also try doing the regression of the scatter plot with the axes reversed. Data with such wide scatter will lead to a falsely low “best unbiased linear estimator”. If radiation is supposed to be the cause of the temperature change it would be logically on the x axis. Climatologists do it the other way around because it gives lesser slope and the greater climate sensitivity.
Just another trick of the trade.

Reply to  Greg
May 25, 2017 11:28 am

That is re. figure 4, it will give a less negative result but will not change the sign.

Reply to  Willis Eschenbach
May 25, 2017 1:38 pm

If you torture the numbers long enough it’s impossible not to find same value zero for sensitivity. The important question is “Is it real?” followed by “Does it matter?”

Reply to  Willis Eschenbach
May 25, 2017 1:41 pm

sorry – fat fingers:
… If you torture the numbers long enough it’s impossible not to find some non zero value for sensitivity.

Reply to  Willis Eschenbach
May 27, 2017 1:22 am

Thanks for the reply Willis. I take your point about the gas burner but I don’t think things are that simple. There is not an on/off switch in the tropics and the tropics acts to moderate changes in the rest of the system but that is not binary either.
You are correct about the way I was looking that rad/temp graph. In this context temp is the independent variable if you can dissociate the two. Unfortunately you can’t, there is a complex back and forth of cause and effect.
The key point is the this is not a lab experiment where you have a ‘controlled variable’ with negligible error; you have two error laden variables. ( In this context ‘error’ means both measurement error and any variability which is not part of the linear relationship one is seeking to find. )
I discuss this in more detail in this article:
simple regression in either orientation will give a biased estimation of the linear relationship. It is probably best to do both and conclude that the correct result lies between the two. That seems to be the way you were approaching this article.
Any attempt to do better than that requires more information about the nature and magnitude of the errors in each variable.

May 25, 2017 11:40 am

Clouds reflect solar radiation (negative SW CRE, global-mean effect of −45 W m−2) and reduce outgoing terrestrial radiation (positive LW CRE, 27 W m−2), with an overall cooling effect estimated at −18 W m−2

I can not understand how anyone can believe such figures. I mean that would put the clouds share of total albedo to 42% or so. Anyone with eyes can clearly see how clouds easily make up for over 2/3 of the albedo, as they are white and bright, as opposed to the dark ocean.
The other part is, that we can tell from night cooling relative to sky condition, that clouds will have a much bigger warming effect. For instance I took these data from Parkersburg, WV. The chart starts with the first measurement within the hour before sunset. Then the cooling is averaged according to the average skycondition. 0 would be all clear sky, 8 all overcast.
One can fairly well see, how strong the impact of cloudiness is on the cooling pattern. Overcast scenarios reduce nocturnal cooling by 85%(!). And if we average all scenarios vs. the only clear sky scenarios, then nocturnal cooling is down by yet around 35%.
So taking the clear sky as a base scenario, where we could assume an emission rate of about 300W/m2 (for the global average), then clouds would reduce this emission rate by roughly 100W/m2.
Now Parkersburg is indeed close to the US average in terms of cloudiness, and the global average will not be very different. So 100W/m2 will be a good approximation the cloud “GHE”. That is allmost 4times as much as the 27W/m2 stated above.
You can find the full story here (starting with page 9):

richard verney
Reply to  Erich
May 25, 2017 1:15 pm

What about humidity? This is the elephant in the room, and it is one reason why the land based thermometer record tells us very little of significance.
In summer on the Spanish Costas usually clear nights are warm nights, and cloudy nights are cool nights. Why is that? the answer is probably that on warm days the humidity is far higher such that the the specific heat capacity of the atmosphere is a lot higher. The warm day leads to an atmosphere that contains far more energy, and hence the energy takes a long time to dissipate, and thus the clear night time temperatures remain warm for longer. But this is a facet of lying on the shores of an ocean (the Mediterranean Sea). In practice, clouds are a rarity on the Spanish Costas
In land the situation is quite different. When you have clouds, I suspect that humidity is higher than normal. Indeed, it may be close to raining. Hence, in land, on a cloudy night the atmosphere has greater specific heat than on clear days, such that on cloudy nights it takes longer to dissipate the energy. I suspect that this is the dominant reason why one perceives cloudy nights to be warm nights.
Without detailed calculations based upon the relative humidity and specific energy contained within the atmosphere on any given day, I am sceptical that warm cloudy nights are driven by radiative effects. I suspect that classic energy transfer, reduced convection and greater humidity leading to greater specific energy that needs longer to dissipate are the dominate drivers.

richard verney
Reply to  richard verney
May 25, 2017 1:17 pm

Something went wrong with the formatting.
I meant to pose the question: What about humidity?
[Fixed. mod.]

Ian W
Reply to  richard verney
May 25, 2017 2:18 pm

The question should be why is everyone using atmospheric temperature as a metric when the enthalpy greatly alter the energy content of the air at the same temperature. The atmospheric heat content should be measured in kilojoules per kilogram. But then the result would probably be less exciting – so expect climate ‘scientists’ to carry on using the wrong metrics.

Reply to  richard verney
May 25, 2017 9:16 pm

If the radiative effect is the primary cooling effect at night, then how does the High Sherriff’s helicopter IR detector find the mary jane grow houses?

Clyde Spencer
Reply to  Erich
May 25, 2017 1:33 pm

You are touching on one of the reasons that “albedo” is a poor choice for energy balance calculations. Clouds are efficient diffuse reflectors that essentially reflect light equally in all directions, thus appearing white not matter how one views them (except directly underneath). On the other hand, water is dominated by specular reflection, which is controlled by the index of refraction and angle of incidence of incoming light. That means that unless one is in the right geometric viewing position, little of no reflected light is observed. Water also has a diffuse component that is correlated with turbidity, but that operates mostly in shallow, coastal waters. The oceans appear dark principally because when viewed with the sun directly overhead, only a ‘hotspot’ of about 2% reflectivity is observable.

Reply to  Clyde Spencer
May 25, 2017 2:32 pm

@Clyde Albedo may be less than perfect, but I did not invent it. As the whole GHE theory is based on a formula that defines absorptivity (A) = 1 – albedo, and puts emissivity (E) = 1, I will refer to it as it is. I am falsifying this theory, and not trying to “improve” it. Also the major mistake here is not the imprecission of the term albedo, but the huge non sense of putting emissivity to 1.
Of course we need to treat A and E equally. If we add clouds to A, which we do as they are part of the albedo, we will also need to count them (or their “GHE”) to E. Now since E of water is already far lower than 1 (~0.84), adding clouds to it will lower it to about ~0.58. So Earth should be about (0.69/0.58)^0.25 * 279.2 = 291.6K. That is warmer than the arithmetic average surface temperature, and does not allow for any GHE or greenhouse gases.
Btw. this is how it works properly:

Reply to  Clyde Spencer
May 26, 2017 11:10 am

Btw.. your problem was, that you could not properly use the Fresnel equation to determine total albedo/reflectivity of a water sphere. Your approach however was completely wrong (sorry!).

The area under the reflectance curve for seawater (Fig. 1, above) is about 9%. Because the sun moves with a constant angular velocity, the value is the time-averaged reflectance of a point on the surface of Waterworld for a 6-hour period. That means that for any particular spot on the surface of the ocean, starting at the terminator at sunrise, on the Equator, it would have an initial instantaneous reflectance of 100%. As the planet rotates under the sun, the percentage of light reflected from the point will decrease until it reaches the minimum of ≈2% at local solar noon. The process will then be reversed until the point again reaches maximum reflectance at the position of the sunset terminator. The total light reflected from that point, over a nominal 12-hour sunlit period, will be about 18% of the incident light. This is more than twice the value listed for the diffuse-reflectance of water and is in the range of values given for vegetation.

It is however pretty simple and your water sphere has a total reflectivity of 0.068 and an absorptivity of 0.932 respectively. So NASA is essentially right. The interesting part, AGAIN, is however emissivity. You can look up the calculations hereto in this excel file (sheet9, the Fresnel equations are in the following sheets). You can also play around with the parameters for n2.

Clyde Spencer
Reply to  Clyde Spencer
May 27, 2017 10:09 am

I’m not about to wade through your spreadsheet, which is effectively undocumented code. I used something similar to create my graph. However, because the total flux is the sum of both the s and p component, I added them instead of taking the average. Is that why our results differ?
As to the quote from my article, I admitted in the comments that my figure of 18% was wrong. I hastily jumped to an unsupported conclusion. However, I stand behind the 9% figure.
However, your statement, “Btw.. your problem was, that you could not properly use the Fresnel equation to determine total albedo/reflectivity of a water sphere. Your approach however was completely wrong (sorry!).”, is an assertion that you do not back up with an explanation. I don’t recollect you weighing in with any objections when I published the article you are quoting. Now is your second chance to explain why my work was wrong.

Reply to  Clyde Spencer
May 27, 2017 2:36 pm


I don’t recollect you weighing in with any objections when I published the article you are quoting

Well I did not read it back then, in fact I did not even know about this blog, and also I am only into the subject for about 3 months. But since you referred to your article in your posts, I checked it.
This is not rocket science! Also you do not need a derivation of the Fresnel function, which I admit, might be tricky. But playing it all in Excel is really simple, although not everyone may have IT expertise. Also the 9% figure is not quite correct, rather it is 10.9% for n2 = 1.34. Which is useless however, as we want to know the hemispheric reflectivity or absorptivity, respectively.
What I did should be pretty obvious, and I thoroughy explained it in my essay. The weighting for one-directional solar radiation will obhere to the formula sin^2 x, with x being the angle of incidence from normal. For every x from 0 to 90, that will give you the accumulated amount of light received, like 0.5 at 45%, or 1 at 90%. Of course, we would rather need the deviation which we could then merge with the deviation of the Fresnel equation. As that part would be complicated, and as we can easily get around it using Excel, I just stick to the base function.
With the “multi directional” emissivity into a hemisphere the weighting must be different. This time we need to use 1-cos x as a base formula. 0.5 will then only be obtained for x = 60°. Those flat angles where reflectivity runs high, will naturally receive a much stronger weighting. This is the major reason, why ultimately emissivity ends up being much lower than absorptivity. And as it is, water drives the planet to (0.934/0.84)^0.25 * 279.2 = 286.7K

Reply to  Clyde Spencer
May 27, 2017 8:03 pm

Sorry .. I was talking non sense, partially. Of course there is no need for a deviation of the Fresnel function. Rather, to obtain to the 2-dimensional reflectivity, we would need to integrate it. Which again is pointless. Anyhow, the math in the excel file is right..

Clyde Spencer
Reply to  Erich
May 27, 2017 9:27 pm

The Excel file might be useful to other than you IF the columns had titles and the graph had notations. It is one thing to slap something together to test an idea, but if you want to convince others that you are onto something you need to make it legible and easily understandable. Have you ever found yourself in a position where an employer asked you to fix a bug in some code written by someone who was no longer with the company, and who never documented anything?
Incidentally, I have a much more sophisticated Excel spreadsheet using complex arithmetic that I use regularly in my research on the optical constants of opaque minerals. I can handle inputs of linear polarization of different azimuths as well as different angles of incidence and complex indexes of refraction. So, I don’t need your spreadsheet, except to perhaps check it against mine. However, I have enough projects going on that I’m not going to play Sherlock Holmes and try to decipher your undocumented spreadsheet.

Reply to  Clyde Spencer
May 28, 2017 5:17 am

Have you ever found yourself in a position where an employer asked you to fix a bug in some code written by someone who was no longer with the company, and who never documented anything?

Well I have been working as a software analyst, and the answer is pretty much YES! The trick is to keep in mind what it is supposed to do, because then allready have an idea how it is done.
Anyhow, the file is not complicated, and there is no “code”. It is only containing the Fresnel equations and the two named formulas for the geometric weighting. The term =(D6+D5)*(G6-G5)/2 merges the average of the Fresnel equation (for each gradient) with the derivation of the weighting. You might want to read it as =(D6+D5)/2 *(G6-G5).
I can lead the horse to the water, but..

Clyde Spencer
Reply to  Erich
May 28, 2017 9:19 am

For undocumented code, it is sometimes easier to start over than to try to understand what the programmer has done, particularly if they were trying to be clever.
It is a mistake to think that spreadsheets are not “code.” They are just a different paradigm than the line-oriented programming that most ‘programmers’ use.
“I can lead the horse to the water, but..”, if the water is tainted the horse knows better than to drink it.

Clyde Spencer
Reply to  Erich
May 28, 2017 9:24 am

You didn’t mention whether you are using the Fresnel amplitudes, intensity, or reflectance. That is what I mean about lacking documentation. It isn’t worth my time to try to work all that out. And, you apparently don’t think it worth your time to make your spreadsheet readable. Therefore, why should I spend my time deciphering what you have written?

Reply to  Erich
May 25, 2017 2:02 pm

It is not about nights being warm or cold per se, it is all about the rate of cooling. And this pattern is very much ubiquitarious, the spanish coast will not be an exception to it.
There is one funny thing though about Spain. The CET zone stretches from (east) Poland to the atlantic shores of Spain. That is 35° in longitude, or 2h20min in daytime, far too much for a reasonable time zone. You could almost fit all of Australia into it, which of course has 3 times zones. And it is specifically Spain which does not quite fit.
So you have a very late sunset and sunrise in Spain. In La Coruna for instance, it only turns dark after 11pm during peak summer. At sunset, of course, temperatures will usually be above daily average, even more so, if there was a clear, hot (summer) day. And that is what you are confusing.

richard verney
Reply to  Erich
May 26, 2017 3:04 am

Thanks your comment.

it is all about the rate of cooling

That is my point. Where I am, on the Spanish Costas, a typical summer day high is about 33degC (late afternoon). At about 1 am/2 am, say 3 to 5 hours after sunset, the temperature is still around 30 degC.
The night time temperatures do not really fall away until after 3am, reaching their coolest about 4am/5am. And then it takes a long time for the day to once again warm up. Morning temperatures, say 10 am (perhaps even 10:30) are still cool, even though the sun has been up for many hours.
I suggest that this is a facet of the relative humidity. The summer atmosphere has high levels of humidity which means that there is a lot of energy entrained in the atmosphere and hence it takes a long time to dissipate that energy and hence to cool at night, not withstanding cloud free skies. Because the atmosphere has high humidity, it takes a long time to once more warm up.
The Spanish Costas are essentially situated opposite to the Sahara desert, and have similar sunny/cloud free profiles. Obviously, the sun is slightly more intense over the Sahara, but with that caveat, the temperature profiles are very different not due to the different intensity of the sun, but rather due to the different humidity parameters.
Under clear sky conditions, the Sahara warms up very quickly, and cools down very quickly. The Spanish Costas warm up slowly and cool down slowly.
My point is simple, namely that one cannot begin to understand the effects of clouds without also having regard to humidity, wind conditions and convectional currents. The way this planet works is complex, and it is very easy to not fully understand its workings.

Reply to  Erich
May 26, 2017 7:03 am

Well first of all I should note that your argument is in line with the consensus model, where not clouds, but vapour is the major greenhouse factor. So even though clouds, if they are present, would reflect some IR back to the surface (which then amounts to 27W/m2 as qouted above), most iIR was reflected by vapour. So the story goes.
Now there are a lot of reasons why this can not be true.
1. Looking at the bigger picture, there are GHGs, and vapour thus can not be one. You would need to read the whole essay to understand.
2. The GHE is at its lowest, where temperatures and vapour are at their highest, which is around the equator. One could argue of course the equator is the warmest region of Earth because of the high prevalence of vapour, but we know better. It is because of the strong solar radiation, about 435W/m2, which would heat a PBB to about 23°C. Oberserved temperatures are 4-5°C higher than that. That compares to a global delta of 288 – 279.2 = ~9K. So, again, the GHE is at its lowest where vapour is at its highest. So vapour can not be GHG.
3. In the Sahara/arabian desert temperatures or even somewhat higher than at the equator, despite less solar input. This region however is extremely dry, and the relative lack of vapour should make it cooler, if vapour was a GHG.
4. As shown there is a strong correlation between cloud cover and nocturnal cooling. An overcast sky will reduce the cooling rate by about 85%. Humidity might be somewhat correlated with the cloud situation, but in no way there would be such a strong variation (in a place like Parkersburg), as to get to a 85% figure.

Clyde Spencer
Reply to  Erich
May 27, 2017 7:34 pm

Before I invest a lot of time to wrap my head around your claims, I want to be sure that you have a consistent story. You originally said, “…your water sphere has a total reflectivity of 0.068…” My claim was that a small patch of water at the Equator, during an equinox, would experience an average reflectivity of at least 9%. You now claim it should be a bit higher. I pointed out that water at higher latitudes would have higher average reflectivities. I have, subsequent to the article, done a spreadsheet calculation of the Water World that gives me an instantaneous hemispherical reflectivity close to 18%, based on slicing the hemisphere into 0.1 degree frustums. The Earth should experience even higher water reflectivities because, besides the specular reflection, there is a diffuse component from suspended particles, particularly near coastlines.
Albedo is typically equated to the relative brightness of a disk the same diameter as a spherical body and misses the reflection changes with angle of incidence and the volume scattering for both water and vegetation canopies. It isn’t “rocket science;” however, it is a little more challenging than balancing one’s check book and seems to be challenging for most climate ‘scientists.’

Reply to  Clyde Spencer
May 27, 2017 7:49 pm

Erich, Clyde Spencer
Albedo of Open Ocean Water at Various Solar Elevation Angles (Solar Zenith Angles)
A complex subject, but the assumptions used in several comments above are incorrect.
Payne, 1972 reported open ocean water albedos of 0.061 (diffuse) to 0.066 (direct – BUT ONLY for high solar elevation angles, angles above 33 degrees over the horizon. Below 33 degrees SEA, the open ocean albedo continuously and predictably increase, going as high as 0.45 at angles below 10 degrees SEA.
Dr Payne’s original abstract, cited over 572 times by others.

An experimental study of the albedo of the sea surface for shortwave solar radiation has been carried out on a fixed platform. Fifteen-minute totals of upward and downward irradiances were recorded continuously for four months over a wide range of atmospheric and sea conditions. The resulting albedo values, the ratio of upward to downward irradiance, are expressed in terms of a particularly convenient pair of parameters, sun altitude and atmospheric transmittance (T). The latter is defined as the ratio of observed downward irradiance to the irradiance at the top of the atmosphere and has not been used before in describing albedo. Examples of albedo values are 0.061±0.005 for heavily overcast skies (0.00.65) of 0.03 for high sun to as large as 0.45 at sun altitudes 25° and increases to 25% for very low sun attitudes. The effect of wind, through surface roughness, is shown to be small but predictable. Effects of whitecaps are not noticeable at wind speeds up to 30 kt, the highest observed in the study.
Application of the results is made to climatological studies of the absorption of solar energy by the surface waters of the ocean. Monthly average albedos, are calculated for the Atlantic Ocean to compare with Budyko’s latitudinally dependent values, and it is shown that although the sets of results agree within 10% at latitudes up to 40°, there are discrepancies at higher latitudes as high as 100%. Finally it is shown with climatological albedo values calculated from the results of this study, that the accuracy of climatological estimates of solar energy absorbed in the ocean are now limited by the accuracy of climatological estimates of downward irradiance.

Payne’s fundamental findings have been duplicated by several other open ocean projects at various wind speeds and latitudes. No find any contradictions. Many find differences below 8 degrees SEA, as Payne summarizes below. There’s a lot of uncertainty that low elevation angles, but ALL measurements show the predicted values of Fresnel equations are invalid.
Pegau and Paulsen measured on the effect of wind speed on open ocean albedo during Dr Curry’s SHEBA months-long Arctic expedition, and their formula axpands on Dr Payne’s to account for wind speeds between 0 and 25-30 meters/sec (essentially a full storm.)

Reply to  RACookPE1978
May 27, 2017 8:01 pm

In Excel terms, for SEA < 33 degrees and WIND as a variable in meters/sec,
Open Ocean Albedo = 0.026/(S9^1.7+(-0.0002*WIND^2+0.0076*WIND+0.0266))+0.15*(S9-0.1)*(S9-0.5)*(S9-1)
Where cell S9 = sin (SEA) = Cos (SZA) for Solar Zenith Angle in radians.
See also the graphics in

Clyde Spencer
Reply to  RACookPE1978
May 27, 2017 9:09 pm

Again, it is my opinion that the use of the term “albedo” is careless and imprecise. It is appropriate for diffuse reflectors such as planets, moons, and asteroids in the solar system that do not have water bodies on them. Albedo is a reasonable first-order approximation for most things on Earth other than water. For the Earth, Total Reflectivity, which includes both specular and diffuse reflectance, should be used for energy-balance calculations. The point in my original article (did your read it?) was that there is a perfectly good theoretical argument that a significant fraction of sunlight is being reflected back out to space that isn’t observed by nadir-viewing satellites. Even the multi-angle viewing satellites have a difficult time observing the outgoing specular reflections at low zenith angles because the satellite has to be looking in the plane normal to the water surface formed by the sun, point of incidence, and the recording sensor. Also, if it has a narrow field of view, then it has to be looking down at the angle of incidence. That means it has to be steered continuously to keep looking at that point on the water and it has to have a huge dynamic range to keep from saturating the sensor. If Payne only recorded a maximum reflectance of 45%, he should have asked himself why it was less than half the theoretical value. The work was a step in the right direction, not relying just on the diffuse reflectance, but appears to be flawed in the design or limited by the instrumentation available to him. It is an interesting commentary on the state of climate science that it has been cited so often, without anyone in the field challenging the results on theoretical grounds. Maybe this is “rocket science!”

Reply to  Clyde Spencer
May 27, 2017 10:53 pm

I will disagree with you there.
See, Payne, and those who follow in his measure-the-actual-ocean field, ARE more accurate than theory. It is not that the perfect physics of the Fresnel equations are wrong. But that they are ONLY limited to the textbook, to the classroom, and to the ideal lab in a sterile environment of pure still water and pure glass and no air motion and at STP (standard temperature and pressure.)
When they get to sea and measure the “albedo” of open water – the total amount of energy reflected from the ocean compared to the energy available to that ocean at that same point in time, clouds, humidity, wind, air mass, and the same proportions of direct and diffuse radiation – they find Fresnel’s equation simply does not work.
It does not calculate the right energy levels. In fact, if you look at the Point Barrow “midnight sun” photo sequecnes of the nighttime sun light reflecting from the Arctic waters at low levels, you see the same thing. 45-50% of the energy is reflected, but there are “black spots” where distance and nearby waves intercept that energy, absorbing it on the front side of the wave, putting the dark side in shadow.
Fresnel is a physicist’s approximation of the real world. The measurements are the real world, complete with messy values and experimental errors and all the “stuff” engineers and field physicists and farmers fight every hour.

Clyde Spencer
Reply to  RACookPE1978
May 28, 2017 9:10 am

Your argument is like someone claiming that the law of gravity only works in the laboratory and does not apply to airplanes. The problem is one of definitions and secondary and tertiary effects. The situation is far more complex than generally acknowledged and Payne apparently doesn’t follow a strict definition of reflectivity. Instead of using the TOA luminance for the sun, he should have measured it in the sky after passing through the long slant-range path. Then the surface reflectance should be measured in the same units. That way, the effects of atmospheric scattering and absorption can be separated from the surface interactions for a true measure of reflectivity.
As to the impact of waves, they too are intricate in their impact. For long swells, with the crests parallel to the sun’s rays, there will be negligible impact. For swells at right angles, the angle of incidence (and reflectivity) will decrease on the side facing the sun, in proportion to the amplitude of the swells. However, unless the angles are such to create total internal reflection, some of the light entering the wave will exit on the far side and experience negligible absorption. Some of the light will just be reflected upward at an angle different than the angle of incidence for calm water. It will therefore be missed if only measured at the gross or average angle of incidence. For increasing wave height, the point will be reached where whitecaps are formed and eventually a frothy surface will be created. At that point, the reflectivity will be dominated by diffuse reflectance and actually have a greater overall reflectivity. However, a direct measurement of the light observed will be much less than the total hemispherical scattering by the whitecaps. So, if it appears that the real world violates theory, then it is probably because there is an attempt to oversimplify very complex phenomena and being careless with definitions.
There is a long, cascading chain of events that interact and they each have to be measured separately to properly understand and model the phenomena.

Reply to  Clyde Spencer
May 27, 2017 11:32 pm

From your original reply,
Clyde Spencer

The point in my original article (did your read it?) was that there is a perfectly good theoretical argument that a significant fraction of sunlight is being reflected back out to space that isn’t observed by nadir-viewing satellites.

But, how have does that account for the atmospheric absorption near the perifery of the globe?
Below 15 degrees Solar Elevation Angle, the inbound light passes through ever-increasing air masses.
1.0 when the sun is directly overhead (Key West for example at noon on June 21-220.
But air mass increases as the sun gets lower: 3-10-17-20 to 35 right on the horizon.
So, at the sea surface, the inbound light is only 5% to 20% (depending on atmospheric clarity) of its top-of-atmosphere levels. Nothing strange, but that 80% of original energy is absorbed into the atmosphere, heating it directly. (NO CO2 or anyother effects, it is absorbed into the dust and pollen and water vapor.
But, once the sunlight hits the ground/sea surface, the reflected light must pass back through the same length of atmosphere before being measurable by the satellite sensors. Only a tiny bit at low solar elevation angles (view angles in this case) makes it through both trips.

Reply to  Erich
May 28, 2017 7:01 am

Before I invest a lot of time to wrap my head around your claims, I want to be sure that you have a consistent story. You originally said, “…your water sphere has a total reflectivity of 0.068…” My claim was that a small patch of water at the Equator, during an equinox, would experience an average reflectivity of at least 9%. You now claim it should be a bit higher. I pointed out that water at higher latitudes would have higher average reflectivities. I have, subsequent to the article, done a spreadsheet calculation of the Water World that gives me an instantaneous hemispherical reflectivity close to 18%, based on slicing the hemisphere into 0.1 degree frustums. The Earth should experience even higher water reflectivities because, besides the specular reflection, there is a diffuse component from suspended particles, particularly near coastlines.
Albedo is typically equated to the relative brightness of a disk the same diameter as a spherical body and misses the reflection changes with angle of incidence and the volume scattering for both water and vegetation canopies. It isn’t “rocket science;” however, it is a little more challenging than balancing one’s check book and seems to be challenging for most climate ‘scientists.’

Ok, I hear you! But first let us pray to the lord to get an edit button. I hate to see my typos without a chance to correct them. I hope to get it all right this time..
You are using the Fresnel function like a sundial. If we had an equatorial day of 12h, the sun would rise at 90°, where reflectivity was 1, and then would climb by 1° for every 4 minutes. So you try to read from this 2 dimensional chart what the reflectivity would be. And indeed that would be higher than 9%, in fact 10.9%, as I stated above.
Now I have already disclosed, that total emispheric reflectivity is only 0.066 (for n = 1.33, 0.068 for n = 1.34). Obviously something does not add up at this point. At the equator reflectivity should be at its lowest, as the most vertical angles of incidence are attained, so how can reflectivity be above global average?
The problem here is the implicit assumption, that your spot of water would receive the same amount of solar radiation throughout the day. That is not so! And I am not talking about reflectivity! If not as a sphere, think of it as half a circle, with your spot moving from one end to the other. Only in the middle, when the sun is in the zenith, it will receive the maximum radiation. Close to sunrise or sunset, it will only be a small fraction. So the high reflectivity at these times will play a much smaller role.
In this case you need to weight the Fresnel equation with sin x, with x again the angle from normal. Doing so gives us the correct result for reflectivity, which then is 2.56% only.

Clyde Spencer
Reply to  Erich
May 28, 2017 9:36 am

You said, “You are using the Fresnel function like a sundial.” Not all readers are as mathematically sophisticated as you. Therefore, I attempted a logical explanation that all could follow that demonstrated that the average reflectivity of seawater should be higher than the ‘albedo’ cited by NASA.
Actually, my hemispherical calculation of instantaneous reflectivity was close to 18%, not the 7% you are claiming.
Reflectivity is a unitless ratio. So, while the actual intensity of sunlight received varies with time and location, the reflectivity doesn’t vary as a result of the intensity. The next step in the energy balance equations is to convert the varying impinging sunlight to an outgoing flux based on the reflectivity and intensity of the incident light.

Reply to  Erich
May 28, 2017 8:27 am

What you bring up is important, but we should also be clear about what we are talking. Sure, the Fresnel equations are merely a theoretical approach, which does not make them wrong. Most of all, they do not tell us why water is blue.
So why is it blue? Sun light reflected by water does not change its color. I mean there might be little variations over the reflectivity within the spectrum of visible light, but that is not the reason.
The “blue light” comes from underneath the water surface. It is Rayleigh scattering. I am not even aware if it comes from water itself, or some particles in it. Anyhow, water is not just water, there is a lot of stuff swimming in it. So unlike what the Fresnel equations suggest, not all light which passes through the surface will be absorbed. Some of it, mostly blue light, will be diffusely scattered back.
What is the overall impact on albedo? I have no clue! There is however a very simple approach, one that Payne probably could not take use of in 1972. The one that I extensively covered in my essay. We can simply look down on earth and see the brightness of water with our own eyes. Ok, that is with the assistence of some satellites or probes.
What we can tell from that, is that water indeed has a very low albedo, and the 0.06% figure seems quite accurate. The picture series where you have earth and moon in one frame is specifically helpful, as we can take the moons albedo for comparison. You can even take it one step further, by looking at it pixel by pixel and determine the color values for earch of it.

Also the direct reflection of the sun (slightly down left from the center) looks very moderate. Of course this might a very different story, if you had different viewing angle, like from the opposite direction..

Reply to  Erich
May 28, 2017 12:47 pm

I do not know what to say anymore. Do you understand, that those areas close to the terminator will only receive little sun light, and therefore the high reflectivity there must be “underweighted”? I think I have told the same simple story now for well a dozen times. It is way too trivial, to make this a major subject.
If, for instance, the sun is in the zenith, then we may have like 1370W/m2. At an angle of 45° it would only be… cos 45 * 1370 = 969W/m2. At 80° then cos 80 * 1370 = 238W/m2, and 89° cos 89 * 1370 = 24W/m2.
Reflectivity only applies to what can be reflected. As there is only little radiation at low angles. You will f****** need to take this into account, if you want to get a valid result. And yes, NASA, who would have thought, got it largerly right. This time.

Clyde Spencer
Reply to  Erich
May 28, 2017 2:04 pm

Your first paragraph and last paragraph are really saying two different things. Your last paragraph is essentially saying what I told you in my last message. You have to take the incoming flux and multiply it by the reflectivity to obtain the outgoing flux. The incoming flux is reduced by the footprint increasing with the angle of incidence, which you take into account, and it also needs to be reduced from atmospheric scattering and absorption through a longer path length than when the sun is directly overhead. We are talking at cross purposes because NASA has improperly been using albedo, which lumps reflectivity and incident flux together, and referring to it as some average reflectivity instead of dealing with them separately.

Clyde Spencer
Reply to  Erich
May 28, 2017 2:16 pm

You raise an important point when you say, “So, at the sea surface, the inbound light is only 5% to 20% (depending on atmospheric clarity) of its top-of-atmosphere levels. Nothing strange, but that 80% of original energy is absorbed into the atmosphere, heating it directly. (NO CO2 or anyother effects, it is absorbed into the dust and pollen and water vapor.”
The reason I have been focusing on the reflectivity at high angles of incidence is that the alarmists have been blaming the “black water” in the ice-free Arctic for warming the water. You have just made a compelling argument that whatever warming is occurring is almost exclusively in the air, and not just in the Arctic, but along the entire path through the atmosphere!

Clyde Spencer
Reply to  Erich
May 28, 2017 2:27 pm

Yes, Rayleigh scattering will preferentially scatter blue light, but there is also the fact that pure water absorbs red light much more strongly so there is less red and green available from deep water to be reflected back.
You said, “We can simply look down on earth and see the brightness of water with our own eyes. …
What we can tell from that, is that water indeed has a very low albedo, and the 0.06% figure seems quite accurate.” This is the fundamental mistake that so many are making. Diffuse reflectors will pretty much look the same from all angles. However, specular reflectors concentrate all the reflected energy into a narrow sheaf of rays that can’t be observed EXCEPT at the angle of incidence. Indeed, the ocean typically has a low ‘albedo’ (Even less than 2% because the ‘hot spot’ can be observed to be brighter than the surroundings.), but that doesn’t mean it has a low reflectance for all angles.

Reply to  Erich
May 28, 2017 7:14 pm

It would be interesting to know the context of the peak at 70°. That could very well be ice. It could just as well have to do with sun light reflected at low angles, although I do not see how it would reach a peak at specifically 70°. Also it could very well be just clouds. I mean they want to give the albedo of the oceon, but from the satellites perspective clouds will stand in the way. So who knows well they filter it.

Keith J
May 25, 2017 11:48 am

Net effect also depends on the position of the sun in the sky.

Clyde Spencer
Reply to  Keith J
May 28, 2017 2:46 pm

I suspect that the bimodal distribution is a direct effect of the shape of the Fresnel specular reflectance curve, which rises strongly after about 60 degrees. Can I assume that the graph is showing that the combined specular and diffuse reflectance of water reaches about 27% at about 5 degrees off nadir? In looking at the moon transit video provided by Erich, I got the impression that the water was maybe brighter than the nominal 15% albedo of the moon, particularly at the solar ‘hot spot’ (2% specular).
The graph explains the apparent brightness of the water surface (albedo), but does it take into account the light scattered away from the sun? That is, is it only showing albedo, or total reflectance, which is different.
I think that there is still work to be done in this area before we can glibly say that the “science is settled.”

Reply to  Clyde Spencer
May 31, 2017 8:20 pm

“In looking at the moon transit video provided by Erich, I got the impression that the water was maybe brighter than the nominal 15% albedo of the moon, particularly at the solar ‘hot spot’ (2% specular).”
Well, then you must be color blind, or blind in some other perceptional way. I did take a closer look at it, and also converted it into a black and white picture. And I can assure you, then the ocean is definitely much darker than the moon.

Clyde Spencer
Reply to  Erich
June 1, 2017 7:21 am

Do you have to work at being rude, or does it just come naturally?
Just for the record, about 10% of men have some form of color blindness. So, your insensitivity in the way you responded does not speak well of your civility. You come across like the warmist trolls who engage in ad hominem attacks when they run out of factual arguments. Maybe you should take some time off work and mellow out.
You might want to read this:
Color vision is very subjective and humans are notoriously poor at judging light intensity, even monochromatic light. That is one reason photometers were invented. As to your having converted the image to B&W, do you have any understanding of how the conversion took place? Did it strip out two of the three channels, average the three channels, or use an RGB to IHS algorithm? Humans are most sensitive to green light and I may be a tetrachromat. Therefore, I am undoubtedly seeing the image differently. In any event, the actual intensity of the light, particularly in the context that we have been discussing it, is the area under the spectral curve. Do you have that information?
Do you know what wavelengths of light were used to make the composite image of the moon transit? It apparently was captured by a weather satellite. Its sensors are optimized for weather and cloud information. Therefore, I suspect that this is a false-color image where the bands chosen were intended for purposes other than capturing a true-color image, as seen by humans. If one or more of the bands used was an IR band, then the water is going to look darker than what an astronaut would see because of the strong IR absorption of water. There is so much you don’t know, and yet you feel justified in being insulting to someone who offers an opinion that is different than yours. You might want to think about that.
I was responding to Willis and you didn’t even bother to answer the questions I asked him.

Reply to  Clyde Spencer
June 1, 2017 1:53 pm

Well .. let us not make too much of a drama over political correctnes.
The pictures were taken by NASAs DSCOVR satellite, positioned at the La Grange point. The colors seem very reasonable, and I see no point in arguing they were false.
Also the conversion to gray is nothing complicated. Every pixel has three color values, for red, green and blue. So technically you just add up these values, devide them by 3, and assign this value to all the three color values. It does not yield wrong results. And the outcome is very clear. I did even use a tool to read those values pixel by pixel (with the original color), again the same outcome.
A reasonable subjection here would be on pictures only giving brightness within their technical limitations. Otherwise a painting the sun would have lightened up the night long before the invention of electrical light. As the clouds shown here reach maximum technical brightness (ie. FF.FF.FF), the picture will understate their brightness by comparison to other surfaces. But that is not what we are discussing.
The spot where there is direct solar reflection may indeed be somewhat brighter than the moon, but it is very small by comparison. And, after all, we can use the Fresnel eqiation to determine that part for all angles of incidence.

Clyde Spencer
Reply to  Erich
June 1, 2017 2:54 pm

“False-color” is a term of art used in the image processing and remote sensing fields where multispectral bands that don’t correspond to the spectral RGB are displayed as though they were RGB. With careful selection and balancing, images can be created that look very natural.
You acknowledge that the hot spot may be equal to or slightly brighter than the moon. That hot spot should be composed of a 2% specular component and an unknown diffuse component. If the Moon has a diffuse albedo of around 12 to 14%, that implies that the water has to have a diffuse component considerably larger than the 6 or 7% you are supporting; additionally, there is a specular component that generally is not observed that should be taken into account for energy balance calculations.

Clyde Spencer
Reply to  Erich
June 1, 2017 3:12 pm

The imagery was as I speculated: ” Exposure time for each of the 10 narrowband channels (317, 325, 340, 388, 443, 552, 680, 688, 764 and 779 nm) is about 40 ms…” “The animation was composed of monochrome images taken in different color filters at 30 second intervals for each frame, resulting in a slight color fringing for the Moon in each finished frame.”

May 25, 2017 11:57 am

Willis, I have spent zero time in the tropics so I’m coming from a place of profound ignorance here.

I could see the daily tropical cycle unfolding most days. Clear at dawn. Then cumulus clouds form usually before noon. Thunderstorms in the afternoon, sometimes lasting into evening or night. link

The idea seemed to be that convection and evaporation would carry humidity up in the atmosphere during the morning. Later in the day, there would be rain that would bring the cooled water back to the sea.
Clouds will block the surface radiation which will be at a wavelength corresponding to 30 or 40 C. If it is raining the condensing water will be giving up heat. The clouds should be radiating heat at some longer wavelength corresponding to a much lower temperature. Do you know if any of the work cited above takes that into account? Is it possible that the clouds that look most like a positive feedback are actually a negative feedback?

Reply to  commieBob
May 25, 2017 4:40 pm

I suggest you apply for a grant to study the effect of global warming on tropical beaches during N.H. winter as represented by toplessness. Race you!!!

Reply to  john harmsworth
May 28, 2017 3:41 am

It’s all yours. My favorite vacation takes me north away from the heat and humidity.

Crispin in Waterloo
Reply to  commieBob
May 26, 2017 6:02 pm

A refinement perhaps on your description:
Water vapour is transparent to visibly light but definitely not to IR. The fact that the air has a huge amount of water vapour in it means it is not really a ‘surface’ and ‘clouds’. The white clouds cool on top by reflecting but in IR they are very dark emitters.
The water vapour between the clouds and the ground works both ways, being an impediment to upwelling and downwelling IR. Ditto above the clouds.
They do not conveniently balance out because of their different temperatures. The air just under a cloud may be very humid, being an effective insulator ‘keeping the cloud bottom warm’.
In short, water vapour doesn’t need to condense to exhibit a large influence on heating and cooling.

May 25, 2017 12:08 pm

The trend in fig. 4 is a joke, R*R must be close to zero! Is this why it’s not stated?

Reply to  Willis Eschenbach
May 25, 2017 1:25 pm

Thanks for the additional info. I perfectly agree with you. I just feel very uncomfortable, whenever I see trend lines in observational data point clouds like the one in fig. 4 – too many people (even ones that are not mathematically illiterate) tend to take the output of Excel, Labview, Diadem etc. like the holy gospels. If the computer says so it must but the ultimate truth. Most don’t realize that while these numbers stated are perfectly correct calculated, they are still significantly divorced from the real world.

Reply to  Willis Eschenbach
May 25, 2017 1:30 pm

… must be the ultimate truth …

Clyde Spencer
Reply to  Willis Eschenbach
May 25, 2017 1:40 pm

I would put my money on net negative cloud feedback. The fact that we are here discussing this is evidence that there has never been a “tipping point” or runaway positive feedback.

Reply to  Willis Eschenbach
May 25, 2017 4:45 pm

I would have thought that the calculation of positive feedback would need to start by removing the cloud albedo from the total available. If it is reflected off the cloud tops back to space it cannot be radiated upward from the Earth or atmosphere to the cloud base. Perhaps I misunderstood the graph?

Reply to  Willis Eschenbach
May 26, 2017 8:22 am

That histogram is interesting since it shows that the cloud feedback is not normally distributed. Climate scientists for some reason almost always assume that climate data follow normal (gaussian) distributions, probably because it makes statistics much simpler. Actual climate data on the other hand are quite often not normally distributed. This is one more case.

Reply to  TheDoctor
May 25, 2017 1:51 pm

I agree with you.
This is a “scatter plot”
It simply shows that there is absolutely no relationship between the two variables, the data are scattered all over the place, no trend.
In order to justify a straight line, the regression coefficient (R squared) must be AT LEAST >0.55. p values are irrelevant in this type of analysis
In this case, without any calculation, a statistician, just looking at the plot, will tell anybody that R squared is <0.1 and will tell anybody that it is pure nonsense of drawing a straight line and pretending a decrease.

Rick C PE
Reply to  rd50
May 25, 2017 3:33 pm

rd50: I am a statistician (well an engineer with a degree in math specialized in prob & stats) and I fully agree. When a scatter plot looks like a shotgun pattern we say there is no association between the variables. There are various statistical tests for association such as a “corner test” but when a regression R-squared is less than 0.1 such tests will invariably just confirm the obvious.

Reply to  rd50
May 25, 2017 5:02 pm

To Rick C PE
I like your description “when a scatter plot looks like a shotgun pattern”
Enough said! Why would anyone believe this nonsense or even try a regression analysis.

Reply to  rd50
May 25, 2017 11:33 pm

So therefore cloud that forms and temperature are independent variables?

Reply to  rd50
May 26, 2017 1:39 am

@ rd50
“… that it is pure nonsense…”
Well, I would not go that far. First of all, if I’m not totally wrong, Willis Eschenbach tried to demonstrate that the often claimed “climate sensitivity” is either not real or at least insignificantly low. Drawing a straight line (=linear regression) through the shot gun pattern makes sense, if you want to show that the data is even consistent with the claim of an negative sensitivity, but at least 93% of the pattern cannot be explained by the alleged sensitivity.
That’s exactly his point.
What I object to is the presentation of -1 +/- 0.3 as a “matter of fact” trend. Many readers of this blog lack the background to properly interpret such pattern. So, to them, this is misleading.

Reply to  TheDoctor
May 25, 2017 2:47 pm

TheDoctor May 25, 2017 at 1:25 pm
… too many people (even ones that are not mathematically illiterate) tend to take the output of Excel, Labview, Diadem etc. like the holy gospels.

Amen brother, amen! Computers allow things like data dredging. Given enough data and enough analysis, spurious correlations are certain (100%) to occur.
One of my favorite bogus papers is a great example of the above. It’s about the health benefits of dark chocolate. It was based on an actual study which was correctly carried out. The bogus part was that it had too few participants and too many parameters. No journalist caught that. The only person who caught it was one blogger.
The above example was a deliberate hoax. There are lots of examples where scientists deceive themselves into thinking they have found something. The result is the replication crisis. The majority of papers are bogus because they can’t be replicated. In the field most likely to be solid (I’m biased), Engineering and Physics, 70% of papers can’t be replicated by others and 50% can’t even be replicated by the original authors. /rant

Reply to  commieBob
May 25, 2017 3:13 pm

The majority of papers are bogus because …

There’s more than one way to read that. Let me be perfectly clear:

Of all the papers published in peer reviewed science journals, the majority are bogus. The reason is …

Reply to  commieBob
May 25, 2017 5:16 pm

To Rick C PE
I like your description “when a scatter plot looks like a shotgun pattern”
Enough said! Why would anyone believe this nonsense or even try a regression analysis.
The same commieBob. The idea that there is a decreasing trend is absurd.

Reply to  TheDoctor
May 26, 2017 5:28 am

To TheDoctor
After saying “R*R must be close to zero!”
You are now saying that drawing a straight line through this shotgun pattern is just fine.
Glad you are not my doctor.

Reply to  rd50
May 26, 2017 8:43 am

@ rd50
“You are now saying that drawing a straight line through this shotgun pattern is just fine.”
Definitely not!
My point is, that in case you want to demonstrate that there is no or only a little temperature depending sensitivity ,applying linear regression to a pattern like the one in Fig. 4 is not nonsense, but quite helpful. The fact that you end up with an R*R-value < 0.1 (0.07) clearly shows that temperature is not a dominating factor!
Claiming that you can calculate a meaningful (!) value for sensitivity based on these data is where it went wrong.
These are not the same things. It depends on what your objective is.

Fischer, Kahlib (Helms School of Government)
May 25, 2017 12:18 pm

Thanks Scott! Consider yourself signed up!
Sent from my Verizon Wireless 4G LTE smartphone

Reply to  Fischer, Kahlib (Helms School of Government)
May 27, 2017 9:56 am

(Commenting only on the goofy autosig)
Bhell Atlantic. That’s not cool.

May 25, 2017 12:20 pm

See Fig 11 from my recent paper.comment image
Fig.11 Tropical cloud cover and global air temperature (29)
The global millennial temperature rising trend seen in Fig11 (29) from 1984 to the peak and trend inversion point in the Hadcrut3 data at 2003/4 is the inverse correlative of the Tropical Cloud Cover fall from 1984 to the Millennial trend change at 2002. The lags in these trends from the solar activity peak at 1991-Fig 10 – are 12 and 11 years respectively. These correlations suggest possible teleconnections between the GCR flux, clouds and global temperatures.
Climate is controlled by natural cycles. Earth is just past the 2004+/- peak of a millennial cycle and the current cooling trend will likely continue until the next Little Ice Age minimum at about 2650.See the Energy and Environment paper at
and an earlier accessible blog version at
Here is the abstract for convenience :
This paper argues that the methods used by the establishment climate science community are not fit for purpose and that a new forecasting paradigm should be adopted. Earth’s climate is the result of resonances and beats between various quasi-cyclic processes of varying wavelengths. It is not possible to forecast the future unless we have a good understanding of where the earth is in time in relation to the current phases of those different interacting natural quasi periodicities. Evidence is presented specifying the timing and amplitude of the natural 60+/- year and, more importantly, 1,000 year periodicities (observed emergent behaviors) that are so obvious in the temperature record. Data related to the solar climate driver is discussed and the solar cycle 22 low in the neutron count (high solar activity) in 1991 is identified as a solar activity millennial peak and correlated with the millennial peak -inversion point – in the RSS temperature trend in about 2004. The cyclic trends are projected forward and predict a probable general temperature decline in the coming decades and centuries. Estimates of the timing and amplitude of the coming cooling are made. If the real climate outcomes follow a trend which approaches the near term forecasts of this working hypothesis, the divergence between the IPCC forecasts and those projected by this paper will be so large by 2021 as to make the current, supposedly actionable, level of confidence in the IPCC .

Dave Yaussy
May 25, 2017 1:06 pm

Ozymandias is a great poem, but it doesn’t have the language you quote:
I met a traveller from an antique land,
Who said—“Two vast and trunkless legs of stone
Stand in the desert. . . . Near them, on the sand,
Half sunk a shattered visage lies, whose frown,
And wrinkled lip, and sneer of cold command,
Tell that its sculptor well those passions read
Which yet survive, stamped on these lifeless things,
The hand that mocked them, and the heart that fed;
And on the pedestal, these words appear:
My name is Ozymandias, King of Kings;
Look on my Works, ye Mighty, and despair!
Nothing beside remains. Round the decay
Of that colossal Wreck, boundless and bare
The lone and level sands stretch far away.

Clyde Spencer
Reply to  Dave Yaussy
May 25, 2017 1:21 pm

Yes, I believe the “Abandon all hope” admonition is from Dante’s Inferno.

Don K
Reply to  Dave Yaussy
May 25, 2017 1:30 pm

“abandon all hope ye who enter here” is Dante — The inferno. Ozymandias is Shelley. (And yes it is quite striking). So endeth my knowledge of classic literature.

Gary Pearse
Reply to  Dave Yaussy
May 25, 2017 2:34 pm

Well, despair is erm… abandonment of all hope!

May 25, 2017 1:20 pm

“I thought, hmmm … I wondered how that compared to the CERES data. Here’s a look at the same thing, net cloud feedback … except theirs is modeled and the CERES satellite data below is observational.”
They are Looking at an ABRUPT 2X c02 case.
Ya cant really compare that to observations.
Unless you could abruptly double c02

Rob Morrow
Reply to  Steven Mosher
May 26, 2017 6:51 am

Ya also cant really compare the value of fantasy data with observations.
Unless you double the hubris and smugness of its generators

May 25, 2017 1:29 pm

Sorry Willis, but the slope of figure 4 does NOT indicate the cloud feedback. Each data point includes the effects of random changes in cloud cover and temperature causing random changes in forcings and feedbacks in unknown quantities. The paper by Spencer&Braswell 2010 shows the problem with this analysis. Dr. Roy Spencer writes here ( )

In the real world, there are always changes in cloud cover (albedo) occurring, which is a forcing. And that “internal radiative forcing” (our term) is what gives the illusion of positive feedback. In fact, feedback in response to internal radiative forcing cannot even be measured. It is drowned out by the forcing itself.

The abstract of the Spencer&Braswell 2011 paper “On the Misdiagnosis of Surface Temperature Feedbacks from Variations in Earth’s Radiant Energy Balance” at

Here we present further evidence that this uncertainty from an observational perspective is largely due to the masking of the radiative feedback signal by internal radiative forcing, probably due to natural cloud variations. That these internal radiative forcings exist and likely corrupt feedback diagnosis is demonstrated with lag regression analysis of satellite and coupled climate model data, interpreted with a simple forcing-feedback model.

The paper shows that the satellite data suggests that climate sensitivity is much less than that of climate models, but the value is difficult to quantify.

May 25, 2017 1:41 pm

Clouds make the night warmer the same way draperies make a house warmer. Draperies/clouds reduce the conductivity and for a given Q, Btu/h, when U goes down dT and the room/surface temperature must increase. It’s Q = U * A * dT that is responsible for the earth’s warm surface, RGHE and 33C are complete nonsense.

Reply to  Nicholas Schroeder
May 26, 2017 7:21 pm

The 33c RGHG effect is complete nonsense , but not for the reasons given . The equilibrium temperature of a gray ball in our orbit is easily calculated to be about 278.5K +- 2.3 from peri- to ap-helion simply from the total energy impinging on a point in our orbit . A conversation with a number of members of the OCO-2 satellite team at the end of this week’s tended to confirm that the temperature of satellite instrument boxes is about that ~ 5c and their sensors must be actively cooled to about 120K .
Our spectrum as seen from the outside apparently reduces our radiative equilibrium closer to that endlessly parroted 255 . I’ve got too many other interests to spend time finding an actual measured spectrum for the earth or mess with digitizing some graph , but give me a spectrum and I’ll give you its computed equilibrium — something which seems to be beyond the education of the journeyman “climate scientist” ( else why the perseveration on the crude 255 meme ) .
What needs to be explained , rather , is why the stratosphere is so much colder than the equilibrium temperature in orbit next to us .
In any case , the fact that our surface temperature is about 3% warmer than the ~ 279 gray body temperature cannot be explained by any spectral effect any more than Venus’s surface temperature , 225% of the gray body temperature in its orbit , can be .

Reply to  Nicholas Schroeder
May 27, 2017 12:24 pm

So what would the earth be like without an atmosphere?
The average solar constant is 1,368 W/m^2 with an S-B BB temperature of 390 K or 17 C higher than the boiling point of water under sea level atmospheric pressure, which would no longer exist. The oceans would boil away removing the giga-tons of pressure that keeps the molten core in place. The molten core would push through the floor flooding the surface with dark magma changing both emissivity and albedo. With no atmosphere a steady rain of meteorites would pulverize the surface to dust same as the moon. The earth would be much like the moon with a similar albedo (0.12) and large swings in surface temperature from lit to dark sides. No clouds, no vegetation, no snow, no ice a completely different albedo, certainly not the current 30%. No molecules means no convection, conduction, latent energy and surface absorption/radiation would be anybody’s guess. Whatever the conditions of the earth would be without an atmosphere, it is most certainly NOT 240 W/m^2 and 255K.
At 100 km, i.e. stratosphere, there are no molecules. The very concepts of energy. heat and temperature get bit flakey.
278.5 K gray body calculation at this orbit – got a link?
390 K BB, 278.5 Gray – emissivity?

The Original Mike M
May 25, 2017 1:43 pm

The bottom line is that despite the IPCC’s admission that –

The climate system is a coupled non-linear chaotic system, and therefore the long-term prediction of future climate states is not possible.
… the climate science cabal shamelessly continues to use their climate models to make scary long-term predictions of impending doom.
As for clouds I can say as a pilot that “I’ve looked at clouds from both sides now.” … and being above them is generally a smoother ride than being under them. (Sorry Joni!)

May 25, 2017 2:22 pm

There are only two “Forcings.” The Sun, Father Sol, irradiates our spherical planet at 1366 W/m2. The surface of Mother Earth radiates to the sky at a flux determined by her surface temperature.
Clouds are not a forcing. Clouds reflect incoming short-wave, and absorb some outgoing LWIR. Any attempt to assign a flux in W/m2 to clouds is fundamentally incorrect.
The energy balance diagrams ubiquitous in “Climate Science” are meaningless because they assume it is legitimate to divide 1366 by four. The heat transfer is directly proportional to the fourth power of the temperature difference, and is ALWAYS from Warmer to Cooler. The energy diagrams were concocted to fool people who do not know that the only way the Earth cools is by radiating to space.
There is an instrument called a pyrGeometer, used to find remote temperatures by comparing measured radiation to a scale. “Climate Scientists” point them at the sky, read a temperature, and claim that the radiation detected is “Downwelling” and warms the surface. Then they run Stefan-Boltzman backwards and claim to have detected a flux of some particular number of W/m2.
I had to look this up as it was not discussed in my engineering Heat Transport class because engineers manage energy which costs money. Radiation from a cooler source, striking a warmer surface, is simply reflected with ZERO transfer of Heat. Radiation is created by vibrating charged particles, namely the protons and electrons of which matter is composed, also neutrons of course. When transferring HEAT, which is ENERGY, the transfer is always from warmer to cooler.
The best proof of this is binary stars. Binary stars with elliptical orbits, basically all of them, have a warmer star approached by and then withdrawing from a cooler star. Guess what? The warmer star does NOT warm up when approached by the cooler star. If it did, the spectrometry would show this, but it NEVER DOES.
CO2 at the Top of Atmosphere absorbs some outgoing LWIR in the proper wavelength around 15 microns. MORE CO2 does some more of this at a slightly higher altitude, thus lowering the temperature at which the Atmosphere radiates to Space, slightly, thus retaining slightly more HEAT in the atmosphere, warming the whole thing slightly. Good luck to anyone trying to calculate this effect, however…

Reply to  Michael Moon
May 25, 2017 3:43 pm

MM, to a first order approximation it HAS been calculated. Have done that math now in two ebooks and several blog posts here and at Climate Etc. What we can find is that a doubling of atmospheric CO2 in the absence of feedbacks will cause an ECS rise of 1.1-1.2C (converted from Kelvin). In fact, using the IPCC AR5 and math that C. Monckton has posted here previously, we can even calculate it is ~1.16C (useless pseudoprecision).
All irrelevant, because what matters are the water vapor and cloud feedbacks (the rest of the possible feedbacks, to a first order IPCC AR4-5 approximation, zero out). Now, WE post here finding clouds weak negative, other guest posts and writings by myself finding water vapor weak positive (e.g. Essay Humidity is still Wet) foot to observational energy budget ECS 1.6-1.7. Tidies up nicely. Even tho WE does not think ECS exists.

A C Osborn
Reply to  Michael Moon
May 26, 2017 5:33 am

MM, Binary Stars, that is the best refutation of cooler bodies warming hotter bodies that I have ever seen.
How accurate though is the spectrometry?
Thank you.

May 25, 2017 2:26 pm

If you want to demonstrate the effects of clouds on TOA fluxes using the ceres data, select region 170 to 270 long and -10 to 10 Lat and subtract total long wave TOA flux from clear sky long wave TOA flux.

May 25, 2017 3:21 pm

Ceppi17 used the same method (clear sky v. all sky) as Dessler 2010. Dessler’s junk paper amplified by NASA found a positive cloud feedback– with an r^2 of 0,02! This paper usingbthe same method finds a negative cloud feedback. Conclusion, maybe the method itself is wrong.
WE also derives a range of observational negative cloud feedbacks. This means all the CMIP5 models are WRONG. They all have significant positive cloud feedback. Essay Cloudy Clouds provides more amusing details.

May 25, 2017 3:43 pm

“[1] We present a new high-resolution model of the Earth’s global atmospheric electric circuit (GEC) represented by an equivalent electrical network. Contributions of clouds to the total resistance of the atmosphere and as current generators are treated more realistically than in previous GEC models. The model of cloud current generators is constructed on the basis of the ISCCP cloud data and the OTD/LIS lightning flash rates and TRMM rainfall data. The current generated and the electric resistance can be estimated with a spatial resolution of several degrees in latitude and longitude and 3 hour time resolution. The resistance of the atmosphere is calculated using an atmospheric conductivity model which is spatially dependent and sensitive to the level of solar activity. An equivalent circuit is constructed assuming the ionosphere and ground are ideal conductors. The circuit solution provides diurnal variations of the ionospheric potential and the GEC global current at the 3 hour time resolution as well as the global distributions and diurnal variations of the air-Earth current density and electric field. The model confirms that the global atmospheric electric activity peaks daily at ∼21 UT. The diurnal variation of the ionospheric potential and the global current have a maximum at 12 and 21–24 UT in July and at 9 and 21 UT in December, and a global minimum at 3–6 UT independent of season. About 80% of the current is generated by thunderstorm convective clouds and 20% by mid-level rain clouds.”

May 25, 2017 3:58 pm

w. ==> “…ample evidence that the climate is a governed system, with a variety of thermoregulatory climate phenomena that work together to constrain the global temperature to a very narrow range (e.g. ± 0.3°C variation over the entire 20th Century). ”
I agree with this almost entirely — it seems obvious to me that the Earth climate is self-regulating, probably due to the stabilizing effects of interdependent chaotic systems operating within their non-chaotic ranges (see Chaos = Stability).
Also obvious are the two states of the climate — the two manifesting “attractors” of the climate system — Ice Ages (glacial episodes) and Interglacials.

Reply to  Kip Hansen
May 25, 2017 4:50 pm

True within limits. The present ‘ice age attractors’ only exist after the Panama isthmus closed~ 2.2mya. The Earth is in a slow tectonic waltz that continues always.
My issue is, that slow waltz, Pleistocene glaciations/interregnums, and resulting paleoclimates say little about the last century, where global population has more than tripled, and fossil fuel consumption has about quintupled. WE’s thermostat assumes neither the house nor the furnace nor the thermostat changes. Possible, not probable.
So, we should look closely at the last century or so only. In that time period, there were two periods of temperature rise; all else is cooling or stable. The first rise period is ~1920-1945. The second is ~1975-2000. IPCC AR4 WG1 SPM fig. 8,2 says the first period cannot have been mainly AGW; not enough increase in atmospheric CO2. Natural variation did not magically stop in 1975. The Achilles heel of warmunism is the AGW attribution problem in the second rise period, confounded by the pause since. An Occam’s razor irrefutable argument.

Reply to  ristvan
May 25, 2017 5:38 pm

ristvan ==> We might better look at the last three thousand years…. that brings in the lesser variations of warm periods and cooler periods.

May 25, 2017 4:00 pm

….. there is NO SUCH THING as “Climate sensitivity”. This concept of SCHNEIDER & MASS (1975) was several times proven wrong, see 1.) Kramm &Dlugi: “Scrutinizing
the atmospheric greenhouse effect and its climatic impact” (2011) and 2.) Gerlich & Tscheuschner:” Falsification of the atmospheric CO2 greenhouse effects within the frame of physics” (2009).
Willis, please check those papers out, they contain the answers to this mistaken concept.

May 25, 2017 4:08 pm

Just confused by what I think is a typo. Where you say:
“the cloud feedback factor of -1.4, -1.0 ± 0/3, and -0.5 W/m2 per °C”
do you mean “-1.0 ± 0.3” or something else?

May 25, 2017 4:34 pm

Maybe the climate changed in Ozymandias’ time as well but he never wrote about it. His civilization probably didn’t originate in a desert I’m thinking. Oh my! Heresy!!

May 25, 2017 4:35 pm

Mmmmm-spherical chickens…

Bill Illis
May 25, 2017 5:02 pm

The cloud radiative feedback value has been carefully chosen by the IPCC and the Hansen’s of the world because if it were 0.0 or -1.0 W/m2/K, then all the catastrophic predictions of global warming theory fall by the way-side.
That is why they refuse to do what Willis has done and actually measure/calculate it.
When you put in -1.0 W/m2 cloud feedback into the calculations, the 3.0C per doubling narrative falls apart and all we get is 1.37C per doubling. Furthermore, there is little feedback on feedback impact and the earth’s atmosphere adjusts to increased CO2/GHGs very quickly – as in within two days. There might still be an ocean uptake lag but it is much smaller.
The IPCC and global warming is OVER if cloud feedback is as negative as -1.0 W/m2/K. It is time for Willis’ numbers to be published.comment image

Reply to  Bill Illis
May 25, 2017 7:53 pm

Willis did not “measure/calculate” the cloud radiative feedback value. All he did was plot monthly change of cloud radiative effect CRE in W/m2 versus the monthly change in temperature, got a near random scatter plot, calculated a best fit slope of -1.0 W/m2/K, and falsely called it net cloud feedback. See my comment of May 25, 2017 at 1:29 pm. This procedure falsely assumes that only temperature change caused the cloud change. The definition of cloud feedback is the change in the top-of-atmosphere radiative flux resulting from the cloud response to a temperature change. But causality also runs the other direction. The cloud changes also cause a temperature change. Each point is a mixture of internal radiative forcing, ie cloud changes causing a temperature change, and cloud feedback, ie temperature change causing cloud changes represented by the change in the cloud radiative effect (CRE).
The Spencer paper shows that the internal radiative forcing contaminated the signal, so the slope is not the cloud feedback, but is greater than the true feedback. That is, the true cloud feedback is less than(more nagative than) -1 on this monthly time scale. But the BIGGER problem is that is only for monthly changes, and there is no way to show that this short-term feedback operates on long time scales relevant to AGW.
The sentence in the top post “Here’s a look at the same thing, net cloud feedback … except theirs is modeled and the CERES satellite data below is observational.” is misleading, as it is not the same thing at all! The modeled net cloud feedback was determined by an abrupt quadrupling of CO2 concentrations, causing a HUGE forcing, so the issue of the internal radiative forcing becomes insignificant. In the models, you can compare the time histories of the resulting warming to the resulting changes in the Earth’s radiative budget, and you can indeed extract an accurate estimate of the feedback in the models. But this tells us nothing about the real world. Spencer’s work shows that the short term cloud feedback could be as much as -6W/m/K. Spencer wrote “Unfortunately, there is no way I have found to demonstrate that this strongly negative feedback is actually occurring on the long time scales involved in anthropogenic global warming.”
It is interesting that the paper referred to, Ceppi et al 2017, shows that in the multi-model mean, almost all of the positive cloud feedback is the long wave cloud feedback of 0.42 W/m2/K, with the short wave, or cloud albedo effect, contributing a tiny 0.02 W/m2/K.

Reply to  Ken Gregory
May 26, 2017 12:20 am

“The definition of cloud feedback is the change in the top-of-atmosphere radiative flux resulting from the cloud response to a temperature change. But causality also runs the other direction. The cloud changes also cause a temperature change.”
As there is no recorded ‘runaway’ temperatures on earth, the above mechanism is a closely coupled system keeping general temperature within narrow limits. AKA negative feedback.

michael hart
May 25, 2017 6:42 pm

“Equilibrium climate sensitivity” indeed does not exist outside of the models, almost at the level of definition. In order to calculate it, the models need to be assured of calculating all possible future climate states, or at least be assured of calculating a truly representative ensemble of climate states. That, in itself, requires a large dollop of faith. Modeled or real climates may end up sitting in a non-equilibrium state for indeterminately long periods of time, skewing results by unknown amounts.
How long should a modeler wait to see? Probably only until funding for CPU cycles runs out, but it may still give significantly different results depending on small changes in input parameters or the day of the week. Back in the real world, reality will choose only one path and there is no reason to think that events will proceed smoothly down a representative path towards a hypothesized equilibrium.
Equilibrium climate sensitivity remains a fancy way of saying “this is what my model says may happen under a certain set of conditions”. To pretend otherwise is to somehow dignify the concept as being on a par with real world measurable physical constants.

May 25, 2017 8:48 pm

Willis, is there anyway you can explain to readers the implications of the sign change in cloud feedback?
How much does that change the global warming estimate in degrees?
My work with old climate models showed this was basically almost all of global warming in this one parameter.

May 25, 2017 9:43 pm

Willis, thank you for the essay.

May 26, 2017 12:18 am

“And because it is decoupled, there is no such thing as “climate sensitivity”
Wouldn’t it be more accurate just to say the sensitivity is effectively zero?
But actually I see your point, within the very idea of ‘climate sensitivity’ is the assumption that it is not only linear, but also that it never gets saturated, buffered or approaches zero. Rather big assumption that any chemist could point out.

May 26, 2017 1:37 am
May 26, 2017 7:08 am

I am trying to understand figure 1. I understand one basic calculation to be
Warming in watts per meter^2 x Sensitivity in K/Wm^2 = warming in K OR
Warming in watts per meter^2 Inverse Sensitivity in Wm^2/K = warming in K
but the values on the y axis are inverse Sensitivity in Wm^2/K . what gives? Feedbacks should be a function of the function greenhouse Gas warming, in wm-2, with gain and also possibly lags, integral, and/or derivative terms, right?

May 26, 2017 7:52 am

Ignore my question I see now that it is w/m-2 feedback for 1 C of surface warming.

May 26, 2017 10:13 am

It is interesting that both the CMIP5 and Ceres data agrees that the cloud feedback is strongly negative in the 40-60 degrees south latitude band. This helps explain the great global climate impact of the opening of the Drake Passage in the Oligocene. It not only isolated Antarctica climatically, it quite possibly also changed the net cloud feedback from positive to negative.

May 26, 2017 4:40 pm

Willis wrote: “On average, Figure 1 shows that for every degree C that the modeled surface warms, the modeled clouds add on another ~ 0.5 W/m2 of additional modeled forcing. Let me say that I find such a large positive cloud feedback to be very doubtful.”
Why? Clouds reflect an average of about 80 W/m2 to space. Why can’t this value decrease by 0.6% per K of surface warming?
If I understand correctly, the data in Figure 3 shows ratios, W/m2/K, not W/m2. So everywhere there was relatively little temperature change (say +0.02 K), even a modest change in TOA flux will produce a large change in feedback – either positive or negative. And if the temperature change were negative (say -0.02 K) the magnitude of the feedback would be just as big, but the sign would be different. This is another way of looking at the problem you discussed with median vs mean.
Figure 4 may be showing us which data points are most robust – those with the largest amount of warming and cooling. However, if you simply plotted dW vs dT, then the climate state in 2000 (a strong La Nina) and in 2015 (a weak? El Nino) could easily be dominating your data. Starting with and El Nino and finishing with a La Nina might make the results look completely different. If you do a linear fit between W and T at each location over the whole period, this problem would be minimized. However that approach doesn’t provide individual dW and dT values to plot. In theory, you have the trend at each locations X+/-Y K/15 yr for both temperature and TOA flux. If you take into account auto-correlation, the error bars are fairly huge. HADCru4 GLOBAL warming at Nick Stoke’s blog is 1.23 +/- 0.71 K/century (0.184 +/- 0.106 K) for 1/2000 to 12/2015. Local variability is far greater than global variability. If you put error bars on the points in Figure 4, the result may not be very convincing.
IIRC, there is combined TOA flux data from many satellites going back to 1979, but it isn’t as homogeneous as CERES. The global warming rate is 1.72 +/- 0.24 K/century (0.67 +/- 0.09 K). With smaller error bars, perhaps the data would be more convincing.
Nic has been discussing these recent papers, which suggest that cloud feedbacks don’t remain constant with time in models.
Gregory (2016) (paywall)
Andrews (2015) (Free)

May 27, 2017 4:22 am

Willis, I know you have said you are not too interested in the peer-review literature debate, but would a reply to Ceppi et al in the journal not open this issue up to further debate?

David Dibbell
May 27, 2017 4:32 am

“Let me say in closing that I don’t think that “climate sensitivity” is a real thing. I say this because of ample evidence that the climate is a governed system, with a variety of thermoregulatory climate phenomena …” In support of this point, I note that the upward heat delivery rate implied by a one-inch-per hour rainfall event is about 16,000 watts/M^2. Even at a tenth of an inch per hour, it is 1,600 W/M^2. So what? It becomes clear that the “climate,” whether defined locally, regionally, or globally, is the composite result of a large number of small-scale heat exchange events of very high power. Any expression of forcings of single-digit W/M^2 or single-digit dec C per doubling are not really meaningful in view of how the atmosphere expresses its operation for all to see. Steam engines rule.

May 28, 2017 12:59 am

Willis has briefly described a ‘flywheel’ aspect of Earth’s climate system that is all too easy to take for granted. Flywheels are essential components in almost all of mans’ gadgets, from gyroscopes, through engines on the Titanic and your bracero’s lawn mower. They work well and are- here’s that word again- Essential. It would be a surprise if a 4+ billion year-old, superbly working machine didn’t have one.

Clyde Spencer
Reply to  Willis Eschenbach
May 28, 2017 4:33 pm

Averages hide a multitude of ‘sins.’ I think that this problem has to be attacked piecemeal, with whatever integration slices or aggregations make sense. Hence my harping on how specular reflection changes with angle of incidence and I suggest that frustums be the first-order sampling and modified where land covers the oceans. Further, the impact of cloud cover over land and ocean has to be taken into account. This is a devilishly complex layer-cake problem where the saying, “The Devil is in the details,” is appropriate!
I haven’t verified RA’s claims about actual values, but clearly, the flux through a cloud-free atmosphere decreases with increasing angles of incidence because the footprint of a light beam spreads out, and a longer slant-range allows for more scattering and absorption.

Reply to  Willis Eschenbach
May 28, 2017 11:26 pm

Willis Eschenbach
Thank you for the courtesy of your reply. I’m working 7×12 night shifts right now, and will get back to you with comments about your feedback as soon as practical.
I do note that your CERES average value of 347 watts/m^2 at TOA would require 1388 Watts/m^2 for TSI. I understand the correct TSI (average value at a nominal earth radius) is much less, only 1362 watts/m^2 TSI now the standard.
(Of course, the original 1988 TSI value of 1368 watts/m^2 was used in the original catastrophic global warming calculations by Hansen and his cronies, so why are those same 1988 calculations and constants and coefficients valid when we “apparently” already “lost” 6 watts/m^2 …. And, if after losing 6 watts/m^2 already, why are we still going to heat up the earth catastrophically due to 3.0 degrees of added warming from CO2? It would seem we would be cooling now, and facing a deficit of 3 watts/m^2 that must be replaced just to keep temperatures the same. But that is a separate question entirely. )

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