Guest Post by Willis Eschenbach
There’s a new study in PNAS, entitled “Observational determination of albedo decrease caused by vanishing Arctic sea ice” by Pistone et al. Let me start by registering a huge protest against the title. The sea ice is varying, it isn’t “vanishing”, that’s just alarmist rhetoric that has no place in science.
In any case, here’s their figure 4B, showing the decrease in albedo from the “vanishing” sea ice:
Figure 1. Graph from Pistone2014 showing CERES albedo data (green, solid line) for the ocean areas of the Arctic.
The authors say:
Using the relationship between SSM/I and CERES measurements to extend the albedo record back in time, we find that during 1979–2011 the Arctic darkened sufficiently to cause an increase in solar energy input into the Arctic Ocean region of 6.4 ± 0.9 W/m2, equivalent to an increase of 0.21 ± 0.03 W/m2 averaged over the globe. This implies that the albedo forcing due solely to changes in Arctic sea ice has been 25% as large globally as the direct radiative forcing from increased carbon dioxide concentrations, which is estimated to be 0.8 W/m2 between 1979 and 2011.
The present study shows that the planetary darkening effect of the vanishing sea ice represents a substantial climate forcing that is not offset by cloud albedo feedbacks and other processes. Together, these findings provide direct observational validation of the hypothesis of a positive feedback between sea ice cover, planetary albedo, and global warming.
So … how are they going about making that case?
Let me start by saying that looking at albedo as they are doing is a very roundabout and inaccurate way of handling the data. The CERES dataset doesn’t have an “albedo” dataset. Instead, they have a dataset for downwelling solar, and another dataset for upwelling solar. The problem is that when the numbers get very small, the values of the calculated albedos get more and more inaccurate. Albedo is reflected solar divided by downwelling solar. So when you get down to where there’s almost no sunshine, you can get things like a gridcell that averages 0.2 W/m2 of incoming sunlight over some month, and reflects 0.4 W/m2 … giving us an impossible albedo of 2.0 …
It’s not clear how Pistone et al. have handled this issue. The way I work around the problem is to calculate the average upwelling reflected sunlight for the Arctic ocean area, and divide that by the average downwelling sunlight for the Arctic ocean area. This gives me an overall average albedo. I get slightly different numbers from theirs, and I am unable to replicate their results. However, I do get about the same trend that they get over the period, a decrease in the albedo of about 1.5% per decade. However, I don’t particularly trust those albedo numbers, averages of ratios make me nervous.
For this reason, I use a different and simpler measure, one which Pistone et al. mention and quantify as well. This is the actual amount of sunlight that makes it into the climate system. The authors call this the “total solar energy input”, and I will follow the practice. And qualitatively, my results agree with theirs—the amount of sunlight absorbed by the arctic has indeed increased over the period of the CERES data, 2000-2013. Figure 2 shows both the clear-sky and the all-sky arctic total energy input:
Figure 2. Increase in solar energy input to the Arctic ocean areas, 2000-2012. Clear sky in black, all sky in red. Units are area-weighted watts/m2.
In addition to the overall trend in all sky solar input (green line), you can see the peak in energy input in 2007, with the high solar input corresponding to very low ice areas. Overall, Figure 2 shows an even greater increase in energy input than Pistone2014 have estimated over the entire period. They report an increase in Arctic solar energy input over the ocean of 0.21 W/m2 over 32 years … and the CERES data shows an increase of 0.3 W/m2 per decade.
So we’ve established that their first claim, of increasing solar energy input to the Arctic ocean area 2000-2012, is true, and perhaps even underestimated. And this is quite reasonable, since we know the sea ice has decreased over the period … but what about their second claim? As you may recall, this was (emphasis mine):
The present study shows that the planetary darkening effect of the vanishing sea ice represents a substantial climate forcing that is not offset by cloud albedo feedbacks and other processes. Together, these findings provide direct observational validation of the hypothesis of a positive feedback between sea ice cover, planetary albedo, and global warming.
The CERES data agrees that the increase in solar energy input from reduced ice cover is not counteracted by Arctic clouds … nor would I have expected it to be counteracted by clouds in the Arctic. As I have discussed, well, more than once, the main climate control system is in the tropics. So if this increase in absorbed energy were counteracted by clouds, my hypothesis is that it would happen be in the tropics. I’ll return to this in a moment.
First, however, they’ve claimed that their results establish the existence of “a positive feedback between sea ice cover [and] planetary albedo”. Since the planetary albedo controls the total solar energy input to the globe, let’s take a look at the same data as Figure 2, total solar energy input, but this time for the entire planet …
Figure 3. Available solar energy at the top of atmosphere (red) and total solar energy input to the globe (blue), 2000-2012. Units are area-weighted watts/m2.
So their claim of increased solar energy input to the Arctic from reduced sea ice is true … but their claim that there is “a positive feedback between sea ice cover [and] planetary albedo” is falsified by the CERES data. The total solar energy input (blue line above), and thus the planetary albedo, is amazingly stable over the time period. There is no feedback at all from the changes in the ice.
To illustrate the stability, Figure 4 shows a breakdown of the total solar input data (blue line above). It’s divided into panels that from top to bottom show the data itself, the seasonal pattern, the trend, and the residuals of the global solar energy input:
Figure 4. Decomposition of the solar energy input signal into trend, seasonal, and residual components. Red scale bars on the right indicate the relative scale of the individual panel. Units are area-adjusted W/m2.
I’ve written before about the amazing stability of the climate system. This is another example. In the past people have objected that the system is forced to be stable, because over time, energy out must generally equal energy in.
But the global solar energy input, the amount of the available solar energy that actually makes it into the climate system, is under no such constraint. There is nothing that it must balance to. Solar energy input is a function of the albedo, which is determined by clouds, snow, ice, vegetation, and wind, and all of these are constantly varying in all parts of the planet … and despite that, the swings of the trend are no greater than ±0.3 w/m2 over the period. The maximum monthly deviation from the seasonal average is a mere one W/m2, and the standard deviation of the residuals (data minus seasonal) is half a watt/m2.
So … how does it happen that we have a strong increase in solar energy input in the Arctic, but the global energy input stays the same?
Well … as I mentioned above, the tropics. Over the period 2000-2012, during which the Arctic received increased solar energy input, here’s what’s happened in the tropics:
Figure 5. Total solar energy input, all skies, tropics. Units are area-adjusted W/m2
As I hypothesized, the control is happening in the tropics. Pistone et al. note that the Arctic solar input is going up because of decreased sea ice … but they did not notice that at the same time, the tropical solar input is going down because of increased clouds. And the net sum of all of the changes, of more energy being absorbed in the extra-tropical areas and less energy being absorbed in the tropics, is … well … no change at all for the globe. It all averages out perfectly, with little change in either the monthly, annual or decadal data.
Coincidence? Hardly.
This is about as neat a demonstration as I can imagine in support of my hypothesis that the system is not ruled by the level of the forcings—instead, it is regulated by a system of interlocking emergent climate phenomena. A number of these phenomena operate in the tropics, and they have a curious property—the warmer the planet gets, the more that they cut down on the incoming solar energy.
So at the end of the day, we find that the claim of the authors that increased solar input to the Arctic is connected to the planetary albedo to be true … except that it is true in exactly the opposite of the direction that they claimed. When more energy is absorbed in the Arctic, less energy is absorbed elsewhere.
In closing, I want to highlight what it was that got me interested in climate science to begin with. I wasn’t interested in finding out why the global temperature had changed by something like ± 0.3°C over the 20th century.
Instead, I was interested in finding out why the global temperature had only changed by ± 0.3°C over the 20th century. I was amazed by the stability of the system, not the fact that it had varied slightly. So let me close with a graph showing the total global solar input residuals, what remains after the seasonal cycle in total solar input is removed.
Figure 5. Residual total solar input after the seasonal cycle is removed. Dotted lines show the inter-quartile range. Smooth curve is the loess trend line.
The monthly deviation from the seasonal cycle is tiny. Half the months are within a third of a W/m2 of the seasonal average … a third of a watt, to me that’s simply amazing.
Now, you might disagree with my hypothesis that the planet is thermoregulated by emergent climate phenomena such as thunderstorms, El Nino, and the PDO.
But the stability shown in the above graphs surely argues strongly for the existence of some kind of regulatory system …
My regards, as usual to everyone.
w.
AVISO: If you disagree with what I or anyone says, please quote the exact words you disagree with. It allows everyone to understand exactly what you are objecting to.
DATA AND CODE: You’ll need the CERES data (227 Mb) , the CERES surface data (117 Mb) and two support files (CERES Setup.R and CERES Functions.R) in your R workspace. The code is Arctic Albedo.R, it should be turnkey.
[UPDATE]
Well, y’all will find this funny, I assume … following up on the question of the net effect of the loss of the sea ice that came up below in the comments, I decided to see what was happening with the upwelling longwave. We’ve established that the loss of the ice increases the total solar energy input … but what about the energy loss via longwave? (Yr. humble author slaps forehead for not thinking of this sooner …)
As you can see, the change in solar energy input is more than offset by increased losses … so the net effect of the melting sea ice is a net energy loss of 0.05 W/m2 per decade.
Note again the stability over time. Note also that this part of the system is not constrained by any need for solar input and longwave output to be stable, or to have the configuration they have. The average solar input is 30 W/m2, and the average longwave loss to space is 63 W/m2 … and despite the marked changes in ice cover over the period, they’ve only changed about 1% per decade over the period …
Gotta love the climate, always more surprises …
w.
[UPDATE II]
Some folks have said that there is a problem with my area-weighting, so let me explain exactly what I did.
The data exists in 180 latitude bands. The center of the bands start at -89/5° (south) and end up at 89.5° (north). To area-weight the data, we want to adjust the results for each gridcell by the area.
What we want to do is adjust the results to give what you would get if they had the size of the average gridcell. Now the area is proportional to the cosine of the mid-latitude. So what we do is multiply each gridcell result by
area of the gridcell / area of the average gridcell
This give each gridcell the value it would have if it were of average size. The effect of tiny gridcells is reduced, and the effect of large gridcells is increased.
Now, what is the average cell size? Well, if we integrate Cos(x) from zero to pi/2, we get 1. So the average gridcell size is 1/(pi/2) = 2/pi ≈ 0.637.
As a result, the weighting factor by which we multiply the gridcell value is, as you recall:
area of the gridcell / area of the average gridcell
which is equal to
cosine of the gridcell midlatitude / 0.637
Once you’ve multiplied the data by those weighting factors, you can compare them directly, as they are all adjusted to the average gridcell size.
The way to test if you’ve done this correctly is to see if the plain vanilla average of the newly-weighted dataset is correct. For example, see the average available solar (~340 W/m2) and solar input (~240 W/m2) values in Figure 3. They are simple averages of weighted data.
Note that there are two ways to do the weighting.
The first is to do all calculations (trends, etc) using the unadjusted variables. Then to get an average, you use what is called an “weighted mean”, which weights the data on the basis of gridcell area as it calculates the average.
The other way to do it is the way I described above, which converts all of the data to what an average sized gridcell would show. Once you’ve done that, you no longer need to do an area-weighted average, because the data itself is weighted. This means that you can use a normal average, and compare things like trends directly.
So … what I do to check my work is to compare the normal mean of the area-weighted data, with the weighted mean of the original data. They should be the same, and that is the case in this analysis.
Finally, an area-weighted mean uses different weights, where the sum of all of the weights is 1. This allows you to calculate the weighted mean as the sum of the product of the data and the weights. These weights are different than the weights I used to area-weight the data itself. These weights which do not sum to 1. However, the end result is the same.
“Pistone et al. note that the Arctic solar input is going up because of decreased sea ice … but they did not notice that at the same time, the tropical solar input is going down because of increased clouds. And the net sum of all of the changes, of more energy being absorbed in the extra-tropical areas and less energy being absorbed in the tropics, is … well … no change at all for the globe. ”
As always, I feel I have received a shot of knowledge concentrate when you post. Terrific. The thermostat is getting more refined – we may soon better understand polar amplification and its unexpected effects elsewhere. Certainly other commenters have a piece of the puzzle in noting that reduced ice also increases LW upwelling, even in the dark with no albedo.
I have a growing suspicion (hypothesis?) that when we talk about non linear, interlocking chaotic behavior of a system, it may be that we just don’t adequately understand it and in frustration appeal to it as such. I think you are reducing the “chaos” considerably. Like John West, I’m amazed that some ambitious climate scientist who wants to differentiate himself from the pack doesn’t glom onto the thermostat hypothesis (they would give it a new name of course) which seems almost free for the stealing.
Willis Eschenbach says:
February 18, 2014 at 6:18 pm
(from your earlier post)
I’ve got that part solved: Inbound diffuse and direct SW solar radiation on an hour-by-hour basis for any day of year at any (arctic or antarctic) latitude.
Have got ice albedo as it varies by day-of-year, open ocean albedo as it varies by hour-of-day (solar elevation angle) and day-of-year and wind speed. So albedo and reflected or absorbed values are ready for you for each day-of-year and latitude for direct and diffuse radiation.
at every day of the year, at every latitude of interst where sea ice is actually present, much more energy is reflected from the Antarctic sea ice than from the Arctic sea ice because the Antarctic sea ice is always between 70 south at minimum extent amd 59.2 south at maximum. It is seeing much higher solar radiation levels at sea ice minimum, and much less air mass every day of the year, and much “cleaner” first year ice (higher albedo) than the “dirty” sea ice in the Arctic with much lower solar elevation angles and uch higher air masses.
So, what is not very clear?
Heat losses.
Specifically, we would need air temperature for the latitudes where the sea ice edge is, or barring that, air temperatures, relative humidity, and wind speed for the latitudes in question for each day of year.
(Forced) Convective heat losses. Convection losses are greater for open water than for ice-covered water. You need hour-by-hour air temperature, wind speed. (From average daily temperature and daily minimum and maximum temperature plots as thay change over the year, you can create an “adequate” hour-by-hour air temperature approximation for the latitudes in question.) And, regardless of ANY “approximations” heat transfer IS instantaneous: It depends on the sea surface temperature at that hour in question, the wind speed (which lets you calculate the heat transfer coefficient for convective losses), and the air temperature.
Conduction heat losses. From sub-surface water temperature and top of water temperature for water, and water temperature and top-of-ice temperature if ice-covered. Heat has to go through the conduction layer to get to the air. More ice means more insulation, and less conductive heat is transferred to the air.
Evaporation heat losses. Ice covers the water, so evaporation phase change energy is NOT released when ice covers the Arctic. Some little bit of ice sublimation energy needs to be calculated when ice covers the sea, but it is less energy/m^2 than evaporation energy – again, phase change energy loss INCREASES when sea ice coverage is lost in the Arctic. (Data needed is relative humidity and air temperature and air pressure.)
Long Wave Radiation heat losses. When ice covers the water, the radiation heat losses are LESS than when open water is exposed to the same “Tsky”. Open water will radiate at 273 K (0 degrees C ) to 277 K (+4 degrees C) into a Tsky of ???? . Ice-covered water will also radiate into that same Tsky, but the surface of the ice will be just warmer than the air temperature: -10 degrees C (263 K) in early September to -25 degrees C near mid-October. Since energy lost to space is proportional (Tsurface^4-Tsky^4), open water loses much more long wave heat energy than does ice-covered Arctic waters. What is unknown? An adequate value for Tsky.
At any given latitude, at one particular day-of-year, every square meter of open Arctic ocean will begin losing more heat every day than can be gained by what little bit is being absorbed from the sun. What specific day-of-year is that? Don’t know yet.
But, under today’s conditions in the Arctic at minimum sea extents, we are already losing more energy from open ocean than we are gaining from the “dark” open ocean. The artful myth of Arctic amplification is wrong.
Bill Illis says:
February 18, 2014 at 4:46 am
While in some cases you’d be right, Bill, in this case I’ve already area-adjusted the data to show the global impact. I had to do that so I could compare e.g. Arctic and tropics and global.
As a separate point, the area involved is only 3.3% of the global surface. Not only is it arctic only, but it’s only the ocean areas, and excludes the land.
w.
KRJ Pietersen says:
February 18, 2014 at 5:32 am
I’ve given some thought to doing the same analysis, but this time masking out the ocean rather than the land … “… but at my back I always hear, time’s wingéd chariot hurrying near …”
w.
O H Dahlsveen says:
February 18, 2014 at 6:08 am
Yeah, I know, it bugs me too. However, the ocean area north of 60°N is what they chose to study, so I had to follow their lead to understand their results.
In fact as you point out the tilt of the earth’s axis is ~ 23.5°, so the tropics of Cancer and Capricorn have that value, and the Arctic and Antarctic circles are at 90° – 23.5° = 66.5°, not 60° … I suspect they picked the line to include the southerly edge of the the ice, but who knows?
w.
Dan says:
February 18, 2014 at 7:45 am
Good question, Dan. There are a couple ways that happens, both over the ocean. First, as wind increases on the ocean, the surface becomes rougher and rougher. This increases the albedo of the water surface with the sun overhead from ~ 2.1% (calm) up to ~13.1% (rough). With the solar angle at 60° instead of 90° (overhead), the albedo goes up from 2.2% (calm) to 3.8% (rough).
Another way that wind directly affects the albedo is also on the ocean. As the wind increases we start getting “whitecaps”. As you might guess from their color, they reflect much more sunlight than plain water. As wind increases from there, the whitecaps become breaking waves, again white, and covering more area.
Next, with high winds you also get “spume”, or “seafoam”. This is the frothy stuff you see on the surface at time, and it also has a higher albedo than calm water.
Finally, with high winds you get spray and “spindrift”. Again this is white in color (think fog), so it also increases the albedo.
As to a link … well, some scientist must have studied this, but me, because I’m a seaman I just studied it in the wild, so to speak, so I fear I don’t have link one …
w.
Billy Liar says:
February 18, 2014 at 8:16 am
The temperature of the planet is about 290 K, and it varied about ± 0.3% over the 20th century … if you don’t call that a highly regulated system, you’ve never tried to regulate a heat engine. Your car’s cruise control is nowhere near that responsive.
w.
RACookPE1978 says:
February 18, 2014 at 8:49 am
Thanks, RA. Worthwhile points. As you say, since diffuse radiation comes from everywhere above the horizon, it’s not affected by rough water … the angles don’t change. Some things do affect the albedo of diffuse radiation, however. One of them is wind-created white water (whitecaps, breaking waves, foam, spindrift, spray), which reflects light from all directions.
Mmm … my bible, the sixth edition of Geiger’s magnificent 1950’s opus “The Climate Near The Ground”, gives the following figures (Table 4.2, p 15):
Geiger says that at solar angles below about 50° above the horizon, wind decreases the albedo as you say. But from 50° to 90° (sun overhead) wind increases the albedo. Geiger also shows a maximum value of 13%, as opposed to your 45% … unless your figures include white water, 45% seems high.
Geiger is an endless amusement of information. Did you know that the albedo of oak tree foliage (0.3 to 2µ) is 18%, but firs are only 10%? Go figure …
w.
Willis Eschenbach says:
February 18, 2014 at 6:54 pm (replying to)
Hmmmn. Check your logic there boss: I think you are mixing energy reflected/energy hitting the surface (albedo) with energy absorbed/energy hitting the surface (1-albedo). [Rev: Nope, you are correct in your definition, but not the values.]
Perfectly calm pure water under lab conditions reflects per Snell’s Law and the ideal Fresnel equations only in the lab under perfect conditions. Open ocean measured albedos – clear days, real water, real winds have to be measured. Dr Curry (and others) use Briegleb’s approximations based only on solar elevation angle (cosine of solar zenith angle in their papers) but Pegau and Paulson corrected that based on their measurements with Curry in the arctic for varying wind speeds during the SHEBA summer ice camps. As wind increases from a dead calm, albedo goes DOWN at all solar elevation angles (not up) as more solar energy is absorbed and less is reflected.
For MU = sin(solar_elevation_angle) or Cosine(solar_zenith_angle)
and WIND = 2 meter wind speed in m/sec
open ocean albedo = (0.026/(MU^1.7 +(-0.0002*WIND^2+0.0076*WIND+0.0266))+0.15*(MU-0.1)*(MU-0.5)*(MU-1)))
Ref: Pegau, Paulson
http://scholarsarchive.library.oregonstate.edu/xmlui/bitstream/handle/1957/27331/PaulsonClaytonAlbedoArcticLeads.pdf?sequence=1
Willis:
You are correct, increasing wind does “spray” water into white caps, but the papers measuring this effect find it does not change the albedo value for clear days and moderate winds (less than 10-12 m/sec), which was my criteria. I am assuming that “clear” days do NOT have storms nor high winds high enough to blow whitecaps. On stormy days, direct radiation is 0,0; ALL radiation is diffuse radiation, BUT those same clouds that create the diffusion also reflect into space 60-70% of the radiation at top of atmosphere. Also, in the Arctic, I am trying to compare the effect of “losing” any given meter of “sea ice” to “open water” … So the presence or absence of storm conditions would mean LESS heat loss under ice-covered waters and even MORE heat loss from storm-tossed wind-blown seas that increase convection and evaporation heat losses greatly.
yes, the open ocean albedo for diffuse radiation is 0.065 for all solar elevation angles, but the amount of radiation actually penetrating those dark and stormy clouds that make high winds to get through to the open storm-tossed ocean surface is only a small fraction of what hits the TOA.
Classic measured open ocean albedos at varying solar elevation angles are Payne, 1972 for the Chesapeake Bay light tower (although he combined direct and indirect radiation together) and Breigleb 1986: which has the “standard” albedo equation without Pegau’s wind correction, but that does separate direct and indirect radiation. Dr Curry uses Briegleb’s equation several times in her various papers, but it ignores wind changes. Olveria measured open ocean albedos several times as he crossed the south Atlantic ocean off of Brazil, and his data confirms Pegua … sort of. His data is mixed with reflections from the ship surface as they steamed south (towards the sun) and contains multiple “tracks” as clouds covered the sun during part of his “clear days” plots.
Best ever “visual” plot of the albedo’s under varying solar elevation angles for direct radiation compared to indirect radiation are the “pretty” multi-color Figure 4 of Rutledge 2004, repeated in 2003 and 2006 from his COVE experiments. Like Payne, Rutledge measured open ocean albedos at varying SEA’s from a fixed tower off the Atlantic coast..
Steven Mosher says:
February 18, 2014 at 9:04 am
Interesting point, Steven.
As I mentioned above, I dislike using albedo because it is a ratio. When the denominator gets small, you get all kinds of strange results. For example, in the CERES dataset, there are 180 latitude x 360 longitude x 156 months ≈ ten million gridcell measurements.
Of these, about three quarters of a million of them, 7.5%, have reflected solar greater than downwelling solar … bad gridcells, no cookies.
Now for most things this isn’t an issue, simply because the mean value of the downwelling solar in those bad gridcells is 0.2 W/m2, with the mean upwelling reflected solar being 0.3 W/m2. In addition, the median downwelling solar value in the bad gridcells is 0, with the median upwelling solar being 0.06. What is the albedo of a gridcell that is reflecting 0.6 W/m2 of solar but not receiving any solar at all?
Now, when I’m adding or subtracting those downwelling and reflected values, there’s no problem. The sum total of the downwelling solar in all of the bad gridcells is 0.005% of the total incoming sunshine. The corresponding figure for the reflected solar is 0.03%, three hundredths of one percent of the total. That’s no problem in any calculation.
But when I calculate the albedo, I have to divide one by the other, and that gives us all kinds of bizarre values.
Finally, bear in mind that all of these bad gridcells are near the poles, where the sunlight goes to zero.
So while cross-checking the arctic numbers is a good thing … I don’t think I’d do it with albedo. I prefer a more well-behaved measure, like the total solar energy input I used in the head post.
Thanks as always,
w.
Amatør1 says: @ur momisugly February 18, 2014 at 9:05 am
I suggest you read this.
By popular demand …
Best to all,
w.
RACookPE1978 just re your comment of at any latitude antarctic reflects more than Arctic at any given equivalent day of the year. The Antarctic is further away from the sun by up to 6% than the Arctic so at the same location, if ice was present at the same stage of the cycle [day 24 say of winter north and south] The arctic sea ice would reflect more energy than the equivalent Antarctic ice.
Your other points are very valid about the sea ice extent being so much further out in general.
You did not comment on the combined albedo of the Antarctic Land ice [presumably equivalent to the Arctic [6% less ]] and the sea ice much higher as so much further out giving much more reflected heat to space down south
Aphan says: @ur momisugly February 18, 2014 at 12:08 pm
Raises hand to ask question-
Why is it, that no one ever just points out that ice sheets melting between glacial periods is NATURAL and PRECEDENTED, and thus EXPECTED? I mean, yeah…they melt. They “vary”. Naturally….
>>>>>>>>>>>>>>>>>>>>>>>>
It is even funnier than that. If you take a longer point of view the glaciers are actually net increasing and all the warmists are doing is crying about minor variations in weather.
A more recent paper looking at glaciers in Norway.
The authors of BOTH papers simply state that most glaciers likely didn’t exist 6,000 years ago, but the highest period of the glacial activity has been in the past 600 years. This is hardly surprising with ~9% less solar energy.
Reblogged this on gottadobetterthanthis and commented:
I always appreciate Willis’ writing, but this is particularly insightful and informative. The fact is, the Arctic is not warming the planet. In fact, the ice isn’t even melting at this point. We shall see, but it seems to me we’ve probably past the peak, and we will be chilling for the next several years. I hope not. Cold kills. Warmer is better.
angech says:
February 18, 2014 at 9:11 pm (Replying to)
Hmmmn.
No. The “further from the sun” phrase requires further amplification.
Let us continue the conversation – as “I” understand, then I need you to show me where I am misunderstanding your phrase.
The earth rotates through space at 23.5 degrees angle, so at various days-of-the-year, each pole (Arctic and Antarctic) ‘rolls” “closer to the sun” on a regular basis. True.
BUT!
On January 3, the earth itself is closer to the sun than on July 3, so on January 3 – mid-summer in the Antarctic – the top-of-atmosphere solar radiation over EVERY square meter of the planet exposed to the sun = 1410 watts/m^2. But the Arctic area has 0.00000 watts/m^2 of sunlight on January 3. And 0.000 on January 4,5,6,7 ….. also.
Now, around that date of January 3, the SOUTH POLE (Antarctic continent and shelf ice and sea ice) has rolled around so IT is exposed to the sun for 24 hours per day for all regions between 67.5 south latitude (all of the remaining sea ice of 3.4 Mkm^2, all of the permanent shelf ice of 3.5 Mkm^2, AND all of the continental ice of 14,0 Mkm^2) for 24 hours per day. (A little bit of the peninsula sticking towards south America does melt off each summer.)
So, when the earth is closest to the sun, 14.0 + 3.5 + 3.4 Mkm^2 = 21.1 Mkm^2 of “some kind of ice” remains exposed to 24 hours per day of sun, right?
Now, on July 3, when the north pole has “rolled” over so IT is exposed to the sun, the entire earth is further from the sun on its elliptical orbit, so the whole earth is exposed to the sun with only 1314 watts/m^2. 1314 is less than 1410, right?
BUT, when the Arctic sea ice is “retreating” in the Arctic summer during this period of low solar radiation levels, it is retreating from 70 north (the north coast of Alaska and Siberia) back towards its minimum of about 80 north in September.
So, when the whole earth’s radiation levels are lower, the little bit of 3.5 Mkm^2 of Arctic ice that remains at minimum sea ice extents is all that can reflect. And that Arctic “reflection” is happening between 78 north latitude and 80 north latitude. (To be clear, on July 3rd, more than the yearly minimum of 3.5 Mkm^2 of Arctic sea remains up north. Assume 6-8 Mkm^2 is up there, but it varies year-to-year.)
Your average “flat plate area” exposed to the sun is going to change based on the sin of the latitude of the area of the sea ice remaining. (If you want the “average” best fit polynomial equation matching sea-ice-edge-to-latitude for 2011-2013 calender years, let me know. It ain’t pretty.)
So, on the same date in mid-September, when the identical “radiation received on a flat plate on the surface” at 80 north is only 1/5 of what is being received at the edge of the Antarctic sea at latitude 60, which is area – Arctic or Antarctic – is more important to the earth’s heat balance?
Regardless of when each area is “closest to the sun” in July or September or January?
Simple tim. Its called a land mask.
If one data set does global and the other does land only.and the land only is validated you check the land versus the land.
If they dont match then you have no reason to trust
The global one that hasnt been through validation.
If they do match then your one step closer.
Next would be to check greenland in both.
Finally would be to check avhrr which the authors
Appear to have started.
Lastly one would check the other global albedo product t g at goes back to 1980s. That would be the SAL product
Or you can stop when you find answer you like
And as feynman observed…fool yourself
angech says: @ur momisugly February 18, 2014 at 9:11 pm
RACookPE1978 just re your comment of at any latitude antarctic reflects more than Arctic at any given equivalent day of the year. The Antarctic is further away from the sun by up to 6% than the Arctic….
>>>>>>>>>>>>>>>>>>>>
?????
In July when the Arctic is in summer (and reflecting sunlight) the earth is 94,500,000 miles from Sun.
In January when the Antarctic is in summer (and reflecting sunlight) the earth is 91,400,000 miles from Sun.
NASA link
Gail! See the comment immediately above your reply. 8<)
I think you are mixing my reply to Angech with my “repeat” of his (incorrect) statement about the 6% yearly change in top-of-atmosphere radiation levels. There IS a 6% change in TOA solar radiation levels. But the “peak” of that 6% change occurs – NOT when the “dirty” Arctic ice is exposed to the sun at 76 north in July 3 – but when the “new first year clean” Antarctic sea ice is exposed to the sun at 65 south on January 3!
It is the Antarctic sea ice which is exposed to 5x the solar radiation levels that the Arctic sea ice is exposed to.
Mosher writes “Simple tim. Its called a land mask.”
And at what time does a land mask mask out the albedo due to clouds?
You’re comparing satellite data that includes clouds with GLASS data that doesn’t. So now you’re making assumptions about the effects of the clouds (and atmospheric effects in general) in your data.
Willis Eschenbach: As I mentioned above, I dislike using albedo because it is a ratio. When the denominator gets small, you get all kinds of strange results. For example, in the CERES dataset, there are 180 latitude x 360 longitude x 156 months ≈ ten million gridcell measurements.
Of these, about three quarters of a million of them, 7.5%, have reflected solar greater than downwelling solar … bad gridcells, no cookies.
Just to elaborate, when the numerator and denominator have random variation, including but not limited to measurement error, the distribution of the ratio is totally non-intuitive. The best studied case, naturally, is independent normal distribution in numerator and denominator (if you want to look this up, I can find you some references, but the best approach has been the bootstrapping approach of Prof. Bradley Efron.) Willis is right in this case: there is no satisfying (e.g. demonstrably accurate) approach.
Some folks have said that there is a problem with my area-weighting, so let me explain exactly what I did.
The data exists in 180 latitude bands. The center of the bands start at -89/5° (south) and end up at 89.5° (north). To area-weight the data, we want to adjust the results for each gridcell by the area.
What we want to do is adjust the results to give what you would get if they had the size of the average gridcell. Now the area is proportional to the cosine of the mid-latitude. So what we do is multiply each gridcell result by
area of the gridcell / area of the average gridcell
This give each gridcell the value it would have if it were of average size. The effect of tiny gridcells is reduced, and the effect of large gridcells is increased.
Now, what is the average cell size? Well, if we integrate Cos(x) from zero to pi/2, we get 1. So the average gridcell size is 1/(pi/2) = 2/pi ≈ 0.637.
As a result, the weighting factor by which we multiply the gridcell value is:
area of the gridcell / area of the average gridcell
which is equal to
cosine of the gridcell midlatitude / 0.637
Once you’ve multiplied the data by those weighting factors, you can compare them directly, as they are all adjusted to the average gridcell size.
The way to test if you’ve done this correctly is to see if the plain vanilla average of the newly-weighted dataset is correct. For example, see the average available solar (~340 W/m2) and solar input (~240 W/m2) values in Figure 3. They are simple averages of weighted data.
Note that there are two ways to do the weighting.
The first is to do all calculations (trends, etc) using the unadjusted variables. Then to get an average, you use what is called an “weighted mean”, which weights the data on the basis of gridcell area as it calculates the average.
The other way to do it is the way I described above, which converts all of the data to what an average sized gridcell would show. Once you’ve done that, you no longer need to do an area-weighted average, because the data itself is weighted. This means that you can use a normal average, and compare things like trends directly.
So … what I do to check my work is to compare the normal mean of the area-weighted data, with the weighted mean of the original data. They should be the same, and that is the case in this analysis.
Finally, an area-weighted mean uses different weights, where the sum of all of the weights is 1. This allows you to calculate the weighted mean as the sum of the product of the data and the weights. These weights are different than the weights I used to area-weight the data itself. These weights which do not sum to 1. However, the end result is the same.
Regards to all,
w.
Willis,
Thanks for the explanation of sea states and their effect on albedo, much appreciated.
Much is being said about heating leading to more water in suspension and greater rainfall. Considering that not all that long ago we were being told to design gardens that conserved water there is strong evidence of fashion and errant surmise conquering all.
Presumably if we do eventually cloud over and receive less light (and heat) from the sun (plus the reflective propensity of clouds) we will have a climate that cools of its own volition.
Today there is much in the news about the glorious increase in butterflies in Britain as a result of a warm summer. Vegetation also benefited and all manner of food crops grew that much better. This can easily be discounted as being an aberrant phenomenon, but how many such summers would it take to change people’s minds about outcomes, answer, you never will.
So many absurd claims have been made now that we are entering the region of personal reputations and people are going to get even more antagonistic towards ‘deniers’ as time passes just to save face.
If there was some acceptance that the earth, for whatever reason, might be warming and all the money spent so fatuously on theorising was put into projects that accepted the fact and moved on to adapt the changed circumstances, how much better we all would be? Just to get rid of this adversarial impasse would concentrate minds wonderfully.
The era of the climate model must come to an end. It is rather like necromancy and mirrors that thing about garbage in-garbage out. It is particularly galling when one considers that Jet Stream science is obviously such a mystic region of understanding and yet without a complete understanding of how the movement of that recalcitrant zephyr works you might as well use that giant network of computers in Exeter to play Tetras.