Guest Post by Willis Eschenbach
Yesterday, I discussed the Shepherd et al. paper, “Recent loss of floating ice and the consequent sea level contribution” (which I will call S2010). I also posted up a spreadsheet of their Table 1, showing the arithmetic errors in their Table.
Today, I’d like to discuss the problems with their method of calculating the loss in the Arctic Ice Pack. To start with, how big is the loss? Here is a graphic showing the change in area of the Arctic ice pack from one year’s loss of ice, with the ice pack area represented by a circle:
Figure 1. One-year change in the area of the Arctic Ice Pack, using the average annual loss which occurred 1996–2007. Note the obligatory polar bears, included to increase the pathos.
OK, so how do they calculate the Arctic ice loss in S2010?
Here is their description from the paper:
We estimated the trend in volume of Arctic sea ice by considering the effects of changes in both area and thickness. According to ERS and Envisat satellite altimeter observations, the 1993-2001 (average wintertime) thickness of Arctic sea ice was estimated to be 273 cm (Laxon et al., 2003), the thickness decreased by 6.7 ± 1.9 cm yr-1 between 1992 and 2001 (Laxon et al., 2003), and the thickness decreased by 4.8 ± 0.5 cm yr-1 between 2003 and 2008 (Giles et al., 2009).
We combined these datasets to produce a new estimate of the 1994-2008 thickness change. Published satellite microwave imager observations (Comiso et al., 2008) show that the 1996-2007 Arctic sea ice area trend was -111 ± 8 x 10^3 km2 yr-1 and, based upon our own analysis of these data, we estimate that the 1990-1999 average wintertime area of Arctic sea was 11.9 x 10^6 km2.
The combined reductions in Arctic sea ice area and thickness amount to a decrease in volume of 851 ± 110 km3 yr-1 during the period 1994 to 2007, with changes in thickness and area accounting for 65 % and 35 % of the overall loss, respectively.
What is the problem with that method?
The problem is that they have assumed that the ice is the same thickness over the entire area. As a result, the reduction in area is causing a large loss of ice, 35% of the loss by their estimate.
But the ice is not all the same thickness. The perimeter of the ice, where the loss occurs, is quite thin. As a result, they have overestimated the loss. Here is a typical example of the thickness of winter ice, from yesterday’s excellent article by Steve Goddard and Anthony Watts:
Figure 2. Ice thickness for May 2010. Note that the thickness of the ice generally tapers, from ~ 3.5 metres in the center to zero at the edges.
So their method will greatly overestimate the loss at the perimeter of the ice. Instead of being 273 cm thick as they have assumed, it will be very thin.
There is another way to estimate the change in ice volume from the melt. This is to use a different conceptual model of the ice, which is a cone which is thickest in the middle, and tapers to zero at the edges. This is shown in Figure 3
Figure 3. An alternative model for estimating Arctic ice pack volume loss.
Upon looking at this drawing, I realized that there is a way to see if my model fits with the facts. This is to use my model to estimate how much of a thickness change would be necessary to create the 111,000 square kilometre loss. It turns out that to get that amount of loss of area, it would require a ~4 cm ice loss over the entire surface … which is a good match to their estimate of ~ 5 cm of loss.
So, what difference does this make in the S2010 estimate of a global loss of 746 cubic kilometres per year? Lets run the numbers. First, I’ll use their method. I have used estimates of their numbers, as their description is not complete enough to give exact numbers.
Thickness loss: (11,900,000 km^2 – 111,000 )* 5 cm / (100,000 cm/km) = 589 cubic km (66 % of total).
Area loss: 111,000 km^2 * 273 cm / (100,000 cm/km) = 303 cubic km (34% of total)
Total: 892 cu km, which compares well with their answer of 851 cubic km. Also, the individual percentages I have calculated (66% and 34%) compare well with their numbers (65% and 35%). The difference is likely due to the decreasing area over the period of the analysis, which I have not accounted for.
So if we use a more realistic conceptual model of the ice (a conical shaped ice pack that is thick in the middle, and thin at the edges), what do we get?
The formula for the volume of a cone is
V (volume) = 1/3 * A (area of base) * h (height)
or
V = 1/3 * A * h
The difference in volume of two cones, therefore, is
V = 1/3 * (A1*h1 – A2*h2)
This means that the volume lost is
V = 1/3 * (11900000 km^2 * 273 cm – 11789000 km2 * 268 cm) / (100000 cm/km)
= 297 cubic km
This is much smaller than their estimate, which was 851 cubic km. And as a result, their estimate of global ice loss, 746 km^3, is reduced by 851 – 297 = 554 km^3, to give a final estimate of global ice loss of 192 cubic kilometres.
FINAL THOUGHTS
1. Is my estimate more accurate than theirs? I think so, because the simplistic assumption that the ice pack is equally thick everywhere is untenable.
2. How accurate is my estimate? I would put the 95% confidence interval (95%CI) at no less than ± 25%, or about ± 75 km^3. If I applied that same metric (±25%) to their estimate of 851 km^3, it would give a 95%CI of ±210 km^3. They give a 95%CI in their paper of ±215 km^3. So we are in agreement that this is not a WAG*, it is a SWAG*.
3. This exercise reveals some pitfalls of this kind of generally useful “back-of-the-envelope” calculation. First, since the final number is based on assumptions rather than data, it is crucial to be very clear about exactly what assumptions were made for the calculations. For example, from reading the paper it is not immediately evident that they are assuming constant thickness for the ice pack. Second, the change in the assumptions can make a huge change in the results. In this case, using my assumptions reduces the final value to a quarter of their global estimate, a value which is well outside their 95%CI estimate.
To close, I want to return to a separate point I made in my last post. This is that the S2010 paper has very large estimates of both gains and losses in the thickness of the Antarctic ice shelves. Now, I’m not an ice jockey, I’m a tropical boy, but it seems unlikely to me that the Venable ice shelf would lose a quarter of a kilometre in thickness in 15 years, as they claim.
Now it’s possible my math is wrong, but I don’t think so. So you colder clime folks out there … does it make sense that an ice shelf would lose 240 metres thickness in fifteen years, and another would gain 130 metres thickness in the same period? Because that is what they are claiming.
As mentioned above, I have posted their table, and my spreadsheet showing my calculations, here. here I’d appreciate anyone taking a look to make sure I have done the math correctly.
PS:
* WAG – Wild Assed Guess, 95%CI = ±100%
* SWAG – Scientific Wild Assed Guess, 95%CI = ±25%
[UPDATE] My thanks to several readers who have pointed out that I should not use 273 cm as the peak thickness of the ice. So following this NASA graphic of submarine measured winter and summer ice, I have recalculated the peak as being about 4 metres.
Using this value, I get arctic ice loss of 344 km^3, and a global ice loss of 239 km^3.
Here is a bit of Propaganda about the Mulholland/LA aqua duct and one reason Mono lake is going away….
http://wsoweb.ladwp.com/Aqueduct/historyoflaa/index.htm
Minor point – should the graphic be changed?
“Inner blue circle is after once year loss”
Drop “once” and add “one” as follows:
Inner blue circle is after one year loss.
Overall, another interesting look at the topic. Thank you
WAG – Wild Assed Guess, 95%CI = ±100%
SWAG – Scientific Wild Assed Guess, 95%CI = ±25%
I’d call it more like a POOMA number. (which stands for Preliminary Order of Magnitude Approximation. Or something.) Especially the confidence interval.
d says:
May 30, 2010 at 6:07 am
the polar bear circle graphic is very confusing
The angles make it a bit hard, but the two young PBs make a radius, so each are about 1,000 Km long (1 Megameter). Mama PB is a step too far ahead, but she looks like she’d make a radius all by herself, so she’s 2 Mm long.
Clearly the PB problem is that their size is increasing and therefore the Arctic can’t support the old populations any more.
The real point of Willis’s circle in a circle is to show that the rate of ice loss posited in the paper may make for a very small change in ice extent.
One of the things to keep in mind about Antarctica is that there is an ice river from East to West, with about a 3000(?) year lag. So the ice falling off the West into the water was made in the East long ago. Thus the only relevant current numbers for ice change are in the East, unless you want to claim that somehow AGW is heating the ice river from underneath and making it slide faster. Quite a trick! I’d be interested (=amused) to hear what mechanism the IPCCrackpots can come up with for that.
Or maybe they think the ocean is pulling on the Western end of the flow, and yanking the whole thing along faster. Or …
Come to think of it, the heavier the ice gets in the East, the quicker it will squeeze out. So accelerated calving might be a reflection of a transcontinental pressure wave resulting from increased ice buildup!
Heh.
” noiv says:
May 30, 2010 at 7:58 am
@gallier2
Thanks for pointing to another evidence. Just a hint: the ice is the white stuff. And are you saying, that you need 30 years of ice-free Arctic to agree that things are changing?”
Illiterate, and uninformed.
There is no such thing as an “evidence”. Piece of evidence, perhaps.
And there have been no years of ice-free Arctic so far. Except for the usual patchy open water in summertime. So what’s your point?
And, to repeat, “things are changing” all the time. Warming has been going on for hundreds of years. It’s up to the Warmists to prove a) it’s accelerating, and b) we did it.
They have failed miserably at both. Despite all pre-emptive claims to the contrary in hopes of preventing anyone from getting access to the data to analyse for themselves.
Willis, you could try using the frustrum of a cone calculation for greater accuracy, but I would suggest a different approach to calculating Arctic melting volume.
I think the ice that survives to the end of summer must be thicker than what melts as a rule, so the area of the difference between March and September can be multiplied by the average thickness of first year ice to get the volume of melt. My guess is this volume is less than both your estimate and the estimate of Sheperd et al.
This does not take into account whatever ice is blown out of the Arctic Ocean to melt in mid-latitudes, which likely includes yearling ice and multi-year ice in possibly random proportions.
David Baigent says:
May 30, 2010 at 3:04 am
The shape of the “ice volume” would approximate that of a “contact lens”
Use V = 4/3¶r3 – V = 4/3¶r3 ( first radius is smaller than the second radius)
Your formula is for the volume of a sphere, – divide the whole works by 2 and get closer to the answer.
Re: Willis model. I know it is back of the envelope but loss of ice thickness in the middle area would proceed slower than the edges if the middle is supposed to be rougly the north pole.
Mike Davis says:
May 30, 2010 at 7:38 am
I do not know about the rest of the world but in my neck of the woods the ice starts forming at the edges of the pond and starts melting towards the center away from the edges. When wind blows it packs ice against a solid object such as the shore.
*
*
Mike,
I agree with your assessment.
However, you must remember that even though ‘thin water’ will experience a quicker freeze point that ‘thick water,’ the water closest to the coldest point will consonantly freeze quicker than that not so.
This is to say that the water closest to that which is already frozen should freeze faster than that which isn’t.
That is seen by water freezing from the shore and thence.
1 forest 1 says:
May 30, 2010 at 5:12 am
That may well be, 1 forest 1 … but an example would be useful. Turn what down? The math? The number of my posts? The style of my posts? My responses to comments?
rw says:
May 30, 2010 at 7:39 am
Along these lines, on another comment thread someone, I forget who, linked to a recent/current page on the GISS site. I guess it makes the point (relentless warming), but at the same time there’s something remarkable about the graphs – the winter of 2009-2010 has disappeared! All those record-breaking (low) temperatures in N. A., Europe and elsewhere apparently mean nothing when it comes to the real temperature record.
*
*
Well, of course!
You see? Its as this: The whole has been ‘homogenized.’
Got that?
” Chris Korvin says:
May 30, 2010 at 3:32 am
What is bad for polar bears is good for penguins.Polar bears supposedly need more ice to hunt and penguins need less to get to the ocean .”
Actually, that’s a Gore-ist crock, too. During warm low-ice periods, the geo-records shows PB populations surge. Because there’s more food. That they are exclusively (amongst bears and land predators) capable of surviving in reasonable numbers on pack ice does not mean that’s what’s best for them. PBs, like all bears, are omnivores. Just ask the towns whose garbage dumps etc. are raided.
mb says:
May 30, 2010 at 4:44 am
mb, thanks for your comments. You say (emphasis mine):
No, that’s not my “main argument”, I see that my writing must not be clear. My main argument is two-fold.
1) By using a constant average height, they have greatly overestimated the effect of the change in area. They calculate the effect of the lost area using the average height, whereas it is lost from the edges where the thickness is minimal.
2) Their estimated average height is unrealistic, because of their incorrect assumptions about the shape of the ice pack.
You also say:
What you say is true. But you are making the same mistake that they made. The undeniable fact that
does not mean that
w.
” Mike Davis says:
May 30, 2010 at 7:38 am
I do not know about the rest of the world but in my neck of the woods the ice starts forming at the edges of the pond and starts melting towards the center away from the edges. When wind blows it packs ice against a solid object such as the shore.”
The air temp in the middle of the pond is not a lot lower than around the edges, as it is near the North Pole. It’s about the sunshine and winds, not to mention the recently discovered huge volcanic ridges on the bottom. Your pond isn’t much more relevant than your bathtub.
But the wind factor has some parallels with what happens. In the Arctic’s case, it’s complicated by the exit ramps around Baffin Island; if the winds blow that way, huge amounts of the ice are forced out into the North Atlantic as icebergs, which then proceed to melt, raising the ocean level by approximately 0%.
2 days ago this paper was not available and not published. Today Willis has read it. WUWT?
Please give us the url so we can see what it says. I see that the estimates Sheppard uses, according to Willis, are based on average ice estimates. The rest of this article goes on the say Sheppard did not use an average. WUWT?
First of all your cone model fails to come close to estimating the area loss for a given height change. The projected area goes as the height squared. In such a geometry a 2% change in height, as measured, would lead to roughly a 4% change in extent, far from the 1% measured. For a 1% change in area a 0.5% in heigh is required. All comparisons to the Shepard paper beyond this point are irrelevant.
One of the problems I see in the S2010 paper is that they appear to be mixing floating glacial ice (ice shelves) and floating sea ice. Ice shelves are driven into the the sea by the mass of the glacier that is their source. Sea ice forms from sea water. Their melting/freezing temperatures are different, and their distributions different, and provenance are different. So why, other than meme, are they mixed. And why was this not pointed out by their peers before publication?
Jbar says:
May 30, 2010 at 7:47 am
“IF the trend continues, then arctic summer sea ice will disappear before the end of this century.The trend is still intact, and in Wall Street parlance, “the trend is your friend”. (I.e., bet on the trend until it actually stops.)”
And so it was said in the 1920’s, then came the 1929 stock market crash and the reversal of the ‘melting arctic trend”.
Smart investors know enough to hedge rather then go ‘all in’ on a trend.
Willis Eschenbach> Thanks for answer!
Well, if you agree that it is an undeniable fact that volume = area * average thickness, there is not much more to discuss. We have simple formulas.
Independent of the shape of the ice, we have:
V1 – V2 = A1 * AverageThickness1 – A2 * AverageThickness2
= A1 *(AverageThickness1 – AverageThickness2)
+AverageThickness2*(A1-A2)
In our case, using the same numbers as you used:
A1 = 11,900,000 km^2
A1 -A2 = 111,000 km^2
AverageThickness1 – AverageThickness2 = 5 /100,000 km
AverageThickness2 = 268/100,000 km
So
V1 – V2 = 5* 11,900,000/100,000 – 111,000 * 268/100,000 km^3
= 892 km^3
Which is quite close to the value in the paper (and of course it agrees with your first, tentative calculation).
mb says:
May 30, 2010 at 6:57 am
I’ve said a number of times that I have no evidence to doubt this, just a sense that it is too large … so why are you asking if I have evidence?
I can see that. But if the velocity is 50 m/year, it seems doubtful that the thinning from that would be 15 metres per year.
Next, it may be that, like taffy when it is pulled, as the mass lengthens it thins. But how much does the volume of taffy decrease when it is stretched?
Well, about zero.
Next, your citation refers to the glaciers feeding the ice shelves, and not to the shelves themselves. This makes sense, because as a glacier slides downhill, the lower end of it can move faster than the upper end, pulling on the and stretching it like taffy.
But no such pulling force exists on the floating ice shelves.
I have looked for other estimates of the thinning of the ice shelves. I find a paper in Science Magazine (subscription required) about the thinning of the Larsen Ice Shelf. It puts the thinning at about 0.8 metres per year, or 12 metres over the period of the study, which seems eminently reasonable. A quarter kilometre of thinning over the period of the study, on the other hand … not so much.
Willis, do you realize what is going to happen thanks to your pointing out such obvious errors right before the paper is officially published?
The journal won’t want to look foolish for accepting it. The journal’s peer reviewers won’t want to look foolish for missing such obvious mistakes. Maybe even the author(s) won’t want to look foolish and fear seeing their mistakes more-officially laid bare.
So they’re going to hurry up and make the changes before publication, resulting in a less-impressive loss number but still sticking to the narrative, There will be no official mention of them, they’ll hope all anyone mentions is the press release, and if anyone notices the differences they’ll blame pre-publication PR hype.
At which point the (C)AGW proponents will all chime in and pile on, calling you a big fat liar who made it all up. This could very likely result in the EPA wanting to fine a bunch of us for excessive environmental lead contamination, among other possible charges.
Before it gets to that point, we really need to consider the alternatives.
😉
jorgekafkazar says:
May 30, 2010 at 9:06 am
…
I’d call it more like a POOMA number. (which stands for Preliminary Order of Magnitude Approximation. Or something.)
_________
That is definitely NOT what POOMA stands for :0
Mike Davis says:
May 30, 2010 at 7:38 am
That’s true, but the Arctic doesn’t seem to melt and freeze from the middle. Here’s what NASA says that the unclassified submarine records of ice thickness show. Upper panel is winter, lower panel is summer.
Looks like my conical model is not far off the mark …
losemal says:
May 30, 2010 at 10:15 am (Edit)
I fear that this sounds interesting, but is totally unclear.
Folks, if you disagree with something that I (or anyone else) has said, QUOTE IT!
In this example, I have no idea which statement of mine you are disputing.
w.
mb: May 30, 2010 at 7:48 am
I fully intended for H to be the apex height, which is what’s relevant for the way in which Willis has set up the calculations. Personally, I think that Willis is being somewhat conservative in his approach to this. I would have used a cusp-shaped profile, which would have given an even smaller number than the one that he obtained for the decrease in volume.
In any case, carry on raving for as long as you need to.
/dr.bill