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.
Shouldn’t the shape be more like a bell curve rather than a triangle. I think this would have the effect of reducing the volume loss substantialy, and also be more accuriate.
Ian H says:
May 31, 2010 at 2:16 am
The problem is not with average depth. We both agree that
The part that you (and the folks who wrote S2010) seem to be missing is that this does not mean that
For example, suppose you have a flat slab and a cone which have the same area “A”, and the same average height “h”. Of course, the volumes are the same.
But if you trim a certain area from the outside edge of the flat slab, you will get a very different volume loss than if you trim the exact same area from the outside edge of the cone.
My friend, you can ask any of the regulars here, when I make a mistake, I am the first one to own up to it. However, I don’t think I made a mistake here.
You say “they made no assumptions about the shape of the ice in writing V=Ah”. This is not true. They show the volumes that they calculate from the loss in thickness (35%) and the loss in area (65%). From this, we can see that they are assuming that the area lost is the same thickness as the rest of the ice.
But this assumption is unjustified, since the area that is lost contains the thinnest of the ice, not thick ice.
I do appreciate that, Ian … just not regarding this situation. I have made mistakes in public before, and I have owned up to them. I can live with that, and have done so quite publicly. But in this case, trimming the edge of a cone removes a different volume than trimming the edge of a slab.
But in this case, you have made the claim that because V = A h, that ∆V = ∆A h … and quite simply, that is not true.
PS – My experience with the peer review process is not that the errors are “gently pointed out” …
But be that as it may, in my opinion, one of the problems with peer review is the anonymity. My feeling is that after the review process, all of the reviews should be made public whether the paper is published or not. This would allow us all to learn from each other’s mistakes and errors. That’s why I do this all in a very public manner … so that when I do make a mistake, it is out there for others to learn from, rather than never coming to light.
Gail Combs says:
May 31, 2010 at 5:01 am …
Gail, I corrected your italics to what I think you meant.
dr.bill says:
May 31, 2010 at 7:47 am
Thanks, Dr. Bill. People are passionate about these questions, as am I. I try to deal with folks reasonably. As I’ve said before, I’m a reformed cowboy, so equanimity is not my ground state … and I do get un-reformed at times. But my assumption is that folks generally operate out of reasonable motives and passions, rather than out of “evil” or “fraud” or “deceit” or the usual kinds of accusations that get tossed around far too lightly.
That said, I don’t do well when people accuse me of lying. In my youth on the cattle ranch, most of the folks around had little of their own except their honor. They guarded this vigorously, and calling a man a liar was the worst insult possible. That was the code of honor I was brought up with, and it has remained with me. While I try to be blase about such accusations, it angrifies my blood mightily, and when accused I have been known to say bad words and to flail mightily. I have done lots of stupid things, and I have made far too many bonehead mistakes, but I tell the truth as I know it.
Other than that, rather than assume that folks are stupid or malevolent, I generally assume that people are either honestly mistaken, or are blinded by their passion for protecting the environment … and as someone who cares deeply for this amazing, mysterious, awesome planet and all of its myriad inhabitants, I can hardly fault anyone for that.
Or at least that’s what I aspire to … but as I said, my cowboy reformation is not 100% solid …
Jeff in Calgary says:
May 31, 2010 at 10:30 am
You are likely right, Jeff. I suspect this is the reason that my simple conical model underestimates the ice volume compared to the PIOMAS values. However, I was not looking for exact numbers. I simply wanted to show that the S2010 calculations, which assume that the ice is the same thickness throughout, were in error.
But if you trim a certain area from the outside edge of the flat slab, you will get a very different volume loss than if you trim the exact same area from the outside edge of the cone.
again: the problem with your claim above is a simple one: you get a much smaller number, because both your cones are much smaller than the ones used by Shepherd. your big cone is much smaller than his starting volume.
the majority of the difference is NOT caused by the different shapes, but by your use of a smaller volume.
this has been explain to you now over a dozen times. you seriously should admit your error finally!
one more explanation:
what Shepperd do is this: they assume that the sea ice has a bucket shape. so they modell the loss by a 20 l bucked and a 15 l bucket. volume loss: 5 l.
you come along, claiming that the ice has beer glass shape, not bucket shape. you use a 0.5 l and a 0.3 l glass to explain it. new difference: 0.2 l
but the difference between 5l and 0.2 l has basically no connection to the shapes that were used. it is mostly caused by the different SIZE.
Total ice volume, S2010: 273*1190 = 324,870 km^3
Reduction in ice volume, S2010: 851 km^3
Percentage reduction in ice volume, S2010: 100*851 / 324,870 = 0.26%
Total ice volume, Willis: (1/3) * 1190 * 475 = 18841.67 km^3
Reduction in ice volume, Willis: 372 km^3
Percentage reduction in ice volume, Willis: 100*372 / 18841.67 = 1.98%
Hmm.
Starting volume: A1H1
Ending volume: A2H2
ΔV=A1H1-A2H2
H2=H1-δH
∴ΔV=A1H1-A2H1+A2δH
ΔV=H1(A1-A2)+A2δH
∴ΔV=H1δA+A2δH
So no matter what shape you have if you know the area before and after and the average height before and after the change in volume is the same. It’s elementary calculus.
Of course if you do what Willis did and start with a different volume you will get a different answer, but that’s because he made an error!
sod says:
May 31, 2010 at 12:47 pm
Shepherd et al. DID NOT USE A CONE. They used a flat slab. This gave them a 35% error in their answer.
Yes, I used a smaller volume. But if I used the Shepherd volume, I would still get a smaller answer than they got. Which was my point. They used a slab, so 35% of their volume was from the reduction in area … which makes no sense, since the area being reduced is at the outer edge of the ice, which is the thinnest ice.
I’m have never insisted that my final answer is correct. It is a BACK OF THE ENVELOPE ESTIMATE done to illustrate the difference between using a cone and using a slab.
But my final answer is much closer to the PIOMAS value for lost ice than is the Shepherd answer. In addition, my value for the volume is much closer PIOMAS volume than is the Shepherd volume.
So which error are you talking about? Because if you want to bust me for whatever error you think I made, you need to bust Shepherd twice as hard.
Sod, here’s one of the underlying problems:
Shepherd Winter Volume, 1990-1999: 11.9 M sq km * 273 cm = 32,487 km^3
PIOMAS Winter Volume, 1990-1999: 22,062 km^3
Shepherd Winter Loss, 1994-2007, 851 km^3
PIOMAS Winter Loss, 1994-2007, 421 km^3
My winter loss estimate, 1994-2007, 354 km^3
Now, the PIOMAS volume value has been verified for a few years at least by the ICESAT values. So the Shepherd volume is about 50% too high.
Because of that, I’m not surprised that my values are smaller than those of Shepherd. You and others keep insisting that my values are in error because I am using both the wrong volume and the wrong shape model. So let me recalculate the figures, using the PIOMAS volume of 22,062 km^3 as my starting point.
Area: (from Shepherd): 11.9M km^2
Peak Height: 5.5 metres
Height loss (from Shepherd): 5 cm
My winter loss estimate, 1994-2007: 402 km^3
Compare this to the PIOMAS volume loss estimate of 421 km^3, and Shepherd’s estimate of 851 km^3, and tell me how wrong my conical model is …
Willis Eschenbach says:
May 31, 2010 at 4:36 pm
Shepherd et al. DID NOT USE A CONE. They used a flat slab. This gave them a 35% error in their answer.
No it didn’t as explained above.
Yes, I used a smaller volume. But if I used the Shepherd volume, I would still get a smaller answer than they got. Which was my point. They used a slab, so 35% of their volume was from the reduction in area … which makes no sense, since the area being reduced is at the outer edge of the ice, which is the thinnest ice.
It makes perfect sense, it’s just HS calculus.
So which error are you talking about? Because if you want to bust me for whatever error you think I made, you need to bust Shepherd twice as hard.
No they seem to have calculated it correctly, had you performed your calculation correctly you would have got the same answer as they did, unfortunately you used inconsistent data which was the source of your error.
Willis,
Do you not agree that the rate of ice loss at the periphery should be much greater than at the apex of your model?
I would think that the annual retreat of winter sea ice relates to first-year ice as a legacy of what has not been restored from the prior summer melt. So, a question to ask is how thick was the first-year ice that has been lost? Well, it’s substantial, according to the NSIDC:
EXTRACT: Estimates based on measurements taken by NASA’s ICESat laser altimeter, first-year ice that formed after the autumn of 2007 had a mean thickness of 1.6 meters. The ice formed relatively late in the autumn of 2007, and NSIDC researchers had actually anticipated this first-year ice to be thinner, but it nearly equaled the thickness of 2006 and 2007.
So, if average first-year ice is around 1.6 metres thick, one might think that the typical thickness would be a great deal more than 5cm (0.05 m) within say 9 km of your modelled periphery. (Re your fig 1) Thus, there has been a much greater thickness loss at the periphery than has been indicated in your conical model. (and probably much less at the apex)
Sorry Willis but your model should not assume uniform thickness change. (but how to find good graded data might be difficult?)
See also this and this
Sod, one last thing. If I use the Shepherd volume of 32,487 km^3, and a conical model, I get an annual loss of 500 km^3/year (vs. Shepherd’s calculation of 851 km^3). So your claim that my “error” is from using the wrong volume doesn’t pass the math test …
Excerpted from: Phil. on May 31, 2010 at 3:09 pm
No, what you have there is basic algebra, not calculus, as found in a basic geometry class.
Oh, you didn’t show all the steps. I had very good teachers, they would’ve taken points off for such sloppiness. One might have chucked a chalkboard eraser at you as well.
Bob_FJ,
I question the NSIDC’s unsupportable statement:
That is an example of why the NSIDC continues to have credibility problems.
According to its own figures, global sea ice is average. The Antarctic rise in ice cover completely offsets the Arctic’s cyclical loss.
Attributing the loss of Arctic ice cover to human emissions of CO2 — which are only about 3% of the total CO2 emitted — does not explain why Antarctic ice is growing at the same time.
To get an idea of how NSIDC adjusts their numbers, do an archive search of: NSIDC. There have been numerous articles here concerning this issue. To have the NSIDC make their official statement above as definitive [and as baseless] as they do, is scientifically unsupportable; it is simply a WAG.
Before making such conjectures, they need to provide some testable evidence showing that the Arctic ice decline is the result of human activity — and then explain why the Antarctic is going in opposite direction.
kadaka (KD Knoebel) says:
May 31, 2010 at 5:09 pm
Excerpted from: Phil. on May 31, 2010 at 3:09 pm
“So no matter what shape you have if you know the area before and after and the average height before and after the change in volume is the same. It’s elementary calculus.”
No, what you have there is basic algebra, not calculus, as found in a basic geometry class.
It is also the result one gets by using calculus, I supplied an algebraic derivation for ease of understanding, however the same result can be arrived at by using calculus as follows:
v(t)=A(t)H(t)
by the Product Rule:
v'(t)=[A(t)H(t)]’=A'(t)H(t)+A(t)H'(t)
Feel free to insert the steps you feel are missing, I note however that you don’t challenge the validity of the statement.
Smokey, Reur May 31, 2010 at 6:03 pm:
Yes, I agree, and I hope I didn’t give the impression that I take the NSIDC seriously. For instance, given some of the pronouncements from their director, Mark Serreze, there is I think a real credibility issue for the group.
However, surely they must get some things right, and my reference sought typical thickness of first-year Arctic sea ice, which is I think, unlikely to be overstated by them? (at 1.6m restoration after the big 2007 melt)
Bob_FJ,
No doubt that we’re on the same page. But please don’t scare me with pics like that! Serreze looks like the guy in my neighborhood pushing his shopping cart.
I was with you up until the end, where you say “If you get something smaller, then you have made a mistake on the way (such as picking a wrong value for the average height.)”
You have left out the other possible option … which is that S2010 has made a mistake along the way (such as picking a wrong value for the average height.) I don’t know if they have, as the numbers are all over the map … but it certainly is possible.
I’m not sure I understand what you mean by that. There are two components to their argument:
1) The data used as input, i.e. A_1 (initial area), A_2 (final area), dh (average decrease in thickness)
2) The model they use.
If you doubt that their data is correct, then you can certainly come up with any result for the decrease in ice volume that you like.
On the other hand, in your initial post you took their data as given, and argued that their model is wrong. You are then certainly free to suggest any other model you like. The point is that no matter what model you choose, the decrease dV will always satisfy
dV >= A_2*dh
If you use their data, this gives the lower bound of 589 km^3 that was mentioned before.
To clear things up, in your initial post you should either state a) that you do not believe their data to be correct (and then you should give reasons for that) or b) that you do not believe their model to be correct. You can then suggest your own model and use it to compute the decrease in ice volume. The answer you get (if you do it correctly) will then be bigger than 589 km^3 (if you use their data) or, more generally, bigger than A_2*dh (if you use your own data). As many people have already pointed out, the way your initial post is written at the moment it contains a rather embarrassing error which you need to correct.
Well Willis, you’ve started quite a little pissing contest here. With all the back and forth WAG’ing going on, we should be grateful that it’s happening in cyber-space, and not at a line of urinals. Could get damp. ☺
/dr.bill
MartinM says:
May 31, 2010 at 1:40 pm
Hmmmm! Missed a decimal place!
273*1190 = 324,870
851/324,870 = 0.262%
1/3*475*1190 =188,417
372/188,417 = 0.197%
Bob_FJ says:
May 31, 2010 at 5:07 pm (Edit)
Bob_FJ, not sure what you mean by the “rate of ice loss”. Do you mean that if the thickness at the periphery reduces by say 10 cm, at the middle it will reduce by 5 cm? Because S2010 assumes a 5 cm loss across the entire surface.
w.
Phil. says:
May 31, 2010 at 5:05 pm
Phil, that’s not clear. It didn’t what? Didn’t use a slab? Didn’t give them an error?
Let me run the numbers for you again.
S2010 assuming a slab
Shepherd Winter Volume, 1990-1999: 11.9 M sq km * 273 cm = 32,487 km^3
Area: 11,900,000 km^2
Reduction in area: -111,000 sq km
Reduction in thickness: 5 cm
Mine assuming a cone, using S2010 figures
My Winter Volume, 1990-1999: 32,487 km^3
Area: 11,900,000 km^2
Reduction in area: -111,000 sq km
Reduction in thickness: 5 cm
Shepherd Winter Loss: 851 km^3
My winter loss using Shepherd’s figures: 500 km^3
The difference is solely from the assumed shape of the ice pack. S2010 assumes it is a slab. I think that assumption is unphysical.
This is borne out by the fact that if I use the PIOMAS volume, I get an answer quite close to the PIOMAS estimate of the ice loss (402 km^3 vs 420 km^3. The S2010 loss estimate of 851 km^3, on the other hand, is twice as large as the PIOMAS estimate.
ck says:
May 31, 2010 at 7:10 pm
One more time. Their data, their slab model.
Their data
Area: 11,900,000 km^2
Thickness: 273 cm
Thickness loss: 5 cm
Area loss: 111,000 km^2
Volume 1: 11,900,000 km^2 * 273 cm / (100000 cm/km) = 32,847 km^3
Volume 2: 11,789,000 km^2 * 265 cm / (100000 cm/km) = 31,595 km^3
Net Loss: 892 km^3 (slightly larger than their answer of 851 km^3, due to changing areas over the period of the calculation)
Now, using my conical model:
Area: 11,900,000 km^2
Thickness: 819 cm
Thickness loss: 5 cm
Volume 1: 1/3 * 11,900,000 km^2 * 819 cm / (100000 cm/km) = 32,847 km^3
Volume 2: 1/3 * 11,789,000 km^2 * 814 cm / (100000 cm/km) = 31,987 km^3
Net Loss: 500 km^3
I think both their model and their data are wrong. Using the PIOMAS volume, my model gives a very good match to the PIOMAS loss.
Not sure what you are calling my “embarassing error” in the initial post … if you mean the height, I updated the initial post some time ago regarding that.