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)
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.
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.
* 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.