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.
mb says:
May 30, 2010 at 10:42 am
mb, you are welcom, thanks for continuing the discussion. Yes, there is more to discuss. You seem to have overlooked my comment, so I’ll repeat it:
But mb, rather than get lost in discussions and debates about average thicknesses, why not just use the formula for the volume of a cone? That way, you can’t go wrong. The formula is:
For one cone we have
For the other we have
Difference = 297
w.
Hi Willis
It’s beyond me, who last took a maths exam at the age of 16 and passed it only by the skin of my teeth, why anyone wouldn’t understand your cone model. However, one thing I wanted to ask is this. Would it be true to say that the centre of the cone, being nearer the pole, would be colder and less liable to melt than at the edge, farther from the pole?
If so, wouldn’t more more area be lost at the edges in your 2-D section of the cone than you show? After all, you assume a constant depth loss over the whole surface of the cone, but if it widens towards the periphery, then for a given volume loss, the periphery loss would be greater, wouldn’t it?
As I say, my maths isn’t good so I could well be wrong, but would welcome your comment.
Willis,
I think you should at least applaud Dr. Shephard for vastly improving the mathematical simplicity of coming up with an incorrect answer.
The norm in the field is to expend millions of trillions of floating point operations to produce it, whereas Shepherd et al produced it in just half a dozen.
Unfortunately, producing a correct answer isn’t quite so economical, unless you take a very long term view and say the trend is zero. 🙂
Willis Eschenbach >
I did not overlook that change in volume is not equal to change in area * average height, where exactly do you think that I overlooked that?
I did compute the difference in volume directly. The result did not depend on the shape of the ice, conical of flat or whatever comes out the same. If you don’t like the result, maybe you can say where my mistake is?
I can say exactly where the mistake is when you use the formula for the volume of a cone to compute ice sheet volume. It goes like this.
If you use the formula for the volume of a cone, for this purpose you are using your model that ice is shaped as a cone. That’s fair enough, as long as we agree on using that model. But if we do agree on that, you cannot later say “Oh wait, its not really a cone, so I use some different values that are more realistic, but I stick them into the formula from the cone model I had before, because I want it both ways.”
So here we are in the cone model, and we are honour bound not to change that choice until we have finished the calculation. Lets talk about V1. As you say, the volume is 1/3 * A1 * h1. The important question is “what is h1?” The mathematical formula assumes that it is the height of the cone. So how do we calculate the height of the cone? We cannot just look at some actual map of ice thickness, and guess what the highest value for the height could be. This would only work if the ice on the map was actually in cone form, but its probably not. And we are honour bound to stay with the cone model!
We do have one value we can use to estimate h, namely the average thickness 273 centimeters. The mistake you make is to say “h=273cm”. The average thickness of the cone is NOT h, it is h/3. So if the average thickness is 273 cm, h has to be 3*273cm=8 meters. Of course, it might be true that the actual ice is not anywhere 8 meter thick. But if that is so, it is because the cone model is not accurate. This cannot impress us, since we are at the moment sworn to uphold that particular model.
Of course you could say “OK, I’m so fed up with this stupid cone-vow, I now want to change the model so that it better reflects the real shape of the ice. I know that the ice is not 8 meter thick”. That’s fine, its a free country, but if you do that, you are not allowed any more to use the formula for the volume of the cone to compute the volume of the ice. You can’t have it both ways.
But you are still allowed to use the formula I gave before. This does not depend on the shape of the ice, so it does not depend on whether the cone model is true or not.
Willis,
I see your point clearly. This seems to narrows down to what exactly do the satellites “see” as they measure the ice “area”, not extent. Is it strictly the area within the waters edge around the thousands of chunk floating, once the ice pack starts to breakup, or are they attempting to adjust the areal value to somehow approximate thousands of little icebergs where most of the volume is underneath the water and the edges are eaten away from the wave action as they break on the edges of these pieces, much like your cone drawing above only thousands of them.
If they only measure strictly what the satellite “sees”, then the volume of ice is almost a magnitude greater than if you are assuming one huge chunk of ice plain like the top example (as a rectangle). If so even your calculations above are underestimating the ice volume. I am guessing this would occur in late April, May, early June during the break up into chunks, not huge plains of ice.
Have you ever read about the strict “area measurement” they display daily?
Good grief, mb, what are you banging on about? Where does Willis say, having hypthesised the cone model, that he wants to change the ground rules? Why can’t you understand that subtracting one cone from the other gives the volume loss and that from that one can estimate the periphery loss? You might not agree the cone model is valid, which is fair enough, but still irrelevant; for purposes of the argument, I fail to see any flaw in Willis’ argument.
@Brian H
… “preventing anyone from getting access to the data to analyse for themselves”
So you’ve spend time on data and found new evidence? How much hours? Ready to listen. Links, please!
mb says:
May 30, 2010 at 11:59 am (Edit)
OK, mb, I see what is happening. You are correct in this way — the difference between your numbers and mine is that you are assuming that the ice is 8.2 metres thick up at the top of the earth, and I am assuming it is 2.73 metres thick at the top of the earth. So your figures basically agree with Shepherds. And we agree on the calculations, we’re just using different ice thicknesses.
Now, you seem to think that we are bound by some kind of “cone-vow” to use the 8 metre figure. I am not bound by any such vow. I’m not in a theoretical dispute. I am not trying to replicate Shepherd’s numbers, as you seem to be doing.
I am trying to get the best estimate possible of what the real world is doing. If you want to be bound to a bad estimate of the average Arctic thickness simply because Shepherd et al. made that estimate, that’s your choice. I’m not bound by that.
Looking at the following figure, I’d say first that the conical shape is a good choice for an estimate. Upper panel is winter, lower is summer.
Second, I’d say that year-round, the thickness at the centre averages about (4.75 + 3.25)/2, or about 4 metres of ice thickness. So if you want my best estimate of the Arctic sea ice loss, using everything that I have learned during this discussion, I would put it at 344 km^3.
This is about 40% of their estimate, and reduces their final figure from 746 km^3 to 239 km^3.
“Why can’t you understand that subtracting one cone from the other gives the volume loss and that from that one can estimate the periphery loss?”
Sorry – badly phrased. I should have said:
“Why can’t you understand that subtracting one cone from the other gives the volume loss? And that the model enables one to estimate the periphery loss?”
I’m glad you spotted what mb was getting wrong, Willis; I was absolutely clueless!
juanita
May 30, 2010 at 7:00 am
I know very OT, but …
Very interesting. I kinda suspected this to be the case considering the distribution of fishes, but never had the time to research it. But what is funny, is that just yesterday I decided to review the history of the Colorado and Sacramento river systems. The site I am getting ready to explore right after writing this post is:
http://jan.ucc.nau.edu/~rcb7/RCB.html
Ed Caryl says:
May 30, 2010 at 7:07 am
So repeat after me, weather is not climate and a local phenomenon is not global.
And two years, or ten, does not a trend make.
And then I get to remembering that the IPCC was created about ten years after the end of a long cooling trend. Guess I shouldn’t mention all the hullabaloo over the coming iceage.
To use some old and tired sayings: What’s good for the goose is good for the gander. Or maybe: you can’t have your cake and eat it too.
My favourite part:
Right. Error bars two orders of magnitude larger than the datum. This is like measuring the thickness of a pencil by pacing it out!
I am sorry I posted too obscurely since I was in a hurry. But the problem and confusion is here”
“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.”
This is really simple geometry. Let me take you through it slowly. If you model the ice as a cone then the covered area goes as the height squared, area being proportional to r^2 and radius proportional to height. If you model the ice this way a 5cm height decrease, or 2%, will decrease the area by ~4%, not the 1% observed. In your calculation of volume lost you have lowered the height of the center by 2%, but only decreased the area by one 1%. You are not preserving the angle of the cone. Your model decreases the thickness by 5cm at the center, but a much smaller amount at the edges. This leads to an average change in thickness much smaller than 5cm for your calculation. I leave you to work out the integral for your average change in thickness. This is the source of confusion for those calculating the volume from average thickness change times area.
Regarding the dispute between mb and Willis:
Does there not exist a simple utility that will perform a numerical integral on a colour-coded graphic in which one can manually assign numeric values to colours in a given spectrum? In an equal-area projection this would be fine as is; in non-equal-area one would also have to map an “area-density function” to account for the distortion.
Such a tool would do away altogether with the need for having a “model”, cone or otherwise, and would permit exact numerical analysis of these and wider data that come by way of shaded maps, without further approximations and assumptions.
If someone knows of such a utility I would be interested, and see the value of it in a broader context; if there isn’t one, analysis we’ve seen here for the last few years argue strongly in favour of the creation of such a tool. In principle it should be extremely simple to construct. Since we’re dealing with pixellated images, one only needs to extract a count of pixels at each colour density or hue, as long as it area is faithfully represented. If not, then one needs to adjust each pixed by some user-definable function of its two coordinates.
Any programmers out there up to this challenge?
“R. Craigen says:
[…]one only needs to[…]Any programmers out there up to this challenge?”
You’re a project manager, right?
Willis,
Update to wayne:
May 30, 2010 at 12:20 pm
Just noticed the graphs you displayed here after my comment above. Seems my comment might only apply to the periphery where total melting occurs and by these maps they show one huge central area never breaks up at all and it is only that area which you are speaking of. Might want to just ignore my comment above (but it should have some effect on ice area during the winter to summer transition).
@Willis
> OK, mb, I see what is happening. You are correct in this way — the difference between your numbers and mine is that you are assuming that the ice is 8.2 metres thick up at the top of the earth, and I am assuming it is 2.73 metres thick at the top of the earth. So your figures basically agree with Shepherds. And we agree on the calculations, we’re just using different ice thicknesses.
And this is why you are wrong.
Shepards give 273cm as an average. You are using it as a peak.
You would best understand this if you looked again at your image comparing their “flat” model versus your “cone” model. You have drawn your cone with a greater height than the flat average – but in actuality they should be the same height, because you use their average as the maximal height of your cone. Then you would be able to see clearly and visually why you have miscalculated and arrived at a much smaller estimate.
Given that they indicate having measured ice with thickness of 5 metres or greater, its odd to think that using calculations based on a maximum thickness of 2.73 metres can be accurate.
Willis Eschenbach says:
Now, you seem to think that we are bound by some kind of “cone-vow” to use the 8 metre figure. I am not bound by any such vow.
————
Dear Willis, this is getting kind of silly. Either you stick with your model of the arctic ice as a cone, or you don’t. I’m happy either way, as long as you make up your mind. If you don’t stick to it, you cannot use the formula for the volume of the cone to compute ice thickness OK?
—————-
Willis Eschenbach further says:
And we agree on the calculations, we’re just using different ice thicknesses.
……..
I am trying to get the best estimate possible of what the real world is doing. If you want to be bound to a bad estimate of the average Arctic thickness simply because Shepherd et al. made that estimate, that’s your choice. I’m not bound by that.
———–
OK, now you flip-flopped, and use the cone model again (that’s the only calculation you make), we are just differing on how tall the cone is. Lets examine if your cone is realistic. You insist that the cone is 273 cm tall. A cone is tallest in the middle and kind of slopes on the sides. So except at the very center of the cone, the thickness is less than 273 cm. Half way towards the perimeter, the thickness will be 137cm. Further out it will be even thinner.
We can compute the average thickness of the cone to be is 91 cm, while the average wintertime thickness of the Arctic ice according to the data we are talking about is around 273 cm. The model you offer is wildly different from what the real world is doing.
Re Ric Werme’s comment on typography
Ah, Ric, a man after my mathematical heart! I agree with most of your comment.
We mathematicians, however, generally dispense with all symbols for multiplication in typography, whether * or other. It’s neither A*B nor AxB, it is AB. As few people know the HTML tags for “cents”, “exponentiation” or other symbols, and don’t care to look them up for a singular comment, between ourselves we tend to use TeX or LaTeX script for anything that can’t be mimicked by ordinary extended alphanumerics or lazy typography that is universally recognizable. Since ^ is TeX superscript, this translates easily for the general public. \times gives the multiplication symbol if it is ever needed, and more arcane things like \equiv for modular congruence, \int for integral, etc. We build matrices with \begin{matrix} … \end{matrix} — there’s basically a construction for any mathematical expression, but it’s specialized knowledge. You gotta know the conventions, but since everyone in math, and most people in physical sciences, use TeX we’re understood as long as we don’t try to speak to a general audience this way.
It would be nice if modern HTML had an “escape to LaTeX” tag that permitted proper typesetting with LaTeX macros, but I’m told it was considered and rejected. Too bad — none of the existing “fixes” really come close.
Inflation, by the way, does not increase your 2 cents to $0.25 — the value of currency is decreased under inflation. Unless of course, you’re charging us this amount, in which case you’re correct.
Either way, the circle chart says it all. After being 100% wrong about ocean acidification, disappearing Himalayan glaciers, impending catastrophes from runaway global warming, toad extinctions, fast rising sea levels and a hundred other fake scares, the alarmists as usual have moved the goal posts once again — this time to Arctic sea ice, simply ignoring the inconvenient and opposite Antarctic trend.
That circle chart has something in common with Dr Roy Spencer’s CO2 chart with the honest y-axis. Can you see the similarity?
mb is right. If a cone had simply sunk by 5 cm, keeping its shape, the area would have reduced by an amount much greater than 111000 km2. By giving the reduced cone the larger, observed area, Willis has ensured that it is a flatter cone, with a height reduction at the apex of 5 cm, but tapering more gradually. Consequently, the thickness change tapers fro 5 cm at the centre, to almost nothing at the periphery. The average is not 5 cm, but about 5/3 cm.
Re: DesertYote (May 30, 2010 at 12:47 PM)
Coincidentally, I was up at Shasta a couple of days ago; learned a bit about the Sacto river.
The reason for the dam is flood control; every few years the sacto river would flood siginficant portions of the central valley. Here’s a current pictures of the dam; note that the reservoir is _FULL_.
Ah, no img tags, I see; try this:
shasta lake
The colour-coded map is, of course, only the unclassified area being reported from a US submarine survey. It says nothing about what the ice is doing on the Siberia side of the Arctic, or close to the Canadian islands and shore. The “cone” visual shape is an artifact of the projection and the selectively reported data. The white areas could be 10 m thick near the shorelines, for all that map tells us. In fact, as WUWT reported last year after the EM-B flew, it was up to 15 m. thick near Ellesmere. http://wattsupwiththat.com/2009/05/17/catlin-artic-ice-survey-an-annie-hall-moment/
I don’t possess one, but there is a small manual tool which allows one to trace any shape on a map and read off the enclosed area. But on a flat map projection of a curved globe, I doubt it would be a meaningful number.