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.
Oh goodie another post about arctic sea ice…
/sarc
Surely the battlefield is bigger than this guys?
Since, as pointed out, we no longer have a “cents” symbol, please let me stick my $0.02 worth in …. 8<)
Assume Floating ice is "smooth" (despite the know, very, very deep pressure ridges below water — not accounted for in the AGW surface "area" hysteria) with 90 % below water, 10 % above water.
A) The Navy data clearly shows a conic-like change in ice thickness – max thickness = 5.0 meter at the center, decreasing linearly towards the edge thickness of 2.0 meter, where it stops abruptly.
mb: Wrong. You (the paper you are defending) cannot use any "average" [ (5+2)/2 = 3.5 for example ] as thickness under any circumstances.
You could, as a bad “approximation” first calculate the true volume of the ice (see below for the implifying sassumptions needed for vol_cyl + vol_cone_above_water + vol_cone_below_water) and then divide that vol_total by the visible surface area (as if the whole ice mass were a cylinder) to get a “volume-adjusted equivilent average thickness” … But that “weighted average ice thickness” is useless.
Others have tried to discuss volumes of spheres (incorrect) and half-spheres. Both are incorrect since the Arctic ice cannot freeze that deep (ocean floor is too shallow); and any spherical model violates the “floating ice” real world.
However, there are some significant things needing improvement in the conic model as well.
B) From the Navy data, the Willis ice model should not be a cone, but it must begin with a cylinder with
volume_cyl_winter = A (t0, t1, t2, t3, etc) * th_cyl (winter, th_cyl = 2.0 appx)
and
volume_cyl_summer = A (t0, t1, t2, t3, etc) * th_cyl (summer, th_cyl = 1.0 appx)
90% of ice is underwater, therefore, the winter ice floats with 0.2 m above water, 1.8 meter below water. Which resembles the photos of submarines surfaced at the pole and Arctic explorers on sleds fairly closely.
Then there are two cones needed:
vol_cone_above_water = A (to, t1, t2, etc) * 1/3 * height_ice_up
vol_cone_below_water = A (t0, t1, t2, etc) * 1/3 * height_ice_down
Probably, one could assume ice_up = 10% of ice_down
However, height_ice_up + height_ice_down + th_cyl = Navy’s total ice for any area or time of year.
Melting is from underneath at a constant rate (if the water below is at a constant temperature and flow rate and salinity – which may be true) across the Arctic.
Freezing must be two ways: In ice-covered area, it is proportional to the “average air temperature above the ice” conducting “cold” down through the insulating ice to the water below. (Heat flow is opposite, from the warm water below to the cold air above, but we’re discussing net melting anyway…)
In open-water, freezing is from the open water to the air directly – and will be MUCH faster than in ice-covered areas. In the three month when arctic air is above freezing, some melting can occur from above. The rest of the year, this is impossible.
This difference between open water and ice covered area is critical to the model volumes and rates:
The ice cone will melt at a constant rate underwater under all parts of the cone. Note that once the ice at the edge reaches the 1.0 meter ice-edge point, melting can proceed very rapidly. This reduces the area of the conic model – and thus its volume – significantly.
At thickness more than 1 meter – unless winds blow whole regions out of the Arctic – melting will be below at a constant rate. There is no change in Area, but a very specific change (decrease) in volume.
Net change in volume (melting) is, of course, the difference between losses at the exposed surfaces to the water, and gains (or losses) from the exposed surfaces to the air (3 months of possible melting, 9 months of definite freezing — BUT freezing gains in those 9 months of below 0.0 C temperatures are NOT limited to the existing ice surface.
mb says:
May 30, 2010 at 1:48 pm
I’m sticking with the conical model, and I have from the beginning. Where did I say I wasn’t? Certainly I didn’t say that in what you quoted. All I said was I am not bound to say that the polar ice is 8 metres thick.
—————-
You really need to learn to read. I don’t insist that the cone is 273 cm tall. I have not flipflopped.
I haven’t a clue what you are calculating. I gave a result for a simple cone that is 4 m tall, which I figure is about the average maximum thickness of the ice. Period.
Dave H says:
May 30, 2010 at 1:33 pm
I agree that my cone should be higher than their slab. And that is what I have done, I have used what seems to be the average maximum thickness for the year-round ice. The graphic I posted above indicates that this is about 4 metres. Which is why I said:
I hope that makes it clear. I have posted an update to the main post indicating this change.
Willis Eschenbach previously said:
The formula is:
V1 = 1/3 A h
For one cone we have
V1 = 1/3 * 11,900,000 * 273 / 100,000 = 10,829
—————-
Got it, the height is 273 cm.
—————
Willis Eschenbach later said:
You really need to learn to read. I don’t insist that the cone is 273 cm tall.
———–
Except that the cone isn’t 273 cm tall.
—————
Willis Eschenbach finally said:
I gave a result for a simple cone that is 325 cm tall, which I figure is about the average maximum thickness of the ice. Period.
—————————
OK, so you suggest that a cone of height 325 cm is the correct model. Does this mean that you suggest that a cone of average thickness 108 cm is a good model for an ice sheet which has a thickness of 273cm?
Willis,
I believe that your calculations using the cone model have an important flaw.
In your model, I think that you assume that the change in ice thickness is a uniform 5cm reduction across the whole surface area of the cone.
With the change in ice area that you use, from 11900000 km^2 to 11789000 km^2, this is not the case. With this change in ice area, the change in thickness is 5cm at the center of the cone, but it tapers down as you go from the center to the edge of the cone. This leads to a substantially lower volume of ice lost, as in your calculation.
Basically, a uniform 5cm reduction in thickness will only occur if the two cones (before and after loss) have the same shape – ie the same angles. This can only happen if the ratio of the heights (273 / 268) is the same as the ratio of the diameters of the cones.
The ratio of the diameters is the ratio of the square roots of the areas – 1.004689 in your calculation. The ratio of the heights is 1.018656.
This difference in ratios basically means that your calculation is for a case where the loss of thickness is 5cm at the center but which tapers away to a much smaller value as you get to the edges of the cone. This is inevitably a much lower figure than for a uniform 5cm thickness loss.
“I’d call it more like a POOMA number. (which stands for Preliminary Order of Magnitude Approximation. Or something.)” –jorgekafkazar
_________
“That is definitely NOT what POOMA stands for :0” –“D”
_________
☻ D wins the Clever Clogs of the Day award! ; ) –jorgekafkazar
Ric Werme says: “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)….”
PostModern Science has lost its bearings.
RACookPE1978 says:
May 30, 2010 at 2:58 pm
RACookPE1978, here’s my 2¢ worth. You have given what is likely a more accurate approximation of the ice volume. And I suspect that your method would give a much better estimate of what is actually happening than mine.
My point was simpler. It was that the S2010 method, of assuming that the ice is a flat slab, and thus that a loss of area around the edges gives a full thickness loss of ice volume, is very unphysical. I merely proposed the next model up in the hierarchy, which is a simple cone.
I realize that my method is overly simplistic. I was not suggesting that it be used as a basis for anything more than a “back of the envelope” calculation. I proposed it to show how foolish it is for the S2010 authors to assume that area loss at the periphery can simply be multiplied by average thickness to give lost volume. You can’t do that.
The proper way to do this calculation would be to use the ICESAT results to get a realistic idea of the actual volume. Second best would be to use a verified and validated model to do the same.
Which is why I was surprised that, after using satellite and airborne radar for the Antarctic calculations, the S2010 authors used such a simplistic and unphysical model for the Arctic ice.
Your interesting analysis is much appreciated,
w.
PS – On the Mac, the cents symbol is “Option-4”.
i am curios whether this comment will pass:
your error is simple Willis, but extremely embarrassing. your big cone (the one based on 11900000 km^2 * 273 cm or *400) is too small already. that is the reason, why you get a smaller reduction in ice.
this cone needs to have the same volume, as their flat model. your cone needs an average thickness of 273 cm of ice. but neither the one with a height of 273 cm, nor the new 400 cm heigh one, has such an average thickness.
you are calculating the difference between two volumes, that have no connection to reality.(both of them contain much too little ice)
this is the simple explanation, for your smaller difference. mb did a very good job, in pointing this out to you.
Ric Werme says:
May 30, 2010 at 6:51 am
shows common mathematical operators, wherein * is the Standard ANSI SQL-92 symbol for multiplication and ^ is the common symbol for exponentiation¢.
Also, OPTION+4 = ALT+4 = ¢ on most modern computer keyboards.
Jbar says:
May 30, 2010 at 7:47 am
“Let’s look at the big picture for a second –>
http://i50.tinypic.com/35iackn.gif
Global and arctic sea ice are slowly diminishing and there is no evidence that this trend has stopped.
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.)”
_________________________________________________________________________
Yes DO look at the big picture. The sun is in a funk. The UV from the sun has dropped 6% and the natural planetary cycle called ENSO or El Nino has gone to the cool cycle. All natural cycles like AMO, PDO, NAO and lunisolar tidal forces are pointing to a near term future of 20-30 years of cooler weather rather than unprecedented warming. Remember the cycles run for 60 to 80 to 200 years, therefore thirty years is only a half cycle. that is why the recent change in terminology from “Global Warming” to “Climate Change”.
Early this year satellite data showed the troposphere was quite warm while land temps were very cold. I asked Dr Spenser about it and he said it was because the oceans were releasing heat. The next month the sea temperature dropped 1C the largest drop since the seventies.
I do not know about you but the weather in my area is decidedly weird. Not only did we get snow five times (we usually do not get snow) but we had a very dry April – no rain at all and a very wet cold May. The highs are 5-8 degrees below normal. However the weirdest thing is the rain storms are now coming out of the east. Except for hurricanes the wind and storms are always out of the west – always. Now they come from the east — weird.
Big picture –
Global sea ice area has been declining steadily for 30 years and there is no evidence that the ice loss is near its end.
http://i50.tinypic.com/35iackn.gif
In that context, why continue with this thread?
In case the link doesn’t show up, my previous should read:
Mathematical, matrix, string array, and string operators shows common operators, wherein * is the standard ANSI SQL-92 symbol for multiplication and ^ is the common symbol for exponentiation. Also, OPTION+4 = ALT+4 = ¢ on most modern computer keyboards.
What mb is saying is correct.
Let’s put it another way. There are two effects that contribute to the decrease:
1) The total area becomes smaller.
2) Average thickness shrinks.
It is true that the contribution of 1) depends on the model of how the ice sheet is shaped, i.e. if it’s conical, flat, or any other arbitrary shape. The contribution of however 2) can always be bounded from below: By definition,
loss of volume through decreased thickness = total Area*average decrease in thickness
This is basically the definition of `average decrease in thickness’, and it is completely independent of the shape of the sheet. Plugging in the values from the original paper, this gives the 589 km^3 that you mentioned above.
It follows that no matter what shape for the sheet you use, the result must always be bigger than 589 km^3. If you get something smaller, then you have made a mistake on the way (such as picking a wrong value for the average height.)
Willis Eschenbach says:
May 30, 2010 at 12:44 pm
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.
mb is making the assumption that you stated, i.e. that the ice is conical and has the stated average thickness, 2.75m. That means the cone must have a peak of ~8.25m, by using the figure you did you get a number 1/3rd of the actual.
I’d appreciate anyone taking a look to make sure I have done the math correctly.
So indeed you have made a mistake, you inadvertently used too thin a cone by a factor of 3 and therefore calculated a volume change too low by a factor of three.
R. Craigen says:
May 30, 2010 at 1:04 pm
My favourite part:
Arctic sea ice area trend was -111 ± 8 x 10^3 km2 yr-1
Right. Error bars two orders of magnitude larger than the datum. This is like measuring the thickness of a pencil by pacing it out!
Unfortunately you have misread it (not entirely your fault because of the prevalent use of the bastardized unit ‘km^2’). This is one of those problems that the SI unit system was designed to prevent, it should be in those units:
-(111 ± 8) x10^3 km^2
or better yet -0.111 ± 0.008 Mm^2
“I have used what seems to be the average maximum thickness for the year-round ice. The graphic I posted above indicates that this is about 4 metres.”
This is little better. What you have now done is taken the max height for the real shape, which is not a cone, and said that must be the height for your cone model. But cones are, well, pointy at the top. The 4m is for a flat-topped distribution. The result is that you still have far less ice volume than Shepherd uses.
But your model, which requires that the cone flattens as it shrinks, is unphysical. It loses volume more rapidly from the centre than the periphery.
For those who were following the penguin/ice question, I think I figured it out.
Most penguins nest on land in summer. The emperor penguins nest on the ice pack in winter. So they have to choose a nesting spot far enough back from the ice edge that they don’t get caught by ice break up in the spring before the chicks are ready to go for a swim. So they start 50 or 60 clicks from the ice edge. The males incubate the egg and the females go to sea. The females return about the time the eggs hatch and the males leave and go to sea. But by that time it is later in the winter and the sea ice has been growing, so now it is 100 to 200 km to the sea. Oddly enough a really cold year is really hard on the emperor penguins because the ice extent is so large, and in a really warm year the ice extent may retreat so rapidly in the spring that the chicks drown.
What an odd bird!
Speaking of computer models, there is a computer model of the Arctic ice mass, called “PIOMAS”. Steve Goddard discusses it here. They give a figure for the Arctic ice volume loss for the 1994-2007 period used by S2010. Remember that the S2010 figure for that period was 851 cubic km (95%CI 635–1065 cu km), and my best estimate was 344 cu km (95%CI 260–430 cu km).
The PIOMAS value for that period is … drumroll please … 426 cu km/yr. (As it is a model result, there is no CI.)
Doesn’t mean too much, it’s a computer model after all … but it’s less than half of the S2010 estimate, and within the CI of my best estimate.
Phil. says:
May 30, 2010 at 4:18 pm (Edit)
Geez, guys, you gotta keep up. I didn’t state that the ice is conical and has an average thickness of 2.75 m. I said it had a peak thickness of 2.73 metres, and based on the feedback and new information, I have increased that peak thickness to 4 metres.
Again I say, if you disagree with something I have said, QUOTE IT.
Willis,
The arguments here about the geometry of the difference between two cones become unimportant in the scale of things. Ignoring global curvature, when drawn to scale the Shepherd model would look something like this, within the limits of screen definition:
____________________________________________________
If we now compare your model, it would also look something like this, (both including thickness change):
____________________________________________________
The lower line emulates your two cones, where the gradient of the cones approaches zero. Not visible is that with any thickness loss, the diameter of the cone base dramatically reduces exponentially with reducing gradient. You can see this in less pronounced fashion in your figure 3, labelled “Area Change” by visualizing a reduced gradient.
Might I suggest that the visible loss of ice around the periphery cannot be explained by either model, but is likely because of higher water and air temperatures, winds, currents, thinner-weaker ice, escape routes to warmer waters, greater exposure to solar radiation per unit area….and maybe a few other things.
BTW Willis FYI Concerning the memory of John Daly, and Marco’s insults of him, (elsewhere), I persisted and he eventually withdrew. My last comment is here: Marco was unable to produce any evidence for his accusations.
EXTRACT to Marco: [4] You should apologise to John Daly for accusing him of nastiness, being ungentlemanly, using strawman circular arguments and so-on, in your various rants. There is no evidence of that in the two articles you cited as demonstrating such.
Nick Stokes says:
May 30, 2010 at 4:57 pm (
Not sure what you mean by “pointy at the top” versus “flat-topped”. Here is a drawing of the cone representing the ice pack, to scale.
As you can see, “pointy” is hardly accurate. You can’t even see the “point” until we increase the vertical scale by a factor of 10,000.
You are right that the real ice is not in the shape of a cone. However, it is also not in the form of a flat slab, as used by S2010. So what? We are trying to estimate the ice loss, not make an accurate 3-D model.
The average thickness of a cone radius r and peak height h is 1/3h . To match a measured average thickness is 273cm you should therefore use a peak height of 3 times this, or 8.19 metres.
This will give a better fit than trying to guess at the peak height by looking at maps. In particular we know that the cone model is going to be least accurate near the point of the cone. This means you don’t want to be using a value of thickness at the point (where the model is least accurate) to set the parameters of the model. It is much better to use a global measure like average thickness to set the appropriate peak height.
scott
May 30, 2010 at 2:13 pm
WOW! I Grew up in Phoenix. There is a joke about the Salt River. Its called a river when there is no water in it. When it has water in it, its called a flood. But its more then a joke because it often really is a flood. That’s because sometimes the lakes are full and can’t hold any more water, but the rain continues to fall. Just like Shasta!
Willis Eschenbach says:
May 30, 2010 at 5:25 pm
Phil. says:
May 30, 2010 at 4:18 pm (Edit)
…
“mb is making the assumption that you stated, i.e. that the ice is conical and has the stated average thickness, 2.75m. That means the cone must have a peak of ~8.25m, by using the figure you did you get a number 1/3rd of the actual.”
Geez, guys, you gotta keep up. I didn’t state that the ice is conical and has an average thickness of 2.75 m. I said it had a peak thickness of 2.73 metres, and based on the feedback and new information, I have increased that peak thickness to 4 metres.
Again I say, if you disagree with something I have said, QUOTE IT.
Well if you say you’re going to do something, in this case compare methods of estimating volume loss, then do it! Instead you suggested using a different model for the ice and comparing the results of the different methods. However you used different initial conditions for the two calculations, not a very clever way to compare them! You did your model calculations for one third of the volume of ice that Shepherd did.