# You cursed brat! Look what you’ve done! I’m melting! Melting!

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.

## 202 thoughts on “You cursed brat! Look what you’ve done! I’m melting! Melting!”

1. wayne says:

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.

Now to gain that 130 meter thickness in 15 year would truly classify as a decadal snowstorm! (and what are we left with, figments, figments everywhere!)

Where do these scientists get their vivid imaginations?

2. now, this is what global warming is telling us! It will melt every ice particle there is.

3. Willis,
I said in another thread that the ice pack should be modeled as a wedge. My point was that growing ice extent late in the ice formation season would result in rapid expansion of the thin edge of the wedge, followed by rapid retreat once the melting season started until the thin edge retreated back to thicker ice, whereupon the retreat would slow down. That’s exactly what we are seeing this year in the Arctic. I’ve lived in a winter climate all my life and have seen fresh water lakes and rivers freeze and break up so often it never ocurred to me that there was any question that it was a wedge shape in the first place!

Would like to see someone do an article on ice and Penguins too. The silly bird brains nest as much as 200 miles from their food source and almost die every year going back and forth across the ice trying to get food back and forth to their chicks and mates. 200 miles! And they waddle! So once has to ask how this came to be? They are just barely adapted enough to manage the treck over and over again, so the question arises, why would they have chosen such a breeding location in the first place? Seems to me the more logical explanation is the breeding grounds were locked into their instincts when there wasn’t a whole lot of ice in the first place and the treck to open sea for food was a short one. As the ice grew, they had to slowly adapt to a longer and longer treck, which in turn implies that at some time in the past there was a lot less than 200 miles of ice between the breeding grounds and the open sea. Even bird brains don’t choose their first breeding ground 200 miles from food.

4. sigh. I meant 1oo miles. A 2 looks just like a 1 when you are so tired that the keyboard starts to get blurry. But Willis posted another article so I had to read it and then a I had to respond to it and then I had this idea about Penguins…. more Ritallin…need more Ritallin….

5. mindbuilder says:

Modeling the thickness with a rounded cross section, like chopping the top off a sphere, might be more accurate than a cone. The result would probably still be much closer to a cone than to a constant thickness.

6. mb says:

In the formula V= 1/3 * A * h, h is height at the apex, not average height.

7. rc says:

Area loss: 111,000 km^2 * 273 cm / (100,000 cm/km) = 303 cubic km (34% of total)

——————

Multiplying the area loss by the average thickness seems like such an obvious overestimation but it sails right through climate science peer review.

Funny how these things always make it “worse than we thought”.

8. 899 says:

davidmhoffer says:
May 30, 2010 at 1:02 am

Would like to see someone do an article on ice and Penguins too. The silly bird brains nest as much as 200 miles from their food source and almost die every year going back and forth across the ice trying to get food back and forth to their chicks and mates. 200 miles!
*
*
A logical deduction is so as to avoid their local predator, the Leopard seal.

How many seals would actually travel that far inland for a meal?

9. @mindbuilder

I think it is so very, very flat that it makes no difference if it is a cone or a hemisphere.

10. mb says:
May 30, 2010 at 2:05 am

In the formula V= 1/3 * A * h, h is height at the apex, not average height.

It seems this is what is being done. As the difference in volume between the two cones is being calculated, the question is moot in any case.

11. 899 says:
May 30, 2010 at 2:23 am
A logical deduction is so as to avoid their local predator, the Leopard seal.

How many seals would actually travel that far inland for a meal>>

Leapord seals don’t hunt on land or ice. They take penguins as prey when the jump off the ice into the sea emaciated, tired and starving from waddling 16o km to the edge of the ice, making them easy prey for the leaport seals. No need to leave your natural habitat for food when they have not choice by to come to yours to eat and are half dead when they get there.

12. rbateman says:

Is climate science calculation a victim of poor imagination, worse than previously imagined?

13. David Baigent says:

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)

14. 899 says:

davidmhoffer says:
May 30, 2010 at 2:34 am

Leopard seals don’t hunt on land or ice. They take penguins as prey when the jump off the ice into the sea emaciated, tired and starving from waddling 160 km to the edge of the ice, making them easy prey for the leopard seals. No need to leave your natural habitat for food when they have not choice by to come to yours to eat and are half dead when they get there.
*
*
David,

Is it your presumption that seals never take birds from the shore, and that they’ve never done so?

I will think that the birds have a ‘species memory’ regarding that matter of where to not make a home.

Alligators and crocodiles have been known to take prey on dry land, even though they are most fitted to do so in the water, or at water’s edge.

15. mb says:

mb says:
In the formula V= 1/3 * A * h, h is height at the apex, not average height.

JER0ME says:
It seems this is what is being done. As the difference in volume between the two cones is being calculated, the question is moot in any case.

———–

Actually, it does matter. If the height at the top of a cone is h, the average height of the cone is h/3. So the correct formula for the volume of the cone, using the average height, is

V = 1/3 * A * h = A * (h/3) = A * average height.

In the example, we have two cones. The difference between the volumes of the cones is

V1-V2= A * (average height1 -average height2)

This gives a difference in ice volume which is three times as big as what we are being told, in excellent agreement with Shepherd el al.

16. Joe Lalonde says:

The growth of Ice is just as complex as the evaporation and precipitation cycle.
Cold temperatures can thicken the ice from below the surface of water(strength of currents is a factor). The precipitation on the surface in snow has to be packed which then there are different types of snowfall that carry different moisture and mass content. Winds also are a factor in packing snow, moving snow and temps, sunlight, cloud cover, night or day which in turn is planetary rotation.

17. David L says:

Why do these climate alarmists love ice so much? Do they have a set amount that they’d be happy with, or would they like to see another ice age? I think we know that in the history of this planet we’ve gone from periods of no ice to much more ice than we currently have. These are the limits of the natural variability. So the loss or gain of a few thousand km2 don’t really concern me. Melt it all for all I care.

18. Chris Korvin says:

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 .

19. Louis Hissink says:

In the mining industry we would go out and measure the ice thickness by in situ sampling. Anything else is intellectual waffle and irrelevant.

Willis, you are spot on with your analysis – it’s like drilling one hole into a mineral deposit and extrapolating that result to the rest of the orebody. OK in theory until you actually have to mine it, and then you discover that the statistical tricks used are basically meaningless.

It’s part of the sample volume variance problem, as well as not knowing how to sample a physical object, here a thin sheet of ice of variable thickness.

But who has the cash to measure ice thickness properly? You need to do that to substantiate the various models proposed here.

One would need to do a drilling traverse over the ice sheet to measure the ice thickness. Or maybe a seismic traverse using percussive methods rather than explosives.

That means spending a long time in winter doing field work.

Can you imagine what OSH departments would impose on this activity?

So the next best effort is to wax intellectually using computer modelling, and if the BS is plausible, converted into “fact”.

20. Well, if Arctic fits on a coaster who would care? Zooming in makes changes obvious:

The region of Jakobshavn Glacier this year (27/05/2009):
http://ice-map.appspot.com/?map=Arc&sat=ter&lvl=7&lat=69.208788&lon=-53.878389&yir=2010&day=147

Same region same day last year (27/05/2009):
http://ice-map.appspot.com/?map=Arc&sat=ter&lvl=7&lat=69.208788&lon=-53.878389&yir=2009&day=147

Looks somewhat different, or not?

Now let’s find the earliest day with comparable circumstances of 2010 in 2009:
http://ice-map.appspot.com/?map=Arc&sat=367&lvl=7&lat=69.208788&lon=-53.878389&yir=2009&day=162

That’s two weeks later (11/06/2009). Let’s assume a melting period of 180 days, then the difference is more than 8 percent.

What’s wrong with this math?

21. Willis Eschenbach says:

mb says:
May 30, 2010 at 3:11 am

mb says:
In the formula V= 1/3 * A * h, h is height at the apex, not average height.

JER0ME says:
It seems this is what is being done. As the difference in volume between the two cones is being calculated, the question is moot in any case.

———–

Actually, it does matter. If the height at the top of a cone is h, the average height of the cone is h/3. So the correct formula for the volume of the cone, using the average height, is

V = 1/3 * A * h = A * (h/3) = A * average height.

In the example, we have two cones. The difference between the volumes of the cones is

V1-V2= A * (average height1 -average height2)

This gives a difference in ice volume which is three times as big as what we are being told, in excellent agreement with Shepherd el al.

mb, many thanks, I’m glad to have somebody take a hard look at the math. However, its a bit more complicated than that.

First, if you wish to claim that we need to use an apex three times as high as the 273 cm ice thickness used in S2010, we would end up with the thickest ice being about 8 metres thick … which just isn’t happening. So the first thing we can see is that by assuming constant ice thickness out to the edge, they are estimating too large a volume of ice.

Next, you have the wrong formula for the difference in the volume of the cones. You have neglected the difference in the areas of the bases of the cones. The correct formula is:

V = 1/3 * (A1*h1 – A2*h2)

Even if we use the unphysical value of 8 metres of ice at the apex, this gives us 500 km^3, still well below their value of 851 km^3.

From looking at Figure 2 above, we can see that the thickest ice is at about 350 cm, much smaller than the 8 metres you propose. Using 350 cm as the apex gives us an ice loss of 326 km^3, rather than the 297 km^3 I had estimated before.

And that give us a total global ice loss of 221 km^3, which is still much smaller than their estimate of 746 km^3. Now we can say “Well, we should use 4 metres for the thickest ice”, but it doesn’t make a whole lot of difference, because we are dealing in estimates and approximations. Under any reasonable assumptions, their estimate of global ice loss is off by at least a factor of two, and likely more. (Remember, they have Antarctic ice shelves dropping a quarter kilometre in thickness …)

22. gallier2 says:
23. mb says:

Willis Eschenbach> I don’t claim that the ice is formed like a cone. On your picture above the cone is clearly marked ” how I model the arctic ice pack”. This means that it is your model, not mine. If you find that this model is ridiculously unrealistic, you are complaining about yourself, please leave me out of that fight. I just discussed the calculations you make inside this model, the model which you proposed.

It’s true that in my previous post I forgot that you also change the area of the two cones A1 is not necessarily equal to A2. Fortunately this is easy to repair: The correct formula is

V1-V2=A1*average height1 – A2*average height2

As you see, this does not change the fact that you underestimate the difference between the volumes by a factor of 3.

Your main argument is that in the paper we are discussing the volume is calculated as area * average height, and that this formula might not be valid for the actual shape of the ice. Then you proceded to give an example of a possible shape (the cone) where the formula would not be true. But the example fails, because it is also true for a cone that volume = area * average height.

It seems that now you argue that the ice has some different shape, and that this for some reason leads to different results on its volume. But as a matter of fact, the formula is true in great generality. I dare you to come up with an even distantly possible shape of the ice which does not satisfy that volume = area * average height!

24. DavidB says:

I was intrigued by the (Emperor) penguin puzzle, so I checked out the Wikipedia entry in search of clues. According to this:

1. The penguins begin mating around March (i.e. in late summer), and in March-April walk to their breeding sites between 50 and 120 kilometers from the edge of the pack ice

2. Although not at the edge of the pack ice, most of the breeding sites are still on pack ice rather than land

3. They lay their eggs in May-June. The females go off to feed while the males look after the eggs

4. The eggs hatch in July-August. Both parents take turns to feed them for several months, which necessitates long treks to the sea. The distance to the sea gradually diminishes as the pack ice melts with the approach of summer.

5. Finally, both adults and young (which now have their adult plumage) walk the considerably shorter distance to the sea in December.

It seems evident from this that the main constraint on the breeding cycle is that they must lay their eggs in a location that will not melt into the sea before the young are hatched and able to walk (or swim in icy waters). However, I don’t understand why they have to walk such a long distance from the sea before breeding. Wouldn’t the Antarctic pack ice be close to its minimum extent by April? I wonder if the Wiki article is accurate on this point. It seems inconsistent that the breeding sites would be closer to the sea in December (mid-summer) than in March-April, after a full summer of melting.

25. Skepshasa says:

Did you mean to say in Figure 1 “…after one year’s ice loss” and NOT “…after once year’s ice loss”? I like the figure but I can’t stand it with the grammar…

26. 1 forest 1 says:

willis

I have been a long time reader of this site and i simply believe you need to turn it down a notch

27. Gail Combs says:

Willis, could you or Steve do a and overlay or side by side using figure 2 and perhaps 2009 or 2007 maximum Ice loss extent so we can do an eyeball comparison on your SWAG. Even better the May 2009 thickness vs the September 2009 extent.

For example here is the September 15th 2009 Ice extent: http://arctic.atmos.uiuc.edu/cryosphere/IMAGES/ARCHIVE/20090915.jpg

An eyeball comparison shows all the purple and blue areas on figure 2 are gone. The ice that is left is the green ice. The compacted ice (red) along the shore lines is also gone except that of Greenland’s. Therefore if you were to use a “constant ice thickness” 1.5 or 1.75 meters would be a better guesstimate.

Since 1 centimeter = 0.01 meters the 273 cm thick they have assumed is 2.73 meters.

And yes I am aware of sublimation and melting from the bottom due to “warmer” arctic waters. However the constant still looks about a half to a full meter too high and that would lead to a very high number in the final calculations.

28. dr.bill says:

re mb: May 30, 2010 at 3:11 am
and Willis Eschenbach: May 30, 2010 at 4:01 am

mb, as Willis has pointed out, you’re making an elementary mistake in geometry by dealing with ‘averages’. Just calculate the before and after cone volumes directly from V = AH/3 and subtract them. This gives: ΔV = (10,532 – 10,829) = -297km^3, as per Willis’ calculations.

/dr.bill

29. beng says:

Ring around the rosie, a pocket full of spheres. Thought you’d outfox me. Well……….

30. d says:

the polar bear circle graphic is very confusing

31. Why do we always have to keep proving to the smug, the-science-is-settled crowd that the earth is not flat? I suppose every few centuries or so it takes a brave, question authority, set of pioneers to rock the boat.

32. 899;
David,
Is it your presumption that seals never take birds from the shore, and that they’ve never done so?>>

My presumption is that an an animal whose main diet is fish, krill, squid and penguins hunts mostly in the ocean and being ungainly on land, rarely pursues prey in that fashion. Brief reading reveals descriptions such as “When hunting penguins, the leopard seal patrols the waters near the edges of the ice, almost completely submerged, waiting for the birds to enter the ocean.” I came across no references regarding hunting of penguins on land or ice.

899
I will think that the birds have a ‘species memory’ regarding that matter of where to not make a home.>>

Think what you want but many of species return repeatedly to the breeding grounds of their birth and this appears to be one of them.

899
Alligators and crocodiles have been known to take prey on dry land, even though they are most fitted to do so in the water, or at water’s edge.>>

Crocodiles in particular are very well fitted to taking prey on land. But they only venture a small distance from water and then only when very hungry. By small distance I’m talking tens of meters not one hundred + kilometers. If the penguins were using the ice expanse to protect their nesting grounds from predation they certainly wouldn’t need 100 + kilometers of it. Or even 10. Or 1. The fact that they can accomplish 100km+ it but only by stretching their physical resources to the absolute limit just to arrive at their food source at maximum vulnerability to their chief predator suggests that it is not natural for them. They evolved with a very narrow ice expanse that was easily crossed and which has built up naturaly over time. I doubt you can propose an evolutionary process that is logical and arrives at the same result.

The reasonable conclusions is that the nesting grounds were established with very narrow ice expanse in place, suggesting that the current ice extent is large by comparison and, given that their nesting grounds are near shore line, the much lower ice extent did not translate to much higher sea levels.

33. Wren says:

The assumption the ice area’s thickness looks like a cone, being highest in the center and tapering toward the ages, seems to conflict with the animation in Steve Goddard’s May 28 article “The great 2007 ice crunch – it wasn’t just melt.”

34. This is sort of OT, but I’d like a few comments about mathematical symbols.
Willis wrote:

V = 1/3 * A * h

If I had written this in grade school math class, my teacher would have corrected my use of asterisk (*) instead of the proper times (×). Skipping several years ahead, ASCII-1968 (okay, it was brand new, but that’s okay) doesn’t have a times sign, nor do typewriters, so programming languages used * instead, and we all found it quite natural. Some used uparrows for exponentiation, e.g. x↑ which worked great until ASCII changed it to be carat, ^, but that worked great until C used it for XOR, which was not so great. Keypunches didn’t have that symbol, so several languages used **, e.g. x**2, which works pretty well.

At any rate, here’s my question. How many people see something like x * y and don’t immediately read it as multiplication? For that matter, how many kids are being taught in school that * is the times operator and × is just an antique that only grandparents use?

Then there’s always ¢ which our typewriter did have, but has never made it into common computer usage. Blame that on inflation, just my 2 ¢, now worth $0.25. 35. AnonyMoose says: Suggestion: Add a vertical line at the ends of the triangle to show the volume (cross-section of the volume) which had to melt to decrease the covered surface area. Maybe color that purple, and color purple the small rectangles which represent the amount of melt needed in the slab model. The difference between the small purple triangles and the end purple rectangles is the difference between the ice retreat in the two models. 36. Justthinkin says: Willis…the ice is melting(which tends to happen when temps get above 0C)May 29th,2010…..Inuvik…Northwest Territories…+29C…Edmonton…Alberta…at -2C…28 hours of snow…but is it weather?Hey.At least I can still ski in Banff! 37. mb says: Willis Eschenbach says> (Remember, they have Antarctic ice shelves dropping a quarter kilometre in thickness …) Yeah, you keep saying that. But do you have any real evidence or reason to doubt this? I understand that you are concerned about the Venable ice shelf, a small glacier on the Antarctic coast, and that the error estimate they give is large. Acording to “Recent Antarctic ice mass loss from radar interferometry and regional climate modelling” ERIC RIGNOT et al. doi:10.1038/ngeo102, “The strong, widespread correlation between ice thinning and ice velocity (>50myr−1), for example, on the Berg, Ferrigno, Venable, Pine Island, Thwaites, Smith and Getz glaciers, indicates that thinning is caused by the velocity of glaciers being well above that required to maintain mass balance, that is, ice stretches longitudinally, which causes it to thin vertically.” This makes sense to me, and it would explain extreme thinning. Any comments? 38. bruce says: damn, Greenland is still covered in white. 39. juanita says: Oh, for cripesakes Gallier2, watch the coordinates, I almost materialized inside of a wall. Did you know, that there used to be so many grizzlies around here, the natives avoided Butte Creek. And my grandfather, who was born here about 115 years ago, told me the Sacramento River would get so full of salmon during the run seasons, the river would turn to foam. Okay pardon me if I’ve told this tired story before. Blame it on Anthony for using a favorite catch phrase to head the post. Over a thousand years ago, the Owens River Valley was full of lakes and swamps and rivers and little creeks fed by enormous glaciers that capped the mountains. There was a virtual forest of cottonwoods and aspen and all those water loving trees. Thousands and thousands of people lived there. We saw a popular chipping ground right after a rain – the ground, as far as the eye could see, glittered with obsidian chips from hunters making arrow heads. There’s rock carvings estimated at over 5,000 years old – the Paiute who live there now say the carvings where there when they came to the valley from Nevada thousands of years ago. Now, of course, everybody knows, water is so scarce in the Owens River Valley, people might shoot you for it. When you look to the east, you can just see the little remnants of the great glaciers, all melted. Those glaciers were formed by the Great Uplift, when water was trapped there, and frozen in the cold. As the years went on, thousands of years, those glaciers melted and formed the lakes and streams. As the glaciers disappeared, the water disappeared. The “old people” (that’s what the Paiute call them) disappeared. The Paiute lived sparingly, watching the trees disappear and the land dry up. Then the white settlers came in and used what was left – compared to what had been there, it was a drop in the bucket. I’m not saying water transfers are okay, I think it’s wrong to take water miles from it’s source to support badly planned subdivisions. But Mono Lake was only a couple of hundred years from drying up on it’s own, the Angelinos just hastened Nature’s process. Earth has it’s plan. And remember what Rick says in Casa Blanca – let me paraphrase – the problems of puny people don’t amount to a hill of beans. We hurt ourselves, and we hurt our fellow creatures, but Earth Abides (great book by the way, George Stewart, probably the dustiest book on your local library shelf). Now, you kids play nice, I’m off to a hockey game. 40. Ed Caryl says: 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. 41. Mike Davis says: Willis: I always thought SWAG 95% CI was +/- 50% When in doubt, Flip a coin! 42. Ian L. McQueen says: Is it safe to assume that melting is uniform over the entire surface of the cone? I’d expect there to be a difference of temperature from edge to center (warmer farther south) so that melting would be more rapid at the edges. Then there is the possible complication of melting from beneath due to water temperature….. IanM 43. gallier2 says: @juanita huh? 44. Mike Davis says: 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. 45. rw says: Although I’m not a climatology buff, it’s fascinating to read these comment threads. Among other things, I see the same tricks tried again and again by those trying to discredit the arguments of the day’s posting. In this case (by noiv), finding some data/figures/whatever that appear to contradict the assertions of the posting, but are often irrelevant. It doesn’t seem to occur to these people that by repeated tricks of this sort, they “give the game away”. 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. 46. Jbar says: Let’s look at the big picture for a second –> 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.) 47. mb says: dr.bill says: mb, as Willis has pointed out, you’re making an elementary mistake in geometry by dealing with ‘averages’. ——— No, I’m not. ———- dr.bill futher says: Just calculate the before and after cone volumes directly from V = AH/3 and subtract them. This gives: ΔV = (10,532 – 10,829) = -297km^3, as per Willis’ calculations. /dr.bill ———- The H you are using here is not the average height of the cone. It is the height of the top point. The average height (thickness) of the cone is not H, it is H/3. When we are dealing with ice sheets, we are measuring the average thickness of the ice sheet, not some abstract, model produced H. If we insist on making the model assumption that the ice forms a circular cone, the height of this cone is three times the average thickness of the cone. 48. @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? 49. Ben says: 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 50. jorgekafkazar says: 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. 51. 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. 52. Brian H says: 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. 53. Brian H says: ” 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. 54. Flask says: 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. 55. Gary Pearse says: 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. 56. 899 says: 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. 57. Willis Eschenbach says: 1 forest 1 says: May 30, 2010 at 5:12 am willis I have been a long time reader of this site and i simply believe you need to turn it down a notch 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? 58. 899 says: 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? 59. Brian H says: ” 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. 60. Willis Eschenbach says: mb says: May 30, 2010 at 4:44 am mb, thanks for your comments. You say (emphasis mine): It’s true that in my previous post I forgot that you also change the area of the two cones A1 is not necessarily equal to A2. Fortunately this is easy to repair: The correct formula is V1-V2=A1*average height1 – A2*average height2 As you see, this does not change the fact that you underestimate the difference between the volumes by a factor of 3. Your main argument is that in the paper we are discussing the volume is calculated as area * average height, and that this formula might not be valid for the actual shape of the ice. Then you proceded to give an example of a possible shape (the cone) where the formula would not be true. But the example fails, because it is also true for a cone that volume = area * average height. 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: It seems that now you argue that the ice has some different shape, and that this for some reason leads to different results on its volume. But as a matter of fact, the formula is true in great generality. I dare you to come up with an even distantly possible shape of the ice which does not satisfy that volume = area * average height! What you say is true. But you are making the same mistake that they made. The undeniable fact that volume = area * average height does not mean that change in volume = change in area * average height w. 61. Brian H says: ” 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%. 62. John McManus says: 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? 63. losemal says: 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. 64. Doug in Seattle says: 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? 65. harrywr2 says: 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. 66. mb says: 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). 67. Willis Eschenbach says: mb says: May 30, 2010 at 6:57 am Willis Eschenbach says> (Remember, they have Antarctic ice shelves dropping a quarter kilometre in thickness …) Yeah, you keep saying that. But do you have any real evidence or reason to doubt this? I understand that you are concerned about the Venable ice shelf, a small glacier on the Antarctic coast, and that the error estimate they give is large. 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? Acording to “Recent Antarctic ice mass loss from radar interferometry and regional climate modelling” ERIC RIGNOT et al. doi:10.1038/ngeo102, “The strong, widespread correlation between ice thinning and ice velocity (>50myr−1), for example, on the Berg, Ferrigno, Venable, Pine Island, Thwaites, Smith and Getz glaciers, indicates that thinning is caused by the velocity of glaciers being well above that required to maintain mass balance, that is, ice stretches longitudinally, which causes it to thin vertically.” This makes sense to me, and it would explain extreme thinning. Any comments? 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. 68. kadaka (KD Knoebel) says: 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. ;-) 69. D says: 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 70. Willis Eschenbach says: 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. 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 … 71. Willis Eschenbach says: losemal says: May 30, 2010 at 10:15 am (Edit) 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. 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. 72. dr.bill says: 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 73. Willis Eschenbach says: mb says: May 30, 2010 at 10:42 am 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. … 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: The undeniable fact that volume = area * average height does not mean that change in volume = change in area * average height 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: V1 = 1/3 A h For one cone we have V1 = 1/3 * 11,900,000 * 273 / 100,000 = 10,829 For the other we have V2 = 1/3 * 11,789,000 * 268 / 100,000 = 10,532 Difference = 297 w. 74. Michael Larkin says: 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. 75. George Turner says: 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. :) 76. mb says: 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. 77. wayne says: 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? 78. Michael Larkin says: 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. 79. @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! 80. Willis Eschenbach says: mb says: May 30, 2010 at 11:59 am (Edit) Willis Eschenbach > 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. 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. 81. Michael Larkin says: “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?” 82. Michael Larkin says: I’m glad you spotted what mb was getting wrong, Willis; I was absolutely clueless! 83. DesertYote says: 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 84. gilbert says: 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. 85. R. Craigen says: 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! 86. losemal says: 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. 87. R. Craigen says: 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? 88. DirkH says: “R. Craigen says: […]one only needs to[…]Any programmers out there up to this challenge?” You’re a project manager, right? 89. wayne says: 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). 90. Dave H says: @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. 91. mb says: 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. 92. R. Craigen says: 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.

93. Smokey says:

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?

94. 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.

95. scott says:

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_.

96. scott says:

Ah, no img tags, I see; try this:

shasta lake

97. Brian H says:

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.

98. jason says:

Oh goodie another post about arctic sea ice…

/sarc

Surely the battlefield is bigger than this guys?

99. 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. 100. Willis Eschenbach says: mb says: May 30, 2010 at 1:48 pm 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? 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. —————- 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. You really need to learn to read. I don’t insist that the cone is 273 cm tall. I have not flipflopped. 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. 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. 101. Willis Eschenbach says: Dave H says: May 30, 2010 at 1:33 pm @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. 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: 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. I hope that makes it clear. I have posted an update to the main post indicating this change. 102. mb says: 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? 103. Mike Edwards says: 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. 104. jorgekafkazar says: “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 105. jorgekafkazar says: 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. 106. Willis Eschenbach says: RACookPE1978 says: May 30, 2010 at 2:58 pm 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. … [much fascinating stuff snipped]

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”.

107. sod says:

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.

108. John Baltutis says:

Ric Werme says:
May 30, 2010 at 6:51 am

This is sort of OT, but…here’s my question. How many people see something like x * y and don’t immediately read it as multiplication? For that matter, how many kids are being taught in school that * is the times operator and × is just an antique that only grandparents use?
Then there’s always ¢ which our typewriter did have, but has never made it into common computer usage. Blame that on inflation, just my 2 ¢, now worth \$0.25.

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.

109. Gail Combs says:

Jbar says:
May 30, 2010 at 7:47 am

“Let’s look at the big picture for a second –>

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.

110. Jbar says:

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.

In that context, why continue with this thread?

111. John Baltutis says:

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.

112. ck says:

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.)

113. Phil. says:

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

114. “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.

115. 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!

116. Willis Eschenbach says:

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.

117. Willis Eschenbach says:

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.

118. Bob_FJ says:

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.

119. Willis Eschenbach says:

Nick Stokes says:
May 30, 2010 at 4:57 pm (

“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.

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.

120. Ian H says:

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.

121. DesertYote says:

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!

122. Phil. says:

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.

123. kadaka (KD Knoebel) says:

From the Update in the article above:

Using this value, I get arctic ice loss of 344 km^3, and a global ice loss of 239 km^3.

Now things are getting interesting. The paper was providing a rate of loss for the previous decade. The terrifying PIOMAS Arctic Sea Ice Volume Anomaly chart gives 340 km^3/yr from 1979 to present, using a straight line incorporating (of course) the last decade. So basically your simple conical model is providing the same result as the “robust” PIOMAS computer model, with only mild ignoring of the time baseline difference. ;-)

Then things break down into discussing area vs extent, do the area numbers used represent only 100% solid ice or can they be less and to what concentration, etc. And if less than 100%, from there the real volume loss total and rate will likely go down from the estimates.

Yup, quite interesting.

124. Willis Eschenbach says:

Nick Stokes says:
May 30, 2010 at 4:57 pm

… The result is that you still have far less ice volume than Shepherd uses.

OK, here’s another way to look at it.

The PIOMAS ice model says that at the height of winter, the ice volume is 28,600 km^3. The NASA submarine model says that the ice is on the order of 4.75 metres at the center during the winter. The average peak ice area per Cryosphere Today is 14,500,000 square km.

Using my estimation of a conical ice mass, we get

Volume = 1/3 * Area * height

= 1/3 * 14,500,000 km^2 * 4.75 metres / (1000 m/km)

= ~ 23,000 cubic m

This is about 20% lower than the modeled value, which is pretty reasonable agreement, within the SWAG 95%CI of ±25%.

Now let’s use the S2010 figures. They don’t give winter volumes, so we’ll use their averages. The PIOMAS info also gives an average annual volume of 21,500 cu km. The volume assumed by S2010, on the other hand, is 273 cm * 11,900,000 sq. km. avg. area / (10,000 cm/km) = 32,500 cu km, about 50% greater than the PIOMAS modeled value … so you are right, Nick, I do have far less ice volume than S2010. And so does PIOMAS … for whatever that’s worth.

125. dr.bill says:

re: Willis Eschenbach: May 30, 2010 at 5:46 pm

I know it’s been a long day and all, but I’m afraid you’re starting to get a bit sloppy. By my calculations (using your new and improved peak height of 4 meters), that magnification factor on the 4th line of your chart should be 12,000, not 10,000.

Tut, tut. ☺☺☺

/dr.bill

126. 899 says:

gilbert says:
May 30, 2010 at 1:01 pm
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.
*
*
It would appear the powers that be had this all figured out at the time.

Remember: The unwashed masses are largely uneducated about matters of history, and so the ‘shamans’ and high priests of the temple of science are able to take advantage of them mercilessly.

Recall from history where not a few times a solar eclipse had been used to scare a people into submission. The ones with the knowledge used it to obtain what they could not otherwise. Solar eclipses happen infrequently enough such that there is no cultural memory for those who don’t record history.

How many people are familiar with the changes in relative climate over time? Most people succumb to the simple notion of the seasons and live with that thinking that not much will change for as long as they live. They essentially live a ‘linear’ existence.

Then along comes Mann-made climate change with the temple high priests making all manner of wild prognostications declaring that should the people not succumb to the temple decrees, then ill fortune shall befall the lot of us.

Remember: Mann-Gore-Pig and company knew they had only a small window of opportunity in which to achieve their master’s goals, before the weather turned down and made them out to be snake oil salesmen and charlatans of the worst sort.

And if the Sol keeps his activity at a consistent low, the truth will hurt, but not nearly as much as would the Mann-Gore-Pig lie: The infliction of misery upon the lot of humanity at the behest of the few intent upon turning us into slaves, to be taxed and starved to death.

127. Willis,
“about 50% greater than the PIOMAS modeled value”
No, S2010 was a wintertime figure
“the 1993-2001 (average wintertime) thickness of Arctic sea ice was estimated to be 273 cm”
So they agree well (32.5 in 1993-2001 vs 28.6 now), and yours is the one out of line.

And your PIOMASS trend that you claim as vindication was from 1979 to 2010. The S2010 figure was for the period 1994-2007. It’s clear from the PIOMASS plot that decline in the S2010 time frame was much greater.

128. 1 forest 1 says:
willis
I have been a long time reader of this site and i simply believe you need to turn it down a notch

What does that mean? Sorry, I have NO respect for anyone who posts fact- or evidence-free criticism. It looks to me like mischief-making, leaving a bad taste for the victim at no cost of producing any substance on your part. Hiding behind an alias while you behave badly doesn’t say much for you either.

129. From Willis, above.

“OK, here’s another way to look at it.

The PIOMAS ice model says that at the height of winter, the ice volume is 28,600 km^3. The NASA submarine model says that the ice is on the order of 4.75 metres at the center during the winter. The average peak ice area per Cryosphere Today is 14,500,000 square km.

Using my estimation of a conical ice mass, we get

Volume = 1/3 * Area * height

= 1/3 * 14,500,000 km^2 * 4.75 metres / (1000 m/km)

= ~ 23,ooo cubic m”

—…—…

“OK, here’s another another way to look at it. 8<)

Assume, from the Navy data, a cylinder of thickness 2 m at the edge, steadily growing in thickness towards the center to a max depth of 4.75 meter. This is conservative (understates the volume) because the Arctic’s “real ice” has a very large “flat” area around the center.
Vol_cylinder = A*t_ave
Vol_cone = A*1/3*t_center
Vol_total of combined conic cylinder = A*(t_edge+1/3*(t_center-t_edge))

Playing the Mission Impossible theme from a cheap tape deck …. “Your mission, Jim, should you decide to accept it, is to determine the difference in “calculated volume loss” for three models (simple flat plate (cylinder), simple cone, and combined conic cylinder) given:

1) a loss of area of 1×10^6 km2
2) a loss of radius of 200 km
3) a loss of thickness of 3/4 meter across the whole surface
4) a loss of radius of 200 km PLUS a loss of thickness of 3/4 meter.

Extra credit assignment: For a ASSUMED volume change of 5,000,000 km^3, determine the final radius and thickness of the three models.

Extra extra credit assignment: For the actual Navy thickness data above for summer and winter, determine the apparent volume loss.

The result is to show how that, despite the apparent “equal” models that all can be displayed as a flat line from a distance, the resultant volume change between apparently very similar models IS very important. And why “modeled” DATA is worthless if the assumptions behind the MATH of the model are not revealed by the “scientist” trying to conceal his assumptions while writing scary headlines in a press release not yet peer-reviewed or printed!

And the 2010 paper being discussed uses a simplistic flat model.

And a simple average thickness model is also, quite simply, dead wrong when 1.3 trillion dollars, and the destruction of the world’s economies by needlessly throttling energy production and the health of billions, is at state.

130. Dr. Dave says:

mb says:
May 30, 2010 at 7:48 am

The H you are using here is not the average height of the cone. It is the height of the top point. The average height (thickness) of the cone is not H, it is H/3.

When we are dealing with ice sheets, we are measuring the average thickness of the ice sheet, not some abstract, model produced H. If we insist on making the model assumption that the ice forms a circular cone, the height of this cone is three times the average thickness of the cone.
=============
mb, you must have received your math training at a government school from a teacher who learned his or her math from a methods book. I can think of no other reason for the above. The volume of a cone is exactly one-third of the volume of a cylinder that has the same base area and the same height as the cone. H/3 is not the average height of the cone. The fraction, 1/3, in the equation for a cone, V = 1/3*A*H comes from the calculus (by the disk method or by the solid of revolution method), it is most emphatically not the “Average Height.”

Sorry about that, folks. I am sure someone has already pointed this out. But after reading MB’s mistake three times, I could not help myself.

Back to lurking! :-)

131. 899 says:

Jbar says:
May 30, 2010 at 4:13 pm
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.

Ya know? If you repeat that 100 times whilst rubbing your tummy in clockwise direction, patting the top of your head and turning in an anti-clockwise direction, one of two things will happen:
[A] You’ll achieve nirvana.
or
[B] You’ll prove that propaganda actually works.

132. R. Craigen says:

Dr. Dave, you’re quite right that the (1/3) in the volume of a cone can be arrived at by an integral (this is not Archimedes’ approach but I digress). However, this says nothing about whether it is the “average height”; mb happens to be right on this point.

Nothing in your comment suggests you have any particular meaning in mind when you say “average height of a cone”. How do you define this? I do teach calculus, and this is an elementary fact, but one can say nothing about whether this or that is “the average” of something until we know what that statement means.

Now there are many kinds of averages and one could come up with some exotic ones here, yadda, yadda. But in point of fact there is only one that serves any useful purpose in this discussion, which happens to coincide with the definition you see in a second-term course in integration:

For a function of one variable (a curve) integrate from a to b and divide by the width b-a of the interval of integration. This gives you an “average height” h of the region such that the rectangle of this height on [a,b] has area h(b-a), which is the same as the area under the curve.

For a function of two variables (a surface above a planar region of area A) integrate over the region and divide by A. The value, h, so obtained is such that a cylinder (more correctly, prism) of height h above the region has the same volume, Ah, as the solid bounded above by the surface.

Since the volume of the cone is (1/3) AH, dividing by A gives average height h=H/3. Thus the cone of height H over A has equal volume with the cylinder of height h=H/3 over the same region.

Sorry, all, for the maths lesson — I think this whole discussion, in any case, has degenerated too badly to be of value. Willis, perhaps you’ve better things to do with your time. Personally I think the data in this paper is flawed and simply doesn’t jive with what we observe in the sea ice. (Currently, for example, global extent is above the 1979-2000 mean and thickness, despite poorly obtained and processed data, looks perfectly “healthy”, whatever that means.) The specifics of the calculations don’t make much difference to me beyond that. You have established that their approach is pretty naive, but all we’ve accomplished beyond that is battling over minutae of how calculations are performed.

133. Lots of great comments here. My two cents is: Something like arctic ice is not unlike a large, okay very, very large tabular ore body. You are absolutely correct one can not use a uniform thickness for something that is not uniform. One should take the elevation of the surface and subtract the elevation of the base and a sufficient number of points, distributed across the entire surface and using standard mapping techniques, a number exist, to calculate an “accurate” volume. When you are looking at such a large area that becomes more difficult. If you start with contour maps of thickness the space the isolines reasonably tight the calculating the volumes between each isoline and adding them up can give a very good estimate, depending on how good the isolines are of course.

134. readingthepapers says:

Can we take for granted that if A is the surface area covered by ice and H is its average thickness then the volume of the ice is V = H*A? This just requires that we understand “average” to mean the average with respect to surface area and taken over the surface covered by ice. Adding a “1” or a “2” to indicate “before” and “after” then the change in volume is V2-V1 = H2*A2-H1*A1. Let’s write dV=V2-V1, dH=H2-H1, dA=A2-A1; then dV = dH*A1 + H1*dA + dH*dA. If the changes are small then, as we learned on the way to Calculus, the term dH*dA is insignificant; but let’s not go there; instead, introduce H=(H1+H2)/2 and A=(A1+A2)/2, then we have
(Eq. 1:) dV = dH*A+H*dA.

So far, so good? And now, supposedly, Shepherd et al. measured the average thickness of the ice and used accepted values for the surface area covered by ice, and they did so “before” and “after” and then they calculated dV=dH*A+H*dA. It is hard to find fault with that. If a picture of a cone is persuading us that the formula is somehow wrong or should be applied differently then that tells us to be wary of the picture; presumably some symbol in the picture is not having its proper algebraic meaning. I think that mb has pointed out quite clearly already that the delta-H of the cone is not the change in average “thickness”, it is the change in the height, which is three times larger.

And yet, there is something to this conversation. Suppose we introduce a fixed surface encompassing the surface of ice both “before” and “after”. Let’s say it has area A0, which is therefore at least as large as A1 and A2. Where ice is absent we assign it a “thickness” 0, and again we can evaluate the average thickness, now averaged over the larger fixed surface, and let’s call this average thickness G. We use G1 and G2 to indicate before and after and dG=G2-G1. Now the volumes are V1=G1*A0 and V2=G2*A0, and dV=dG*A0. Better give that a label:
(Eq. 2:) dV = dG*A0.

Same left hand side, dV, in Eqs. 1 and 2; different expression on the right hand side; both are correct. If one is measuring dH then it is a good idea to use Eq. 1, and if one is measuring dG then one should rather use Eq. 2. The picture of the cone and the associated computations serve as a warning that one may be wrong by a factor 3 or so if one uses the inappropriate equation.
Now, finally, concerning changes in the thickness of the ice sheet, I would think it a lot more straightforward to measure dG than to measure dH. To measure dG I take an appropriate set of sample locations and measure the local change in thickness (using a thickness of 0 where there is no ice), then do the averaging (suitably weighted if the sample points weren’t suitably uniform). To measure dH we pretty much have to calculate the volume along the way and then evaluate the difference between two volumes. I would rather be measuring differences between local thicknesses and do the averaging afterwards, so obtaining dG instead of dH. But if Shepherd et al. measured changes in average thickness, dH, and then applied Eq. 1, good for them.

135. Willis Eschenbach says:

Nick Stokes says:
May 30, 2010 at 6:47 pm

Willis,
“about 50% greater than the PIOMAS modeled value”
No, S2010 was a wintertime figure
“the 1993-2001 (average wintertime) thickness of Arctic sea ice was estimated to be 273 cm”
So they agree well (32.5 in 1993-2001 vs 28.6 now), and yours is the one out of line.

Nick, thanks for that. First, the 28.6 from PIOMAS is not “now”, it is the average.

However, S2010 are doing something curious. They are using the average wintertime thickness, but they don’t give a time frame for the area (winter? summer?). I don’t understand why. It is this kind of thing that makes comparisons difficult.

In any case, using S2010 wintertime thickness and S2010 who knows when area, they get a number which is larger than the wintertime PIOMAS volume. If we use peak wintertime area (~ 14 million km^2) and the S2010 winter thickness figure, we get 38,000 cubic km. In both cases, they are much larger than the PIOMAS figure. And unfortunately, they give no definition of “winter”.

And your PIOMASS trend that you claim as vindication was from 1979 to 2010. The S2010 figure was for the period 1994-2007. It’s clear from the PIOMASS plot that decline in the S2010 time frame was much greater.

Nope. My PIOMASS trend was from 1994-2007. I had said:

The PIOMAS value for that period [1994-2007] is … drumroll please … 426 cu km/yr. (As it is a model result, there is no CI.)

The PIOMAS trend for the entire period is given here, and it is 340 km^3/yr.

I didn’t just squint at the plot and guess. I calculated the value from the actual data.

136. Willis Eschenbach says:

Jbar says:
May 30, 2010 at 4:13 pm

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.

I’m sorry, Jbar, but that’s just AGW propaganda.

Big picture. Global sea ice area is right now at the 1979-2008 average value, with no long-term decline at all. Look at the red line in my Figure 3 here, go to the “Cryosphere Today” web site for more information.

137. 1 forest 1 says:

willis

how about less childish titles for your future Articles
might be a good place to start :-)

138. Bob_FJ says:

Willis,
Further to my comment above, and to yours somewhat akin that crossed but 6 minutes later; perhaps I should elaborate with some numbers on mine;
Using your Fig.1, the base radius of your concept cone is 1946 Km at baseline year, and 1937 km at year 2. (Winter loss per year). If we use your refined cone height of 4m, the tangent for the gradient at baseline year would be: Tan = 4/ 1,000 * 1,946 = ~0.00000206

The annual winter reduction of base radius is 1,946 – 1937 = 9 km or 9,000 metres.

The uniform reduction in thickness T in metres to amount to that would be:
T = 9,000 * Tan, or T = 9,000 * 0.00000206 = ~0.019m or 1.9 cm

I think this confirms that the concept of uniform thickness loss, (apex to perimeter), is wrong, because it is unreasonable to imagine that 9 Km of sea-ice could totally retreat by virtue of a mere 1.9 cm, (or about ¾ of an inch), of thickness reduction.

I previously put forward some reasons in my link above, as to why sea-ice loss around the periphery is evidently much greater than inland towards the colder apex of much greater thermal inertia.

An additional complication in all this is that sea ice area is not the same as sea ice extent, which in some source definitions may be as low as 15% of the sea area, or as high as 100%! Ho hum!

BTW, per “National Geographic”:
July is the North Pole’s warmest month, when the mean temperature rises to a freezing 32 degrees Fahrenheit (0 degrees Celsius). (not much melting going on, even if there were an effective downhill for it to run)

Willis, let me congratulate you on your very good thought starter!

139. Willis Eschenbach says:

ck says:
May 30, 2010 at 4:18 pm

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.)

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.

Their estimate (2.73 m) comes from here. It is worth noting that the area that paper measured is 3.1 million km^2 … while the area of the winter ice is about four times that. So there is a lot of estimation going on. Which is not a bad thing … but you have to allow that their 2.73 figure may not be representative of the full area in either time or space.

It is also worth noting that the difference between the submarine and satellite measurements for the same area often differ by up to two metres, and that for the 1997 time series, the submarine and satellite datasets are strongly negatively correlated (slope = -.81, r^2=0.29, p=2e-9). So it is far from an exact science.

That reference also says an interesting thing:

Our results show that errors are present in current simulations of Arctic sea ice, either in the radiative forcing, or in the parameterization of the effect of surface melt on the absorption of shortwave radiation. The observed variability of Arctic sea ice thickness, which shows that the sea ice mass can change by up to 16% within one year, contrasts with the concept of a slowly dwindling ice pack, produced by greenhouse warming. This variability must be taken into account when determining the significance of trends derived from intermittent submarine measurements of ice draught.

Words of wisdom …

140. Ian H says:

I’m sorry Willis, but if a distribution has area A and average depth h then V = Ah regardless of what cross section you use. You could use a cylindrical cross section or a conical cross section or a cross section shaped like one of Angelina Jolie’s breasts and you’d still have exactly the same relationship between the three quantities. It is true by definition assuming that the words “Area”, “Volume” and “Average depth” are used normally. The only reason you got a different result is because your model didn’t have the correct average depth.

There is absolutely no error that I can see in the original calculation. In particular they made no assumptions about the shape of the ice cap in writing V=Ah. Not to put too fine a point on it, you’ve simply made a mistake. How about owning up to the error. It would be the honorable thing to do.

The peer review process has many faults. It also has a few advantages. One of these is being able to make a fool of yourself in front of only one or two anonymous peers who will then gently point out your error and stop you before you can make a fool of yourself in front of the entire world. As I think you now appreciate, this can be a very good thing.

141. Geoff Sherrington says:

Permit a quick whip to the Antarctic. Here are some figures:

“The site is at an elevation of approximately 2850 meters (9300 ft), most of which is measured in ice thickness. The mean annual temperature is -49.3 degrees C (-56.7 F). The lowest recorded temperature is -82.8 degrees C (-117.0 F) and the highest temperature is -13.6 C (+7.5 F).

An extremely arid environment limits annual snowfall; however, a relatively constant wind speed of 5-15 knots compounds the accumulation and accounts for the heavy snow drifting common to inland antarctic stations. The surrounding terrain is completely flat, featureless snow.
http://quest.nasa.gov/antarctica/background/NSF/sp-stay.html

Now, I’m interested in that word “snowfall”, while admitting that I’m a tropical person.

Seems to me that the accumulation at the South Pole could be largely particles of ice broken off the surroundings and swirled around by the frequent winds. It’s 1,300 km to the nearest sea at present, so the question arises about the source and fate of oxygen isotopes. The hypothesis here is that the “snow” accumulation carries mostly a mixture of isotopes that happened to be residing on the ice plateau before being blown to a fixed place at the Pole. Ice has not melted at the Pole, it’s too cold, so classical snow cannot form there. How far away from the Pole to you have to go to get Arctic style snow conditions? Such fractionation as is postulated for oxygen isotopes as water evaporates from the sorrounding oceans might not be relevant if very little snow comes directly from the sea. This could help explain the lag between water ages in ice bubbles and ice ages as determined by isotopes, a severe, unresolved problem.

Indeed, it places caution on the interpretation of Vostok core (even colder, even higher) as well as South Pole core. Anyone able to shoot down the hypothesis that oxygen isotopes in South Pole and Vostok core are related more to local circumstances of abrasion/relocation/accumulation on the ice plateau and not much to global events?

142. Gail Combs says:

Geoff Sherrington says:
May 31, 2010 at 3:37 am

“Permit a quick whip to the Antarctic. Here are some figures:

…..An extremely arid environment limits annual snowfall; however, a relatively constant wind speed of 5-15 knots compounds the accumulation and accounts for the heavy snow drifting common to inland antarctic stations. The surrounding terrain is completely flat, featureless snow.
http://quest.nasa.gov/antarctica/background/NSF/sp-stay.html

…Seems to me that the accumulation at the South Pole could be largely particles of ice broken off the surroundings and swirled around by the frequent winds. It’s 1,300 km to the nearest sea at present, so the question arises about the source and fate of oxygen isotopes. The hypothesis here is that the “snow” accumulation carries mostly a mixture of isotopes that happened to be residing on the ice plateau before being blown to a fixed place at the Pole. Ice has not melted at the Pole, it’s too cold, so classical snow cannot form there. How far away from the Pole to you have to go to get Arctic style snow conditions? Such fractionation as is postulated for oxygen isotopes as water evaporates from the sorrounding oceans might not be relevant if very little snow comes directly from the sea. This could help explain the lag between water ages in ice bubbles and ice ages as determined by isotopes, a severe, unresolved problem.

Indeed, it places caution on the interpretation of Vostok core (even colder, even higher) as well as South Pole core. Anyone able to shoot down the hypothesis that oxygen isotopes in South Pole and Vostok core are related more to local circumstances of abrasion/relocation/accumulation on the ice plateau and not much to global events?
__________________________________________________________________________

Boy that sure could explain the difference between the Vostok core CO2 measurements and the plant stomata CO2 measurements. They are not measuring what they think they are measuring. The oxygen isotopes from the “snow” and the CO2 from the atmosphere would not necessarily be correlated as thought.

143. Dr. Dave says:
May 30, 2010 at 9:23 pm

The volume of a cone is exactly one-third of the volume of a cylinder that has the same base area and the same height as the cone. H/3 is not the average height of the cone. The fraction, 1/3, in the equation for a cone, V = 1/3*A*H comes from the calculus (by the disk method or by the solid of revolution method), it is most emphatically not the “Average Height.”

The only definition for average height I’m aware of is volume ÷ area, which for cones works out to be H/3.

Please complete your comment by providing the proper definition for “Average Height” and how to compute it.

144. Phil. says:

Dennis Nikols says:
May 30, 2010 at 10:10 pm
Lots of great comments here. My two cents is: Something like arctic ice is not unlike a large, okay very, very large tabular ore body. You are absolutely correct one can not use a uniform thickness for something that is not uniform.

You certainly can, in calculus it’s called the Intermediate Value Theorem.

145. 899 says:

Willis,

Have you seen this?

Quote:
Now on to calculating the volume. That calculation is straightforward :
volume = (A1 * 0.5) + (A2 * 1.5) + (A3 * 2.5) + (A4 * 3.5) + (A5 * 4.5)
Where A1 is the area of ice less than one metre, A2 is the area of ice less than two metres, etc. The 2010/2008 volume ratio came out to 1.24, which means there has been approximately a 25% increase in volume over the last two years. The average thickness has increased from about 2.0 metres to 2.5 metres. That means an extra 20 inches of ice will have to melt this summer. So far, this seems unlikely with the cold Arctic temperatures over the last couple of weeks.

146. 899 says:

Geoff Sherrington says:
May 31, 2010 at 3:37 am
-snip-
Indeed, it places caution on the interpretation of Vostok core (even colder, even higher) as well as South Pole core. Anyone able to shoot down the hypothesis that oxygen isotopes in South Pole and Vostok core are related more to local circumstances of abrasion/relocation/accumulation on the ice plateau and not much to global events?
*
*
Only if you’re willing to say that the Earth’s air always stays in one place and absolutely never moves around.

Last I checked, the atmosphere is in such a state of flux that any such hypothesis as you are wont to remark of would be sheer science fiction fantasy.

147. dr.bill says:

Willis, pardon my presumption, but I would like to commend you on your equanimity in responding to some of the comments you have been fielding.

The semi-pro attack dogs are to be expected, and I’m sure they cause you no problems, but the (possibly) well-meaning ‘forest for trees boneheads’ who completely missed the point of the exercise are another thing altogether.
Your manner of responding to them is an edification.

/dr.bill

148. Jeff in Calgary says:

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.

149. Willis Eschenbach says:

Ian H says:
May 31, 2010 at 2:16 am

I’m sorry Willis, but if a distribution has area A and average depth h then V = Ah regardless of what cross section you use. You could use a cylindrical cross section or a conical cross section or a cross section shaped like one of Angelina Jolie’s breasts and you’d still have exactly the same relationship between the three quantities. It is true by definition assuming that the words “Area”, “Volume” and “Average depth” are used normally. The only reason you got a different result is because your model didn’t have the correct average depth.

The problem is not with average depth. We both agree that

V = A h

The part that you (and the folks who wrote S2010) seem to be missing is that this does not mean that

∆V = ∆A h

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.

There is absolutely no error that I can see in the original calculation. In particular they made no assumptions about the shape of the ice cap in writing V=Ah. Not to put too fine a point on it, you’ve simply made a mistake. How about owning up to the error. It would be the honorable thing to do.

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.

The peer review process has many faults. It also has a few advantages. One of these is being able to make a fool of yourself in front of only one or two anonymous peers who will then gently point out your error and stop you before you can make a fool of yourself in front of the entire world. As I think you now appreciate, this can be a very good thing.

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.

150. Willis Eschenbach says:

Gail Combs says:
May 31, 2010 at 5:01 am …

Gail, I corrected your italics to what I think you meant.

151. Willis Eschenbach says:

dr.bill says:
May 31, 2010 at 7:47 am

Willis, pardon my presumption, but I would like to commend you on your equanimity in responding to some of the comments you have been fielding.

The semi-pro attack dogs are to be expected, and I’m sure they cause you no problems, but the (possibly) well-meaning ‘forest for trees boneheads’ who completely missed the point of the exercise are another thing altogether.
Your manner of responding to them is an edification.

/dr.bill

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 …

152. Willis Eschenbach says:

Jeff in Calgary says:
May 31, 2010 at 10:30 am

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.

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.

153. sod says:

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!

154. sod says:

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.

155. MartinM says:

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.

156. Phil. says:

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!

157. Willis Eschenbach says:

sod says:
May 31, 2010 at 12:47 pm

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.

Shepherd et al. DID NOT USE A CONE. They used a flat slab. This gave them a 35% error in their answer.

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!

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.

158. Willis Eschenbach says:

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 …

159. Phil. says:

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.

160. Bob_FJ says:

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

161. Willis Eschenbach says:

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 …

162. kadaka (KD Knoebel) says:

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.

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.

163. Smokey says:

Bob_FJ,

I question the NSIDC’s unsupportable statement:

Greenhouse gases emitted through human activities and the resulting increase in global mean temperatures are the most likely underlying cause of the sea ice decline…

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.

164. Phil. says:

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.

165. Bob_FJ says:

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)

166. Smokey says:

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.

167. ck says:

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.

168. dr.bill says:

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

169. John Baltutis says:

MartinM says:
May 31, 2010 at 1:40 pm

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%

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%

170. Willis Eschenbach says:

Bob_FJ says:
May 31, 2010 at 5:07 pm (Edit)

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?

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.

171. Willis Eschenbach says:

Phil. says:
May 31, 2010 at 5:05 pm

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.

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.

172. Willis Eschenbach says:

ck says:
May 31, 2010 at 7:10 pm

… 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.

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.

173. Phil. says:

Willis Eschenbach says:
May 31, 2010 at 9:21 pm
Phil. says:
May 31, 2010 at 5:05 pm

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.

Phil, that’s not clear. It didn’t what? Didn’t use a slab? Didn’t give them an error?

Doesn’t give them an error, as explained above the shape of the object makes no difference.

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

Correct answer: -50×11.9-2.73×111= 898km^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.

You get the same answer irrespective of geometry, actually the cone is the less physical as it implies a very thin sheet round the edge, something like a mexican hat would be more physical. However calculus says that if we know the initial area and average thickness and rates of change the geometry doesn’t matter.

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.

No this calculation uses different data and it doesn’t have any relevance to the comparison of two different models. The PIOMAS data shows a slope of ~1000km^3/yr over the last decade.

174. Phil. says:

Willis Eschenbach says:
May 31, 2010 at 9:49 pm
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.

That you divided by 3 when you shouldn’t have, your conical model using the parameters consistent with Shepherd gives 891km^3 i.e. equal to the slab model 898km^3 (allowing for rounding error). As calculus says it should.

175. ck says:

Volume 1: 1/3 * 11,900,000 km^2 * 819 cm / (100000 cm/km) = 32,847 km^3

I assume this is a typo; should be 32,487 km^3.

This model is indeed closer to agreeing with the data, but the average decrease in thickness dh in your model is still smaller than you claim: around 4.2 cm, to be precise. (By making your cone pointier, the slopes of the two cones have become ‘more parallel’, but they’re still not quite parallel, so dh <= 5 cm.)

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.

Even in the update to your original post you still claim to get a decrease of only 344 km^3. For the reasons given above, this cannot be correct if you take the S2010 data (as you do in your computation).

176. Bob_FJ says:

Willis, Reur May 31, 2010 at 9:05 pm

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.

First of all Willis, I’d like to say that as a fellow sceptic, I’ve admired your frequent very active work, but on this occasion, I cannot agree with you. Both Shepherd’s and your models are flawed, maybe explained below?
Meanwhile, returning to your question:

No, I have no data for the “gradient” of melting from periphery to apex, but I’m postulating things like:
1) The north pole region barely reaches the melting point of ice during the hottest month of July, and since you assume this area, (close to your model apex), has the greatest ice depth, it must also have the greatest thermal inertia. Thus, there is logically virtually no surface melting there. (but maybe some other minor losses)
2) On the other hand, at the peripheries, it is very clearly evident that during the summer there is major loss of ice attributed to many factors, for at least the reasons suggested here.
3) The maximum ice coverage in winter, as modelled, is mostly a legacy of what happened in the previous summer.
4) It is a flawed concept to use an average thickness reduction over the radial extent of the area of ice. To elaborate, if your circular model is divided into annuli and the ice loss is actually greater in an outer annulus, then the net effect there is very much greater than for any inner annulus with lesser ice thickness loss and smaller area.
5) Additionally, IF the NSIDC are correct in saying that the 2007 winter recovery resulted in a mean first-year ice depth of 1.6 metres, should your cone of height 4 metres not have a base datum somewhere around 1.6 metres rather than zero? (that may be over simplifying it a bit)
6) I don’t know if Shepherd’s average thickness change is calculated per unit area or radially. Do you?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
There is another way of calculating volume change assuming uniform thickness change t,
by using the surface area of a cone formula: Sloping area = ‘Pie’ * r * s, where s = the slope length. (then multiply by t) If you work that through, you may spot something.

All the best Willis!

177. Willis Eschenbach says:

Phil. says:
May 31, 2010 at 9:59 pm

… Doesn’t give them an error, as explained above the shape of the object makes no difference.

If you trim a certain area from the edges of a slab, you remove more volume than if you trim the same area from the edges of a cone. How is it, then, that the shape makes no difference?

You get the same answer irrespective of geometry, actually the cone is the less physical as it implies a very thin sheet round the edge, something like a mexican hat would be more physical. However calculus says that if we know the initial area and average thickness and rates of change the geometry doesn’t matter.

Phil, you can keep repeating that all you want. Let me give you a physical example. Here are two objects that have the exact same area and the exact same average thickness. They are 10 cm in diameter, so they have a base area of 25 * pi. The cylinder is 2 cm tall, and the cone is 6 cm tall, so they both have a volume of 50 * pi and both have an average thickness of 2 cm. Here they are:

Now, I’ll reduce the area of both of them by the exact same amount. You keep claiming that “geometry” or “calculus” means that if we reduce the area by the same amount, it will reduce the volume by the same amount. Here is that situation:

As you can see, despite my having trimmed the exact same area from each object, and despite the fact that the average thickness of the two objects are identical at 2 cm, neither geometry nor calculus can force these two trimmed objects to have the same volume. Much more volume has been lost from the cylinder. So the shape indeed does matter, it matters very much. As I have said before, the fact that

V = A * average h

doesn’t mean that

∆V = ∆A * average h

Finally, you say the cone would “imply a very thin sheet round the edge”. The algorithms that calculate whether each small ocean gridcell is ice-covered count out until the point where there is 15% ice in a gridcell. Less ice than that and the gridcell is not counted, and above that the gridcell is counted as ice covered.

Now, what is the average thickness of the ice in that final gridcell at the edge of the ice? I’d say “very thin” …

Yes, an “S” or “Mexican hat” shaped model makes more sense. But a simple cone provides an answer that is within about 5% of the PIOMAS volume loss, so it is certainly good enough for the back of the envelope type of calculation used in S2010.

178. dr.bill says:

re Willis Eschenbach: June 1, 2010 at 2:33 am

An interesting corollary to Willis’ demonstration with the cylinder and cone is that as the base area shrinks, the ‘average height’ of the cone actually increases, and continues to ‘climb’ steadily as more base area is removed. This clearly shows the perils of using mindless averages for such purposes, to say nothing of the cognitive disconnect. I can just hear it now: “That cone is melting like crazy, and wow, look at it get thicker by the minute!”

Just find the damned volumes and subtract them.

/dr.bill

179. Phil. says:

Willis Eschenbach says:
June 1, 2010 at 2:33 am
Phil. says:
May 31, 2010 at 9:59 pm

“… Doesn’t give them an error, as explained above the shape of the object makes no difference.”

If you trim a certain area from the edges of a slab, you remove more volume than if you trim the same area from the edges of a cone. How is it, then, that the shape makes no difference?

Pay attention, although in that example ΔA is the same but ΔH is not, look at the equation I gave you.

“You get the same answer irrespective of geometry, actually the cone is the less physical as it implies a very thin sheet round the edge, something like a mexican hat would be more physical. However calculus says that if we know the initial area and average thickness and rates of change the geometry doesn’t matter.”

Phil, you can keep repeating that all you want.
And I will until you understand it!

Let me give you a physical example. Here are two objects that have the exact same area and the exact same average thickness. They are 10 cm in diameter, so they have a base area of 25 * pi. The cylinder is 2 cm tall, and the cone is 6 cm tall, so they both have a volume of 50 * pi and both have an average thickness of 2 cm. Now, I’ll reduce the area of both of them by the exact same amount.

But what have you done to the thickness?

You keep claiming that “geometry” or “calculus” means that if we reduce the area by the same amount, it will reduce the volume by the same amount.

No I don’t! I keep saying:
ΔV=HΔA+AΔH

180. dr.bill says:

Phil.: June 1, 2010 at 6:23 am
No I don’t! I keep saying: ΔV=HΔA+AΔH

As a theoretical construct, there’s nothing wrong with your formula, either in algebraic or differential form. The problem is that it can’t always be used in actual practice. The reason for that is the need to know the change, not in the height of the object, but in its average height, before your formula can be used. In all but the simplest of cases, you need to already know the new volume so that you can calculate the new average height, and thus the change in the average height. I will use Willis’ example to illustrate this.
——————
Willis’ Method:
Start with a cone of height 6, and base radius 5. Decrease the radius to 4, giving a cylinder of height 1.2 with a cone of height 4.8 sitting on top of it. This gives:
V1 = (1/3)π·5²·6 = 50π
V2 = (1/3)π·4²·(4.8) + π4²·(1.2) = 44.8π
ΔV = V2 – V1 = -5.2π

——————
Your Method:
Same case, but using an initial average height of 2 units. All references to H in your formulas are for average heights. So here’s what we get:
V1 = A1H1
V2 = A2H2
ΔV = H1ΔA + A2ΔH
H1 = 2.0
H2 = unknown
A1 = π·5² = 25π
A2 = π·4² = 16π
ΔA = -9π
ΔV = (2.0)·(-9π) + (16π)·ΔH

——————
So now what do you do? ΔH is unknown, and in the present case, it turns out to be a positive change, not a negative one. If you use Willis’ results, you can figure out that the new average height is 2.8, so the change is +0.8, and if you use that in your formula, you will get the correct result. In order to find the +0.8, however, you need to already know the result that you are looking for. And in this simple example, only the base was changed.

/dr.bill

181. kadaka (KD Knoebel) says:

From: Phil. on May 31, 2010 at 6:17 pm

Feel free to insert the steps you feel are missing, I note however that you don’t challenge the validity of the statement.

Didn’t see the point. V=Ah is valid for the volume enclosed by an orthogonal projection of an area between two parallel planes, so nothing right there to argue about. However in your new post:

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)

Now that just looks silly, and I don’t only mean the questioanable notation. We’re not working with functions using the same variable that yield the values. We have measurements, hard numbers. You know, constant terms. And the first derivative of a constant term is zero. Plus your result as shown is an equation for the rate of change of volume with respect to t, whatever “t” represents (time? thickness?), which ain’t what we are going after.

182. Phil. says:

dr.bill says:
June 1, 2010 at 9:10 am
Phil.: June 1, 2010 at 6:23 am
No I don’t! I keep saying: ΔV=HΔA+AΔH

As a theoretical construct, there’s nothing wrong with your formula, either in algebraic or differential form. The problem is that it can’t always be used in actual practice. The reason for that is the need to know the change, not in the height of the object, but in its average height, before your formula can be used.

Which Shepherd had which is why he calculated it that way. Willis thinks that method is wrong, not that it is inconvenient, and in that he is incorrect.

183. Phil. says:

kadaka (KD Knoebel) says:
June 1, 2010 at 12:00 pm
From: Phil. on May 31, 2010 at 6:17 pm

Feel free to insert the steps you feel are missing, I note however that you don’t challenge the validity of the statement.

Didn’t see the point. V=Ah is valid for the volume enclosed by an orthogonal projection of an area between two parallel planes, so nothing right there to argue about. However in your new post:

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)

Now that just looks silly, and I don’t only mean the questioanable notation. We’re not working with functions using the same variable that yield the values. We have measurements, hard numbers. You know, constant terms. And the first derivative of a constant term is zero. Plus your result as shown is an equation for the rate of change of volume with respect to t, whatever “t” represents (time? thickness?), which ain’t what we are going after.

All three variables are changing with respect to time (t), still with all your excuses you haven’t been able to refute the basic formula. While I personally prefer the Leibnitz notation the Lagrange notation is easier to produce in this context. The idea that volume, thickness and extent aren’t continuous functions of time is rather novel.
H’=-50
A=11.9
H=2.73
A’=-111
Giving V’= 898km^3/yr

184. dr.bill says:

re Phil.: June 1, 2010 at 12:47 pm

Hi Phil,

I guess we can leave it at that. You might not have noticed the topic drift, but at some point in the proceedings we weren’t talking about Shepherd any more, just geometry. kadaka (KD Knoebel) (June 1, 2010 at 12:00 pm) does have a point, though. If the shape has ‘re-entrant’ or ‘lack of single-valuedness’ features, there’s a chance of getting into trouble with that expression. Given some of the other fast-and-loose things going on in climate calculations, however, that might well be classified as just a ‘venial’ sin.

/dr.bill

185. Bob_FJ says:

Dr. Bill, Reur June 1, 2010 at 2:54 pm:

Some aspects of Geometry go a bit wobbly when any variables approach zero or infinity. For instance, is the circumference of a circle where the radius approaches infinity a straight line in a real sense, or should you call it well; erh um; sort of a straight line?

Here’s a scale drawing again of Willis’s model within the screen definition available. (ignoring global curvature)
_____________________________________________________

Note too that if typical first-year ice is 1.6m thick per NSIDC, then the height of the “cone”, sitting on a 1.6m “pancake” is not 4m, but more like 2.4m.

Oh and BTW, the surface area of the slope on a cone is in real terms “Pi” * r^2, because s (= slope) approaches r. So multiply the area A1 by the real average (?) thickness change, and you are almost there. Then do A2 to pick up Willis’s small outer annulus loss.

(?) = Whatever that is

186. Bob_FJ says:

Dr. Bill,
Sorry, I should refine the wording in my last para above:

Oh and BTW, the surface area of the slope on Willis’s cone is in real terms “Pi” * r^2, because s, (= slope), approaches r. So simply multiply the base area A1 by the real average thickness change, and you are almost there. Then do A2 to pick up Willis’s small outer annulus loss.

187. kadaka (KD Knoebel) says:

Excerpt from: Phil. on June 1, 2010 at 1:58 pm

All three variables are changing with respect to time (t)…

Fine. Show the three equations for volume, area, and height where time is the variable. That should help advance climate science tremendously. I’m surprised no one else has discovered them yet, maybe too many climatologists were focused too strongly on CO2 and ignored time as being important. Oh, and specify if time is expressed as a cycling number, restarts at the same point every year, or if it comes from some definitive historical origin point t₀.

188. dr.bill says:

Bob_FJ: June 1, 2010 at 5:41 pm

You lost me there Bob, but that’s OK. I think we’ve beaten the life out of this by now anyway. I do, however, like that scale model of the ice-sheet. It’s reminiscent, on an even smaller scale by far, of the model of Earth’s atmosphere portrayed as the thickness of the skin on an apple. :-)

/dr.bill

189. Bob_FJ says:

Dr. Bill, Reur June 1, 2010 at 6:28 pm
Sorry if I didn’t explain it well, but if you are not interested in an elaboration, just skip to the bit below the line

Here is another example of wobbly geometry at the physical limits; The sine and tangent for any given included angle are clearly two different thingies right? Well not always in reality, because as the angle approaches zero they both also approach zero. Thus, in Willis’s cone, in the scale of things, in reality, the sine and tangent are the same. Or, if you like, the base radius is the same as the slope.

An alternative method of calculation is firstly to take the sloping surface area of cone1 and multiply by the “average” thickness (?) change. There may be an argument to subtract A2, (base area) from A1 and apply some other thickness, but whatever, it would appear to be a small value. The problem is to define thickness; see below the line.

The formula for the sloping surface area of a cone is “Pi” * r * s, where s = the slope. However, ‘s’ goes wobbly and becomes ‘r’, so in other words the surface area of Willis’s cone is the same as the base, (“pi” * r^2), in real terms. This is beginning to look like the Shepherd model, but gee, that thickness issue:

Willis‘s cone = __________________________________________________

The sea-ice shape is complex with complex dynamics atop the Beaufort Gyre currents etc. Take for instance the Catlin jolly; they spent a lot of time travelling backwards and on an island. (first trip). There have been reports of open water areas closing in a matter of hours. Pray how is the average thickness of such a dynamic measured? And how do you compare it with something different next year? There is also the question of varying snow, and how the radar copes with that. Also, if average first-year sea-ice according to NSIDC is typically 1.6 m thick, I find it difficult to imagine that it feathers to very thin at the peripheries. Surely there is a big step-down reduction partly from mechanical effects.

That is why I put (? Whatever that is) against “average thickness” previously.
Both models are flawed.

190. dr.bill says:

Bob_FJ: June 2, 2010 at 3:54 pm

Hi Bob. I ended up here again by accident, and noticed your comment

You’re correct in what you say about cones when they get very short. There’s not much difference between the slant-length and the radius, and using one or the other would give much the same outcome.

As far as averages are concerned, I try to stay away from using them whenever possible, because they sometimes change in non-intuitive ways, and you often need to know the answer before you can find the average. With Willis’ cone, for example, once you slice a piece off the perimeter, the average height actually increases, and what you have is a disk with a cone sitting on top of it (which is pretty much like the geometry that you suggested, with a drop-off at the edge). If you slice off another piece, the average height increases again, and has to be found by ‘knowing the answer’. The way Willis’ approached the ice-volume calcualtion (eventually) was to include the fact that the edge might not be zero height (as you correctly note), and then model the thing as a disk plus cone, which is what I also would have done.

In any case, I find it a lot easier to just deal with volumes using regular expressions for each object, and then I don’t have to keep the complications in mind. If I know the formula for the volume of something, why not just use that, instead of constructing averages that don’t really tell me much, if anything?

For example, if you had a sphere sitting on the ground, its volume is easy to calculate just knowing the radius. I don’t see the point of turning this into an artificial area and height combination. For one thing, what would you use as an area? There are several ‘defensible’ choices, and each one would then give a different average height. You could just as easily do it the other way, i.e. pick something and call it the height, and then calculate the corresponding average area. Seems like a lot of work for no purpose.

If someone just hands me a pair of area and height values, and if I don’t have any other information, then sure, I’d just multiply them and get a volume, but if I know something about the shape of the thing, it’s better to use all I know.

/dr.bill

191. Bob_FJ says:

Dr.bill Reur June 6, 2010 at 3:12 pm

Hi Bill, thanks for your comments. Like you I guess, I never cease to lament how climatologists can bandy around averages like they do. Quite often they are either meaningless or virtually impossible to calculate because the data often contain complex parameters that are not even linear and may also be incomplete in coverage, spatially and/or temporally.

One of my favourites is the IPCC/Trenberth “Earth’s Energy Budget” diagram, where there is this huge vertical (not hemispherical!) up-welling EMR from the surface apparently based on an S-B calculation on the so-called earth’s average temperature. But, I’d better stop there to avoid going too far off topic. (BTW, I’m toying with writing an article on this for Anthony to consider posting.)

Coming back to Willis’s two cones volume method, it cannot work using average thickness loss. The recent annual winter loss is logically small at the apex, and large at the periphery, so the slope on the second smaller cone is consequently steeper, giving different geometry. On the other hand, IF the quoted average thickness loss is real, which I doubt, the surface area x thickness could be used.

OH, BTW, just for fun I calculated the difference in slope versus radius. For simplicity, Cone height = 4m, and radius = 19,000Km.
Slope = 19,000.00000000042 !

192. Willis Eschenbach says:

I guess I was not clear in my exposition, big surprise.

My point in proposing a conical model was not to get an exact answer. It was to demonstrate that using a slab model, as was done in S2010, gives an answer that is way high. This error in S2010 greatly changes their final result, since Arctic ice is by far the largest item in their equation.

193. dr.bill says:

re Willis Eschenbach: June 10, 2010 at 1:50 pm

Nah, you were perfectly clear (and correct), Willis.
Bob and I are just chatting.

/dr.bill

194. Phil. says:

dr.bill says:
June 6, 2010 at 3:12 pm
Bob_FJ: June 2, 2010 at 3:54 pm

As far as averages are concerned, I try to stay away from using them whenever possible, because they sometimes change in non-intuitive ways, and you often need to know the answer before you can find the average.

But in the case of the paper above you have measurements of the average height so you can use the exact solution given by calculus and multiply the area by the average height. Using a model, such as Willis’s cone will always be an approximation and not as accurate.

195. dr.bill says:

Phil.: June 12, 2010 at 8:08 am

dr.bill says:
June 6, 2010 at 3:12 pm
Bob_FJ: June 2, 2010 at 3:54 pm

As far as averages are concerned, I try to stay away from using them whenever possible, because they sometimes change in non-intuitive ways, and you often need to know the answer before you can find the average.

But in the case of the paper above you have measurements of the average height so you can use the exact solution given by calculus and multiply the area by the average height. Using a model, such as Willis’s cone will always be an approximation and not as accurate.

C’mon Phil, up to now I’ve had the idea that you’re a pretty smart guy, but
you have measurements” (!), and “you can use the exact solution” (!!).
That’s not even wrong. :-(

/dr.bill

196. Phil. says:

dr.bill says:
June 12, 2010 at 11:59 am
Phil.: June 12, 2010 at 8:08 am

dr.bill says:
June 6, 2010 at 3:12 pm
Bob_FJ: June 2, 2010 at 3:54 pm

As far as averages are concerned, I try to stay away from using them whenever possible, because they sometimes change in non-intuitive ways, and you often need to know the answer before you can find the average.

But in the case of the paper above you have measurements of the average height so you can use the exact solution given by calculus and multiply the area by the average height. Using a model, such as Willis’s cone will always be an approximation and not as accurate.

C’mon Phil, up to now I’ve had the idea that you’re a pretty smart guy, but
“you have measurements” (!), and “you can use the exact solution” (!!).
That’s not even wrong. :-(

Actually you are wrong, I suggest you read up on the mean theorem of calculus.
And yes there were measurements of the average thickness. Shepherd et al. used the correct method and got the correct result Willis’s error prone analysis notwithstanding.

197. dr.bill says:

Phil.: June 12, 2010 at 1:18 pm

Actually you are wrong, I suggest you read up on the mean theorem of calculus.
And yes there were measurements of the average thickness. Shepherd et al. used the correct method and got the correct result Willis’s error prone analysis notwithstanding.

Thanks for the tip, Phil. It’s really good to have a helpful and knowledgeable person like yourself keeping me on the path to righteousness. ☺ ☺

Actually, one of the big problems with Calculus is that people remember the results without remembering the conditions and limitations. Ahem…

So tell me then, using your enviable grasp of such things,
what is the average thickness of a baseball?

O/T to Moderator: Just checking, but is the word ‘dickhead’ acceptable for use in comments, or is it frowned upon?

/dr.bill

198. Bob_FJ says:

Phil, Reur June 12, 2010 at 8:08 am to Dr. Bill

But in the case of the paper above you have measurements of the average height so you can use the exact solution given by calculus and multiply the area by the average height.

Ah yes, that’s right, we have measurements! So why was it necessary to launch the ESA’s Cryosat 2 last April? (replacing Cryosat 1 that failed on launch 4 years ago). Cryosat 2 has new sophisticated radar that gives altimetry of the ice and underlying surfaces. Here are a couple of ESA release extracts:

EXTRACT: “… the CryoSat-2 satellite will monitor precise changes in the thickness of the polar ice sheets and floating sea ice. CryoSat’s original objective was to determine if there was a trend towards diminishing ice cover. There now seems little doubt that there are indeed trends – the challenge now is to characterise them…”
http://www.esa.int/esaLP/ESAOMH1VMOC_LPcryosat_0.html

EXTRACT:CryoSat-2’s sophisticated instruments will measure changes at the margins of the vast ice sheets that lie over Greenland and Antarctica and in the marine ice floating in the polar oceans. By accurately measuring thickness change in both types of ice, CryoSat-2 will provide information critical to scientists’ understanding of the role ice plays in the Earth system.
http://www.esa.int/esaLP/SEMTMB9MT7G_LPcryosat_0.html

Note that altimetry of the top surface of sea ice only gives freeboard of the ice, which reportedly has varying density laminally, especially perennial ice.

Even so, I remain puzzled how the complex dynamics of the ice, (such as measuring a constantly moving 3 dimensional target,) and comparing it a year later are handled. (Re my Bob_FJ says: June 2, 2010 at 3:54 pm )

199. Willis Eschenbach says:

Phil. says:
June 12, 2010 at 1:18 pm (Edit)

Actually you are wrong, I suggest you read up on the mean theorem of calculus.
And yes there were measurements of the average thickness. Shepherd et al. used the correct method and got the correct result Willis’s error prone analysis notwithstanding.

Phil, for the Shepherd et al. analysis to be right, the edge of the ice pack would have to be the same thickness as the rest of the ice pack. That’s why their analysis finds that 35% of the volume loss comes from the loss of area.

Now, you can wave your hands and repeat “calculus” as though it were a mystical incantation as much as you want, but that doesn’t make the edge of the ice pack the same thickness as the middle of the ice pack. They explicitly claim the 35% number. I have repeated their calculations to verify that they are using a flat slab model for the ice pack. Sorry … it’s thinner on the edges, Phil. Look at any of the thickness images and it is very clear. Middle thicker. Edges thinner.

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