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

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?

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.

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.

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”.

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

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.

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)

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.

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.

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”.

Sorry, wrong coordinates:
http://ice-map.appspot.com/?map=Arc&sat=ter&lvl=7&lat=67.5&lon=-35.0&yir=2009&day=147
http://ice-map.appspot.com/?map=Arc&sat=ter&lvl=7&lat=67.5&lon=-35.0&yir=2010&day=147

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.

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

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

Willis et al
Very interesting, unable to comment on your maths,
Do any of these papers help with your question on ice thickness? [google search ‘defence research thickness of ice’]
http://www.igsoc.org/journal/3/28/igs_journal_vol03_issue028_pg727-732_724-726.pdf
http://www.igsoc.org/journal/12/66/igs_journal_vol12_issue066_pg417-421.pdf
http://www.nasa.gov/topics/earth/features/seaice_skinny.html
sea ice thickness section in wiki does nor provide reference http://en.wikipedia.org/wiki/Measurement_of_sea_ice

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.

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.”

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?

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.

Willis:
I always thought SWAG 95% CI was +/- 50%
When in doubt, Flip a coin!

@juanita
huh?

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.

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.

Here is a bit of Propaganda about the Mulholland/LA aqua duct and one reason Mono lake is going away….
http://wsoweb.ladwp.com/Aqueduct/historyoflaa/index.htm

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.

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.

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.

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?

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?

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?

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).

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.
😉

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

Hi Willis
It’s beyond me, who last took a maths exam at the age of 16 and passed it only by the skin of my teeth, why anyone wouldn’t understand your cone model. However, one thing I wanted to ask is this. Would it be true to say that the centre of the cone, being nearer the pole, would be colder and less liable to melt than at the edge, farther from the pole?
If so, wouldn’t more more area be lost at the edges in your 2-D section of the cone than you show? After all, you assume a constant depth loss over the whole surface of the cone, but if it widens towards the periphery, then for a given volume loss, the periphery loss would be greater, wouldn’t it?
As I say, my maths isn’t good so I could well be wrong, but would welcome your comment.

Willis 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.

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.

I’m glad you spotted what mb was getting wrong, Willis; I was absolutely clueless!

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.

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.

“R. Craigen says:
[…]one only needs to[…]Any programmers out there up to this challenge?”
You’re a project manager, right?

@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.

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.

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

Oh goodie another post about arctic sea ice…
/sarc
Surely the battlefield is bigger than this guys?

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?

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.

Big picture –
Global sea ice area has been declining steadily for 30 years and there is no evidence that the ice loss is near its end.
http://i50.tinypic.com/35iackn.gif
In that context, why continue with this thread?

What mb is saying is correct.
Let’s put it another way. There are two effects that contribute to the decrease:
1) The total area becomes smaller.
2) Average thickness shrinks.
It is true that the contribution of 1) depends on the model of how the ice sheet is shaped, i.e. if it’s conical, flat, or any other arbitrary shape. The contribution of however 2) can always be bounded from below: By definition,
loss of volume through decreased thickness = total Area*average decrease in thickness
This is basically the definition of `average decrease in thickness’, and it is completely independent of the shape of the sheet. Plugging in the values from the original paper, this gives the 589 km^3 that you mentioned above.
It follows that no matter what shape for the sheet you use, the result must always be bigger than 589 km^3. If you get something smaller, then you have made a mistake on the way (such as picking a wrong value for the average height.)

Willis,
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.

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.

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.

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.

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! 🙂

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.

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.

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!

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.

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.

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.

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.

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.

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

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.

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

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!

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

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

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

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.

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

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