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.
From the Update in the article above:
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.
Nick Stokes says:
May 30, 2010 at 4:57 pm
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.
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
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.
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.
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.
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.
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! 🙂
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.
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.
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.
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.
Nick Stokes says:
May 30, 2010 at 6:47 pm
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”.
Nope. My PIOMASS trend was from 1994-2007. I had said:
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.
Jbar says:
May 30, 2010 at 4:13 pm
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.
willis
how about less childish titles for your future Articles
might be a good place to start 🙂
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!
ck says:
May 30, 2010 at 4:18 pm
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:
Words of wisdom …
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.
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?
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.
Dr. Dave says:
May 30, 2010 at 9:23 pm
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.
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.
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.
Source:
http://www.prisonplanet.com/arctic-ice-volume-has-increased-25-since-may-2008.html
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?
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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.
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