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.
Willis Eschenbach says:
May 31, 2010 at 9:21 pm
Phil. says:
May 31, 2010 at 5:05 pm
Willis Eschenbach says:
May 31, 2010 at 4:36 pm
Shepherd et al. DID NOT USE A CONE. They used a flat slab. This gave them a 35% error in their answer.
No it didn’t as explained above.
Phil, that’s not clear. It didn’t what? Didn’t use a slab? Didn’t give them an error?
Doesn’t give them an error, as explained above the shape of the object makes no difference.
Let me run the numbers for you again.
S2010 assuming a slab
Shepherd Winter Volume, 1990-1999: 11.9 M sq km * 273 cm = 32,487 km^3
Area: 11,900,000 km^2
Reduction in area: -111,000 sq km
Reduction in thickness: 5 cm
Correct answer: -50×11.9-2.73×111= 898km^3
The difference is solely from the assumed shape of the ice pack. S2010 assumes it is a slab. I think that assumption is unphysical.
You get the same answer irrespective of geometry, actually the cone is the less physical as it implies a very thin sheet round the edge, something like a mexican hat would be more physical. However calculus says that if we know the initial area and average thickness and rates of change the geometry doesn’t matter.
This is borne out by the fact that if I use the PIOMAS volume, I get an answer quite close to the PIOMAS estimate of the ice loss (402 km^3 vs 420 km^3. The S2010 loss estimate of 851 km^3, on the other hand, is twice as large as the PIOMAS estimate.
No this calculation uses different data and it doesn’t have any relevance to the comparison of two different models. The PIOMAS data shows a slope of ~1000km^3/yr over the last decade.
Willis Eschenbach says:
May 31, 2010 at 9:49 pm
Not sure what you are calling my “embarassing error” in the initial post … if you mean the height, I updated the initial post some time ago regarding that.
That you divided by 3 when you shouldn’t have, your conical model using the parameters consistent with Shepherd gives 891km^3 i.e. equal to the slab model 898km^3 (allowing for rounding error). As calculus says it should.
Volume 1: 1/3 * 11,900,000 km^2 * 819 cm / (100000 cm/km) = 32,847 km^3
I assume this is a typo; should be 32,487 km^3.
This model is indeed closer to agreeing with the data, but the average decrease in thickness dh in your model is still smaller than you claim: around 4.2 cm, to be precise. (By making your cone pointier, the slopes of the two cones have become ‘more parallel’, but they’re still not quite parallel, so dh <= 5 cm.)
Not sure what you are calling my “embarassing error” in the initial post … if you mean the height, I updated the initial post some time ago regarding that.
Even in the update to your original post you still claim to get a decrease of only 344 km^3. For the reasons given above, this cannot be correct if you take the S2010 data (as you do in your computation).
Willis, Reur May 31, 2010 at 9:05 pm
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!
Phil. says:
May 31, 2010 at 9:59 pm
If you trim a certain area from the edges of a slab, you remove more volume than if you trim the same area from the edges of a cone. How is it, then, that the shape makes no difference?
Phil, you can keep repeating that all you want. Let me give you a physical example. Here are two objects that have the exact same area and the exact same average thickness. They are 10 cm in diameter, so they have a base area of 25 * pi. The cylinder is 2 cm tall, and the cone is 6 cm tall, so they both have a volume of 50 * pi and both have an average thickness of 2 cm. Here they are:
Now, I’ll reduce the area of both of them by the exact same amount. You keep claiming that “geometry” or “calculus” means that if we reduce the area by the same amount, it will reduce the volume by the same amount. Here is that situation:
As you can see, despite my having trimmed the exact same area from each object, and despite the fact that the average thickness of the two objects are identical at 2 cm, neither geometry nor calculus can force these two trimmed objects to have the same volume. Much more volume has been lost from the cylinder. So the shape indeed does matter, it matters very much. As I have said before, the fact that
doesn’t mean that
Finally, you say the cone would “imply a very thin sheet round the edge”. The algorithms that calculate whether each small ocean gridcell is ice-covered count out until the point where there is 15% ice in a gridcell. Less ice than that and the gridcell is not counted, and above that the gridcell is counted as ice covered.
Now, what is the average thickness of the ice in that final gridcell at the edge of the ice? I’d say “very thin” …
Yes, an “S” or “Mexican hat” shaped model makes more sense. But a simple cone provides an answer that is within about 5% of the PIOMAS volume loss, so it is certainly good enough for the back of the envelope type of calculation used in S2010.
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
Willis Eschenbach says:
June 1, 2010 at 2:33 am
Phil. says:
May 31, 2010 at 9:59 pm
“… Doesn’t give them an error, as explained above the shape of the object makes no difference.”
If you trim a certain area from the edges of a slab, you remove more volume than if you trim the same area from the edges of a cone. How is it, then, that the shape makes no difference?
Pay attention, although in that example ΔA is the same but ΔH is not, look at the equation I gave you.
“You get the same answer irrespective of geometry, actually the cone is the less physical as it implies a very thin sheet round the edge, something like a mexican hat would be more physical. However calculus says that if we know the initial area and average thickness and rates of change the geometry doesn’t matter.”
Phil, you can keep repeating that all you want.
And I will until you understand it!
Let me give you a physical example. Here are two objects that have the exact same area and the exact same average thickness. They are 10 cm in diameter, so they have a base area of 25 * pi. The cylinder is 2 cm tall, and the cone is 6 cm tall, so they both have a volume of 50 * pi and both have an average thickness of 2 cm. Now, I’ll reduce the area of both of them by the exact same amount.
But what have you done to the thickness?
You keep claiming that “geometry” or “calculus” means that if we reduce the area by the same amount, it will reduce the volume by the same amount.
No I don’t! I keep saying:
ΔV=HΔA+AΔH
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
From: Phil. on May 31, 2010 at 6:17 pm
Didn’t see the point. V=Ah is valid for the volume enclosed by an orthogonal projection of an area between two parallel planes, so nothing right there to argue about. However in your new post:
Now that just looks silly, and I don’t only mean the questioanable notation. We’re not working with functions using the same variable that yield the values. We have measurements, hard numbers. You know, constant terms. And the first derivative of a constant term is zero. Plus your result as shown is an equation for the rate of change of volume with respect to t, whatever “t” represents (time? thickness?), which ain’t what we are going after.
dr.bill says:
June 1, 2010 at 9:10 am
Phil.: June 1, 2010 at 6:23 am
No I don’t! I keep saying: ΔV=HΔA+AΔH
As a theoretical construct, there’s nothing wrong with your formula, either in algebraic or differential form. The problem is that it can’t always be used in actual practice. The reason for that is the need to know the change, not in the height of the object, but in its average height, before your formula can be used.
Which Shepherd had which is why he calculated it that way. Willis thinks that method is wrong, not that it is inconvenient, and in that he is incorrect.
kadaka (KD Knoebel) says:
June 1, 2010 at 12:00 pm
From: Phil. on May 31, 2010 at 6:17 pm
Feel free to insert the steps you feel are missing, I note however that you don’t challenge the validity of the statement.
Didn’t see the point. V=Ah is valid for the volume enclosed by an orthogonal projection of an area between two parallel planes, so nothing right there to argue about. However in your new post:
It is also the result one gets by using calculus, I supplied an algebraic derivation for ease of understanding, however the same result can be arrived at by using calculus as follows:
v(t)=A(t)H(t)
by the Product Rule:
v’(t)=[A(t)H(t)]‘=A’(t)H(t)+A(t)H’(t)
Now that just looks silly, and I don’t only mean the questioanable notation. We’re not working with functions using the same variable that yield the values. We have measurements, hard numbers. You know, constant terms. And the first derivative of a constant term is zero. Plus your result as shown is an equation for the rate of change of volume with respect to t, whatever “t” represents (time? thickness?), which ain’t what we are going after.
All three variables are changing with respect to time (t), still with all your excuses you haven’t been able to refute the basic formula. While I personally prefer the Leibnitz notation the Lagrange notation is easier to produce in this context. The idea that volume, thickness and extent aren’t continuous functions of time is rather novel.
H’=-50
A=11.9
H=2.73
A’=-111
Giving V’= 898km^3/yr
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, Reur June 1, 2010 at 2:54 pm:
Some aspects of Geometry go a bit wobbly when any variables approach zero or infinity. For instance, is the circumference of a circle where the radius approaches infinity a straight line in a real sense, or should you call it well; erh um; sort of a straight line?
Here’s a scale drawing again of Willis’s model within the screen definition available. (ignoring global curvature)
_____________________________________________________
Note too that if typical first-year ice is 1.6m thick per NSIDC, then the height of the “cone”, sitting on a 1.6m “pancake” is not 4m, but more like 2.4m.
Oh and BTW, the surface area of the slope on a cone is in real terms “Pi” * r^2, because s (= slope) approaches r. So multiply the area A1 by the real average (?) thickness change, and you are almost there. Then do A2 to pick up Willis’s small outer annulus loss.
(?) = Whatever that is
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.
Excerpt from: Phil. on June 1, 2010 at 1:58 pm
Fine. Show the three equations for volume, area, and height where time is the variable. That should help advance climate science tremendously. I’m surprised no one else has discovered them yet, maybe too many climatologists were focused too strongly on CO2 and ignored time as being important. Oh, and specify if time is expressed as a cycling number, restarts at the same point every year, or if it comes from some definitive historical origin point t₀.
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
Dr. Bill, Reur June 1, 2010 at 6:28 pm
Sorry if I didn’t explain it well, but if you are not interested in an elaboration, just skip to the bit below the line
Here is another example of wobbly geometry at the physical limits; The sine and tangent for any given included angle are clearly two different thingies right? Well not always in reality, because as the angle approaches zero they both also approach zero. Thus, in Willis’s cone, in the scale of things, in reality, the sine and tangent are the same. Or, if you like, the base radius is the same as the slope.
An alternative method of calculation is firstly to take the sloping surface area of cone1 and multiply by the “average” thickness (?) change. There may be an argument to subtract A2, (base area) from A1 and apply some other thickness, but whatever, it would appear to be a small value. The problem is to define thickness; see below the line.
The formula for the sloping surface area of a cone is “Pi” * r * s, where s = the slope. However, ‘s’ goes wobbly and becomes ‘r’, so in other words the surface area of Willis’s cone is the same as the base, (“pi” * r^2), in real terms. This is beginning to look like the Shepherd model, but gee, that thickness issue:
Willis‘s cone = __________________________________________________
The sea-ice shape is complex with complex dynamics atop the Beaufort Gyre currents etc. Take for instance the Catlin jolly; they spent a lot of time travelling backwards and on an island. (first trip). There have been reports of open water areas closing in a matter of hours. Pray how is the average thickness of such a dynamic measured? And how do you compare it with something different next year? There is also the question of varying snow, and how the radar copes with that. Also, if average first-year sea-ice according to NSIDC is typically 1.6 m thick, I find it difficult to imagine that it feathers to very thin at the peripheries. Surely there is a big step-down reduction partly from mechanical effects.
That is why I put (? Whatever that is) against “average thickness” previously.
Both models are flawed.
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
Dr.bill Reur June 6, 2010 at 3:12 pm
Hi Bill, thanks for your comments. Like you I guess, I never cease to lament how climatologists can bandy around averages like they do. Quite often they are either meaningless or virtually impossible to calculate because the data often contain complex parameters that are not even linear and may also be incomplete in coverage, spatially and/or temporally.
One of my favourites is the IPCC/Trenberth “Earth’s Energy Budget” diagram, where there is this huge vertical (not hemispherical!) up-welling EMR from the surface apparently based on an S-B calculation on the so-called earth’s average temperature. But, I’d better stop there to avoid going too far off topic. (BTW, I’m toying with writing an article on this for Anthony to consider posting.)
Coming back to Willis’s two cones volume method, it cannot work using average thickness loss. The recent annual winter loss is logically small at the apex, and large at the periphery, so the slope on the second smaller cone is consequently steeper, giving different geometry. On the other hand, IF the quoted average thickness loss is real, which I doubt, the surface area x thickness could be used.
OH, BTW, just for fun I calculated the difference in slope versus radius. For simplicity, Cone height = 4m, and radius = 19,000Km.
Slope = 19,000.00000000042 !
I guess I was not clear in my exposition, big surprise.
My point in proposing a conical model was not to get an exact answer. It was to demonstrate that using a slab model, as was done in S2010, gives an answer that is way high. This error in S2010 greatly changes their final result, since Arctic ice is by far the largest item in their equation.
re Willis Eschenbach: June 10, 2010 at 1:50 pm
Nah, you were perfectly clear (and correct), Willis.
Bob and I are just chatting.
/dr.bill
dr.bill says:
June 6, 2010 at 3:12 pm
Bob_FJ: June 2, 2010 at 3:54 pm
As far as averages are concerned, I try to stay away from using them whenever possible, because they sometimes change in non-intuitive ways, and you often need to know the answer before you can find the average.
But in the case of the paper above you have measurements of the average height so you can use the exact solution given by calculus and multiply the area by the average height. Using a model, such as Willis’s cone will always be an approximation and not as accurate.
C’mon Phil, up to now I’ve had the idea that you’re a pretty smart guy, but
“you have measurements” (!), and “you can use the exact solution” (!!).
That’s not even wrong. 🙁
/dr.bill
dr.bill says:
June 12, 2010 at 11:59 am
Phil.: June 12, 2010 at 8:08 am
dr.bill says:
June 6, 2010 at 3:12 pm
Bob_FJ: June 2, 2010 at 3:54 pm
As far as averages are concerned, I try to stay away from using them whenever possible, because they sometimes change in non-intuitive ways, and you often need to know the answer before you can find the average.
But in the case of the paper above you have measurements of the average height so you can use the exact solution given by calculus and multiply the area by the average height. Using a model, such as Willis’s cone will always be an approximation and not as accurate.
C’mon Phil, up to now I’ve had the idea that you’re a pretty smart guy, but
“you have measurements” (!), and “you can use the exact solution” (!!).
That’s not even wrong. 🙁
Actually you are wrong, I suggest you read up on the mean theorem of calculus.
And yes there were measurements of the average thickness. Shepherd et al. used the correct method and got the correct result Willis’s error prone analysis notwithstanding.
Thanks for the tip, Phil. It’s really good to have a helpful and knowledgeable person like yourself keeping me on the path to righteousness. ☺ ☺
Actually, one of the big problems with Calculus is that people remember the results without remembering the conditions and limitations. Ahem…
So tell me then, using your enviable grasp of such things,
what is the average thickness of a baseball?
O/T to Moderator: Just checking, but is the word ‘dickhead’ acceptable for use in comments, or is it frowned upon?
/dr.bill