This is a nice nit to pick! I think that, although the area at the top of the frustrum could get large, that the corresponding air density is so low that it doesn't matter. However, I've been wondering about what happens when one calculates the pressure near the center of the earth. If one uses a column, one gets a nice, finite pressure, integrating the weight of the rock/magma/stuff upward. However, if one considers that the surface area of the sphere around the center of the earth becomes arbitrarily small, it seems that all of that weight is borne by a tiny area--the pressure should approach infinity. This is, I think, equivalent to using the frustrum instead of the column. So, is there a compensating force (the walls of the frustrum are radial, so there might be a net upward force from the pressure around it?), or is there something wrong with the model of using a frustrum, or is it something else? Bill -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com] On Behalf Of Daniel Asimov Sent: Tuesday, August 01, 2006 4:05 PM To: math-fun Subject: Re: [math-fun] A modest proposal for carbon sequestration in space << If I remember my physics correctly, air pressure is approx. 14.7#/sq in, which is the weight of the entire column of air 1 inch square sitting on the square inch.

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Not to be a nitpicker, but since "up" is not given by a parallel vector field, wouldn't the shape of this "air column" be more like a frustum of a 4-sided pyramid -- with a spherical "one-square-inch" top of radius R ~ Earth's radius, and a spherical base, of radius R+h (where h = maximum distance of relevant air from Earth) ? --Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun