I think rwg meant what you get if you take one of the "spherical squares" — each of which is just 1/6 of a sphere — and then pick a vertex anywhere on the line perpendicular to it at its center point . . . at a distance h from the center point . . . and then connect this vertex to each point of the spherical square by a line segment. (Also note that negative h can be taken to mean the vertex is outside the original sphere.) This results in a "right spherical square cone". —Dan
On Mar 8, 2016, at 3:40 AM, David Makin <makinmagic@tiscali.co.uk> wrote:
I assume you mean round-topped pyramids - and that the sides are chopped with a vertical section like a rectangle but with one side a curve ?
On 8 Mar 2016, at 10:00, rwg wrote:
My impiety stems from measuring height from the (presumably planar) rim of the base of the cone, which would give the same volume whether the base was concave or convex. Eavesdropper Gary Snethen observes that measuring height from the center of a rotationally symmetric base apparently gives the right answer for spherical cones, among others. --rwg Problem: Project a (unit, say) cube onto its circumsphere, producing six "spherical squares". What is the volume of the round-bottomed "pyramid" whose apex lies at height h above the center of one such square?
On 2016-03-07 04:13, Dan Asimov wrote:
The only misapplication I find is the part about Jesus' mother. —Dan
On Mar 6, 2016, at 9:41 PM, Bill Gosper <billgosper@gmail.com> wrote: Where is the misapplication of integral calculus in the following?: While preserving its edge, deform the base of a circular, say, cone to make it convex or concave, with area A. Then the volume formula should remain V = A h/3, because you can decompose it into a tapered series of curved plates dV = A (z/h)^2 dz. And I am the Virgin Mary. --rwg