The solid nappe of that cone intersects the sphere in a region that includes one semicircle, in fact a semicircle one of whose endpoints is exactly at the cusp ("corner") you mention. If the cone has its apex at (0, 0, -1) and is tangent to the plane z = -1, then this semicircle will have its endpoints at (0, 0, -1) and (0, 0, 1), with that cusp at (0, 0, -1). —Dan
On Jan 2, 2016, at 1:56 AM, rwg <rwg@sdf.org> wrote:
On 2016-01-01 12:44, Dan Asimov wrote:
Isn't there also a "circular" height at the other extreme -- where one edge of the conical beam is tangent to the sphere, at least for the case of the cone (one nappe) being pi/4 from its axis.
I think not. (Therefore I aren't) If the curve lies on the cone, it would develop a corner to fit the apex as it approached, no? gosper.org/sphere+cone.png [1] --rwg
On Jan 1, 2016, at 8:31 AM, rwg <rwg@sdf.org> wrote:
Intuitive picture: You're inside a big sphere on the equator. You run the center of your conical flashlight beam along the equator. By the 2D thm, the spot has constant width. But the height varies, being circular only at the antipode. --rwg
On 2015-12-31 06:34, Gareth McCaughan wrote:
On 31/12/2015 04:56, James Propp wrote:
Is there a version of the central angle theorem that applies to spheres and solid angles?
I don't think so. Draw a picture like this. You have the unit sphere, centre O. You have an area element dA somewhere on it, say at P. You have a point Q somewhere else on the sphere. Let alpha be angle OPQ (= angle OQP). The solid angle subtended at O is (by definition) dA. The solid angle subtended at Q is smaller because of two factors: - The distance is 2 cos alpha instead of 1. - The area element's normal isn't along PQ but at an angle alpha to it. This means that the ratio of subtended angles is 1/(2 cos alpha)^2 from the first factor, times cos alpha from the second factor. The cosine factors don't cancel the way they do in two dimensions; you get 1 / 4 cos alpha.
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