User bob - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:08:14Z http://mathoverflow.net/feeds/user/3766 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14032/finding-a-minimum-bounding-sphere-for-a-frustum Finding a minimum bounding sphere for a frustum Bob 2010-02-03T19:35:40Z 2010-02-04T01:31:16Z <p>I have a frustum (truncated pyramid defined by six planes) and I need to compute a bounding sphere for this frustum that's as small as possible.</p> <p>I can choose the centre of the sphere to be right in the centre of the frustum and the radius be the distance to one of the "far" corners (the base of the pyramid), but that usually leaves quite a lot of slack around the narrow end of the frustum. There must be a better way!</p> <p>This seems like simple geometry, but I can't seem to figure it out. Any ideas?</p> http://mathoverflow.net/questions/14032/finding-a-minimum-bounding-sphere-for-a-frustum Comment by Bob Bob 2010-02-03T19:55:45Z 2010-02-03T19:55:45Z It's not square no, however I know that one of the sides will always be longer than the other. I think I've got it, actually. Simplifying it into 2D on the longest axis, and stipulating that the distance to the far vertex on the base and the far vertex on the truncated top should be equal (i.e. the bounding sphere touches the outside vertices), and the maths just kind of works out from there I think. I got that the distance from the base should be (Wn^s + Wf^2 +x^2)/2x, where Wn is the half-width of the narrow side, Wf is the half-width of the fat end, and x is distance from base to top.