Path: utzoo!utgpu!jarvis.csri.toronto.edu!cs.utexas.edu!samsung!munnari.oz.au!bruce!dbrmelb!davidp From: davidp@dbrmelb.dbrhi.oz (David Paterson) Newsgroups: comp.graphics Subject: Re: smallest sphere enclosing a set of points Message-ID: <679@dbrmelb.dbrhi.oz> Date: 7 Dec 89 06:22:51 GMT References: <28@ <658@cditi.UUCP> <4893@skinner.nprdc.arpa> Organization: CSIRO, Div. Building Constr. and Eng'ing, Melb., Australia Lines: 22 In article <4893@skinner.nprdc.arpa>, malloy@nprdc.arpa (Sean Malloy) writes: > > Then generalize it. Find the largest distance between any two points. > Take all the pairs of points with that separation, and average their > coordinates. This will give you the center of the sphere; the distance > from that point to any of the equidistant points will give you the > radius. > > Each point in the set of points with largest separation gets counted > twice in the averaging, but since you divide by the number of endpoints, > not the number of separations, the averaging still works. > Sorry, still doesn't work, consider, for instance the circle surrounding points (-1,0),(1,0),(0,2). My solution was (see earlier posting) to solve simultaneously three quadratic equations using Newton's method. Does anyone know a quicker way? ----------------------------------------------------------------------- David Paterson, CSIRO, Highett, 3190, Victoria, Australia