Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Max distance from point on ellipsoid to surface Date: Tue, 22 May 2012 19:37:07 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 14 Message-ID: <jpgpt3$c6s$1@newscl01ah.mathworks.com> References: <jpgh3q$c1$1@newscl01ah.mathworks.com> <jpgnlq$1fp$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-02-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1337715427 12508 172.30.248.47 (22 May 2012 19:37:07 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Tue, 22 May 2012 19:37:07 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 3499673 Xref: news.mathworks.com comp.soft-sys.matlab:768661 My surface is determined by a mathematical equation 3x+4y^2+6z+6=0. So basically I need to find tangent plane and normal line(to that tangent plane) at each point of surface and see if that normal line (extended) orthogonally intersects tangent plane of ellipsoid at any point. And Lagrange multipliers would lead this? I'm not sure if I got it right. Thanks. "Roger Stafford" wrote in message <jpgnlq$1fp$1@newscl01ah.mathworks.com>... > "stefaneli" wrote in message <jpgh3q$c1$1@newscl01ah.mathworks.com>... > > I want to find a point on an given ellipsoid that is the farthest from a given surface. (The distance between a point on ellipsoid and the surface should be max). So how can I do this? I was thinking about some kind of Lagrange multipliers. But what do you think? Thanks. > - - - - - - - - - - - > How is your surface determined? Is it given by a set of discrete points or is it determined by a mathematical equation? > > With a mathematical equation, what you want to find is a point on the surface such that if the normal to the surface at that point is extended, it will intersect the ellipsoid orthogonally. That is what Lagrange multipliers would lead to. That will give you two equations to be satisfied. Together with the requirement of the point being on the surface, that is three equalities altogether. Of course you would have to exclude the multiple solutions that give the closest distance rather than the furtherest. Also if your surface has a bounding edge, a non-orthogonal solution might occur on this edge. > > How hard such a problem would be depends entirely on the particular surface equation. > > Roger Stafford