Path: news.mathworks.com!not-for-mail From: "Bruno Luong" <b.luong@fogale.findmycountry> Newsgroups: comp.soft-sys.matlab Subject: Re: Normal vector of circles Date: Sun, 24 Oct 2010 08:14:04 +0000 (UTC) Organization: FOGALE nanotech Lines: 16 Message-ID: <ia0psc$6qe$1@fred.mathworks.com> References: <i9vb60$eu5$1@fred.mathworks.com> <i9vi1d$plf$1@fred.mathworks.com> <i9vkph$i54$1@fred.mathworks.com> <i9vlht$6qq$1@fred.mathworks.com> <i9voa0$15e$1@fred.mathworks.com> <i9vq61$b5$1@fred.mathworks.com> <i9vtb8$jov$1@fred.mathworks.com> Reply-To: "Bruno Luong" <b.luong@fogale.findmycountry> NNTP-Posting-Host: webapp-03-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1287908044 6990 172.30.248.38 (24 Oct 2010 08:14:04 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Sun, 24 Oct 2010 08:14:04 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 390839 Xref: news.mathworks.com comp.soft-sys.matlab:680853 "Paulo " <paulofreitas7@portugalmail.pt> wrote in message <i9vtb8$jov$1@fred.mathworks.com>... > > The initial and original problem is: Given a sphere to you, you can take the axes and any other information from it, then you close your eyes while I rotate it. Now, how can you calculate the rotation? You get confused by the person who poses the problem. The sphere by itself has little to do with the rotation (all we know is the *mapping image* as the sphere unchanged). You can imagine selecting n points on the sphere, i.e., n vectors having the same norm = r. After the rotation it maps to n new points. The question is what is the rotation that maps the old to new points. This problem can be formalized as vectors as following: P := { x1, x2, ... , xn in R^3 : |xi| = r }, Q := rotation of P, find the rotation from P and Q. I claim that if you select P having two points (n=2) not opposite (x1 different +/- x2) and their transformations, it is enough to find the rotation. Bruno