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Approved-By: [log in to unmask] User-Agent: Thunderbird 2.0.0.16 (Windows/20080708) References: <[log in to unmask]> <001901c90179$f1271230$d3753690$@[log in to unmask]> Content-Type: text/plain; charset=ISO-8859-1; format=flowed Message-ID: <[log in to unmask]> Date: Thu, 21 Aug 2008 16:19:58 +0100 Reply-To: ImageJ Interest Group <[log in to unmask]> Sender: ImageJ Interest Group <[log in to unmask]> From: Michael Doube <[log in to unmask]> Subject: Re: Transform stack using eigenvectors In-Reply-To: <001901c90179$f1271230$d3753690$@[log in to unmask]>To answer my own question: The aligned position is the 3 dot products of the original coordinate (x,y,z) and the 3 eigenvectors (the principal axes). Each eigenvector is a normal to the plane that contains the other 2 eigenvectors because they are orthogonal. A plane is defined by the vector of its normal and the position of the plane on that normal. The dot product of a plane and a point is the distance between plane and point, assuming that (0,0,0) is a solution for the plane. So if you subtract the centroid from the point and work out the 3 dot products corresponding to the 3 eigenplanes, you have the distance along each of the eigenvectors which defines the point in the principal axis coordinate frame, and hence gives you 'aligned' coordinates. Then the aligned coordinates can be drawn in ordinary image coordinates and the object is aligned according to its principal axes. If that makes no sense or is blatantly wrong, please email me, otherwise it's going in a plugin tonight... Mike > -----Original Message----- > From: ImageJ Interest Group [mailto:[log in to unmask]] On Behalf Of > Michael Doube > Sent: Friday, August 15, 2008 5:57 PM > To: [log in to unmask] > Subject: Transform stack using eigenvectors > > Hi all > > I have written a plugin that calculates the 3D moments of inertia of a > bone imaged in CT, using the Jama package and eigen decomposition. > > Now that I have a set of eigenvalues and eigenvectors I want to align my > object such that its principal axes are parallel with the x,y,z axes of > a new stack. > > Can anyone shed some light as to how to best achieve this? > > Mike > >
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