Vectors on ellipsoid surface
in [Matlab]
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From: Bruno Luong on 23 May 2010 10:38 "Els " <y.e.t.reeuwijk(a)student.utwente.nl> wrote in message <htba3v$iq3$1(a)fred.mathworks.com>... > But what I am not getting is how to give in P1 and P2 (the two points between which C lies), and is the projected C in 90 degrees of this line? Sorry? What are P1 and P2? I quote your post #1 [ ...I plotted an ellipsoid [x,y,z]=ellipsoid (standard in Matlab) Now I have a point C(x,y,z) which lays within this surface. I am looking for the projection of C on the surface of the ellipsoid,called C', where the normal vector perpendicular to the surface is in line with the vector between C and C'. ] My understanding is C is given, and C' is the projection on C on the surface of the ellipsoid to be found. In my code, C is called "c" (lower case); and C' is "x(:,imin)". > > What should I do if my origin is not on [0,0,0], just change [X Y Z] = ellipsoid(0,0,0,radii(1),radii(2),radii(3),N); ? No, translate it. Simply subtract all your coordinates by the center of the ellipsoid, then the new coordinates are centered about the origin. > > And what should I do if my axis are not straigt but rotated? In my code the axis ellipsoid is not straight, but oriented in the column vectors of matrix U. I wrote in the comments that I assumed the ellipsoid is E = { x in R^n : x'*Q*x = 1 }, where Q is a (n x n) sdp matrix, this is general ellipsoid *centered* about the origin. The most general form of ellipsoid is E = { x0 + x in R^n : x'*Q*x = 1 }, x0 in R^n and Q is a (n x n) sdp matrix. If you know 1) the radii: radii(1), radii(2), and radii(3) (three scalars in a vector) 2) the corresponding orientation U1, U2, U3 (three vectors, they must be orthogonal and normed to 1) 3) and the center x0 (a vector) of your ellipsoid Then define: U = [U(:,1) U(:,2) U(:3)] S = diag(1./radii.^2); Q = U*S*U' Before calling my code, just subtract x0 to c, then add it back to the result. c = c-x0 .... % Do my code % cprojected = x(:,imin)+x0 Bruno
From: Els on 23 May 2010 13:42 > > But what I am not getting is how to give in P1 and P2 (the two points between which C lies), and is the projected C in 90 degrees of this line? > > Sorry? What are P1 and P2? I quote your post #1 > > My understanding is C is given, and C' is the projection on C on the surface of the ellipsoid to be found. In my code, C is called "c" (lower case); and C' is "x(:,imin)". > What you suggest is correct, only c lies on a line which runs from P1 to P2. This is a line through the ellipsoid. C' is the projection of c on the surface of the ellipsoid, where the angle between the line P1(or P2) to c and c to C' is 90 degrees. So I think here has been a misunderstanding. > > > > What should I do if my origin is not on [0,0,0], just change [X Y Z] = ellipsoid(0,0,0,radii(1),radii(2),radii(3),N); ? > If you know > 1) the radii: radii(1), radii(2), and radii(3) (three scalars in a vector) > 2) the corresponding orientation U1, U2, U3 (three vectors, they must be orthogonal and normed to 1) > 3) and the center x0 (a vector) of your ellipsoid > > Then define: > U = [U(:,1) U(:,2) U(:3)] > S = diag(1./radii.^2); > Q = U*S*U' Is Q even necessary? After this code, I let the engine start. But I have the feeling somethings going wrong. What I have now is this: C=[1.0150 -1.5380 -6.0150] cx=1; cy=-5;cz=1; % Center of my ellipsoid ax=12; ay=12; az=20; % radii of my ellipsoid radii(1)=ax; radii(2)=ay; radii(3)=az; U(:,1) = [0.1;0.2;1]; U(:,2) = [0.5;0.3;0.7]; U(:,3) = [0.6;0.6;0.4] % example of possible axis of my ellipsoid % Data, generate ellipse centered at origin, { x: x'*Q*x = 1 } c = C'-[cx; cy; cz] % c = 1*randn(n,1) U = [U(:,1) U(:,2) U(:,3)] S = diag(1./radii.^2) s = diag(S) % Q = U*s*U' A = U*diag(radii);
From: Bruno Luong on 24 May 2010 03:08 "Els " <y.e.t.reeuwijk(a)student.utwente.nl> wrote in message <htbpdd$e9i$1(a)fred.mathworks.com>... > > U(:,1) = [0.1;0.2;1]; U(:,2) = [0.5;0.3;0.7]; U(:,3) = [0.6;0.6;0.4] % example of > > possible axis of my ellipsoid > Axis of an ellipsoid MUST be orthogonal. You can't just put a random orientation like you did. Bruno
From: Els on 24 May 2010 04:55 Ok, you are correct. The orthogonal axis problem I solved now. Still my question about the P1 and P2 remains. Where can I give in these points? So that a line between P1 and P2 comes into existance, where c lies on. And C' is a projection of the c point perpendicular to this line. Really hope you can help me to solve this final problem.
From: Bruno Luong on 24 May 2010 05:28
"Els " <y.e.t.reeuwijk(a)student.utwente.nl> wrote in message <htdetc$c98$1(a)fred.mathworks.com>... > Ok, you are correct. The orthogonal axis problem I solved now. > > Still my question about the P1 and P2 remains. Where can I give in these points? So that a line between P1 and P2 comes into existance, where c lies on. And C' is a projection of the c point perpendicular to this line. Project the ellipsoid boundary on the plane perpendicular to P1-P2, then use 2D version of the algorithm. Bruno |