Ellipsoid and Plane Intersection EquationDate: 04/20/2000 at 09:07:09 From: Martin Sheen Subject: Solid Geometry (Ellipsoids) I need the equation for the intersection of a plane surface with an ellipsoid. For example, if the ellipsoid has half-axes a, b, and c, and the plane is normal to the vector [i,j,k] and also passes through the point (i,j,k), what are the half-axes and orientation of the ellipse of intersection? If (i,j,k) lies on an axis the answer is easy enough, but I'm stuck for the general case. I'm not a student, but a physicist doing research in health care. I would be most grateful for any pointers. Date: 04/20/2000 at 14:04:17 From: Doctor Rob Subject: Re: Solid Geometry (Ellipsoids) Thanks for writing to Ask Dr. Math, Martin. The intersection you describe will be a closed curve in space. As such, it doesn't have a single equation. Single equations describe surfaces, not curves. You will either need two equations in x, y, and z, or else three equations for x, y, and z in terms of some parameter. Of course, you already have two equations: those of the ellipsoid and the plane: x^2/a^2 + y^2/b^2 + z^2/c^2 = 1 (x-i)*i + (y-j)*j + (z-k)*k = 0 Apparently that is not what you want, however. You can find the center of the ellipse using the following technique. Find the projection of the ellipse on the xy-plane by solving the second equation for z and substituting that into the first equation. Bring this equation into general form A*x^2 + 2*B*x*y + C*y^2 +2*D*x + 2*E*y + F = 0 by expanding and collecting like terms. Then if the center of the projection has coordinates (H,K), you can find them by solving the simultaneous equations A*H + B*K = -D B*H + C*K = -E These will be the x- and y-coordinates of the center of the ellipse before projection. Now repeat this procedure with the projection of the ellipse on the xz-plane, by solving for y and substituting, and finding the center of that projection. It will have coordinates (x,z) = (H,L). Then the center of the ellipse in space will be at (H,K,L). If I did the calculations correctly, H = a^2*i*(i^2+j^2+k^2)/(a^2*i^2+b^2*j^2+c^2*k^2) K = b^2*j*(i^2+j^2+k^2)/(a^2*i^2+b^2*j^2+c^2*k^2) L = c^2*k*(i^2+j^2+k^2)/(a^2*i^2+b^2*j^2+c^2*k^2) (a rather pleasing set of formulae, I think). To find the direction and magnitude of the major axis, you can use the following. Let the square distance from a point (x,y,z) on the ellipse from the center (H,K,L) be f(x,y,z) = (x-H)^2 + (y-K)^2 + (z-L)^2 You want to maximize this value subject to the constraints of the two equations of the ellipsoid and plane. To do this, solve the ellipsoid equation for z and substitute that in the last equation above, and the plane equation. Then solve the plane equation for y, and substitute that into the above equation. This will eliminate both y and z, so that f is a function of x alone (but a rather complicated one). Now find its derivative df/dx, set that equal to zero, and solve for x. You should get four values of x. Substituting these into earlier equations, you can solve for the corresponding values of y and z. These four points will be the endpoints of the major and minor axes of the ellipse in space. Then it is a simple matter to obtain the lengths and directions of the half-axes. To do all this symbolically is quite a mess, but with actual numbers this should be quite feasible. Try it and see. I hope this satisfies your request. If not, write again. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 05/02/2000 at 12:41:55 From: Martin Sheen Subject: Re: Solid Geometry (Ellipsoids) Very many thanks for your extremely prompt reply. You did invite me to write again; may I trespass on your time further? Sorry; I wasn't clear quite about what I wanted - I was trying to be brief. I was actually thinking of the equation of the two-dimensional ellipse in the coordinate frame of the intersecting plane. Easy enough to transform, of course. What I'm doing is viewing a section through an ellipsoidal eye from a viewpoint normal to the intersecting plane, and displaying the intersection on that plane along with a projection of the eye structures and isodose lines for planning radio-therapy treatments. From your help I've got the ellipse center (H,K,L) okay (in eye coordinates), but I'm a bit stuck with the half-axes. Solving the ellipsoid equation for z and substituting into f(x,y,z) gives an f(x,y) with terms not only in x^2 and y^2 but also a Sqrt[1 - (x/a)^2 - (y/b)^2], and substituting into the plane equation will give terms in x and y and the same Sqrt[...]. If we get rid of the square roots we'll get simultaneous equations with fourth-order and second-order terms respectively, and I don't know how to cope with that. If instead we solve the plane equation for z and substitute that into f(x,y,z) and the ellipsoid equations, we get two simultaneous equations of the form A*x^2 + 2*B* x*y + C*y^2 + ..., which may be simpler but I'm still stuck. So; sorry to bother you again, but is there an alternative? The difference between the ellipsoid axes is not going to be extreme, and absolute accuracy is not required, so if there's a simpler approximation assuming only first-order difference from a sphere, that would probably do. Very many thanks again for your time. Martin Sheen Date: 05/02/2000 at 16:32:39 From: Doctor Rob Subject: RE: Solid Geometry (Ellipsoids) Thanks for the further explanation. You can rotate coordinates so that the x'y'-plane is parallel to the plane in question. This you can do as follows. Let (i,j,k) be direction numbers for the z' axis in the xyz-system. Then (-k,0,i) is a perpendicular direction (rather arbitrarily chosen among all such), which we'll make the direction numbers for the x' axis. Then we can normalize these to direction cosines lambda3 = i/sqrt(i^2+j^2+k^2) mu3 = j/sqrt(i^2+j^2+k^2) nu3 = k/sqrt(i^2+j^2+k^2) lambda1 = -k/sqrt(i^2+k^2) mu1 = 0 nu1 = i/sqrt(i^2+k^2) To find the direction cosines of the y' axis, they are lambda2 = i*j/[sqrt(i^2+k^2)*sqrt{i^2+j^2+k^2)] mu2 = -(i^2+k^2)/[sqrt(i^2+k^2)*sqrt{i^2+j^2+k^2)] nu2 = j*k/[sqrt(i^2+k^2)*sqrt{i^2+j^2+k^2)] You can verify that the three direction vectors are mutually orthogonal unit vectors, so will represent a rotation of coordinates which will not change distances (an *isometry* or *orthogonal transformation*). Now make the substitution x = lambda1*x' + lambda2*y' + lambda3*z' y = mu1*x' + mu2*y' + mu3*z' z = nu1*x' + nu2*y' + nu3*z' This will transform the equation of the ellipsoid into a rather general second-degree equation in x, y, and z, and the equation of the plane will be transformed into z' = sqrt(i^2+y^2+z^2) Then you can substitute this value for z' in the equation of the ellipsoid. This will yield the x'y'-equation of the ellipse in the general form A*x'^2 + 2*B*x'*y' + C*y'^2 + 2*D*x' + 2*E*y' + F = 0 Next you'll want to rotate these coordinates in order to remove the x'*y' term. This is done by determining any angle t such that tan(2*t) = 2*B/(A-C) Then the transformation is x' = cos(t)*x" - sin(t)*y" y' = sin(t)*x" + cos(t)*y" z' = z" Now it is just a matter of completing the square on x" and y", moving the constant term to the other side of the equation, and dividing by it to bring the equation of the ellipse into the standard form (x"-h')^2/a'^2 + (y"-k')^2/b'^2 = 1 Then you can read off the semi-major and semi-minor axis lengths a' and b' from this form, and the center (h',k',sqrt(i^2+j^2+k^2)) in the x"y"z"-coordinate system. Finally you can transform this back into the old coordinate system using the substitution equations given above, if desired. Alternatively, you can translate the coordinate system via the substitution X = x" - h' Y = y" - k' Z = z" - sqrt(i^2+j^2+k^2) and the equation of the ellipse and the plane become X^2/a'^2 + Y^2/b'^2 = 1 Z = 0 I am not sure which of the coordinate systems above may have been what you had in mind when you wrote, "the equation of the two-dimensional ellipse in the coordinate frame of the intersecting plane." You pick. If you would like me to work an example, I can oblige. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 05/08/2000 at 09:20:27 From: Martin Sheen Subject: RE: Solid Geometry (Ellipsoids) Again, I'm most grateful for your time and effort. This is just what I need. Coding it without any mistakes is another matter, of course, but that's a familiar problem. All the best, Martin Sheen |
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