0
$\begingroup$

Having an oblate spheroid (S): (x/a)^2+(y/a)^2+(z/b)^2=1 eqt(1) with a>b

and a plane not passing through the origin (P): ux+vy+wz+D=0 eqt(2)

The result of this intersection is an ellipse (except special cases circle or a point). I would get the semi-axis length (major and minor) and the coordinates of the center of this ellipse.

I tried:

from eqt(2) I get z = -(ux+vy+D)/w eqt(3) then I subtitute z in eqt(1) to get an equation in general conic quadratic form Ax^2+Bxy+Cy^2+Dx+Ey+F=0

With

A = (b*w)^2 + (a*u)^2
B = 2*u*v*a^2
C = (b*w)^2 + (a*v)^2
D = 2*u*a^2
E = 2*v*a^2
F = (a*D)^2 - (a*b*w)^2

Numerical example: (S): (x/20)^2 + (y/20)^2 + (z/16)^2 =1 and (P): 3x -2y +4z -8 =0

that yields:

A = 7696
B = -710740992
C = 5696
D = 2400
E = -1600
F = -1612800

Somehow there is an error and probably by B and F

The derivation of semi-axis major and minor lengths, the coordinates of the center from the general quadratic conic form, I get from Osmund's answer 1 and answer 2 But it seems that there is a problem by him in the equation where he derives the distance center-focus, isn't?

$\endgroup$
  • $\begingroup$ You haven’t derived the equation of the orthogonal projection of the intersection curve onto the $x$-$y$ plane, not that of the curve itself. The center of this projected curve will have the same $x$- and $y$- coordinates as the center of the actual intersection, but the projection will be foreshortened relative to the actual intersection curve, so regardless of the correctness of the linked formulas, you’re going to have to make some adjustments to the values that you get from them. $\endgroup$ – amd Mar 20 '18 at 1:10
  • $\begingroup$ What will you do when $w=0$? $\endgroup$ – amd Mar 20 '18 at 1:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.