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I have an oblate spheroid (E): (x/a)^2+(y/a)^2+(z/b)^2 = 1 and a plane (P): ux+vy+wz+d=0 / d=0 (passing by origin)

I have plugged z = - (ux+vy)/w from (P) into (E), then I became this equation:

[(wb)^2+(au)^2]x^2 + [(wb)^2+(av)^2]y^2 + (2*uva^2)xy -(wab)^2 = 0

That is the equation of conics (here in this case, an ellipse, bcz intersection of an plane passing by origin with an oblate spheroid is always an ellipse except when it is parallel to plane xy is a circle)

General equation of conics:

Ax^2 + Bxy + Cy^2 + F = 0 [Dx and Ey vanish bcz the ellipse centered always at O(0,0,0)]

I have rotated and everything done, the I got this form:

A'x^2 + C'y^2 - F' = 0 with A' = a^2*[u^2 + v^2] + (bw)^2 C' = (bw)^2 F' = (abw)^2

When I continue to find the semi-axis major a' and semi-axis minor b' of this ellipse, I get:

a' = a and b' = (abw)/sqrt[a^2*(u^2 + v^2) + (b*w)^2]

My problem, is: the semi-axis major a' shouldn't equal the semi-major of the oblate spheroid a at each case!!! What is wrong?!!

Thanks for any help, regards!

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The equation you got is the projection of the ellipse (intersection between spheroid and plane) on the $(x,y)$ plane. As your plane cuts the "equator" of the spheroid along a diameter, then yes: the semi-axis major $a'$ of the projection is equal to the semi-major of the oblate spheroid. And the same is true for the intersection, of course.

You can see below what happens: the plane $ACF$ through the origin (light gray) cuts the $(x,y)$ plane (dark gray) along line $AC$. But of course $AC$ is a diameter of the spheroid equator (black) and is the major axis of the intersection ellipse (blue). Line $EF$ lies on the plane but has no special meaning: it is there just to help you visualize the situation.

enter image description here

To convince you that $AC$ is indeed the major axis, here's the same diagram seen from "above".

enter image description here

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  • $\begingroup$ @Arentino thanks, you are right but what I look for, to calculate a' in general for each ellipse as an intersection and of course also b', passing by Origin O(0,0,0) and not only in this case when it's at equator $\endgroup$ – user412889 Feb 3 '17 at 19:36
  • $\begingroup$ If the plane passes by the origin then its intersection with the $(x,y)$ plane is a line through the origin, and the intersection of the spheroid with the $(x,y)$ plane is a circle (which I called "equator" for short) centered at the origin and of radius $a$. That line always cuts the equator along a diameter, which is also the major axis of the intersection. $\endgroup$ – Aretino Feb 3 '17 at 19:42
  • $\begingroup$ I don't know if my question is well asked and/or answered. But there is infinty planes passing with center O(0,0,0) but they all are not parallel or perpendiculat to plane xy, at this case, there is semi-axis major and minor $\endgroup$ – user412889 Feb 3 '17 at 19:45
  • $\begingroup$ Would be of help to make you see a diagram to convince you? Or should I suggest a simpler algebric approach? $\endgroup$ – Aretino Feb 3 '17 at 19:49
  • $\begingroup$ Could you see this? gnu.org/software/3dldf/graphics/elpsd_17.png $\endgroup$ – user412889 Feb 3 '17 at 19:52

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