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I'm currently looking for a projection of (1) the oblate spheroid onto (2) the slanted plane.

(1) Having the oblate spheroid: $\frac {x^2}{a^2}+\frac {y^2}{b^2}+\frac {z^2}{c^2}=1$ with $a=b>c$ and $\varphi_{ca}=\frac {c}{a}$

Its current position could be represented by a specific direction (azimuth angle $\alpha$ and elevation angle $\epsilon$).

enter image description here

(2) and the slanted plane: $tan(\theta)\cdot x+z=0$ with $0\leq\theta<\pi/2$

I succeeded in finding the length of semi-major and semi-minor axis of the projection (ellipse), if $\theta=0$ (i.e., projection onto the x-y plane).

I referred to the following link: enter link description here

enter image description here

$\therefore\bar{OG}=a$ and $\bar{OH}=a\cdot \sqrt{cos^2(\epsilon)+\varphi_{ca}^2\cdot sin^2(\epsilon)} $

All points above the ellipse can be represented by $\bar{OG}$, $\bar{OH}$, and azimuth angle $\alpha$.

Likewise, I want to get the distance between the origin and a particular point on the projection when $\theta\neq0$.

Is there anyone who can refer to the material that explains this concept? I'd be very grateful if anyone could help me with this.

Thanks for any help, regards!

Jiseob Kim
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  • You can find the angle $\phi$ between the z-axis of the ellipsoid and the normal to the plane. Then the new $\epsilon$ will be $\frac{\pi}{2} -\phi$. Then you can use the same equations of the semi-minor and semi-major axes as in your question. –  Jul 20 '21 at 08:49
  • @GeometryLover Thank you for your concern for my question. I applied the new $\epsilon$ according to your suggestion. This way is definitely valid when azimuth angle $\alpha=0$. But, if the ellipsoid rotates in the azimuth direction ($\alpha\neq0$), it becomes more complicated. I still don't know how to get the answer. – Jiseob Kim Jul 20 '21 at 09:37
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    My understanding is that the $z$-axis of the spheroid (corresponding to the $c$ semi-axis) is pointing in the direction (unit vector) $v = (\cos \epsilon \cos \alpha, \cos \epsilon \sin \alpha, \sin \epsilon)$, and the normal to the plane is pointing in the direction $n = (\sin \theta, 0, \cos \theta)$. Using dot product between $v$ and $n$, you get $\cos \phi = v \cdot n$, from which $\epsilon = \frac{\pi}{2}- \phi$ –  Jul 20 '21 at 10:29
  • In the above comment, at the end, I should have written $\epsilon' = \frac{\pi}{2} - \phi $ where $\epsilon'$ is the new $\epsilon$ corresponding to a plane of projection other than the $xy$ plane. –  Jul 20 '21 at 10:55
  • Once you find the semi-major and semi-minor axes of the ellipse of projection, you need to find their directions in the plane of projection. And for that, you need to project the $z$-axis of the spheroid onto the plane, and this will give the direction of the semi-minor axis, then the direction of the semi-major axis will be perpendicular to that. The projection of the $z$ axis onto the plane is given by $v' = [I - n n^T] v $ where $v$ and $n$ are defined in my previous comment. –  Jul 20 '21 at 11:01
  • @GeometryLover Sorry for the delay in my replying. I was very impressed by your detailed response. I think I understand now. I'll try to solve this matter again, and maybe your idea will help me a lot. I appreciate your effort! – Jiseob Kim Jul 21 '21 at 02:56

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