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I am searching the spherical coordinates for the circular edge that are obtained when a sphere is cut at a certain position with a plane. The sphere has herby a radius $r$ and is focused at the center of a coordinate system. The plane cut is performed at a certain $x, y,$ or $z$ position (see an exemplary cut in the linked image).

https://i.stack.imgur.com/7MFgW.png

What I am now interested in is finding the parametrization of the cutting edge, however not as parametrization of a circle, but instead in spherical coordinates of the sphere. This means I want to find the coordinates of every point on the cut, expressed in the spherical coordinate system. For a cut through the z-plane the solutions looks trivial with a azimuth angle changing between $0$ and $2\pi$ and a fixed elevation angle, as seen in the exemplary image. However the solution is not trivial for a cut through the $x$, or $y$ plane.

Does anyone know the solution for it?

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  • $\begingroup$ The cut is performed either perpendicular to the x, y or z axis. For a cut perpendicular to the z-axis the solution is trivial. So, the main challenge lies in finding the solution for the cut through the x or y axis or beyond that a generalized formula valid for a perpendicular cut through any of the 3 planes. $\endgroup$
    – pffelix
    Nov 9, 2019 at 19:36
  • $\begingroup$ You mean the plane of small circle is not perpendicular to any of the three ${x,y,z}$ axes? $\endgroup$
    – Narasimham
    Nov 10, 2019 at 9:31
  • $\begingroup$ in my case the plane is perpendicular to one of the axes, however, a generalized solution might be of interested for some others. $\endgroup$
    – pffelix
    Nov 11, 2019 at 6:20
  • $\begingroup$ You need to apply two Euler angle matrix rotations on rigid body (sphere). $\endgroup$
    – Narasimham
    Nov 11, 2019 at 7:48
  • $\begingroup$ @Narasimham Good idea for a second solution. As far as I understand the topic , Euler angle transformation should be connected to Rodrigues' rotation formula. Do you think such a solution will lead to a different result than the formula below? If yes, do you see a possiblity to derive it as second solution? $\endgroup$
    – pffelix
    Nov 13, 2019 at 14:10

1 Answer 1

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The equations of the cut at some $x_p$ are given by $$x^2+y^2+z^2=r^2$$ and $$x=x_p$$ We can the write $$y^2+z^2=r^2-x_p^2$$ Now plug in the expression for $y$ and $z$ in polar coordinates:$$\begin{align}z&=r\cos\theta\\y&=r\sin\theta\sin\phi\end{align}$$ You get $$\cos^2\theta+\sin^2\theta\sin^2\phi=1-\frac{x_p^2}{r^2}$$ To get the limits, note that $\cos\theta$ varies between $\pm\sqrt{1-\frac{x_p^2}{r^2}}$. Then you can write the expression for $\sin\phi$ in terms of $\theta$.

If you do a cut at $y_p$ instead, just replace $x_p$ with $y_p$ and use $\cos\phi$ instead of $\sin\phi$.

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  • $\begingroup$ Thanks i tried out the solution and it works ;) Do you see a possiblity to solve the last equation also for $\theta$ in terms of $\phi$? $\endgroup$
    – pffelix
    Nov 13, 2019 at 13:58
  • $\begingroup$ To the solutions should be noted that the the squared cosine and sine terms and the squared cut $x_p$ lead to a paramterized quarter circle, however the other part of the circle can be easily concluded by symmetry. $\endgroup$
    – pffelix
    Nov 13, 2019 at 14:14
  • $\begingroup$ To answer your first question, I don't have an easy way of saying what the limits for $\phi$ should be $\endgroup$
    – Andrei
    Nov 13, 2019 at 15:42

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