How to find inverse coordinates?

[Rick16]

TL;DR

given are x(u, v, w), y(u, v, w), z(u, v, w), and I want to find u(x, y, z), v(x, y, z), w(x, y, z)

Is there a systematic way to do it? In particular, I have the coordinates $x=au \sin v \cos w$, $y=bu\sin v\sin w$, $z=cu\cos v$, where a, b, c are constants, and I want to find $u(x,y,z)$, $v(x,y,z)$, $w(x,y,z)$. I could solve the three equations for u, v, and w and then try to insert the resulting equations into each other, and with a lot of fumbling around I may get the solutions that I want. But this does not seem to be the most rational approach.

 

[fresh_42]

I would first get rid of the constants and introduce $x'={(au)}^{-1}x\, , \,y'={(bu)}^{-1}y\, , \,z'={(cu)}^{-1}z$ and then operate with them like the usual polar coordinates, e.g. $x'^2+y'^2=\sin^2(v)$ etc.

 

[Rick16]

Thank you very much. I was able to solve it quickly with these substitutions.

 

[Gavran]

I would first get rid of the constants and introduce $x'={(au)}^{-1}x\, , \,y'={(bu)}^{-1}y\, , \,z'={(cu)}^{-1}z$ and then operate with them like the usual polar coordinates, e.g. $x'^2+y'^2=\sin^2(v)$ etc.

You lost the coordinate $u$ and you have to operate with the usual spherical coordinates.

 

[fresh_42]

You lost the coordinate $u$ and you have to operate with the usual spherical coordinates.

I did not.
\begin{align*}
x'^2+y'^2&=(au)^{-2}x^2+(bu)^{-2}y^2\\
&=(au)^{-2}(au \sin v \cos w)^2+(bu)^{-2}(bu\sin v\sin w)^2\\
&=\sin^2 v\left(\cos ^2w+\sin^2w\right)\\
&=\sin^2 v
\end{align*}
And, yes, my substitutions make a sphere out of the ellipsoid. This has to be reversed at the end of the calculations.

 

[Gavran]

I did not.
\begin{align*}
x'^2+y'^2&=(au)^{-2}x^2+(bu)^{-2}y^2\\
&=(au)^{-2}(au \sin v \cos w)^2+(bu)^{-2}(bu\sin v\sin w)^2\\
&=\sin^2 v\left(\cos ^2w+\sin^2w\right)\\
&=\sin^2 v
\end{align*}
And, yes, my substitutions make a sphere out of the ellipsoid. This has to be reversed at the end of the calculations.

I mean that your transformation from xyz-Cartesian coordinate system to x’y’z’-Cartesian coordinate system is wrong because in the equations you got the spherical coordinate $u$ can not be seen.
“get rid of the constants”, what you said and what is correct, does not mean to get rid of the constants and the coordinate $u$.

 

[fresh_42]

I mean that your transformation from xyz-Cartesian coordinate system to x’y’z’-Cartesian coordinate system is wrong because in the equations you got the spherical coordinate $u$ can not be seen.
“get rid of the constants”, what you said and what is correct, does not mean to get rid of the constants and the coordinate $u$.

This is not wrong. It might only be more work to do. I showed a path, not a solution. $u$ is a radius and scaling it back again is no problem.

 

[Gavran]

This is not wrong. It might only be more work to do.

So it is better to use the next transformations: $x'=a^{-1}x$, $y'=b^{-1}y$ and $z'=c^{-1}z$.