# map on a unit sphere with polar coordinates

My brother, who is in hospital atm and cannot verify by himself asked me to post the following question, thank you in advance, and sorry if the topic has already been covered, i do not have the math knowledge to look for it in the forums.

He wants to map, on the unit sphere with polar coordinates, the square with vertices $(0,1)$, $(1,0)$. $(-1,0)$, $(0,-1)$ with centre $(0,0)$ having $(0,1)$ and $(0,-1)$ as poles.

He wants to know if the map $(a,b)\mapsto (a\pi , \arcsin(b))$ is adequate or at least it works.

He would also like to know if there's a more clever way to do that.

Thank you.

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This would map the point $(0,1)$ to the point $\left(0,\frac{\pi}{2}\right)$, which is not on the unit circle. – Joe Johnson 126 Oct 8 '12 at 10:56
I don't see why people are voting to close. All of this personal information may be irrelevant, but the question is perfectly valid. – Alexander Gruber Mar 12 '13 at 21:01

In my understanding, a polar coordinate system is usually used to describe points in the plane. You appear to be looking for something describing points on the (unit) sphere. So I guess you're actually referring to a geographic coordinate system.

The geographic coordinate system has a mostly rectangular structure: you have latitudes from $-\frac\pi2$ to $\frac\pi2$ and for every latitude you have longitudes from $-\pi$ to $\pi$. You could get your square into this shape using

$$\begin{pmatrix}a\\b\end{pmatrix}\mapsto \begin{pmatrix}\frac{a}{1-\lvert b\rvert}\pi\\ b\frac\pi2\end{pmatrix}$$

There is a slight problem, however: At the poles, the longitude is undefined. So there are many different coordinates representing the same point. In the above formula, this undefinedness manifests itself as a division by zero. You could make a case distinction and define that in those cases where you would have to divide by zero, you simply assume the latitude to be zero, as it doesn't really matter.

You also have an overlap at the $\pm\pi$ meridian, where the two edges of your input triangle represent the same points on the sphere. If this matters, you'd have to remove two edges from your input square, except for the corners corresponding to the poles.

On the whole, I doubt that any of the above will be particularly useful for practical applications, as the transformation defined in this way probably won't fulfill any of the requirements you'd expect from a projection. Except perhaps being simple to calculate.

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