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March 21

Mapping the sphere to the plane

Suppose I have a unit sphere, described using spherical polar coordinates. A surface area element has area $\mathrm {d} A=\sin \theta \mathrm {d} \theta \mathrm {d} \phi$ .

Next, suppose I want to project this to a plane in such a way that distances from the point with $\theta =0$ are correct. This requires that my projection, if described in plane polar coordinates, has radius $r=\theta$ .

Further, I want areas on the sphere to correspond to areas on the plane. Since an area element on the plane is given by $\mathrm {d} A=r\mathrm {d} r\mathrm {d} \Phi$ , by comparing area elements it seems irresistible to conclude that one way of achieving this is by $\Phi ={\frac {\sin \theta }{\theta }}\phi$ , or perhaps $\Phi ={\frac {\sin \theta }{\theta }}(\phi -\pi /2)+\pi /2$ . Obviously, this won't fill the whole plane, but this is not my intention.

Does this work, or have I gone wrong somewhere?--Leon (talk) 21:10, 21 March 2016 (UTC)[reply]

Various possibilities are discussed at Map_projection#Equal-area. Sławomir
Biały 21:17, 21 March 2016 (UTC)[reply]

I think that's the Werner projection (which does have the properties you want). -- BenRG (talk) 07:02, 22 March 2016 (UTC)[reply]