I wanted an equal-area projection of the earth onto a disk. This is a simple one. We’ll use spherical coordinates, with polar angle \phi and azimuthal angle \theta. This means that latitude is, in radians, \pi/2-\phi, and longitude is \theta. We’ll say that the sphere has a radius of R_s.

We’ll have the North pole at the center of the circle, and the South pole around the perimeter. So we can have the angle on the flat map be \theta_f, the distance from the center is r_f, and the radius of the circle is R_f.

This simple projection means that we just need to find r_f(\phi). We can do that by matching the relative area covered from the North pole to a certain latitude further south:

\frac{1}{4\pi R_s^2} \int_0^{\phi} \int_0^{2\pi} R_s^2 \sin(\Phi) d\theta d\Phi = \frac{1}{\pi R_f^2} \int_0^{r_f} \int_0^{2\pi} r d\theta dr .This lets us solve for \phi:

\cos(\phi) = 1-2 \left( \frac{r_f}{R_f} \right)^2 .

And now we can sample the map at the proper coordinates to make our result:

This really distorts a lot of the globe. Especially as we get to the edge: Antarctica is unrecognizable.

Now, you may wonder, how do we know that this preserves area? Things look quite distorted. Well, we can do a bit of math, of course. Basically, we’ll see how the areal element changes in the mapping:

\begin{aligned} A_{ratio} &= \left( \frac{d\phi}{dx} \frac{d\theta}{dy} - \frac{d\phi}{dy} \frac{d\theta}{dx} \right) \sin(\phi) R_s^2 \\ &= 4 \left( \frac{R_s}{R_f} \right)^2 \end{aligned}

So if R_f = R_s/2, then this is an equal-area mapping. But since there’s no dependence on the coordinates, that means that the area ratio is equal at every point. Because of that, we can scale down the plate, and each point will be equally proportional to the original. So if we want a plate with an areal ratio of one to one quadrillion (1 square millimeter of the plate yields 1000 square kilometers of Earth), the plate will have a diameter of 80.6 cm. Neat.