Categories
map

Variable-Radius Flat Earth

No, not the conspiracy theory. For example, here is an Earth-splat:

And here is an Earth-square:

And both are equal-area projections! How did I do this? Math. Math math math.

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 can have the angle on the flat map be \theta_f, the distance from the center is r_f, and the radius of the shape at each angle is R(\theta_f).

Since the radius changes, we can’t just say that \theta_f = \theta. Instead, use integrals to match the total area, finding a relation between the two:
\begin{aligned} A_{ratio} \int_0^{\theta_f} \int_0^{R(t)} r dr dt &= \int_0^{\theta} \int_0^{R(t)} R_s^2 \sin(\phi) d\phi dt \\ A_{ratio} \int_0^{\theta_f} \frac{1}{2} R^2(t) dt &= 2 \theta R_s^2 \\ \int_0^{\theta_f} \frac{R^2(t) A_{ratio}}{4 R_s^2 } dt &= \theta \end{aligned}

We want the area ratio to be a constant, and we know \frac{d\theta}{d\theta_f} = \frac{R^2(\theta_f)A_{ratio}}{4 R_s^2 } from the integral, and that \frac{d\theta}{dr} = 0 . This yields

\begin{aligned} A_{ratio} &= \left( \frac{d\phi}{dr} \frac{d\theta}{d\theta_f} - \frac{d\phi}{d\theta_f} \frac{d\theta}{dr} \right) \frac{\sin(\phi) R_s^2}{r} \\ &= \frac{d\phi}{dr} \frac{R^2(\theta_f)A_{ratio} }{4 }\frac{\sin(\phi) }{r} \end{aligned}

We can separate variables and integrate to find \cos(\phi) = 1-2\frac{r^2}{R^2(\theta_f)} .

Leave a Reply

Your email address will not be published.

This site uses Akismet to reduce spam. Learn how your comment data is processed.