Categories

# Equal Area Map Projections

I’m most interested in making equal-area projections of spheres onto different shapes. For example, turning a spherical map of Earth into regular polyhedra. I’ll talk about the math I’ve used for these, and show some of the results. And exellent reference for map projections is Map Projections: A Working Manual.

I’ll be using this equirectangular map of the Earth as the basis for my projections:

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 $R_s$.

Since I want to project a sphere onto flat surfaces, we can use the Jacobian determinant to figure out the local areal distortion. If we project the sphere onto a plane using cartesian $(x,y)$ or polar $(r,\theta_f)$ coordinates, we find the area ratio:

\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 \\ &= \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} \end{aligned}

For an equal-area projection, we want to find $\phi(x,y)$ and $\theta(x,y)$ such that $A_{ratio} =1$ at all points.

As a first demonstration, let’s make a rectangular equal-area projection. Let’s say that the two spherical coordinates only depend on one axis each: $\theta= a x$ and $\phi(y)$, with $\phi(y=0)=\pi$, $\phi(y=H)=0$, $\theta(x=0)=0$, $\theta(x=W)=2\pi$ where the width of the image is $w$ :

\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 \\ &= - \frac{d\phi}{dy} a \sin(\phi) R_s^2 \end{aligned}

$\cos(\phi) = \frac{2 y}{H}-1$

$A_{ratio} = \frac{4 \pi R_s^2}{W H}$

And as a result, here is an equal-area square map:

That’s quite distorted, however. So how can we minimize the distortion? The answer, of course, is math. But we need to change variables from $\phi$ to latitude $\Phi$, $\Phi = \pi/2-\phi$. From the book cited above, the maximum angular deformation, $\omega$, is:

\begin{aligned} \sin(\omega/2) &= b'/a' \\ a' &= \sqrt{ h^2+k^2 + 2 h k \sin(\theta') } \\ b' &= \sqrt{ h^2+k^2 - 2 h k \sin(\theta') } \\ h &= \sqrt{ (dx/d\Phi)^2+(dy/d \Phi)^2}/R_s \\ k &= \sqrt{ (dx/d\theta)^2+(dy/d \theta)^2}/(R_s \cos(\Phi)) \end{aligned}

With a lot of manipulation, we find

$\sin(\omega/2) = \frac{| \pi \cos(\Phi)^2 - (W/H) |}{\pi \cos(\Phi) ^2 + (W/H) }$

I’d like to find the value of $W/H$ that minimizes the average distortion. It turns out that the integral is exactly solvable, but horrible. Via numerics, the optimal aspect ratio is $W/H \approx 2.4445$. Take a look:

It does look better!

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