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So, I want to project a sphere onto a square with the poles at two opposite corners, preserving areas (long story, but I have a practical application in mind).

Wikipedia and other online sources lists the Collignon Projection as an equal-area pseudocylindrical projection that maps the sphere onto a triangle or diamond, which is close enough for me. However, I wasn't able to find any sources that went into any more detail than that.

Does anyone know the formula for the equal-area diamond Collignon Projection? Or could anyone help me re-derive it? (I'm unsure what needs to be done mathematically in order to ensure equal areas.)

This is the projection I want.

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I found myself in a similar situation. This website was the only derivation I could find though it's written in French. I've reviewed the article and now think I have enough of a grip on their derivation that I can attempt my own, hopefully simpler.

This type of projection was first derived for the case of a square by Collignon in 1865, then it was generalized to any rhombus by Tissout in 1881. I will provide the derivation for a square first then show how to adapt that for rhombuses.

enter image description here

So there's a square (NESW) whose area is equal to the area of a unit sphere, $4\pi$. A triangle like NEW has the area of a unit hemisphere, $a=2\pi$. By the Pythagorean theorem, the triangle legs (like NO) have a length of $b=\sqrt{2\pi}$. Since the projection is area preserving, if we draw a horizontal line like AC, we decide the triangle NAC should have the same area as a spherical cap, which is $2\pi(1-sin\theta)$ for some latitude $\theta$. We notice that NAC is just a scaled down version of NEW: it's area is scaled down by $\frac{2\pi(1-sin\theta)}{2\pi} = 1-\sin\theta$, and its legs are scaled down by the square root of that, $s = \sqrt{1-\sin\theta}$. So the length of NB is $sb$, and the cartesian coordinate $y = b-sb = (1-s)b$.

That just leaves us with finding $x$. To find $x$, we do a similar thing to another triangle, NOD. We decide that triangle NOD has the same area as a unit hemispherical wedge, or $c=2\pi \frac{\phi-\phi_0}{2\pi} = \phi-\phi_0$ for some longitude $\phi$, where $\phi_0$ is the reference longitude that's represented by NO. The length along OD must be $d=\frac{2c}{b}$, and if we scale $d$ by $s$ we find the length along FB. That leaves us with $x = sd$.

An easy way to generalize the projection to rhombuses might just be to apply an area preserving linear transformation to $x$ and $y$:

$$ \left[ {\begin{array}{cc} \sigma & 0 \\ 0 & 1/\sigma \\ \end{array} } \right] $$

for some value of $\sigma$. This maintains the equal area property of the projection while smooshing it to whatever ratio you like!

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