Two-point equidistant projection of the sphere According to this Wikipedia article, there is a projection from the $2$-sphere to a region in the plane that preserves distances to two given points. The article says that the projection was first described by someone called Hans Maurer in 1919 and that it has been used in cartography. I have been trying to find the equations describing the projection, but I cannot find them. Does anyone know about this projection?
 A: You can reconstruct it by yourself.
Choose two reference points and obtain their Cartesian coordinates. Now take any other point and convert to Cartesian coordinates. The dot product of the vectors give the cosines of the terrestrial distances (on a unit sphere). Now you have the three sides of a flat triangle on the projected map.
The process can be inverted, but for now I don't see a better way than by solving the spherical triangle.

The terrestrial distances are given by
$$d_{ij}=\arccos(\cos\theta_i\sin\phi_i\cos\theta_j\sin\phi_j+\sin\theta_i\sin\phi_i\sin\theta_j\sin\phi_j+\cos\phi_i\cos\phi_j).$$

To solve the flat triangle in a Cartesian system, you can compute an angle formed with the base by the cosine law and use polar coordinates (angle and length of the side). Denoting $a,b$ the reference points,
$$\psi=\hat{bap}=\arccos\frac{d_{ap}^2+d_{bp}^2-d_{ab}^2}{2d_{ap}d_{bp}}$$
and
$$\rho=d_{ap}.$$

I cannot resist giving the complete expression in terms of the spherical coordinates of the point to be projected.
$$\begin{cases}
\psi_p=\arccos\dfrac{\arccos^2(\cos\theta_p\sin\phi_px_a+\sin\theta_p\sin\phi_py_a+\cos\phi_pz_a)+\arccos^2(\cos\theta_p\sin\phi_px_b+\sin\theta_p\sin\phi_py_b+\cos\phi_pz_b)-d_{ab}^2}{2\arccos(\cos\theta_p\sin\phi_px_a+\sin\theta_p\sin\phi_py_a+\cos\phi_pz_a)\arccos(\cos\theta_p\sin\phi_px_b+\sin\theta_p\sin\phi_py_b+\cos\phi_pz_b)},\\
\rho_p=\arccos(\cos\theta_p\sin\phi_px_a+\sin\theta_p\sin\phi_py_a+\cos\phi_pz_a)
.\end{cases}$$
In Cartesian coordinates,
$$\begin{cases}
x_p=\dfrac{\arccos^2(\cos\theta_p\sin\phi_px_a+\sin\theta_p\sin\phi_py_a+\cos\phi_pz_a)+\arccos^2(\cos\theta_p\sin\phi_px_b+\sin\theta_p\sin\phi_py_b+\cos\phi_pz_b)-d_{ab}^2}{2\arccos(\cos\theta_p\sin\phi_px_b+\sin\theta_p\sin\phi_py_b+\cos\phi_pz_b)},\\
y_p=\sqrt{\arccos^2(\cos\theta_p\sin\phi_px_a+\sin\theta_p\sin\phi_py_a+\cos\phi_pz_a)-x_p^2}.\end{cases}$$
