Airy Projection

A Map Projection. The inverse equations for $\phi$ are computed by iteration. Let the Angle of the projection plane be $\theta_b$. Define

$$a=\cases{ 0 & for $\theta_b={\textstyle{1\over 2}}\pi$\cr ... ...ver 2}}({\textstyle{1\over 2}}\pi-\theta_b)]} & otherwise.\cr}$$ (1)

For proper convergence, let $x_i=\pi/6$ and compute the initial point by checking

$$x_i=\left\vert{\mathop{\rm exp}\nolimits [-(\sqrt{x^2 + y^2} + a\tan x_i)\tan x_i]}\right\vert.$$ (2)

As long as $x_i>1$, take $x_{i+1}=x_i/2$ and iterate again. The first value for which $x_i<1$ is then the starting point. Then compute

$$x_i=\cos^{-1}\{\mathop{\rm exp}\nolimits [-(\sqrt{x^2+y^2} + a\tan x_i)\tan x_i]\}$$ (3)

until the change in $x_i$ between evaluations is smaller than the acceptable tolerance. The (inverse) equations are then given by

$$\phi = {\textstyle{1\over 2}}\pi-2x_i$$ (4)

$$\lambda = \tan^{-1}\left({-{x\over y}}\right).$$ (5)