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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

0 & for $\theta_b={\textstyle{1\over 2}}\pi$\cr
...ver 2}}({\textstyle{1\over 2}}\pi-\theta_b)]} & otherwise.\cr}
\end{displaymath} (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.
\end{displaymath} (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]\}
\end{displaymath} (3)

until the change in $x_i$ between evaluations is smaller than the acceptable tolerance. The (inverse) equations are then given by
$\displaystyle \phi$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\pi-2x_i$ (4)
$\displaystyle \lambda$ $\textstyle =$ $\displaystyle \tan^{-1}\left({-{x\over y}}\right).$ (5)

© 1996-9 Eric W. Weisstein