Numerical Examples for Miller Cylindrical Projection #

$$ \DeclareMathOperator{\arcsinh}{arcsinh} $$

$$ \eqalign{ x &= 1.0\times[-75^\circ-0^\circ]\times\pi/180^\circ \cr &= -1.3089969\;\text{units} } $$

$$ \eqalign{ y =& 1.0\times[\ln{\tan(45^\circ+0.4\times50^\circ)}]/0.8 \cr =& (\ln{\tan 65^\circ)}/0.8 \cr =& 0.9536371\;\text{units} } $$

$$ \eqalign{ y &= 1.0\times\{\arcsinh[\tan(0.8\times50^\circ)]\}/0.8 \cr &= 1.0\times\arcsinh{0.8390996}/0.8 \cr &= 0.9536371\;\text{units} } $$

$$ h = \sec(0.8\times50^\circ) = 1/\cos40.0000000^\circ = 1.3054073 $$

$$ k = \sec50^\circ = 1/\cos50^\circ = 1.5557238 $$

$$ \eqalign{ \sin{1/2\omega} &= [\cos(0.8\times50^\circ) - \cos50^\circ]/[\cos(0.8\times50^\circ) + \cos50^\circ] \cr &= 0.0874887 } $$

$$ \omega=10.0382962^\circ $$

$$ \eqalign{ \phi &= 2.5\arctan \mathrm{e}^{0.8\times0.9536371/1.0}-(5\pi/8)\times180^\circ/\pi \cr &= 2.5\arctan \mathrm{e}^{0.7629097} - 1.9634954\times180^\circ/\pi \cr &= 2.5\arctan (2.1445070) - 1.9634954\times180^\circ/\pi \cr &= 2.5\times65.0000006^\circ - 112.5000000^\circ \cr &= 50.0000015^\circ } $$

$$ \eqalign{ \phi &= \arctan[\sinh(0.8\times(-0.8428443)/1.0)]/0.8 \cr &= \arctan (-0.7265425)/0.8 \cr &= -44.9999978^\circ } $$

$$ \eqalign{ \lambda &= -5^\circ + (-1.2217305/1.0)\times 180^\circ/\pi \cr &= -5^\circ + (-70.0000014^\circ) \cr &= -75.0000014^\circ } $$