Numerical Examples - Van der Grinten Projection

Numerical Examples for Van der Grinten Projection #

SPHERE #

Forward Equations #

Given

 Radius of sphere: $R=\;\;$ unit Central meridian: $\lambda_0=\;$° Point: $\phi=\;$° $\lambda=\;$°
Find $x, y$.

From equations (29-6), (29-3), (29-4), (29-5), and (29-6a) in order,

\eqalign{ \theta &= \arcsin |2\times(-50^\circ)/180^\circ| \cr &= 33.7489886^\circ }
\eqalign{ A &= ½|180^\circ/(-160^\circ-(-85^\circ))-(-160^\circ-(-85^\circ))/180^\circ| \cr &= ½|(-2.4000000) - (-0.4166667)| \cr &= 0.9916667 }
\eqalign{ G &= \cos33.7489886^\circ/(\sin33.7489886^\circ+\cos33.7489886^\circ-1) \cr &= 2.1483315 }
\eqalign{ P &= 2.1483315/(2/\sin33.7489886^\circ - 1) \cr &= 5.5856618 }
\eqalign{ Q &= 0.9916667^2+2.1483315 \cr &= 3.1317343 }

From equation (29-1),

\eqalign{ x =& -\pi\times1.0\{ 0.9916667\times(2.1483315-5.5856618^2) \cr & +[0.9916667^2\times(2.1483315 - 5.5856618^2)\cr & -(5.5856618^2 + 0.9916667^2)\times(2.1483315^2 - 5.5856618^2)]^{1/2}\}/ \cr & (5.5856618^2+0.9916667^2)\cr =& -1.1954154\text{ units} }
taking the initial “—” sign because $(\lambda-\lambda_0)$ is negative. Note that $\pi$ is not converted to $180^\circ$ here, since there is no angle in degrees to offset it. From equation (29-2),
\eqalign{ y =& -\pi\times1.0\{ 5.5856618\times3.1317343 - 0.9916667\cr & \times[(0.9916667^2+1)\times(7.0000000+0.9916667^2) \cr & -3.1317343^2]^{1/2} \}/(5.5856618^2 + 0.9916667^2) \cr =& -0.9960733\text{ units} }

taking the initial “—” sign because $\phi$ is negative.

Inverse Equations #

Inversing forward example:

Given: $R, \lambda_0$, for forward example

 $x=\;$ units $y=\;$ units
Find $\phi, \lambda$.

Using equations (29-9) through (29-19) in order,

\eqalign{ X &= -1.1954154/(\pi\times1.0) \cr &= -0.3805125 }
\eqalign{ Y &= -0.9960733/(\pi\times1.0) \cr &= -0.3170600 }
\eqalign{ c_1 &= -|-0.3170600|\times(1+(-0.3805125)^2+(-0.3170600)^2)\cr &= -0.3948401 }
\eqalign{ c_2 &= -0.3948401-2\times(-0.3170600)^2+(-0.3805125)^2 \cr &= -0.4511044 }
\eqalign{ c_1 &= -|-0.3170600|\times[1+(-0.3805125)^2+(-0.3170600)^2]\cr &= -0.3948401 }
\eqalign{ d =& (-0.3170600)^2/2.0509147+[2\times(-0.4511044)^3/2.0509147^3 \cr & - 9\times(-0.3948401)\times(-0.4511044)/2.0509147^2]/27 \cr =& 0.0341124 }
\eqalign{ a_1 =& [(-0.3948401)-(-0.4511044)^2/(3\times2.0509147)]/2.0509147\cr =& -0.2086455 }
\eqalign{ m_1 &= 2\times[-(-0.2086455)/3]^{1/2} \cr &= 0.5274409 }
\eqalign{ \theta_1 &= ⅓\arccos(3\times0.0341124/((-0.2086455)\times0.5274409)) \cr &= ⅓\arccos(-0.9299322) \cr &= 52.8080829^\circ }
\eqalign{ \phi =& -180^\circ\times[(-0.5274409)\cos(52.8080829^\circ+60^\circ) \cr & (-0.4511044)/(3\times0.5274409)] \cr =& -49.9999985^\circ }
taking the initial “—” sign because $y$ is negative.