Numerical Examples - Sinusoidal Projection

# Numerical Examples for Sinusoidal Projection #

## SPHERE #

### Forward Equations #

Given

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

From equations (30-1) through (30-5) in order,

\eqalign{ x &= 1.0\times[-75^\circ-(-90^\circ)]\cos(-50^\circ) \cr &= 0.1682814\text{ units} }
\eqalign{ y &= 1.0\times(-50^\circ)\times\pi/180^\circ \cr &= -0.8726646\text{ units} }
\eqalign{ h &= \{1+ [-75^\circ-(-90^\circ)]^2\times(\pi/180^\circ)^2\times\sin^2(-50^\circ) \}^{1/2} \cr &= 1.0199119 }
\eqalign{ k &= 1.0 }
\eqalign{ \theta' &= \arcsin(1/1.0199119) \cr &= 78.6597719^\circ }
\eqalign{ \omega &= 2\arctan[(1/2)[-75^\circ-(-90^\circ)]\times(\pi/180^\circ)\times\sin(-50^\circ)] \cr &= 11.4523842^\circ }

### Inverse Equations #

Inversing forward example:

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

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

From equations (30-6) and (30-7),

\eqalign{ \phi &= (-0.8726646/1.0)\times180^\circ/\pi \cr &= -49.9999985^\circ }
\eqalign{ \lambda &= -85^\circ+0.1682814/[1.0\cos(-49.9999985^\circ)]\times180^\circ/\pi \cr &= -70.0000007^\circ }

## ELLIPSOID #

### Forward Equations #

Given

 Clarke 1866WGS-84 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00676866 Central meridian: $\lambda_0=$ ° Point: $\phi=$ ° $\lambda=$ °
Find: $x, y$.

Using equations (30-8), (3-21), and (30-9) in order,

\eqalign{ x =& 6378206.4\times[-75^\circ-(-90^\circ)]\times(\pi/180^\circ)\cos(-50^\circ)/ \cr & [1-0.0067687\sin^2(-50^\circ)]^{1/2} \cr =& 1075471.54\text{ m} }
\eqalign{ M =& 6378206.4\times[(1-0.0067687/4-3\times0.0067687^2/64 \cr & -5\times0.0067687^3/256)\times (-50^\circ)\times \pi/180^\circ \cr & -(3\times0.0067687/8+3\times0.0067687^2/32 \cr & +45\times0.0067687^3/1024)\times\sin(2\times(-50^\circ)) \cr & +(15\times0.0067687^2/256+45\times0.0067687^3/1024) \cr & \times \sin(4\times(-50^\circ)) -(35\times0.0067687^3/3072) \cr & \times\sin(6\times(-50^\circ))] \cr =& -5540628.03\;\text{m} }
$$y = -5540628.03\text{ m}$$

### Inverse Equations #

Inversing forward example:

Given

 $x=\;$m $y=\;$m
Find: $\phi, \lambda$

Using equations (30-10), (7-19), (3-24), (3-26), and (30-11) in order,

$$M = -5540628.03$$
\eqalign{ \mu =& -5540628.03 /[6378206.4\times(1-0.0067687/4 \cr & -3\times0.0067687^2/64 - 5\times0.0067687^3/256)] \cr =& -0.8701555 \;\text{radians} = -49.8562392^\circ }
\eqalign{ e_1 &= [1-(1-0.0067687)^{1/2}]/[1+(1-0.0067687)^{1/2}] \cr &= 0.001697916 }
\eqalign{ \phi =& -49.8562392^\circ+[(3\times0.0016979/2-27\times0.0016979^3/32)\sin(2\times(-49.8562392^\circ)) \cr &+(21\times0.0016979^2/16-55\times0.0016979^4/32)\sin(4\times(-49.8562392^\circ)) \cr &+(151\times0.0016979^3/96)\sin(6\times(-49.8562392^\circ)) \cr &+(1097\times0.0016979^4/512)\sin(8\times(-49.8562392^\circ))]\times180^\circ/\pi \cr =&-50.0000001^\circ }
\eqalign{ \lambda =& -90^\circ+\{1075471.54\times[1-0.0067687\sin^2(-50.0000001^\circ)]^{1/2}/ \cr & [6378206.4\cos(-50.0000001^\circ)]\}\times(180^\circ/\pi) \cr =& -75^\circ }