Numerical Examples - Polyconic Projection

# Numerical Examples for Polyconic Projection #

## SPHERE #

### Forward Equations #

Given

 Radius of sphere: $R=\;\;$ units Origin: $\phi_0=\;$ ° $\lambda_0=\;$ ° Point: $\phi=\;$ ° $\lambda=\;$ °

Find $x, y, h$

From equations (18-2) through (18-4),
\eqalign{ E &= (-75^\circ - (-96^\circ))\sin 40^\circ \cr &= 13.4985398^\circ }

\eqalign{ x &= 1.0\cot 40^\circ\sin 13.4985398^\circ \cr &= 0.2781798\;\text{units} }
\eqalign{ y &= 1.0\times[40^\circ \times \pi/180^\circ- 30^\circ \times \pi/180^\circ + \cot40^\circ(1-\cos13.4985398^\circ)]\cr &= 0.2074541\;\text{units} }

From equations (18-6) and (18-5),
\eqalign{ D &= \arctan[(13.4985398^\circ\times\pi/180^\circ - \sin13.4985398^\circ)/(\sec^240^\circ-\cos13.4985398^\circ)] \cr &= 0.1701833^\circ }
\eqalign{ h &= (1-\cos^240^\circ\cos13.4985398^\circ)/(\sin^240^\circ\cos0.1701833^\circ) \cr &= 1.0392385 }

### Inverse Equations #

Inversing the forward example:

Given $R, \phi_0, \lambda_0$ for forward example

 Point: $x=\;$ units $y=\;$ units

Find $\phi, \lambda$

Since $y \ne -R\times \phi_0$, use equations (18-7) and (18-8):

\eqalign{ A &= 30^\circ\times \pi/180^\circ+0.2074541/1 \cr &= 0.7310529 }
\eqalign{ B &= 0.2781798^2/1^2 + 0.7310529^2 \cr &= 0.6118223 }
Assuming an initial $\phi_n = A = 0.7310529$ radians, it is simplest to work with equation (18-9) in radians:
\eqalign{ \phi_{n+1} =& 0.7310529 - [{\bf 0.7310529}\times (0.7310529\tan 0.7310529+1) \cr & - 0.7310529 -½(0.7310529^2 + 0.6118223)\tan 0.7310529]/ \cr & [(0.7310529 - {\bf 0.7310529})/\tan 0.7310529 - 1] \cr =& 0.6963533 }
Using $0.6963533$ in place of $0.7310529$ (except that the boldface retains the value of $A$) a new $\phi_{n+1}$ of $0.6981266$ radian is obtained. Again substituting this value, $0.6981317$ radian is obtained. The fourth iteration results in the same answer to seven decimal places. Therefore,
\eqalign{ \phi = 0.6981317 \times 180^\circ/\pi = 40.0000012^\circ }
From equation (18-10),
\eqalign{ \lambda &= [\arcsin(0.2781798\tan 40.0000012^\circ/1)]/\sin 40.0000012^\circ + (-96^\circ) \cr &= -75.0000010^\circ }

## ELLIPSOID #

### Forward Equations #

Given:

 Clarke 1866WGS-84 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00676866 Origin: $\phi_0=$ ° $\lambda_0=$ ° Point: $\phi=$ ° $\lambda=$ °
Find $x, y, h$.

From equation (3-21),

\eqalign{ M =&6378206.4[(1-0.0067687/4-3\times 0.0067687^2/64 - 5\times 0.0067687^3/256)\times 40^\circ\times\pi/180^\circ \cr &-(3\times 0.0067687/8+3\times 0.0067687^2/32 +45\times0.0067687^3/1024)\sin(2\times40^\circ) \cr &+(15\times 0.0067687^2/256 +45\times 0.0067687^3/1024)\sin(4\times 40^\circ) \cr &-(35\times 0.0067687^3/3072)\sin(6\times40^\circ)] \cr =& 4429318.91\;\text{m} }
Using $30^\circ$ in place of $40^\circ$ ,
$$M_0 = 3319933.29\;\text{m}$$

From equation (4-20),

\eqalign { N &= 6378206.4/(1-0.0067687\sin^2 40^\circ)^{1/2} \cr &= 6387143.95\;\text{m} }

From equations (18-2), (18-12), and (18-13),

\eqalign{ E &= (-75^\circ - (-96^\circ))\sin40^\circ \cr &= 13.4985398^\circ }
\eqalign{ x &= 6387143.95\cot 40^\circ\sin 13.4985398^\circ \cr &= 1776774.54\;\text{m} }
\eqalign{ y =& 4429318.90 - 3319933.29 + 6387143.95\cot 40^\circ \cr & (1-\cos 13.4985398^\circ) \cr =& 1319657.78\;\text{m} }
To calculate scale factor $h$, from equations (18-16) and (18-15),
\eqalign{ D =& \arctan\{(13.4985398^\circ\times\pi/180^\circ-\sin13.4985398^\circ)/[\sec^240^\circ \cr & - \cos13.4985398^\circ - 0.0067687\sin^240^\circ/(1-0.0067687\sin^240^\circ)] \} \cr =& 0.1708381^\circ }
\eqalign{ h =& [1-0.0067687 - 2(1-0.0067687\sin^2 40^\circ)\sin^2 ½(13.4985398^\circ)/ \cr & \tan^2 40^\circ]/(1-0.0067687)\cos 0.1708381^\circ \cr =& 1.0393954 }

### Inverse Equations #

Inversing forward example:

Given

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

First calculating $M_0$, from equation (3-21), as in the forward example,



Since $y \ne M_0$, from equations (18-18) and (18-19),

\eqalign{ A &= (3319933.29+1319657.78)/6378206.4 \cr &= 0.7274131 }
\eqalign{ B &= 1776774.54^2/6378206.4^2 + 0.7274131^2 \cr &= 0.6067309 }
Assuming an initial value of $\phi_n = 0.7274131$ the following calculations are made in radians from equations (18-20), (3-21), (18-17), and (18-21):
\eqalign{ C &= (1-0.0067687\sin^20.7274131)^{1/2}\tan0.7274131 \cr &= 0.8889365 }
$$M_n = 4615626.09\;\text{m}$$
\eqalign{ M'_n =& 1-0.0067687/4 - 3\times0.0067687^2/64 -5\times0.0067687^3/256 \cr & - 2\times (3\times0.0067687/8 + 3\times 0.0067687^2/32 \cr & + 45 \times 0.0067687^3/1024)\cos(2\times0.7274131) \cr & + 4\times(15\times0.0067687^2/256 + 45\times0.0067687^3/1024)\cos(4\times0.7274131) \cr & - 6\times(35\times0.0067687^3/3072)\cos(6\times0.7274131) \cr =& 0.9977068 }
$$M_a = 4615626.09/6378206.4 = 0.7236558$$
\eqalign{ \phi_{n+1} =& 0.7274131 - [{\bf 0.7274131}\times(0.8889365\times0.7236558+1) \cr & -0.7236558 - ½(0.7236558^2 + 0.6067309)\times0.8889365]/ \cr & [0.0067687\sin(2\times0.7274131)\times(0.7236558^2 + 0.6067309 \cr & + 2\times {\bf 0.7274131} \times 0.7236558) / (4\times 0.8889365) \cr & + ({\bf 0.7274131} - 0.7236558)\times(0.8889365 \times 0.9977068 \cr & -2/\sin(2\times 0.7274131)) - 0.9977068] \cr =& 0.6967280\;\text{radians} }
Substitution of $0.6967280$ in place of $0.7274131$ in equations (18-20), (3-21), (18-17), and (18-21), except for boldface values, which are $A$, not $\phi_n$, a new $\phi_{n+1}$ of $0.6981286$ is obtained. Using this in place of the previous value results in a third of $0.6981317$ , which is unchanged by recalculation to seven decimals. Thus,
\eqalign{ \phi = 0.6981317 \times 180^\circ/\pi = 40^\circ }
From equation (18-22), using the finally calculated $C$ of $0.8379255$
\eqalign{ \lambda &= [\arcsin(1776774.54\times0.8379255/6378206.4)]/\sin40^\circ + (-96^\circ) \cr &= -75^\circ }