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:
| ellipsoid | $a=$ | 6378206.4 m |
| $e^2=$ | 0.00676866 | |
| Origin: | $\phi_0=$ | ° |
| $\lambda_0=$ | ° | |
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
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 |
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 $$$
\eqalign{
\lambda &= [\arcsin(1776774.54\times0.8379255/6378206.4)]/\sin40^\circ + (-96^\circ) \cr
&= -75^\circ
}
$$