Numerical Examples for Stereographic Projection #
SPHERE #
Forward Equations #
Given
| Radius of sphere: | $R=\;\;$ units |
| Center: | $\phi_1=\;$ ° |
| $\lambda_0=\;$ ° | |
| Central scale factor: | $k_0=\;$ |
| Point: | $\phi=\;$ ° |
| $\lambda=\;$ ° | |
Find $x, y, k$
Using equations (21-4), (21-2), and (21-3) in order,
$$
\eqalign {
k &= 2\times 1.0/[1+\sin40^\circ\sin30^\circ+\cos40^\circ\cos30^\circ\cos(-75^\circ-(-100^\circ))] \cr
&= 1.0402304
}
$$
$$
\eqalign{
x &= 1.0\times1.0402304\cos30^\circ\sin(-75^\circ-(-100^\circ)) \cr
&= 0.3807224\;\text{units}
}
$$
$$
\eqalign{
y &= 1.0\times1.0402304[\cos40^\circ\sin30^\circ - \sin40^\circ\cos30^\circ\cos(-75^\circ-(-100^\circ))] \cr
&= -0.1263802\;\text{units}
}
$$
Examples of other forward equations are omitted, since the above equations are general.Inverse Equations #
Inversing forward example:
Given $R, \phi_1, \lambda_0, k_0$ for forward example
| Point: | $x=\;$ units |
| $y=\;$ units | |
Using equations (21-18) and (21-19),
$$
\rho = [0.3807224^2 + (-0.1263802)^2]^{1/2} = 0.4011502\;\text{units}
$$
$$
\eqalign{
c &= 2 \arctan[0.4011502/(2\times1.0\times1.0)] \cr
&= 22.6832261^\circ
}
$$
Using equations (21-14) and (21-15),$$
\eqalign{
\phi =& \arcsin[\cos22.6832261^\circ\sin40^\circ + (-0.1263802)\sin22.6832261^\circ \cr
& \cos40^\circ/0.4011502] \cr
=& 29.9999991^\circ
}
$$
$$
\eqalign{
\lambda =& -100^\circ + \arctan[-0.1503837\sin12.9082572^\circ/(0.2233906 \cr
&\cos40^\circ\cos12.9082572^\circ - (-0.1651911)\sin40^\circ \sin12.9082572^\circ)] \cr
=& -109.9999978^\circ
}
$$
ELLIPSOID #
Oblique Aspect #
Forward Equations #
Given:
| ellipsoid | $a=$ | 6378206.4 m |
| $e^2=$ | 0.00676866 | |
| $e=$ | 0.0822719 | |
| Center: | $\phi_1=$ | ° |
| $\lambda_0=$ | ° | |
| Central scale factor: | $k_0=$ | |
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
From equation (3-1),
$$
\eqalign{
\chi_1 =& 2\arctan\{ \tan(45^\circ + 40^\circ/2)[(1-0.0822719\sin40^\circ)/ \cr
& (1+0.0822719\sin40^\circ)]^{0.0822719/2} \} -90^\circ \cr
=& 2\arctan2.1351882-90^\circ \cr
=& 39.8085922^\circ
}
$$
$$
\eqalign{
\chi =& 2\arctan\{ \tan(45^\circ + 30^\circ/2)[(1-0.0822719\sin30^\circ)/ \cr
& (1+0.0822719\sin30^\circ)]^{0.0822719/2} \} -90^\circ \cr
=& 2\arctan1.7261956-90^\circ \cr
=& 29.8318339^\circ
}
$$
From equation (14-15),$$
\eqalign{
m_1 &= \cos40^\circ/(1-0.0067687\sin^240^\circ)^{1/2} \cr
&= 0.7671179
}
$$
$$
\eqalign{
m &= \cos30^\circ/(1-0.0067687\sin^230^\circ)^{1/2} \cr
&= 0.8667591
}
$$
From equation (21-27),$$
\eqalign{
A =& 2\times6378206.4\times0.9999\times0.7671179/\{ \cos39.8085922^\circ \cr
& [1+\sin39.8085922^\circ\sin29.8318339^\circ + \cos39.8085922^\circ \cr
& \cos29.8318339^\circ\cos(-90^\circ-(-100^\circ))]\} \cr
=& 6450107.68\;\text{m}
}
$$
From equations (21-24), (21-25), and (21-26),$$
\eqalign{
x &= 6450107.68\cos29.8318339^\circ\sin(-90^\circ-(-100^\circ)) \cr
&= 971630.79\;\text{m}
}
$$
$$
\eqalign{
y =& 6450107.68[\cos39.8085922^\circ\sin29.8318339^\circ \cr
& - \sin39.8085922^\circ\cos29.8318339^\circ\cos(-90^\circ-(-100^\circ))] \cr
=& -1063049.26\;\text{m}
}
$$
$$
\eqalign{
k &= 6378206.40\cos29.8318339^\circ/(6378206.40\times0.8667591) \cr
&= 1.0121248
}
$$
Inverse Equations #
Inversing forward example:
Given
| $x=\;$m |
| $y=\;$m |
From equation (14-15),
$$
\eqalign{
m_1 &= \cos40^\circ/(1-0.0067687\sin^240^\circ)^{1/2} \cr
&= 0.7671179
}
$$
From equation (3-11), as in the forward oblique example,$$
\chi_1 = 39.8085922^\circ
$$
From equations (20-18) and (21-38),$$
\eqalign{
\rho &= [971630.79^2 + (-1063049.26)^2]^{1/2} \cr
&= 1440187.53\text{ m}
}
$$
$$
\eqalign{
c_e =& 2\arctan[1440187.57\cos39.8085922^\circ/(2\times6378206.40 \cr
& \times0.9999)\times0.7671179] \cr
=& 12.9018251^\circ
}
$$
From equation (21-37),$$
\eqalign{
\chi =& \arcsin[\cos12.9018251^\circ\sin0.6947910040690184+(-1063049.3)\sin12.9018251^\circ\cr
& \cos39.8085922^\circ/1440187.57] \cr
=& 29.8318335^\circ
}
$$
Using $\chi$ as the first trial $\phi$ in equation (3-4),$$
\eqalign{
\phi =& 2\arctan\{ \tan(45^\circ + 29.8318335^\circ/2)[(1-0.0822719\sin29.8318335^\circ)/ \cr
& (1+0.0822719\sin29.8318335^\circ)]^{0.0822719/2} \} -90^\circ \cr
=& 29.9991438^\circ
}
$$
Using this new trial value in the same equation for $\phi$, not for $\chi$,$$
\eqalign{
\phi =& 2\arctan\{ \tan(45^\circ + 29.8318335^\circ/2)[(1-0.0822719\sin29.9991438^\circ)/ \cr
& (1+0.0822719\sin29.9991438^\circ)]^{0.0822719/2} \} -90^\circ \cr
=& 29.9999953^\circ
}
$$
Repeating with
$ 29.9999953^\circ $
in place of
$ 29.9991438^\circ $
, the next trial $\phi$ is$$
\phi = 29.9999996^\circ
$$
The next trial calculation produces the same $\phi$ to seven decimals. Therefore,
this is $\phi$.Using equation (21-36),
$$
\eqalign{
\lambda =& -100^\circ+\arctan[971630.8\sin12.9018251^\circ \cr
& (1440187.57\cos39.8085922^\circ\cos12.9018251^\circ \cr
& -(-1063049.30)\sin39.8085922^\circ\sin12.9018251^\circ)] \cr
=& -100^\circ+\arctan(216946.86/1230366.77) \cr
=& -100^\circ+10.0000000^\circ \cr
=& -90.0000000^\circ
}
$$
Instead of the iterative equation (3-4), series equation (3-5) may be used (omitting terms with $e^8$ here for simplicity):
$$
\eqalign{
\phi =& 29.8318335^\circ\times \pi/180^\circ + (0.0067687/2 + 5\times0.0067687^2/24 \cr
& 0.0067687^3/12)\sin(2\times29.8318335^\circ)+(7\times0.0067687^2/48 \cr
& + 29\times0.0067687^3/240)\sin(4\times29.8318335^\circ) \cr
& + (7\times0.0067687^3/120)\sin(6\times29.8318335^\circ) \cr
=& 0.5235988\text{ radian} \cr
=& 29.9999995^\circ
}
$$
Polar Aspect With Known $k_0$ #
Forward Equations #
Given:
| ellipsoid | $a=$ | 6378206.4 m |
| $e^2=$ | 0.00672267 | |
| $e=$ | 0.0819919 | |
| Center: | $\phi_1=$ | |
| $\lambda_0=$ | ° | |
| Central scale factor: | $k_0=$ | |
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
Since this is the south polar aspect, for calculations change signs of $x, y, \phi_1, \lambda_1$, and $\lambda_0$:
$ \lambda_0=100^\circ $
,
$ \phi=75^\circ $
,
$ \lambda=-150^\circ $
Using equations (15-9) and (21-33),
$$
\eqalign{
t &= \tan(45^\circ-75^\circ/2)/[(1-0.0822719\sin 75^\circ)/(1+0.0822719\sin 75^\circ)]^{0.0822719/2} \cr
&= 0.1325179
}
$$
$$
\eqalign{
\rho =& 2\times6378388.0\times0.994\times0.1325120/[(1+0.0819919)^{(1+0.0819919)} \cr
& \times(1-0.0819919)^{(1-0.0819919)}] \cr
=& 1674638.30\text{ m}
}
$$
Using equations (21-30) and (21-31),$$
\eqalign{
x &= 1674638.31\sin(-150^\circ-100^\circ) \cr
&= 1573645.26\text{ m}
}
$$
$$
\eqalign{
y &= -1674638.31\cos(-150^\circ-100^\circ) \cr
&= 572760.03\text{ m}
}
$$
Changing signs of x and y for the south polar aspect,
$$
x = -1573645.26\text{ m}
$$
$$
y = -572760.03\text{ m}
$$
From equation (14-15),
$$
\eqalign{
m &= \cos 75^\circ/(1-0.0067227\sin^275^\circ)^{1/2}\cr
&= 0.2596346
}
$$
From equation (21-32),$$
\eqalign{
k &= 1674638.31/(6378388.0\times0.2596346) \cr
&= 1.0112244
}
$$
Inverse Equations #
Inversing forward example:
Given
| $x=\;$m |
| $y=\;$m |
Since this is the south polar aspect, for calculations change signs as stated in text:
$ \lambda_0=100^\circ $
,
$ x=1573645.3\text{ m} $
,
$ y=572760.0\text{ m} $
.
From equations (20-18) and (21-39),
$$
\eqalign{
\rho &= (1573645.3^2 + 572760.0^2)^{1/2} \cr
&= 1674638.33\text{ m}
}
$$
$$
\eqalign{
t =& 1674638.33\times[(1+0.0819919)^{(1+0.0819919)} \cr
& (1-0.0819919)^{(1-0.0819919)}]^{1/2}/(2\times6378388.0\times0.994) \cr
=& 0.1325120
}
$$
To iterate with equation (7-9), use as the first trial $\phi$,
$$
\eqalign{
\phi &= 90^\circ - 2\arctan 0.1325120 \cr
&= 74.9031986^\circ
}
$$
Substituting in (7-9),$$
\eqalign{
\phi =& 90^\circ - 2\arctan\{0.1325120\times[(1-0.0819919\sin74.9031986^\circ)/ \cr
& (1+0.0819919\sin74.9031986^\circ)]^{0.0819919/2} \} \cr
=& 74.9999558^\circ
}
$$
Using this second trial $\phi$ in the same equation instead of
$ 74.9031986^\circ $
,$$
\phi = 74.9999997^\circ
$$
The third trial gives the same value to seven places.From equation (20-16), using ATAN2 function and after adjusting the result to $(-180^\circ,\;+180^\circ]$ range,
$$
\eqalign{
\lambda &= 100^\circ + \arctan[1573645.3/(-572760.0)] \cr
&= -150.0000016^\circ
}
$$
The sign of $\phi$ and $\lambda$ must be reversed for the south polar aspect. Finally,
$$
\phi = -74.9999997^\circ
$$
$$
\lambda = 150.0000016^\circ
$$
Polar Aspect With Known $\phi_c$ #
Forward Equations #
Given:
| ellipsoid | $a=$ | 6378206.4 m |
| $e^2=$ | 0.00672267 | |
| $e=$ | 0.0819919 | |
| Standard parallel: | $\phi_c=$ | ° |
| $\lambda_0=$ | ° | |
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
Since this is the south polar aspect, for calculations change signs of $x, y, \phi_c, \lambda_1$, and $\lambda_0$:
$ \phi_c=71^\circ $
,
$ \lambda_0=100^\circ $
,
$ \phi=75^\circ $
,
$ \lambda=-150^\circ $
Using equation (15-9),
$$
\eqalign{
t &= \tan(45^\circ-75^\circ/2)/[(1-0.0819919\sin 75^\circ)/(1+0.0819919\sin 75^\circ)]^{0.0819919/2} \cr
&= 0.1325120
}
$$
For $t_c$ substitute
$ 71^\circ $
in place of
$ 75^\circ $
in (15-9), and$$
\eqalign{
t_c &= \tan(45^\circ-71^\circ/2)/[(1-0.0819919\sin 71^\circ)/(1+0.0819919\sin 71^\circ)]^{0.0819919/2} \cr
&= 0.1684118
}
$$
From equations (14-15) and (21-34),$$
\eqalign{
m_c &= \cos 71^\circ/(1-0.0067227\sin^271^\circ)^{1/2}\cr
&= 0.3265509
}
$$
$$
\eqalign{
\rho &= 6378388.0\times0.3265509\times0.1325120/0.1684118 \cr
&= 1638869.54\text{ m}
}
$$
Equations (21-30), (21-31), and (21-32) are used as in the preceding south polar example,$$
\eqalign{
x &= 1638869.54\sin(-150^\circ-100^\circ) \cr
&= 1540033.61\text{ m}
}
$$
$$
\eqalign{
y &= -1638869.54\cos(-150^\circ-100^\circ) \cr
&= 560526.39\text{ m}
}
$$
Changing signs of x and y for the south polar aspect,
$$
x = -1540033.61\text{ m}
$$
$$
y = -560526.39\text{ m}
$$
$$
\eqalign{
m &= \cos 75^\circ/(1-0.0067227\sin^275^\circ)^{1/2}\cr
&= 0.2596346
}
$$
$$
\eqalign{
k &= 1638869.54/(6378388.0\times0.2596346) \cr
&= 0.9896256
}
$$
Inverse Equations #
Inversing forward example:
Given
| $x=\;$m |
| $y=\;$m |
Since this is the south polar aspect, for calculations change signs as stated in text:
$ \phi_c=71^\circ $
,
$ \lambda_0=100^\circ $
,
$ x=1540033.6\text{ m} $
,
$ y=560526.4\text{ m} $
.
From equations (15-9) and (14-15), as calculated in the corresponding forward example,
$$
\eqalign{
t_c &= \tan(45^\circ-71^\circ/2)/[(1-0.0819919\sin 71^\circ)/(1+0.0819919\sin 71^\circ)]^{0.0819919/2} \cr
&= 0.1684118
}
$$
$$
\eqalign{
m_c &= \cos 71^\circ/(1-0.0067227\sin^271^\circ)^{1/2}\cr
&= 0.3265509
}
$$
From equations (20-18) and (21-40),
$$
\eqalign{
\rho &= [1540033.6^2 + 560526.4^2]^{1/2} \cr
&= 1638869.53\text{ m}
}
$$
$$
\eqalign{
t &= 1638869.53\times0.1684118/(6378388.0\times0.3265509) \cr
&= 0.1325120
}
$$
For the first trial $\phi$ in equation (7-9)$$
\eqalign{
\phi &= 90^\circ - 2\arctan 0.1325120 \cr
&= 74.9031989^\circ
}
$$
Substituting in (7-9),$$
\eqalign{
\phi =& 90^\circ - 2\arctan\{0.1325120\times[(1-0.0819919\sin74.9031989^\circ)/ \cr
& (1+0.0819919\sin74.9031989^\circ)]^{0.0819919/2} \} \cr
=& 74.9999561^\circ
}
$$
Replacing
$ 74.9031989^\circ $
with
$ 74.9999561^\circ $
, the next trial $\phi$ is$$
\phi = 75.0000001^\circ
$$
The next iteration results in the same $\phi$ to seven places.From equation (20-16), using ATAN2 function and after adjusting the result to $(-180^\circ,\;+180^\circ]$ range,
$$
\eqalign{
\lambda &= 100^\circ + \arctan[1540033.6/(-560526.4)] \cr
&= -149.9999997^\circ
}
$$
The sign of $\phi$ and $\lambda$ must be reversed for the south polar aspect. Finally,
$$
\phi = -75.0000001^\circ
$$
$$
\lambda = 149.9999997^\circ
$$