Numerical Examples - Stereographic Projection

# 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
Find $\phi, \lambda$

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:

 Clarke 1866WGS-84 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00676866 $e=$ 0.0822719 Center: $\phi_1=$ ° $\lambda_0=$ ° Central scale factor: $k_0=$ Point: $\phi=$ ° $\lambda=$ °
Find $x, y, k$

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
Find: $\phi, \lambda$

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:

 InternationalWGS-84 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00672267 $e=$ 0.0819919 Center: $\phi_1=$ -90° (South Pole)90° (North Pole) $\lambda_0=$ ° Central scale factor: $k_0=$ Point: $\phi=$ ° $\lambda=$ °
Find $x, y, k$

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
Find: $\phi, \lambda$

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:

 InternationalWGS-84 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00672267 $e=$ 0.0819919 Standard parallel: $\phi_c=$ ° $\lambda_0=$ ° Point: $\phi=$ ° $\lambda=$ °
Find $x, y, k$

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
Find: $\phi, \lambda$

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,
$$\phi = -75.0000001^\circ$$
$$\lambda = 149.9999997^\circ$$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}