Numerical Examples - Azimuthal Equidistant Projection

# Numerical Examples for Azimuthal Equidistant Projection #

## SPHERE #

### Forward Equations #

Given

 Radius of sphere: $R=\;\;$ unit Center: $\phi_1=\;$° $\lambda_0=\;$° Point: $\phi=\;$° $\lambda=\;$°
Find $x, y$

Using equations (5-3) and (25-2),

\eqalign{ \cos c &= \sin40^\circ\sin(-20^\circ) + \cos 40^\circ\cos(-20^\circ)\cos (100^\circ - (-100^\circ))\cr &= -0.8962806 }
$$c = 153.6733925^\circ$$
\eqalign{ k' &= (153.6733925^\circ\times\pi/180^\circ)/\sin 153.6733925^\circ \cr &= 6.0477621 }

Using equations (22-4) and (22-5),

\eqalign{ x &= 3.0\times6.0477621\cos(-20^\circ)\sin(100^\circ-(-100^\circ)) \cr &= -5.8311398\text{ units} }
\eqalign{ y &= 3.0\times6.0477621[\cos40^\circ\sin(-20^\circ)-\sin40^\circ\cos(-20^\circ)\cos(100^\circ-(-100^\circ))]\cr &= 5.5444634\text{ units} }

Since the above equations are general, examples of other forward formulas are not given.

### Inverse Equations #

Inversing forward example:

Given: $R, \phi_1, \lambda_0$, for forward example

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

Using equations (20-18) and (25-15),

\eqalign{ \rho &=[(-5.8311398)^2+5.5444634^2]^{1/2} \cr &= 8.0463200\text{ units} }
\eqalign{ c &= 8.0463200/3.0 \cr &= 2.6821067\text{ radians} \cr &= 153.6733925^\circ }
Using equation (20-14),
\eqalign{ \phi &= \arcsin(\cos153.6733925^\circ\sin40^\circ-5.5444634\sin153.6733925^\circ\cos40^\circ/8.0463200) \cr &= -19.9999999^\circ }
Using equation (20-15), using ATAN2 function and after adjusting the result to $(-180^\circ,\;+180^\circ]$ range,
\eqalign{ \lambda =& -100^\circ + \arctan[(-5.8311398)\sin153.6733925^\circ/(8.0463200\cos40^\circ\cr & \cos153.6733925^\circ - 5.5444634\sin40^\circ\sin153.6733925^\circ)] \cr =& 99.9999999^\circ }

## ELLIPSOID #

### Polar Aspect #

#### Forward Equations #

Given:

 InternationalWGS-84 ellipsoid $a=$ 6378388. m $e^2=$ 0.00672267 $e=$ 0.0819919 Center: $\phi_1=$ -90° (South Pole)90° (North Pole) $\lambda_0=$ ° Point: $\phi=$ ° $\lambda=$ °
Find $x, y, k$

Using equation (3-21),

\eqalign{ M =&6378388.0\times[(1-0.0067227/4-3\times 0.0067227^2/64 - 5\times 0.0067227^3/256)\times 80^\circ\times\pi/180^\circ \cr &-(3\times 0.0067227/8+3\times 0.0067227^2/32 +45\times0.0067227^3/1024)\sin(2\times80^\circ) \cr &+(15\times 0.0067227^2/256 +45\times 0.0067227^3/1024)\sin(4\times 80^\circ) \cr &-(35\times 0.0067227^3/3072)\sin(6\times80^\circ)] \cr =& 8885403.07\text{ m} }
Using the same equation (3-21), but with $90^\circ$ in place of $80^\circ$ ,
$$M_p = 10002288.30\text{ m}$$
Using equation (14-15),
\eqalign{ m &= \cos80^\circ/(1-0.0067227\sin^280^\circ)^{1/2} \cr &= 0.1742171 }
Using equations (25-16), (21-30), (21-31), and (21-32),
\eqalign{ \rho &= 10002288.30-8885403.07 \cr &= 1116885.23\text{ m} }
\eqalign{ x &= 1116885.23\sin[5^\circ-(-100^\circ)]\cr &= 1078828.29\text{ m} }
\eqalign{ y &= -1116885.23\cos[5^\circ-(-100^\circ)]\cr &= 289071.17\text{ m} }
\eqalign{ k &= 1116885.23/(6378388.0\times0.1742171)\cr &= 1.0050946 }

#### Inverse Equations #

Inversing forward example:
Given

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

Using equation (3-21), as in the corresponding forward example,

$$M_p = 10002288.30\text{ m}$$
Using equations (20-18), (25-28) or (25-29) for south polar case, and (7-19),
\eqalign{ \rho &= [1078828.3^2 + 289071.2^2]^{1/2} \cr &= 1116885.25\text{ m} }
\eqalign{ M &= 10002288.30 - 1116885.25 \cr &= 8885403.05\text{ m} }
\eqalign{ \mu =& 8885403.05/[6378388.0(1-0.0067227/4 - 3\times0.0067227^2/64 \cr & -5\times0.0067227^3/256)] \cr =& 79.9503324^\circ }
Using equations (3-24) and (3-26),
\eqalign{ e_1 &= [1-(1-0.0067227)^{1/2}]/[1+(1-0.0067227)^{1/2}] \cr &= 0.0016863 }
\eqalign{ \phi =& 1.3953965\text{ radians}+(3\times0.0016863/2 - 27\times0.0016863^3/32)\sin(2\times79.9503324^\circ)+ \cr & (21\times0.0016863^2/16-55\times0.0016863^4/32)\sin(4\times79.9503324^\circ)+ \cr & (15\times0.0016863^3/96)\sin(6\times79.9503324^\circ)+ \cr & (1097\times0.0016863^4/512)\sin(8\times79.9503324^\circ) \cr =& 79.9999998^\circ }
Using equation (20-16) or (20-17) for the south polar case,
\eqalign{ \lambda &= -100^\circ + \arctan[1078828.3/(-289071.2)] \cr &= 5.0000014^\circ }

### Oblique Aspect (Guam) #

#### Forward Equations #

Given: 1

 Clarke 1866 ellipsoid $a=$ 6378206.4 m $e^2=$ 0.00676866 Center: $\phi_1=$ 13°28'20.87887" $\lambda_0=$ 144°44'55.50254" False origin: $x_0=$ 50000 m $y_0=$ 50000 m Point: $\phi=$ ° $\lambda=$ °
Find $x, y$

Using equation (25-18), after converting angles to degrees and decimals: ($\phi_1=13.472466353^\circ, \lambda_0=144.748750706^\circ$ $\phi=13.339038461^\circ$ , $\lambda=144.635331292^\circ$ ),

\eqalign{ x =& 6378206.4\times(144.635331292^\circ-144.748750706^\circ)\times \pi/180^\circ\cr & \times\cos13.339038461^\circ/(1-0.0067687/\sin^213.339038461^\circ) \cr =& -12287.519\text{ m} }
Since $50000$ m is added to the origin for the Guam projection,
\eqalign{ x &= -12287.519 + 50000.0 \cr &= 37712.48\text{ m} }
From equation (3-21),
\eqalign{ M =&6378206.4\times[(1-0.0067687/4-3\times 0.0067687^2/64 - 5\times 0.0067687^3/256)\times 13.3390385^\circ\times\pi/180^\circ \cr &-(3\times 0.0067687/8+3\times 0.0067687^2/32 +45\times0.0067687^3/1024)\sin(2\times13.3390385^\circ) \cr &+(15\times 0.0067687^2/256 +45\times 0.0067687^3/1024)\sin(4\times 13.3390385^\circ) \cr &-(35\times 0.0067687^3/3072)\sin(6\times13.3390385^\circ)] \cr =& 1475127.96\text{ m} }
Substituting $\phi_1=13.472466353^\circ$ in place of $13.339038461^\circ$ in the same equation,
$$M_1 = 1489888.76\text{ m}$$
Using equation (25-19), and using the x without false origin,
\eqalign{ y =& 1475127.96-1489888.76+(-12287.52)^2\tan13.339038461^\circ \cr & \times(1-0.0067687\sin^213.339038461^\circ)^{1/2}/(2\times6378206.4) \cr =& -14757.999\text{ m} }
Adding 50000 meters for the false origin,
\eqalign{ y &= -14758.00 + 50000.0 \cr &= 35242.00\text{ m} }

#### Inverse Equations #

Inversing forward example:
Given

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

First subtract 50,000 m from $x$ and $y$ to relate them to actual projection origin: $x = -12287.52\text{ m}$ , $y = -14758.00 \text{ m}$ . Calculation of $M_1$ from equation (3-21) is exactly the same as in the forward example, or

$$M_1 = 1489888.76\text{ m}$$
From equation (25-30), the first trial M is found from an assumed $\phi = \phi_1$:
\eqalign{ M =& 1489888.76+(-14758.00)-(-12287.52)^2\tan13.472466353^\circ \cr & \times(1-0.0067687\sin^213.472466353^\circ)^{1/2}/(2\times6378206.4) \cr =& 1475127.93\text{ m} }
Using equation (7-19) and the above trial M,
\eqalign{ \mu =& 1475127.93/[6378206.4(1-0.0067687/4 - 3\times0.0067687^2/64 \cr & -5\times0.0067687^3/256)] \cr =& 0.2316688\text{ radians} }
Using equation (3-24),
\eqalign{ e_1 &= [1-(1-0.0067687)^{1/2}]/[1+(1-0.0067687)^{1/2}] \cr &= 0.0016979 }
Using equation (3-26) in radians, although it could be converted to degrees,
\eqalign{ \phi =& 0.2316688+(3\times0.0016979/2 - 27\times0.0016979^3/32)\sin(2\times0.2316688)+ \cr & (21\times0.0016979^2/16-55\times0.0016979^4/32)\sin(4\times0.2316688)+ \cr & (15\times0.0016979^3/96)\sin(6\times0.2316688)+ \cr & (1097\times0.0016979^4/512)\sin(8\times0.2316688) \cr =& 0.2328101\text{ radians} \cr =& 0.2328101 \times 180^\circ/\pi = 13.3390382^\circ }
If this new trial value of $\phi$ is used in place of $\phi_1$, in equation (25-30), a new value of M is found:
$$M = 1475127.96\text{ m}$$
This in turn, used in (7-19), gives
$$\mu = 0.2316688 \text{ radians} = 13.2736455^\circ$$
and from (3-26),
$$\phi = 13.3390385^\circ$$
The third trial, through the above equations and starting with this value of $\phi$, produces no change to seven decimal places. Thus, this is the final value of $\phi$. Converting to degrees, minutes, and seconds,
$$\phi = 13^\circ20'20.5384''$$

Using equation (25-31) for longitude,

\eqalign{ \lambda =& 144.7487507^\circ + [(-12287.52)\times(1-0.0067687\sin^213.3390385^\circ)^{1/2}/ \cr & (6378206.4\cos13.3390385^\circ)]\times 180^\circ/\pi \cr =& 144.6353313^\circ \cr =& 144^\circ38'07.1926'' }

### Oblique Aspect (Micronesia) #

#### Forward Equations #

Given:

 Clarke 1866WGS-84 ellipsoid $a=$ 6378206.4 $e^2=$ 0.00676866 Center: $\phi_1=$ $\lambda_0=$ False origin: $x_0=$ m $y_0=$ m Point: $\phi=$ ° $\lambda=$ °
Find $x, y$

First convert angles to degrees and decimals:

\eqalign{ \phi_1 &= 15.1849119^\circ \cr \lambda_0 &= 145.7416589^\circ \cr \phi &= 15.2465258^\circ \cr \lambda &= 145.79303^\circ }

From equations (4-20a), (4-20), (25-20), and (25-21) in order,

\eqalign{ N_1 &= 6378206.4/(1-0.0067687\sin^215.1849119^\circ)^{1/2} \cr &= 6379687.93\text{ m} }
\eqalign{ N &= 6378206.4/(1-0.0067687\sin^215.2465258^\circ)^{1/2} \cr &= 6379699.69\text{ m} }
\eqalign{ \psi =& \arctan[(1-0.0067687)\tan15.2465258^\circ+0.0067687\times \cr & 6379687.93\times\sin15.1849119^\circ/(6379699.69\times\cos15.2465258^\circ)] \cr =& 15.2461374^\circ }
\eqalign{ Az =& \arctan\{ \sin(145.79303^\circ-145.7416589^\circ)/[\cos15.1849119^\circ \cr & \tan15.2461374^\circ- \sin15.1849119^\circ\cos(145.79303^\circ-145.7416589^\circ)]\} \cr =& 38.9881344^\circ }

Since $\sin Az \ne 0$, from equation (25-22a),

\eqalign{ s =& \arcsin[\sin(145.79303^\circ-145.7416589^\circ)\cos15.2461374^\circ/ \cr & \sin38.9881344^\circ] \cr =& 0.001374913\text{ radians} }

From equations (25-23) through (25-27) in order,

\eqalign{ G &= 0.0067687^{1/2}\sin15.1849119^\circ/(1-0.0067687)^{1/2}\cr &= 0.021623198 }
\eqalign{ H &= 0.0067687^{1/2}\sin15.1849119^\circ/(1-0.0067687)^{1/2}\cr &= 0.061925215 }
\eqalign{ c =& 6379687.93\times0.001374913\times\{1-0.001374913^2\times0.061925215^2\times(1-0.061925215^2)/6 \cr & +(0.001374913^3/8)\times0.021623198\times0.061925215\times(1-2\times0.061925215^2) \cr & +(0.001374913^4/120)[0.061925215^2\times(4-7\times0.061925215^2)-3\times0.021623198^2\cr & \times(1-7\times0.061925215^2)]-(0.001374913^5/48)\times0.021623198\times0.061925215 \} \cr =& 8771.52\text{ m} }
\eqalign{ x &= 8771.52\sin38.9881344^\circ + 28657.52 \cr &= 34176.20 \text{ m} }
\eqalign{ y &= 8771.52\cos38.9881344^\circ + 67199.99 \cr &= 74017.88 \text{ m} }

#### Inverse Equations #

Inversing forward example:

Given

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

From equations (25-32) through (25-41) in order,

\eqalign{ c &= [(34176.2-28657.52)^2 + (74017.88-67199.99)^2]^{1/2} \cr &= 8771.51\text{ m} }
\eqalign{ Az &= \arctan[(34176.2-28657.52) / (74017.88-67199.99)] \cr &= 38.9881292^\circ }
\eqalign{ N_1 &= 6378206.4/(1-0.0067687\sin^215.1849119^\circ)^{1/2} \cr &= 6379687.93\text{ m} }
\eqalign{ A =& -0.0067687\cos^215.1849119^\circ\cos^238.9881292^\circ/ \cr & (1-0.0067687) \cr =& -0.003834730 }
\eqalign{ B =& 3\times0.0067687\times(1-(-0.003834730))\cos15.1849119^\circ \cr & \sin15.1849119^\circ\cos38.9881292^\circ/(1-0.0067687) \cr =& 0.004032465 }
\eqalign{ D &= 8771.51/6379687.93 \cr &= 0.0013749 }
\eqalign{ E =& 0.0013749-(-0.003834730)\times0.0013749^3/6\cr & -0.004032465\times(1+3\times(-0.003834730))\times0.0013749^4/24 \cr =& 0.001374913 \text{ radians} \cr =& 0.0787767^\circ }
\eqalign{ F =& 1-(-0.003834730)\times0.001374913^2/6-0.004032465 \cr & \times0.001374913^3/6 \cr =& 1.000000004 }




1. Parameters for this projection are fixed ↩︎