Numerical Examples for Azimuthal Equidistant Projection #
SPHERE #
Forward Equations #
Given
| Radius of sphere: | $R=\;\;$ unit |
| Center: | $\phi_1=\;$° |
| $\lambda_0=\;$° | |
| Point: | $\phi=\;$° |
| $\lambda=\;$° | |
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 |
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:
| ellipsoid | $a=$ | 6378388. m |
| $e^2=$ | 0.00672267 | |
| $e=$ | 0.0819919 | |
| Center: | $\phi_1=$ | |
| $\lambda_0=$ | ° | |
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
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 |
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: | $\phi=$ | ° | $\lambda=$ | ° |
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 |
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:
| ellipsoid | $a=$ | 6378206.4 | |
| $e^2=$ | 0.00676866 | ||
| Center: | $\phi_1=$ | ||
| $\lambda_0=$ | |||
| False origin: | $x_0=$ | m | |
| $y_0=$ | m | ||
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
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 |
From equations (25-32) through (25-41) in order,
$$
\eqalign{
c &= [(34176.2-28657.52)^2 + (74017.88-67199.99)^2]^{1/2} \cr
&= 8771.51\text{ m}
}
$$
$$
\eqalign{
Az &= \arctan[(34176.2-28657.52) / (74017.88-67199.99)] \cr
&= 38.9881292^\circ
}
$$
$$
\eqalign{
N_1 &= 6378206.4/(1-0.0067687\sin^215.1849119^\circ)^{1/2} \cr
&= 6379687.93\text{ m}
}
$$
$$
\eqalign{
A =& -0.0067687\cos^215.1849119^\circ\cos^238.9881292^\circ/ \cr
& (1-0.0067687) \cr
=& -0.003834730
}
$$
$$
\eqalign{
B =& 3\times0.0067687\times(1-(-0.003834730))\cos15.1849119^\circ \cr
& \sin15.1849119^\circ\cos38.9881292^\circ/(1-0.0067687) \cr
=& 0.004032465
}
$$
$$
\eqalign{
D &= 8771.51/6379687.93 \cr
&= 0.0013749
}
$$
$$
\eqalign{
E =& 0.0013749-(-0.003834730)\times0.0013749^3/6\cr
& -0.004032465\times(1+3\times(-0.003834730))\times0.0013749^4/24 \cr
=& 0.001374913 \text{ radians} \cr
=& 0.0787767^\circ
}
$$
$$
\eqalign{
F =& 1-(-0.003834730)\times0.001374913^2/6-0.004032465 \cr
& \times0.001374913^3/6 \cr
=& 1.000000004
}
$$
$$
$$
$$
$$
$$
$$
Parameters for this projection are fixed ↩︎