Numerical Examples for Lambert Azimuthal Equal-Area #
SPHERE #
Forward Equations #
| Radius of sphere: | $R=\;\;$ units |
| Center: | $\phi_1=\;$ ° |
| $\lambda_0=\;$ ° | |
| Point: | $\phi=\;$ ° |
| $\lambda=\;$ ° | |
Find $x, y$
Using equation (24 -2),
$$
\eqalign{
k' &= \{2/[1+\sin40^\circ\sin(-20^\circ)+\cos40^\circ\cos(-20^\circ)\cos(100^\circ-(-100^\circ))] \}^{1/2} \cr
&= 4.3912175
}
$$
Using equations (22-4) and (22-5),$$
\eqalign{
x &= 3.0\times4.3912175\cos(-20^\circ)\sin(100^\circ-(-100^\circ)) \cr
&= -4.2339303\text{ units}
}
$$
$$
\eqalign{
y &= 3.0\times4.3912175[\cos40^\circ\sin(-20^\circ)-\sin40^\circ\cos(-20^\circ)\cos(100^\circ-(-100^\circ))] \cr
&= 4.0257775\text{ units}
}
$$
Examples for the polar and equatorial reductions, equations (24-3) through (24-14), are omitted, since the above general equations give the same results.Inverse Equations #
Inversing forward example:
Given $R, \phi_1, \lambda_0$ for forward example
| Point: | $x=\;$ units |
| $y=\;$ units | |
Using equations (20-18) and (24-16),
$$
\eqalign{
\rho &= [(-4.2339303)^2 + 4.0257775^2]^{1/2} \cr
&= 5.8423497 \text{ units}
}
$$
$$
\eqalign{
c &= 2\arcsin[5.8423497/(2\times3.0)] \cr
&= 153.6733917^\circ
}
$$
From equation (20-14),$$
\eqalign{
\phi =& \arcsin[\cos153.6733917^\circ\sin40^\circ+4.0257775\sin2.6821067\cr
& \cos40^\circ/5.8423497] \cr
=& -19.9999993^\circ
}
$$
From equation (20-15), using the ATAN2 function and adjusting longitude to be between $-180^\circ$ and $+180^\circ$,$$
\eqalign{
\lambda =& -100^\circ + \arctan[-4.2339303\sin153.6733917^\circ / \cr
& (5.8423497\cos40^\circ\cos153.6733917^\circ \cr
& -4.0257775\sin40^\circ\sin153.6733917^\circ)] \cr
=& -100^\circ + \arctan[-1.8776951/(-5.1589246)] \cr
=& -100^\circ + (-159.9999996^\circ) \cr
=& 100.0000004^\circ
}
$$
In polar spherical cases, the calculation of $\lambda$ from equations (20-16) or (20-17) is simpler than the above, but the quadrant adjustment follows the same rules.ELLIPSOID #
Oblique Aspect #
Forward Equations #
Given:
| ellipsoid | $a=$ | 6378206.4 m |
| $e^2=$ | 0.00676866 | |
| $e=$ | 0.0822719 | |
| Center: | $\phi_1=$ | ° |
| $\lambda_0=$ | ° | |
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
Using equation (3-12),
$$
\eqalign{
q =& (1-0.0067687)\{\sin 30^\circ/(1-0.0067687) \cr
& -[1/(2\times0.0822719)\ln[(1-0.0822719\sin^2 30^\circ)/ \cr
& (1+0.0822719\sin^2 30^\circ)] \} \cr
=& 0.9943535
}
$$
Inserting
$ \phi_1 = 40^\circ $
in place of
$ 30^\circ $
in the same equation,$$
q_1 = 1.2792602
$$
Inserting $90^\circ$ in place of
$ 30^\circ $
,$$
q_p= 1.9954814
$$
Using equation (3-11),$$
\eqalign{
\beta &= \arcsin(0.9943535/1.9954814) \cr
&= 29.8877622^\circ
}
$$
$$
\eqalign{
\beta_1 &= \arcsin(1.2792602/1.9954814) \cr
&= 39.8722878^\circ
}
$$
Using equation (3-13),$$
\eqalign{
R_q &= 6378206.4\times(1.9954814/2)^{1/2} \cr
&= 6370997.24\text{ m}
}
$$
Using equation (14-15),$$
\eqalign{
m_1 &= \cos40^\circ/(1-0.0067687\sin^240^\circ)^{1/2} \cr
&= 0.7671179
}
$$
Using equations (24-19) and (24-20),$$
\eqalign{
B =& 6370997.24\times\{ 2/[1+\sin39.8722878^\circ\sin29.8877622^\circ \cr
& + \cos39.8722878^\circ\cos29.8877622^\circ\cos(-110^\circ-(-100^\circ))]\}^{1/2} \cr
=& 6411606.09\text{ m}
}
$$
$$
\eqalign{
D &= 6378206.4\times0.7671179/(6370997.24\cos39.8722878^\circ) \cr
&= 1.0006653
}
$$
Using equations (24-17) and (24-18),$$
\eqalign{
x &= 6411606.09\times1.0006653\cos29.8877622^\circ\sin(-110^\circ-(-100^\circ)) \cr
&= -965932.11\text{ m}
}
$$
$$
\eqalign{
y =& (6411606.09/1.0006653)[\cos39.8722878^\circ\sin29.8877622^\circ \cr
& - \sin39.8722878^\circ\cos29.8877622^\circ\cos(-110^\circ-(-100^\circ))]\cr
=& -1056814.93\text{ m}
}
$$
Inverse Equations #
Inversing forward example:
Given
| $x=\;$m |
| $y=\;$m |
Since these are the same map parameters as those used in the forward example, calculations of map constants not affected by $\phi$ and $\lambda$ are not repeated here.
$$
q_p= 1.9954814
$$
$$
\beta_1 = 39.8722878^\circ
$$
$$
R_q = 6370997.24\text{ m}
$$
$$
D = 1.0006653
$$
Using equations (24-28), (24-29) and (24-27)$$
\eqalign{
\rho &= [(-965932.1/1.0006653)^2 + (-1056814.9\times1.0006653)^2]^{1/2} \cr
&= 1431827.12\text{ m}
}
$$
$$
\eqalign{
c_e &= 2\arcsin[1431827.12/(2\times6370997.24)]\cr
&= 12.9039908^\circ
}
$$
$$
\eqalign{
q =& 1.9954814[\cos12.9039908^\circ\sin39.8722878^\circ\cr
& +1.0006653\times(-1056814.9)\sin12.9039908^\circ\cr
& \cos39.8722878^\circ/1431827.12] \cr
=& 0.9943535
}
$$
For the first trial $\phi$ in equation (3-16),$$
\eqalign{
\phi &= \arcsin(0.9943535/2) \cr
&= 29.8133914^\circ
}
$$
Substituting into equation (3-16),$$
\eqalign{
\phi = & 29.8133914^\circ+[(1-0.0822719\sin^229.8133914^\circ)^2/ \cr
& (2\cos29.8133914^\circ)]\times \{ 0.9943535/(1-0.0067687) \cr
& - \sin29.8133914^\circ/(1-0.0067687\sin^229.8133914^\circ) \cr
& +[1/(2\times0.0822719)]\ln[(1-0.0822719\sin29.8133914^\circ)/ \cr
& (1+0.0822719\sin29.8133914^\circ)] \}\times 180^\circ/\pi \cr
=& 29.9998293^\circ
}
$$
Substituting
$ 29.9998293^\circ $
in place of
$ 29.8133914^\circ $
in the same equation, the new trial $\phi$ is found to be$$
\phi = 30.0000002^\circ
$$
The next iteration results in no change to seven decimal places.Using equation (24-26),
$$
\eqalign{
\lambda =& -100^\circ + \arctan\{ (-965932.1)\sin12.9039908^\circ/[1.0006653 \cr
& \times 1431827.12\cos39.8722878^\circ\cos12.9039908^\circ \cr
& - 1.0006653^2\times(-1056814.9)\sin39.8722878^\circ\sin12.9039908^\circ] \} \cr
=& -109.9999999^\circ
}
$$
using the ATAN2 function.Polar Aspect #
Forward Equations #
Given:
| ellipsoid | $a=$ | 6378206.4 m |
| $e^2=$ | 0.00672267 | |
| $e=$ | 0.0819919 | |
| Center: | $\phi_1=$ | |
| $\lambda_0=$ | ° | |
| Point: | $\phi=$ | ° | $\lambda=$ | ° |
From equation (3-12),
$$
\eqalign{
q =& (1-0.0067227)\{\sin 80^\circ/(1-0.0067227) \cr
& -[1/(2\times0.0819919)\ln[(1-0.0819919\sin^2 80^\circ)/ \cr
& (1+0.0819919\sin^2 80^\circ)] \} \cr
=& 1.9649283
}
$$
Using the same equation with $90^\circ$ in place of
$ 80^\circ $
,$$
q_p = 1.9955122
$$
From equation (14-15),$$
\eqalign{
m &= \cos80^\circ/(1-0.0067227\sin^280^\circ)^{1/2} \cr
&= 0.1742171
}
$$
Using equations (24-23), (21-30), (21-31), and (21-32),$$
\eqalign{
\rho &= 6378388.0\times(1.9955122-1.9649283)^{1/2} \cr
&= 1115468.35\text{ m}
}
$$
$$
\eqalign{
x &= 1115468.35\sin[5^\circ-(-100^\circ)] \cr
&= 1077459.69\text{ m}
}
$$
$$
\eqalign{
y &= -1115468.35\cos[5^\circ-(-100^\circ)] \cr
&= 288704.45\text{ m}
}
$$
$$
\eqalign{
k &= 1115468.35/(6378388.0\times0.1742171) \cr
&= 1.0038196
}
$$
$$
\eqalign{
h &= 1/1.0038196 \cr
&= 0.9961950
}
$$
Inverse Equations #
Inversing forward example:
Given
| $x=\;$m |
| $y=\;$m |
First $q_p$ is found to be $ 1.9955122 $ from equation (3-12), as in the corresponding forward example for the polar aspect. From equations (20-18) and (24-31),
$$
\eqalign{
\rho &= (1077459.7^2 + 288704.5^2)^{1/2} \cr
&= 1115468.37\text{ m}
}
$$
$$
\eqalign{
q &= +[1.9955122-(1115468.37/6378388.0)^2] \cr
&= 1.9649283
}
$$
Iterative equation (3-16) may be used to find $\phi$. The first trial $\phi$ is$$
\eqalign{
\phi &= \arcsin(1.9649283/2) \cr
&= 79.2542240^\circ
}
$$
When this is used in equation (3-16) as in the oblique inverse example, the next trial $\phi$ is found to be$$
\phi = 79.9744266^\circ
$$
Using this value instead, the next trial is$$
\phi = 79.9999675^\circ
$$
and the next,$$
\phi = 79.9999998^\circ
$$
From equation (20-16),$$
\eqalign{
\lambda &= -100^\circ + \arctan[1077459.7/(-288704.5)] \cr
&= 5.0000022^\circ
}
$$
using the ATAN2 function and after adjusting the result to $(-180^\circ,\;+180^\circ]$ range.