Numerical Examples - Lambert Azimuthal Equal-Area

# 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
Find $\phi, \lambda$

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:

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

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
Find: $\phi, \lambda$

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:

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

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
Find: $\phi, \lambda$

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.