Numerical Examples - Albers Equal-Area Conic Projection

# Numerical Examples for Albers Equal-Area Conic Projection #

## SPHERE #

### Forward Equations #

Given

 Radius of sphere: $R=\;\;$ unit Standard parallels: $\phi_1=\;$° $\phi_2=\;$° Origin: $\phi_0=\;$° $\lambda_0=\;$° Point: $\phi=\;$° $\lambda=\;$°
Find $\rho, \theta, x, y, k, h, \omega$

From equations (14-6), (14-5), (14-3), (14-3a), and (14-4) in order

\eqalign { n &= (\sin29.5^\circ +\sin45.5^\circ)/2 \cr &= 0.6028370 }
\eqalign{ C &= \cos^2 29.5^\circ+2\times0.6028370\sin29.5^\circ \cr &= 1.3512213 }
\eqalign{ \rho &= 1.0\times(1.3512213-2\times0.6028370\sin35.0^\circ)^{1/2}/0.6028370 \cr &= 1.3473026 \;\text{units} }
\eqalign{ \rho_0 &= 1.0\times(1.3512213-2\times0.6028370\sin23^\circ)^{1/2}/0.6028370 \cr &= 1.5562263 \;\text{units} }
\eqalign{ \theta &= 0.6028370\times[(-75.0^\circ)-(-96^\circ)] \cr &= 12.6595771^\circ }
From equations (14-1) (14-2) and (14-7) in order,
\eqalign{ x &= 1.3473026\sin 12.6595771^\circ \cr &= 0.2952720 \;\text{units} }
\eqalign{ y &= 1.5562263 - 1.3473026\cos 12.6595771^\circ \cr &= 0.2416774 \;\text{units} }
\eqalign{ h &= \cos 35.0^\circ/(1.3512213-2\times 0.6028370\sin 35.0^\circ) \cr &= 1.0085547 }
and
$$k=1/1.0085547=0.9915178$$
From equation (4-9),
\eqalign{ \sin½\omega &= |1.0085547-0.9915178|/(1.0085547+0.9915178) \cr \omega &= 0.9761175^\circ }

### Inverse Equations #

Inversing forward example:

Given: $R, \phi_1, \phi_2, \phi_0, \lambda_0$ for forward example

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

As in the forward example, from equations (14-6), (14-5), and (14-3a) in order

\eqalign { n &= (\sin29.5^\circ +\sin45.5^\circ)/2 \cr &= 0.6028370 }
\eqalign{ C &= \cos^2 29.5^\circ+2\times0.6028370\sin29.5^\circ \cr &= 1.3512213 }
\eqalign{ \rho_0 &= 1.0\times(1.3512213-2\times0.6028370\sin23^\circ)^{1/2}/0.6028370 \cr &= 1.5562263 \;\text{units} }
From equations (14-10), (14-11), (14-8), and (14-9) in order,
\eqalign{ \rho &=[0.2952720^2 + (1.5562263-0.2416774)^2]^{1/2} \cr &= 1.3473027 \;\text{units} }
\eqalign{ \theta =& \arctan[0.2952720/(1.5562263-0.2416774)] \cr =& 12.6595763^\circ }
\eqalign{ \phi &=\arcsin\{[1.3512213-(1.3473027\times0.6028370/1)^2]/(2\times0.6028370) \} \cr &=\arcsin 0.5735764 \cr &=34.9999974^\circ }
\eqalign{ \lambda &= 12.6595763^\circ/0.6028370 + (-96^\circ) \cr &= 20.9999988^\circ + (-96^\circ) \cr &= -75.0000012^\circ }

## ELLIPSOID #

### Forward Equations #

Given:

 Clarke 1866WGS-84 ellipsoid $a=6378206.4\,\text{m}$ $e^2=0.00676866$ or: $e=0.0822719$ Standard parallels: $\phi_1=\;$ $\phi_2=\;$ Origin: $\phi_0=\;$° $\lambda_0=\;$° Point: $\phi=$° $\lambda=$°
Find $\rho, \theta, x, y, k, h, \omega$.

From equation (14-15),

\eqalign{ m1 &=\cos 29.5^\circ/(1-0.0067687\sin^2 29.5^\circ)^{1/2} \cr &=0.8710708 }
\eqalign{ m2 &=\cos 45.5^\circ/(1-0.0067687\sin^2 45.5^\circ)^{1/2} \cr &=0.7021191 }
From equation (3-12),
\eqalign{ q_1 =& (1-0.0067687)\{\sin 29.5^\circ/(1-0.0067687) \cr & -[1/(2\times0.0822719)\ln[(1-0.0822719\sin^2 29.5^\circ)/ \cr & (1+0.0822719\sin^2 29.5^\circ)] \} \cr =& 0.9792529 }
Using the same formula for $q_2$, (with $\phi_2$, instead of $\phi_1$),
$$q_2=1.4201080$$
Using the same formula for $q_0$ (with $\phi_0$ instead of $\phi_1$),
$$q_0=0.7767080$$
From equations (14-14), (14-13), and (14-12a) in order,
\eqalign{ n &= (0.8710708^2 - 0.7021191^2)/(1.4201080-0.9792529) \cr &= 0.6029035 }
\eqalign{ C &= 0.8710708^2+0.6029035\times0.9792529 \cr &= 1.3491594 }
\eqalign{ \rho_0 &= 6378206.4\times(1.3491594-0.6029035\times0.7767080)^{1/2}/0.6029035 \cr &= 9929079.57 \;\text{m} }
These are the constants for the map.

Using equation (3-12), but with $\phi$ in place of $\phi_1$:

$$q=1.1410831$$

From equations (14-12), (14-4), (14-1), and (14-2) in order,

\eqalign{ \rho &= 6378206.4\times(1.3491594-0.6029035\times1.1410831)^{1/2}/0.6029035 \cr &= 8602328.23 \;\text{m} }
$$\theta = 0.6029035\times[-75^\circ-(-96^\circ)] = 12.6609735^\circ$$
$$x = 8602328.23\sin 12.6609735^\circ = 1885472.73\;\text{m}$$
\eqalign{ y &=9929079.57 - 8602328.23\cos 12.6609735^\circ \cr &=1535925.00\;\text{m} }
From equations (14-15), (14-16), (14-18), and (4-9) in order,
\eqalign{ m &=\cos 35^\circ/(1-0.0067687\sin^2 35^\circ)^{1/2} \cr &=0.8200656 }
\eqalign{ k &=8602328.23\times0.6029035/(6378206.4\times0.8200656) \cr &= 0.9915546 }
$$h=1/0.9915546=1.0085173$$
\eqalign{ \sin ½\omega &= |1.0085173-0.9915546|/(1.0085173+0.9915546) \cr \omega &= 0.9718683^\circ }

### Inverse Equations #

Inversing forward example:

Given

 $x=\;$m $y=\;$m
Find $\rho, \theta, \phi, \lambda$.

The same constants $n, C, \rho_0$, are calculated with the same equations as those used for the forward example. For the particular point:

From equation (14-10),

\eqalign{ \rho &=[1885472.73^2+(9929079.57 - 1535925.00)^2]^{1/2} \cr &=8602328.23\;\text{m} }
From equation (14-11),
\eqalign{ \theta =& \arctan[1885472.73/(9929079.57-1535925.00)] \cr =& 12.6609735^\circ \text{Since the denominator is positive,} \cr & \text{there is no adjustment to } \theta }
From equation (14-19),
\eqalign { q &= [1.3491594-(8602328.23\times0.6029035/6378206.4)^2]/0.6029035 \cr &= 1.1410831 }

Using for the first trial $\phi$ the arcsin of $(1.1410831/2)$ or $34.7879968^\circ$calculate a new $\phi$ from equation (3-16),

\eqalign{ \phi =& 34.7879968^\circ+[(1-0.0067687\sin^234.7879968^\circ)^2/(2\cos 34.7879968^\circ)] \cr &\times\{1.1410831(1-0.0067687)-\sin 34.7879968^\circ/(1-0.0067687\sin^234.7879968^\circ) \cr &+[1/(2\times0.0822719)]\ln[(1-0.0822719\sin34.7879968^\circ) \cr &/(1+0.0822719\sin34.7879968^\circ)]\}\times 180^\circ/\pi \cr =& 34.9990287^\circ }
Note that $180^\circ/\pi$ is included to convert to degrees. Replacing $34.7879968^\circ$ by $34.9990287^\circ$ for the second trial, the calculation using equation (3-16) now provides a third $\phi$ of $34.9999967^\circ$. A third iteration yields $34.9999999^\circ$ and the fourth iteration produces:
$$\phi = 35.0000000^\circ$$
For the longitude use equation (14-9)
\eqalign{ \lambda &= (-96^\circ)+12.6609735^\circ/0.6029035 \cr &= -75.0000000^\circ }