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: | $\phi=\;$° |
| $\lambda=\;$° | |
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 |
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:
| 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: | $\phi=$° | $\lambda=$° |
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 |
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)$$
\eqalign{
\lambda &= (-96^\circ)+12.6609735^\circ/0.6029035 \cr
&= -75.0000000^\circ
}
$$