Numerical Examples for Lambert Conformal Conical 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 $\rho, \theta, x, y, k$
From equations (15-3), (15-2), and (15-1a) in order
$$
\eqalign{
n &= \ln(\cos33^\circ/\cos45^\circ)/\ln[\tan(45^\circ+45^\circ/2)/\tan(45^\circ+33^\circ/2)] \cr
&= 0.6304777
}
$$
$$
\eqalign{
F &=[\cos33^\circ\tan^{0.6304777}(45^\circ+33^\circ/2)]/0.6304777 \cr
&=1.9550002
}
$$
$$
\eqalign{
\rho_0 &= 1.0\times1.9550002/\tan^{0.6304777}(45^\circ+23^\circ/2) \cr
&= 1.5071429\;\text{units}
}
$$
The above constants apply to the map generally. For the specific $\phi$ and $\lambda$, using equations (15-1), (14-4), (14-1), and (14-2) in order,$$
\eqalign{
\rho &= 1.0\times1.9550002/\tan^{0.6304777}(45^\circ+35^\circ/2) \cr
&= 1.2953636\;\text{units}
}
$$
$$
\eqalign{
\theta &= 0.6304777\times((-75^\circ)-(-96^\circ)) \cr
&= 13.2400316^\circ
}
$$
$$
\eqalign{
x &= 1.2953636\sin13.2400316^\circ \cr
&= 0.2966785\;\text{units}
}
$$
$$
\eqalign{
y &= 1.5071429-1.2953636\cos13.2400316^\circ\cr
&= 0.2462112\;\text{units}
}
$$
From equation (15-4),$$
\eqalign{
k =& \cos33^\circ\tan^{0.6304777}(45^\circ+33^\circ/2)/ \cr
&[\cos35^\circ\tan^{0.6304777}(45^\circ+35^\circ/2)] \cr
=& 0.9970040
}
$$
or from equation (4-5)$$
\eqalign{
k &= 0.6304777\times1.2953636/(1.0\times\cos35^\circ) \cr
&= 0.9970040
}
$$
Inverse Equations #
Inversing forward example:
Given: $R, \phi1, \phi2, \phi_0, \lambda_0$, for forward example
| $x=\;$ units |
| $y=\;$ units |
After calculating $n, F,$ and $\rho_0$ as in the forward example, obtaining the same values, equation (14-10) is used:
$$
\eqalign{
\rho &= [0.2966785^2 + (1.5071429-0.2462112)^2]^{1/2} \cr
&= 1.2953636
}
$$
From equation (14-11),$$
\eqalign{
\theta &= \arctan[0.2966785/(1.5071429-0.2462112)] \cr
&= 13.2400331^\circ\;\text{Not adjusted since denominator is positive.}
}
$$
From equation (14-9),$$
\eqalign{
\lambda &= 13.2400331^\circ/0.6304777 + (-96^\circ)\cr
&= -74.9999977^\circ
}
$$
From equation (15-5),$$
\eqalign{
\phi &= 2\arctan[(1.0\times1.9550002/1.2953636)^{1/0.6304777}] - 90^\circ \cr
&= 34.9999978^\circ
}
$$
ELLIPSOID #
Forward Equations #
Given:
| ellipsoid | $a=$ | 6378206.4 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),
$$
\eqalign{
m_1 &= \cos33^\circ/(1-0.0067687\sin^233^\circ)^{1/2} \cr
&= 0.8395138
}
$$
$$
\eqalign{
m_2 &= \cos45^\circ/(1-0.0067687\sin^245^\circ)^{1/2} \cr
&= 0.7083064
}
$$
From equation (15-9),
$$
\eqalign{
t_1 &= \tan(45^\circ-33^\circ/2)/[(1-0.0822719\sin 33^\circ)/(1+0.0822719\sin 33^\circ)]^{0.0822719/2} \cr
&= 0.5449623
}
$$
$$
t_2= 0.4162031\text{, using above with \(45^\circ\) in place of \(33^\circ\),}
$$
$$
t_0= 0.6636390\text{, using above with \(23^\circ\) in place of \(33^\circ\),}
$$
From equations (15-8),(15-10), and (15-7a) in order,
$$
\eqalign{
n &= \ln(0.8395138/0.7083064)/\ln(0.5449623/0.4162031) \cr
&= 0.6304965
}
$$
$$
\eqalign{
F &= 0.8395138/(0.6304965\times0.5449623^{0.6304965}) \cr
&= 1.9523837
}
$$
$$
\eqalign{
\rho_0 &= 6378206.4\times1.9523837\times0.6636390^{0.6304965} \cr
&= 9615955.20\;\text{m}
}
$$
The above are constants for the map. For the specific $\phi, \lambda$, using equation (15-9),
$t= 0.5225935$, using above calculation with $35^\circ$ in place of $33^\circ$.
From equations (15-7), (14-4), (14-1) and (14-2) in order,
$$
\eqalign{
\rho &= 6378206.4\times1.9523837\times0.5225935^{0.6304965} \cr
&= 8271173.83\;\text{m}
}
$$
$$
\eqalign{
\theta &= 0.6304965\times[(-75^\circ)-(-96^\circ)] \cr
&= 13.2404257^\circ
}
$$
$$
\eqalign{
x &= 8271173.83\sin 13.2404257^\circ \cr
&= 1894410.90\;\text{m}
}
$$
$$
\eqalign{
y &= 9615955.20-8271173.83\cos13.2404257^\circ \cr
&= 1564649.47\;\text{m}
}
$$
From equations (14-15), (14-16),
$$
\eqalign{
m &= \cos35^\circ/(1-0.0067687\sin^235^\circ)^{1/2} \cr
&= 0.8200656
}
$$
$$
\eqalign{
k &= 8271173.83\times0.6304965/(6378206.4\times0.8200656) \cr
&= 0.9970171
}
$$
Inverse Equations #
Inversing forward example:
Given
| $x=\;$m |
| $y=\;$m |
$$
\eqalign{
\rho &= [1894410.90^2 + (9615955.20 - 1564649.47)^2]^{1/2} \cr
&= 8271173.84\;\text{m}
}
$$
From equation (14-11),$$
\eqalign{
\theta &= \arctan[1894410.90/(9615955.20-1564649.47)]\cr
&= 13.2404257^\circ\;\text{The denominator is positive; therefore \(\theta\) is not adjusted.}
}
$$
From equation (15-11),$$
\eqalign{
t &= [8271173.84/(6378206.4\times1.9523837)]^{1/0.6304965} \cr
&= 0.5225935
}
$$
To use equation (7-9), an initial trial $\phi$, is found as follows:$$
\eqalign{
\phi &= 90^\circ - 2\arctan0.5225935\cr
&= 34.8174476^\circ
}
$$
Inserting this into the right side of equation (7-9),$$
\eqalign{
\phi =& 90^\circ - 2\arctan\{0.5225935\times[(1-0.0822719\sin 34.8174476^\circ)/ \cr
& (1+0.0822719\sin 34.8174476^\circ)]^{0.0822719/2} \} \cr
=& 34.9991681^\circ
}
$$
Replacing $34.8174476^\circ$ with $34.9991681^\circ$ for the second trial, a $\phi$, of $34.9999962^\circ$ is obtained. Recalculation with the new $\phi$, results in $$
\phi=35.0000000^\circ
$$
From equation (14-9),$$
\eqalign{
\lambda &= 13.2404257^\circ/0.6304965+(-96^\circ) \cr
&= -75.0000000^\circ
}
$$
Examples using equations (3-5) and (7-13) are omitted here, since comparable examples for these equations have been given above.