Numerical Examples - Lambert Conformal Conical Projection

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
Find $\rho, \theta, \phi, \lambda$.

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:

 Clarke 1866WGS-84 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: $\phi=$ ° $\lambda=$ °
Find $\rho, \theta, x, y, k$.

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
The map constants $n, F$, and $\rho_0$, are calculated as in the forward example, obtaining the same values. Then, from equation (14-10),
\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 $\phi = 35.0000000^\circ$, which does not change to seven decimals with a fourth trial. Therefore,
$$\phi=35.0000000^\circ$$
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.