Mercator Projection

# Numerical Examples for Mercator Projection #

## SPHERE #

### Forward Equations #

Given

 Radius of sphere: $R=\;\;$ unit Central meridian: $\lambda_0=\;$° Point: $\phi=\;$° $\lambda=\;$°
Find $x, y, k$

Using equations (7-1) , (7-2), and (7-3),

\eqalign { x &= \pi\times1.0\times[(-75.0^\circ) - (-180.0^\circ)]/180^\circ \cr &= 1.8325957\;\text{units} }
\eqalign { y &= 1.0\times\ln\tan(45^\circ+35.0^\circ/2) = 1.0\times\ln\tan(62.5^\circ) \cr &= 1.0\times \ln 1.9209821 = 0.6528366\;\text{units} }
$$h = k = \sec 35.0^\circ = 1/\cos 35.0^\circ = 1/0.8191520 = 1.2207746$$

### Inverse Equations #

Given: $R, \lambda_0$ for forward example

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

Using equations (7-4) and (7-5),

\eqalign{ \phi &= 90^\circ - 2\arctan(\mathrm{e}^{-0.6528366/1.0}) \cr &= 90^\circ - 2\arctan(0.5205670) \cr &= 90^\circ - 2\times27.5^\circ = 35.0^\circ }
$$\lambda = (1.8325957/1.0)\times180^\circ/\pi+(-180.0^\circ)= -75.0^\circ$$

## ELLIPSOID #

Given:
 Clarke 1866 WGS-84 ellipsoid $a=6378206.4\,\text{m}$ $e^2=0.00676866$ or: $e=0.0822719$ Central meridian: $\lambda_0=\;$°

### Forward Equations #

 Point: $\phi=$° $\lambda=$°
Find $x, y, k$. Using equations (7-6), (7-7), and (7-8),

$$x = 6378206.4\times[(-75.0^\circ) - (-180^\circ)] \times \pi/180^\circ = 11688673.7$$
\eqalign{ y &= 6378206.4\ln\left[\tan(45^\circ+35.0^\circ/2)\left(\frac{1-0.0822719\sin35.0^\circ}{1+0.0822719\sin35.0^\circ}\right)^{0.0822719/2}\right] \cr &= 6378206.4\ln[1.9209821\times0.9961223] \cr &= 6378206.4\ln 1.9135332 = 4139145.66 }
\eqalign { k &= (1-0.0067687\sin^2 35.0^\circ)^{1/2}/\cos 35.0^\circ \cr &= 1.2194146 }

### Inverse Equations #

Inversing forward example:

Given: $a, e, \lambda_0$ for forward example,

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

Using equation (7-10),

$$t = \mathrm{e}^{-4139145.6/6378206.4} = 0.5225935$$
From equation (7-11), the first trial $\phi$
$$\phi = 90^\circ - 2\arctan 0.5225935 = 34.8174474^\circ$$
Using this value on the right side of equation (7-9),
\begin{align} \phi =& 90^\circ - 2\arctan\{0.5225935[(1-0.0822719\sin34.8174477^\circ)/ \cr & (1+0.0822719\sin34.8174477^\circ)]^{0.0822719/2} \} \cr =& 34.9991681^\circ \end{align}
Replacing $34.8174477^\circ$ with $34.9991681^\circ$ for the second trial, recalculation using (7-9) gives $34.9999963^\circ$. The third trial gives $35.0000000^\circ$, which does not change (to seven places) with recalculation.

For $\lambda$, using equation (7-12),

$$\lambda = 11688673.72/6378206.4\times 180^\circ/\pi+(-180^\circ) = -75.0000000$$

Using equations (7-13) and (3-5) instead, to find φ,

\eqalign { \chi &= 90^\circ - 2\arctan 0.5225935 \cr &= 90^\circ - 55.1825523^\circ \cr &= 34.8174477^\circ }
using $t$ as calculated above from (7-10). Using (3-5), $\chi$ is inserted as in the example given above for inverse auxiliary latitude $\chi$
$$\phi = 34.9999999^\circ$$