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:

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),

$$ x = 6378137.0000000\times[-75^\circ - (-180^\circ)] \times \pi/180^\circ= 11688546.53 $$

$$ \eqalign{ y &= 6378206.4\ln\left[\tan(45^\circ+35.0^\circ/2)\left(\frac{1-0.0822719\sin35.0^\circ}{1+0.0822719\sin35.0^\circ}\right)^{0.0822719/2}\right] \cr &= 6378206.4\ln[1.9209821\times0.9961223] \cr &= 6378206.4\ln 1.9135332 = 4139145.66 } $$

$$ \eqalign { k &= (1-0.0067687\sin^2 35.0^\circ)^{1/2}/\cos 35.0^\circ \cr &= 1.2194146 } $$

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 $\chi$

$$ \phi = 34.9999999^\circ $$