Transverse Mercator Projection

# Numerical Examples for Transverse Mercator Projection #

## SPHERE #

### Forward Equations #

Given

 Radius of sphere: $R=\;\;$ unit Origin: $\phi_0=\;$° $\lambda_0=\;$° Central scale factor: $k_0=\;$° Point: $\phi=\;$° $\lambda=\;$°
Find $x, y, k$

Using equations (8-5), (8-1), (8-3), and (8-4) in order

\eqalign { B &= \cos 40.5^\circ \sin[(-73.5^\circ)-(-75.0^\circ)] \cr &= \cos 40.5^\circ \sin 1.5^\circ = 0.0199051 }
\eqalign{ x &= \unicode{xbd} \times 1.0 \times 1.0 \ln[(1+0.0199051)/(1-0.0199051)] \cr &= 0.0199077 \; \text{units} }
\eqalign{ y &= 1.0\times1.0\times[\arctan[\tan 40.5^\circ/\cos1.5^\circ) - 0^\circ] \cr &= 1.0\times1.0\times40.5096980^\circ\pi/180^\circ \cr &= 0.7070276\; \text{units} }
$$k = 1.0/(1-0.0199051^2)^{1/2} = 1.0001982$$

### Inverse Equations #

Inversing forward example:

Given: $R, \phi_0, \lambda_0, k_0$ for forward example

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

Using equation (8-8)

$$D = 0.7070276/(1.0\times 1.0) + 0 = 0.7070276\;\text{radians}$$
For the hyperbolic functions of $(x/Rk_0)$, the relationships $$\sinh x =(\mathrm{e}^x-\mathrm{e}^{-x})/2$$ and $$\cosh x =(\mathrm{e}^x+\mathrm{e}^{-x})/2$$ are recalled if the function is not directly available on a given computer or calculator. In this case,
\eqalign { \sinh(x/Rk_0) &= \sinh[0.0199077/(1.0\times1.0)] \cr &= (\mathrm{e}^{0.0199077} - \mathrm{e}^{0.0199077})/2 \cr &= 0.0199090 }
\eqalign { \cosh(x/Rk_0) &= \cosh[0.0199077/(1.0\times1.0)] \cr &= (\mathrm{e}^{0.0199077} + \mathrm{e}^{0.0199077})/2 \cr &= 1.0001982 }
From equation (8-6), with D in radians, not degrees,
\eqalign{ \phi &= \arcsin(\sin 0.7070276/1.0001982) = \arcsin(0.6495767/1.0001982) \cr &= 40.4999995^\circ }
From equation (8-7),
\eqalign{ \lambda &= -75.0^\circ + \arctan[0.0199090/0.7602960] \cr &= -75.0^\circ + \arctan 0.0261859 = -75.0^\circ + 1.4999972^\circ \cr &= -73.5000028^\circ }
If more decimals were supplied with the x and y calculated from the forward equations, the $\phi$ and $\lambda$ here would agree more exactly with the original values.

## ELLIPSOID #

Given:
 Clarke 1866 WGS-84 ellipsoid $a=6378206.4\,\text{m}$ $e^2=0.00676866$ Origin: (UTM zone 18) $\phi_0=\;$° $\lambda_0=\;$° Central scale factor $k_0=\;$

### Forward Equations #

 Point: $\phi=$° $\lambda=$°
Find $x, y, k$.

Using equations (8-12) through (8-15) in order,

$$e'^2 = 0.0067687/(1-0.0067687) = 0.0068148$$
$$N = 6378206.4/(1-0.00676866\sin^240.5^\circ)^{1/2} = 6387330.52\;\text{m}$$
$$T = \tan^2 40.5^\circ = 0.7294538$$
$$C = 0.0068148\cos^2 40.5^\circ = 0.0039404$$
$$A = (\cos40.5)\times[(-73.5^\circ) - (-75^\circ)]\pi/180^\circ = 0.0199074$$
From equation (3-21)1
\begin{align} M =&6378206.4[(1-0.0067687/4-3\times 0.0067687^2/64 - 5\times 0.0067687^3/256)\times 40.5^\circ\times\pi/180^\circ \cr &-(3\times 0.0067687/8+3\times 0.0067687^2/32 +45\times0.0067687^3/1024)\sin(2\times40.5^\circ) \cr &+(15\times 0.0067687^2/256 +45\times 0.0067687^3/1024)\sin(4\times 40.5^\circ) \cr &-(35\times 0.0067687^3/3072)\sin(6\times40.5^\circ)] \cr =& 4484837.66\;\text{m} \end{align}
\begin{align} M_0 =&6378206.4[(1-0.0067687/4-3\times 0.0067687^2/64 - 5\times 0.0067687^3/256)\times 0^\circ\times\pi/180^\circ \cr &-(3\times 0.0067687/8+3\times 0.0067687^2/32 +45\times0.0067687^3/1024)\sin(2\times0^\circ) \cr &+(15\times 0.0067687^2/256 +45\times 0.0067687^3/1024)\sin(4\times 0^\circ) \cr &-(35\times 0.0067687^3/3072)\sin(6\times0^\circ)] \cr =& 0.00\;\text{m} \end{align}
Equations (8-9) and (8-10) may now be used:
\begin{align} x =& 0.9996\times6387330.52\times[0.0199074+(1-0.7294538+0.0039404) \cr &\times0.0199074^3/6+(5-18\times0.7294538+0.7294538^2+72\times0.0039404 \cr &-58\times0.006814784946237987)\times0.0199074^5/120] \cr =& 127106.47\;\text{m} \end{align}
\begin{align} y =& 0.9996\times\{4484837.66-0.00+6387330.52\times0.8540807\times[0.0199074^2/2 \cr & +(5-0.7294538+9\times0.0039404+4\times0.0039404^2)\times0.0199074^4/24 \cr & +(61-58\times0.7294538+0.7294538^2+600\times0.0039404-330 \cr & \times0.0068148)\times0.0199074^6/720]\} \cr =& 4484124.43\;\text{m} \end{align}

These values agree exactly with the UTM tabular values, except that 500,000.0m must be added to x for “false eastings.” To calculate k, using equation (8-11),

\begin{align} k =& 0.9996\times[1+(1+0.0039404)\times0.0199074^2/2+(5-4\times0.7294538+42 \cr & \;\times0.0039404+13\times0.0039404^2-28\times0.0068148)\times0.0199074^4/24 \cr & \;+(61-148\times0.7294538+16\times0.7294538^2)\times0.0199074^6/720] \cr =& 0.9997989 \end{align}
\begin{align} k =&0.9996\times[1+(1+0.0068148\cos^2 40.5^\circ)\times127106.47^2 \cr &(2\times0.9996^2\times6387330.52^2)] \cr =&0.9997989 \end{align}

### Inverse Equations #

Inversing forward example:

Given

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

Calculating $M_0$ from equation (3-21),
see above

From equations (8-12), (8-20), (3-24), and (7-19) in order,

$$e'^2 = 0.0067687/(1-0.0067687) = 0.0068148$$
$$M = 0 + 4484124.43/0.9996 = 4485918.80\;\text{m}$$
\eqalign{ e_1 &= [1-(1-0.0067687)^{1/2}]/[1+(1-0.0067687)^{1/2}] \cr &= 0.0016979 }
\begin{align} \mu =& 4485918.80/[6378206.4\times(1-0.0067687/4-3\times0.0067687^2/64 \cr &\;-5\times0.0067687^3/256)] \cr =&0.7045135\;\text{radian} \end{align}
From equation (3-26), using $\mu$ in radians, omitting the last term,
\begin{align} \phi_1 =& 0.7045135 + (3\times0.0016979/2-27\times0.0016979^3/32)\sin(2\times0.7045135) \cr & +(21\times0.0016979^2/16-55\times0.0016979^4/32)\sin(4\times0.7045135) \cr & +(151\times0.0016979^3/96)\sin(6\times0.7045135) \cr =& 0.7070283\;\text{radian} \cr =& 0.7070283\times180^\circ/\pi \cr =& 40.5097362^\circ \end{align}
Now equations (8-21) through (8-25) may be used:
$$C_1 = 0.0068148\cos^2(40.5097362^\circ) = 0.0039393$$
$$T_1=\tan^2(40.5097362^\circ) = 0.7299560$$
$$N_1=6378206.4/(1-0.0067687\sin^240.5097362^\circ)^{1/2} = 6387334.16\;\text{m}$$
\eqalign{ R_1 &=6378206.4\times(1-0.0067687)/(1-0.0067687\sin^240.5097362^\circ)^{3/2} \cr &= 6362271.37\;\text{m} }
$$D=127106.47/(6387334.16\times0.9996) = 0.0199077$$
Returning to equation (8-17),
\begin{align} \phi =& 40.5097362^\circ -(6387334.16\times \tan40.5097362^\circ/6362271.37)\times [0.0199077^2/2 \cr & -(5+3\times0.7299560+10\times0.0039393-4\times0.0039393^2-9\times0.0068148) \cr & \times 0.0199077^4/24 +(61+90\times0.7299560+298\times0.0039393+45\times0.7299560^2\cr & -252\times0.0068148-3\times0.0039393^2)\times 0.0199077/720]\times180^\circ/\pi \cr =& 40.4999996^\circ \end{align}
From equation (8-18),
\begin{align} \lambda =& -75^\circ + [0.0199077 - (1 + 2\times0.7299560 + 0.0039393)0.0199077^3/6 \cr & +(5 - 2\times0.0039393 +28\times0.7299560 -3\times0.0039393^2 + 8\times0.0068148 \cr & +24\times0.7299560^2)/0.0199077^5/120]/\cos40.5097362^\circ \cr =& -73.5000000^\circ \end{align}

1. For lengths of meridional arcs, $M$ and $M_0$, Snyder uses equation (3-22) that is specific for Clarke-1866 ellipsoid. Here I changed to use the equation (3-21) that applies to any ellipsoid. ↩︎