Numerical Examples - 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

$$ \eqalign { B &= \cos 40.5^\circ \sin[(-73.5^\circ)-(-75.0^\circ)] \cr &= \cos 40.5^\circ \sin 1.5^\circ = 0.0199051 } $$
$$ \eqalign{ x &= \unicode{xbd} \times 1.0 \times 1.0 \ln[(1+0.0199051)/(1-0.0199051)] \cr &= 0.0199077 \; \text{units} } $$
$$ \eqalign{ y &= 1.0\times1.0\times\{\arctan[\tan 40.5^\circ/\cos1.5^\circ - 0^\circ]\} \cr &= 1.0\times1.0\times0.7070276^\circ\pi/180^\circ \cr &= 0.7070276\; \text{units} } $$
$$ k = 1.0/(1-0.0199051^2)^{1/2} = 1.0001982 $$

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 #

Forward Equations #

Given:

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

$$ \begin{align} k =& 0.9996\times[1+(1+0.0039404)\times0.0199074^2/2+(5-4\times0.7294538+42 \cr & \;\times0.0039404+13\times0.0039404^2-28\times0.0068148)\times0.0199074^4/24 \cr & \;+(61-148\times0.7294538+16\times0.7294538^2)\times0.0199074^6/720] \cr =& 0.9997989 \end{align} $$
Using equation (8-16) instead,
$$ \begin{align} k =&0.9996\times[1+(1+0.0068148\cos^2 40.5^\circ)\times127106.47^2 \cr &(2\times0.9996^2\times6387330.52^2)] \cr =&0.9997989 \end{align} $$

Inverse Equations #

Inversing forward example:

Given

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

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

$$ \begin{align} M_0 =&6378206.4[(1-0.0067687/4-3\times 0.0067687^2/64 - 5\times 0.0067687^3/256)\times 0^\circ\times\pi/180^\circ \cr &-(3\times 0.0067687/8+3\times 0.0067687^2/32 +45\times0.0067687^3/1024)\sin(2\times0^\circ) \cr &+(15\times 0.0067687^2/256 +45\times 0.0067687^3/1024)\sin(4\times 0^\circ) \cr &-(35\times 0.0067687^3/3072)\sin(6\times0^\circ)] \cr =& 0.00\;\text{m} \end{align} $$

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