Numerical Examples - Oblique Mercator Projection

# Numerical Examples for Oblique Mercator Projection #

## SPHERE #

### Forward Equations #

Given

 Radius of sphere: $R=\;\;$ unit Central scale factor: $k_0=\;$° Central line through: $\phi_1=\;$° $\lambda_1=\;$° $\phi_2=\;$° $\lambda_2=\;$° Point: $\phi=\;$° $\lambda=\;$°
Find $x, y, k$

Using equation (9-1),

\begin{align} \lambda_p =& \arctan[(\cos45^\circ\sin0^\circ\cos0^\circ - \sin45^\circ\cos0^\circ\cos(-90^\circ))/ \cr &(\sin45^\circ\cos0^\circ\sin(-90^\circ) - \cos45^\circ\sin0^\circ\sin0^\circ] \cr =&180^\circ \end{align}
Result is adjusted to be in the $(-180^\circ, +180^\circ]$ interval.

From equation (9-2),

\eqalign{ \phi_p &= \arctan[-\cos(180^\circ-0^\circ)/\tan45^\circ] \cr &= 45^\circ }
The other pole is then at $\phi= -45^\circ, \lambda = 0^\circ$. From equation (9-6a),
$$\lambda_0 = 180^\circ + 90^\circ = -90^\circ$$
From equation (9-6),
\eqalign{ A &= \sin45^\circ\sin(-30^\circ)-\cos45^\circ\cos(-30^\circ)\sin(120^\circ-(-90^\circ)) \cr &= 0.7071068\times(-0.5000000)-0.7071068\times0.8660254\times(-0.5000000) \cr &= -0.0473672 }
From equation (9-3)
\eqalign{ x &=1.0\times1.0\arctan\{[\tan(-30^\circ)\cos45^\circ+\sin45^\circ\sin(120^\circ-(-90^\circ))]/\cos(120^\circ-(-90^\circ))\} \cr &=-2.4201335 \;\text{units} }
From equation (9-4),
\eqalign { y &= (1/2)\times1.0\times1.0\ln[(1-0.0473672)/(1+0.0473672)] \cr &= -0.0474026 \;\text{units} }
From equation (9-5),
$$k=1.0/[1-(-0.0473672)^2]^{1/2} = 1.0011237$$
If the parameters are given in terms of a central point (for equations (9-7) and (9-8)), we shall assume certain artificial parameters (calculated with different formulas) which give the same pole as above:*1

Given

 Azimuth of central line: $\beta=\;\;$ ° Center: $\phi_c=\;$° $\lambda_c=\;$°
Using equations (9-7) and (9-8),
\eqalign{ \phi_p &= \arcsin(\cos20.0^\circ\sin48.806299^\circ) \cr &= 45^\circ }
\eqalign{ \lambda_p &= \arctan[\cos48.806299^\circ/(-\sin20.0^\circ\sin48.806299^\circ)] + (-68.6557771^\circ) \cr &= 180^\circ \;\;\text{(computation uses the ATAN2 function)} }

### Inverse Equations #

Inversing forward example:

Given:

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

First, $\phi_p$ and $\lambda_p$ are determined, exactly as for the forward example, so that $\lambda_0=-90^\circ$, and $\phi_c = 45^\circ$. Determining hyperbolic functions, if not readily available,

\eqalign{ y/Rk_0 &= -0.0474026/(1\times1) = -0.0474026 \cr e^{-0.0474026} &= 0.9537034 \cr \sinh(y/Rk_0) &= (0.9537034-1/0.9537034)/2 \cr &= -0.0474204 \cr \cosh(y/Rk_0) &= (0.9537034+1/0.9537034)/2 \cr &= 1.0011237 \cr \tanh(y/Rk_0) &= (0.9537034-1/0.9537034)/(0.9537034+1/0.9537034) \cr &= -0.0473671 }
From equation (9-9),
\eqalign { \phi =& \arcsin\{\sin45^\circ\times(-0.0473671)+\cos45^\circ\sin[(-2.4201335)/(1\times1)]/1.0011237 \} \cr =& \arcsin (-0.5) = -29.999998^\circ }
From equation (9-10),
\eqalign{ \lambda =& (-90^\circ)+\arctan\{[\sin0.7853982^\circ\sin[(-2.4201335)/(1\times1)] \cr & -\cos0.7853982^\circ\times(-0.0474204)]/\cos[(-2.4201335)/(1\times1)]\} \cr =& (-90^\circ)+(-149.9999981^\circ) \cr =& -239.9999981^\circ = 120.0000019^\circ }

## ELLIPSOID (Alternate A) #

### Forward Equations #

Given:

 Clarke 1866WGS-84 ellipsoid $a=6378206.4\,\text{m}$ $e^2=0.00676866$ or $e=0.0822719$ Central scale factor $k_0=\;$ Center: $\phi_0=\;$° Central line through: $\phi_1=\;$° $\lambda_1=\;$° $\phi_2=\;$° $\lambda_2=\;$° False coordinates: $x_0=\;$° $y_0=\;$° Point: $\phi=$° $\lambda=$°
Find $x, y, k$.

Following equations (9-11) through (9-24) in order:

\eqalign{ B &=[1+0.00676866\cos^440^\circ/(1-0.00676866)]^{1/2} \cr &= 1.0011727 }
\eqalign{ A =& 6378206.4\times1.0011727\times0.9996\times(1-0.00676866)^{1/2} \cr & (1-0.00676866\sin^240^\circ) \cr =& 6379333.23; }
\eqalign{ t_0 =&\tan(45^\circ-40^\circ/2)[(1-0.0822719\sin40^\circ)/(1+0.0822719\sin40^\circ)]^{0.0822719/2} \cr =&0.4683428 }
\eqalign{ t_1 =&\tan(45^\circ-47.5^\circ/2)[(1-0.0822719\sin47.5^\circ)/(1+0.0822719\sin47.5^\circ)]^{0.0822719/2} \cr =&0.3908266 }
\eqalign{ t_2 =&\tan(45^\circ-25.7^\circ/2)[(1-0.0822719\sin25.7^\circ)/(1+0.0822719\sin25.7^\circ)]^{0.0822719/2} \cr =&0.6303639 }
\eqalign{ D =&1.0011727\times(1-0.00676866)^{1/2}/[\cos40^\circ\times \cr &(1-0.00676866\sin^240^\circ)^{1/2}] \cr =& 1.3043327 }
\eqalign { E &=[1.3043327 + (1.3043327^2-1)]\times0.4683428^{1.0011727} \cr &=1.0021857 \cr &\text {using the "+" sign since $\phi_0$ is north or positive.} }
$$H = 0.3908266^{1.0011727} = 0.3903963$$
$$L = 0.6303639^{1.0011727} = 0.6300229$$
$$F= 1.0021857/0.3903963 = 2.5670986$$
$$G= (2.5670986- 1/2.5670986)/2$$
\eqalign { J &= (1.0021857^2-0.6300229\times0.3903963)/(1.0021857^2+0.6300229\times0.3903963) \cr &= 0.6065716 }
$$P=(0.6300229-0.3903963)/(0.6300229+0.3903963) = 0.2348315$$
\eqalign { \lambda_0 =& [(-122.3^\circ)+(-80.2^\circ)]/2-\arctan\{0.6065716\tan[1.0011727 \cr & \times((-122.3^\circ)-(-80.2^\circ))/2]/0.2348315\}/1.0011727 \cr =& -101.25^\circ - \arctan (-0.9953887)/1.0011727 \cr =& -56.4349627^\circ }
\eqalign{ \gamma_0 &= \arctan \{\sin[1.0011727\times((-122.3^\circ)-(-56.4349627^\circ))]/1.0887769\} \cr &= -39.985883^\circ }
\eqalign{ \alpha_c &= \arcsin[1.3043327\sin(-39.985883^\circ)] \cr &= -56.9466071^\circ }
These are constants for the map. For the given $\phi$, and $\lambda$, following equations (9-25) through (9-34) in order:
\eqalign{ t =&\tan(45^\circ-40.8^\circ/2)[(1-0.0822719\sin40.8^\circ)/(1+0.0822719\sin40.8^\circ)]^{0.0822719/2} \cr =&0.4598671 }
$$Q=1.0021857/0.4598671^{1.0011727} = 2.1812805$$
$$S=(2.1812805-1/2.1812805)/2 = 0.8614171$$
$$T=(2.1812805+1/2.1812805)/2 = 1.3198634$$
$$V = \sin[1.0011727\times((-74^\circ) - (-56.4349627^\circ))] = -0.3021309$$
\eqalign{ U =& [0.3021309\cos(-39.985883^\circ) + 0.8614171\sin(-39.985883^\circ)]/1.3198634 \cr =& -0.2440041 }
\eqalign{ v &= 6379333.23\ln[(1+(-0.2440041))/(1-(-0.2440041))]/(2\times1.0011727) \cr &= 1586767.31 \;\text{m} }
\eqalign{ u =& 6379333.23\arctan\{ [0.8614171\cos (-39.985883^\circ) \cr & + (-0.3021309)\sin(-39.985883^\circ)]/\cos[1.0011727\times ((-74^\circ)-(-56.4349627^\circ))]\} \cr & / 1.0011727 \cr =& 4655443.69 \;\text{m} }
\eqalign{ k =& 6379333.23 \cos[1.0011727\times4655443.6913226/6378206.4] \cr & \times(1-0.0067687\sin^240.8^\circ)/\{6378206.4\cos40.8^\circ \cr & \cos[1.0011727\times((-74^\circ)-(-56.4349627^\circ))] \} \cr =& 1.0307554 }
\eqalign{ x &= 1586767.31\cos (-56.9466071^\circ) + 4655443.69\sin (-56.9466071^\circ) + 4000000 \cr &= 963436.08\;\text{m} }
\eqalign{ y &= 4655443.69\cos (-56.9466071^\circ) + 1586767.31\sin (-56.9466071^\circ) + 500000 \cr &= 4369142.8\;\text{m} }

### Inverse Equations #

The above example for alternate A will be inverted,*2 first using equations (9-11) through (9-24), then using equations (9-40) through (9-48). Since no new equations are involved for inverse alternate B, an example of the latter will be omitted. As stated with the inverse equations, the constants for the map are chosen as in the forward examples.

Inversing forward example for alternate A:

Given:

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

Using equations (9-11) through (9-24) in order, again gives the following constants:

$$B=1.0011727$$
$$A=6379333.23\;\text{m}$$
$$E=1.0021857$$
$$\lambda_0=-56.4349627^\circ$$
$$\gamma_0=-39.985883^\circ$$
$$\alpha_c=-56.9466071^\circ$$

Following equations (9-40) through (9-48) in order:

\eqalign{ v =& (963436.08 - 4000000)\cos(-56.9466071^\circ) \cr &-(4369142.8-500000)\sin(-56.9466071^\circ) \cr =& 1586767.31\;\text{m} }
\eqalign{ u =& (4369142.8 - 500000)\cos(-56.9466071^\circ) \cr &-(963436.08-4000000)\sin(-56.9466071^\circ) \cr =& 4655443.69\;\text{m} }
\eqalign{ Q' =& \mathrm{e}^{-(1.0011727\times1586767.31/6379333.23)} = \mathrm{e}^{(-0.2490273)} \cr =& 0.7795587 }
$$S' = (0.7795587-1/0.7795587) = -0.2516092$$
$$T' = (0.7795587+1/0.7795587)/2 = 1.0311679$$
\eqalign{ V' =& \sin(1.0011727\times4655443.69/6379333.23) \cr =& \sin 41.8617536^\circ = 0.6673356 }
\eqalign{ U' =& [0.6673356\cos(-39.985883^\circ)+(-0.2516092)\sin(-14.4161438^\circ)]/ \cr & 1.0311679 \cr =& 0.6526562 }
\eqalign{ t &= \{1.0021857/[(1+0.6526562)/(1-0.6526562)]^{1/2} \}^{1/1.0011727} \cr &= 0.4598671 }
The first trial $\phi$ for equation (7-9) is
$$90^\circ - 2\arctan 0.4598671 = 40.6077094^\circ$$
Calculating a new trial $\phi$:
\eqalign{ \phi =& 90^\circ - 2\arctan\{ 0.4598671\times[(1-0.0822719\sin 40.6077094^\circ)/ \cr & (1+0.0822719\sin 40.6077094^\circ)]^{0.0822719/2}\} \cr =& 40.7992509^\circ }
A second iteration gives $40.7999971^\circ$. Third iteration gives:
$$\phi = 40.8^\circ$$
For longitude $\lambda$:
\eqalign{ \lambda =& -56.4349627^\circ-\arctan\{[(-0.2516092)\cos (-39.985883^\circ) \cr & -0.6673356\sin (-39.985883^\circ)]\cos(1.0011727 \cr & \times 4655443.69/6379333.23)\} /1.0011727 \cr =& -74^\circ }

Using series equation (3-5) with (7-13), to avoid itemtion of (7-9), and beginning with equation (7-13),

\eqalign{ \chi &= 90^\circ - 2\arctan 0.4598671 \cr &= 40.6077094^\circ }
Since equation (3-5) is used in an example under Auxiliary latitudes, the calculation will not be shown here.

## ELLIPSOID (Alternate B) #

### Forward Equations #

Given:

 Clarke 1866WGS-84 ellipsoid $a=6378206.4\,\text{m}$ $e^2=0.00676866$ or $e=0.0822719$ Central scale factor $k_0=\;$ Center: $\phi_0=\;$° $\lambda_c=\;$° Azimuth of central line: $\alpha_c=\;$° Point: $\phi=\;$° $\lambda=\;$°
Find: $u, v$ (example uses center of Zone 2, Path 16, Landsat mapping, with Hotine Oblique Mercator).

Using equations (9-11) through (9-39) in order,

\eqalign{ B &=[1+0.0067687\cos^436^\circ/(1-0.0067687)]^{1/2} \cr &= 1.0014586 }
\eqalign{ A =& 6378206.4\times1.0014586\times1\times(1-0.0067687)^{1/2} \cr & (1-0.0067687\sin^236^\circ) \cr =& 6380777.05\;\text{m} }
\eqalign{ t_0 =&\tan(45^\circ-36^\circ/2)[(1-0.0822719\sin36^\circ)/(1+0.0822719\sin36^\circ)]^{0.0822719/2} \cr =&0.5115582 }
\eqalign{ D =&1.0014586\times(1-0.0067687)^{1/2}/[\cos36^\circ\times \cr &(1-0.0067687\sin^236^\circ)^{1/2}] \cr =& 1.2351194 }
\eqalign{ & F = 1.2351194 + (1-1.2351194)^{1/2} = 1.9600471 \cr & \text{using the "+" sign since $\phi_0$ is north or positive.} }
$$E = 1.9600471\times 0.5115582^{1.0014586} = 1.0016984$$
$$G= (1.9600471- 1/1.9600471)/2 = 0.7249276$$
\eqalign{ \gamma_0 &= \arcsin [(\sin 14.3394883^\circ)/1.2351194] \cr &= 11.5673996^\circ }
\eqalign{ \lambda_0 &= -77.7610558^\circ - [\arcsin(0.7249276\tan 11.5673996^\circ)]/1.0014586 \cr &= -86.28148^\circ }
\eqalign{ u_{36^\circ,-76.87^\circ\dots} =& (6380777.05/1.0014586)\arctan[(1.2351194^2-1)^{1/2}/\cr & \cos14.3394883^\circ] = 4092868.93 }
These are constants for the map. The calculations of $u, v, x$, and $y$ for $(\phi, \lambda)$ follow the same steps as the numerical example for equations (9-25) through (9-34) for alternate A. For the given point, it is found that
$$u=4414439.01\;\text{m}$$
$$v=-2356.25\;\text{m}$$