Auxiliary Latitudes

# Numerical Examples for Auxiliary Latitudes #

For all examples under this heading, the ellipsoid will be used.
 $a=6378206.4\,\text{m}$ a is not needed here $e^2=0.00676866$ or: $e=0.0822719$

Auxiliary latitudes will be calculated for geodetic latitude $\phi =$ °

## Conformal latitude #

Using closed equation (3-1):

\begin{align} \chi =& 2\arctan\{\tan{(45^\circ + 40^\circ/2)}[(1-0.0822719\sin{40^\circ})/(1+0.0822719 \cr & \sin{40^\circ})]^{0.0822719/2}\}-90^\circ \cr =& 2\arctan\{ 2.1445069 [0.8995456]^{0.0822719/2} \} -90^\circ \cr =& 2\arctan(2.1351882) - 90^\circ \cr =& 2\times 64.9042961^\circ - 90^\circ \cr =& 39.8085922^\circ \end{align}

Using series equation (3-2), obtaining $\chi$ first in radians, and omitting terms with $e^8$ for simplicity:

\begin{align} \chi =& 40^\circ\times\pi/180^\circ-(0.0067687 + 5 \times 0.0067687^2/24 + 3 \times 0.0067687^3/32)\times \sin(2\times 40^\circ)+ \\ & (5 \times 0.0067687^2/48 + 7 \times 0.0067687^3/80)\times \sin(4\times 40^\circ)- \\ & (13 \times 0.0067687^3/480) \times \sin(6\times 40^\circ) \\ =& 0.6981317 - (0.0033891)\times 0.9848078 \\ & (0.0000048)\times 0.3420201 - (0.0000000)\times -0.8660254 \\ =& 0.6947910\;\textrm{radian} = 39.8085923^\circ \end{align}

For inverse calculations, using closed equation (3-4) with iteration and given
$\chi$=° , find $\phi$:

First trial:

\begin{align} \phi =& 2\arctan\{\tan (45^\circ+ 39.8085923^\circ/2) [(1+0.0822719\sin 39.8085923^\circ)/ \\ & (1-0.0822719\sin 39.8085923^\circ)]^{0.0822719/2} \} - 90^\circ \\ =& 2\arctan\{2.1351882 [1.1112023]^{0.0411359} \} - 90^\circ \\ =& 129.9992367^\circ - 90^\circ \\ =& 39.9992367^\circ \end{align}

Second trial:

\begin{align} \phi =& 2\arctan\{ \tan (45^\circ+ 39.8085923^\circ/2) [(1+0.0822719\sin{39.9992367^\circ}/ \\ & (1-0.0822719\sin{39.9992367^\circ})]^{0.0411359}\}-90^\circ \\ =& 2\arctan(2.1445068)-90^\circ = 39.9999971^\circ \end{align}

The third trial gives $40.0000001^\circ$also given by the fourth trial.

Using series equation (3-5):

\begin{align} \phi =& 39.8085923^\circ \times \pi/180 + (0.0822719^2/2 + 5\times 0.0822719^4/24 + 0.0822719^6/12)\sin(2\times 39.8085923^\circ) \\ &+ (7\times 0.0822719^4/48 + 29\times0.0822719^6/240)\sin(4\times 39.8085923^\circ) \\ &+ (7\times 0.0822719^6/120) \sin(6\times39.8085923^\circ) \\ =& 0.6947910 + (0.0033939)\times 0.9836256 \\ &+ (0.0000067)\times 0.3545461 \\ &+ (0.0000000)\times -0.8558300 \\ =& 0.6981317\;\text{radian} = 40.0000001^\circ \end{align}

## Isometric latitude #

Using equation (3-7):

\begin{align} \psi =& \ln\{\tan(45^\circ+40^\circ/2)[ (1-0.0822719\sin(40^\circ)/(1+0.0822719\sin(40^\circ)]^{0.0822719/2} \} \\ =& \ln(2.1351882) \\ =& 0.7585548 \end{align}

Using equation (3-8) with the value of $\chi$ resulting from the above examples:

\begin{align} \psi =& \ln\tan(45^\circ + 39.8085923^\circ/2) \\ =& \ln(2.1351882) \\ =& 0.7585548 \end{align}

For inverse calculations, using equation (3-9) with
$\psi$= :

\begin{align} \chi &= 2\arctan\mathrm{e}^{0.7585548} - 90^\circ \\ &= 2\arctan(2.1351882) - 90^\circ \\ &= 39.8085933^\circ \end{align}

From this value of $\chi, \phi$; may be found from (3-4) or (3-5) as shown above.

Using iterative equation (3-10), to find $\phi$:

First trial:

\begin{align} \phi &= 2\arctan\mathsf{e}^{0.7585548} - 90^\circ \\ &= 39.8085933^\circ \text{, just as above} \end{align}

Second trial:

\begin{align} \phi =& 2\arctan\mathrm{e}^{0.7585548}[(1+0.0822719\sin 39.8085933^\circ)/(1-0.0822719 \\ & \sin 39.8085933^\circ)]^{0.0822719/2} - 90^\circ \\ =& 2\arctan (2.1351882\times 1.0058368) - 90^ \circ \\ =& 39.9992376^\circ \end{align}

Third trial:

\begin{align} \phi =& 2\arctan\mathrm{e}^{0.7585548}[(1+0.0822719\sin 39.9992376^\circ)/(1-0.0822719 \\ & \sin 39.9992376^\circ)]^{0.0822719/2} - 90^\circ \\ =& 2\arctan (2.1351882\times 1.0050660) - 90^ \circ \\ =& 39.9999981^\circ \end{align}

Fourth trial, substituting: $39.9999981^\circ$ in place of $39.9992376^\circ$

$$\phi=40.0000011^\circ$$

## Authalic latitude #

Using equations (3-11) and (3-12):

\begin{align} q =& (1-0.0067687)\{ \sin40^\circ/(1-0.0067687\sin^240^\circ)- \\ & [1/(2\times0.0822719)]\ln[(1-0.0822719\sin40^\circ)/(1+0.0822719\sin40^\circ)]\} \\ =& 0.9932313\;(0.6445903 - 6.0774092\ln 0.8995456) \\ =& 1.2792602 \end{align}
\begin{align} q_p =& (1-0.0067687)\{ \sin 90^\circ/(1-0.0067687\sin^2 90^\circ)- \\ & [1/(2\times0.0822719)]\ln[(1-0.0822719\sin 90^\circ)/(1+0.0822719\sin 90^\circ)]\} \\ =& 1.9954814 \end{align}
\begin{align} \beta =& \arcsin(1.2792602/1.9954814) \\ =& 39.8722859^\circ \end{align}

Determining $\beta$ from series equation (3-14) involves the same pattern as the example for equation (3-5) given above.

For inverse calculations, using equation (3-17) with iterative equation (3-16), given
$\beta$=°,

and $q_p = 1.9954814$ as determined above:

\begin{align} q &= 1.9954814\sin39.8722878^\circ \\ &= 1.2792603 \end{align}

First trial:*1

\begin{align} \phi &= \arcsin(1.2792603/2) \\ &= 39.7642444^\circ \end{align}

Second trial:

\begin{align} \phi =& 39.7642444^\circ + (180^\circ/\pi)\{[(1-0.0067687\sin^2 39.7642444)^2/(2\cos 39.7642444^\circ)] \\ & [1.2792603/(1-0.0067687) - \sin 39.7642444^\circ/ (1-0.0067687\sin^2 39.7642444^\circ) \\ &+[1/(2\times0.0822719)]\ln[(1-0.0822719\sin39.7642444^\circ) \\ &/(1+0.0822719\sin39.7642444^\circ))]]\} \\ =& 39.9996022^\circ \end{align}

Third trial, substituting $39.9996022^\circ$ in place of $39.7642444^\circ$

$$\phi=40.0000000^\circ$$

Finding φ from β by series equation (3-18) involves the same pattern as the example for equation (3-5) given above.

## Rectifying latitude #

Using equations (3-20) and (3-21)

\begin{align} M =&a[(1-0.0067687/4-3\times 0.0067687^2/64 - 5\times 0.0067687^3/256)\times 40^\circ\times\pi/180^\circ \\ &-(3\times 0.0067687/8+3\times 0.0067687^2/32 +45\times0.0067687^3/1024)\sin(2\times40^\circ) \\ &+(15\times 0.0067687^2/256 +45\times 0.0067687^3/1024)\sin(4\times 40^\circ) \\ &-(35\times 0.0067687^3/3072)\sin(6\times40^\circ)] \\ =&a[0.9983057\times 0.6981317 - 0.0025426\times\sin 80^\circ] + 0.0000027\sin 160^\circ \\ &-0.0000000\times \sin240^\circ \\ =& 0.6944458a \end{align}
$$M_p = 1.5681349a, \text{using $90^\circ$ in place of $40^\circ$ in the above example.}$$
$$\mu = 90^\circ\times 0.6944458a/1.5681349a = 39.8563451^\circ$$

Calculation of $\mu$ from series (3-23), and the inverse $\phi$ from (3-26), is similar to the example for equation (3-5) except that $e_1$ is used rather than $e$.

From equation (3-24),

\begin{align} e_1 &= [1-(1-0.0067687)^{1/2}]/[1+(1-0.0067687)^{1/2}] \\ &= 0.001697916 \end{align}

## Geocentric latitude #

Using equation (3-28),

\begin{align} \phi_g &= \arctan[(1-0.0067687)\tan 40^\circ] \\ &= 39.8085032^\circ \end{align}

## Reduced latitude #

Using equation (3-31),

\begin{align} \eta &= \arctan[(1-0.0067687)^{1/2}\tan 40^\circ] \\ &= 39.9042229^\circ \end{align}