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Next: Symbol Glossary Up: Geodesics on a spheroid Previous: Geodesic arc length

Longitude difference

Again from Fig. 3 and Eqn. (17) we see that:
\begin{displaymath}
\tan\alpha=\frac{\cos\theta}{(1-e^2\cos^2\theta)^{1/2}}\frac{\mathrm{d}L}{\mathrm{d}\theta}
\end{displaymath} (23)

which, in terms of the arc-length on the auxiliary sphere, $\sigma$, using the geodesic condition (17) becomes:


\begin{displaymath}
\frac{\mathrm{d}L}{d\sigma}=\sin\alpha_0\frac{(1-e^2(1-\cos^...
...ha_0 \sin^2\sigma))^{1/2}}
{(1-\cos^2\alpha_0 \sin^2\sigma)}
\end{displaymath} (24)

In the spherical limit, $e=0$, this is readily integrated to give $L=\lambda$, where $\tan\lambda=\sin\alpha_0 \tan\sigma$. $\lambda$ is the longitude difference, corresponding the arc-length $\sigma$ on the auxiliary sphere.

Expanding in powers of $e^2$ and integrating term by term, we thus obtain:

\begin{displaymath}
L =\lambda-e^2\sin\alpha_0( J_0\sigma+e^2 J_2\sin 2\sigma+e^4 J_4\sin
4\sigma+e^6 J_6 \sin 6\sigma+O(e^8))
\end{displaymath} (25)

where


$\displaystyle J_0$ $\textstyle =$ $\displaystyle \frac{1}{2}\left(1+\frac{e^2}{8}(2-\mu)+\frac{e^4}{64}(8-8\mu+3\mu^2)+
\frac{5e^6}{1024}(16-24\mu+18\mu^2-5\mu^3)\right)$  
$\displaystyle J_2$ $\textstyle =$ $\displaystyle \frac{\mu}{32}\left(1+\frac{e^2}{2}(2-\mu)+
\frac{15e^4}{256}(16-16\mu+5\mu^2)\right)$  
$\displaystyle J_4$ $\textstyle =$ $\displaystyle \frac{\mu^2}{512}\left(1+\frac{15e^2}{16}(2-\mu)\right)$  
$\displaystyle J_6$ $\textstyle =$ $\displaystyle \frac{5\mu^3}{24576}$  
$\displaystyle \mu$ $\textstyle =$ $\displaystyle \cos^2\alpha_0$ (26)

Vincenty[5] has again rearranged a subset of the resulting equations into nested forms more suitable for computation:


\begin{displaymath}
L=\lambda-(1-C)f\sin\alpha_0(\sigma+C\sin\sigma(\cos 2\sigma_m
+C\cos\sigma(-1+2\cos^2 2\sigma_m)))+ O(f^4)
\end{displaymath} (27)

where
\begin{displaymath}
C=f \cos^2\alpha_0(4+f(4-3\cos^2\alpha_0))/16
\end{displaymath} (28)

As in Eqn. (22), the origins of $\sigma$,$\lambda$ and $L$ have been shifted from the equator to the initial point (1). $\lambda$ is then given by

\begin{displaymath}
\tan\lambda=\frac{\sin\sigma\sin\alpha_1}
{\cos\phi_1\cos\sigma-\sin\phi_1\sin\sigma\cos\alpha_1}
\end{displaymath} (29)

on solving the spherical triangle on the auxiliary sphere.


next up previous
Next: Symbol Glossary Up: Geodesics on a spheroid Previous: Geodesic arc length
Ed Williams
2002-03-21