Longitude difference

Again from Fig. 3 and Eqn. (17) we see that:

$$\tan\alpha=\frac{\cos\theta}{(1-e^2\cos^2\theta)^{1/2}}\frac{\mathrm{d}L}{\mathrm{d}\theta}$$ (23)

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

$$\frac{\mathrm{d}L}{d\sigma}=\sin\alpha_0\frac{(1-e^2(1-\cos^... ...ha_0 \sin^2\sigma))^{1/2}} {(1-\cos^2\alpha_0 \sin^2\sigma)}$$ (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:

$$L =\lambda-e^2\sin\alpha_0( J_0\sigma+e^2 J_2\sin 2\sigma+e^4 J_4\sin 4\sigma+e^6 J_6 \sin 6\sigma+O(e^8))$$ (25)

where

$$J_0 = \frac{1}{2}\left(1+\frac{e^2}{8}(2-\mu)+\frac{e^4}{64}(8-8\mu+3\mu^2)+ \frac{5e^6}{1024}(16-24\mu+18\mu^2-5\mu^3)\right)$$

$$J_2 = \frac{\mu}{32}\left(1+\frac{e^2}{2}(2-\mu)+ \frac{15e^4}{256}(16-16\mu+5\mu^2)\right)$$

$$J_4 = \frac{\mu^2}{512}\left(1+\frac{15e^2}{16}(2-\mu)\right)$$

$$J_6 = \frac{5\mu^3}{24576}$$

$$\mu = \cos^2\alpha_0$$ (26)

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

$$L=\lambda-(1-C)f\sin\alpha_0(\sigma+C\sin\sigma(\cos 2\sigma_m +C\cos\sigma(-1+2\cos^2 2\sigma_m)))+ O(f^4)$$ (27)

where

$$C=f \cos^2\alpha_0(4+f(4-3\cos^2\alpha_0))/16$$ (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

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

on solving the spherical triangle on the auxiliary sphere.