56.6: Vincenty’s Formulæ- Inverse Problem

In the inverse problem, we're given two points on the Earth's surface \(\left(\phi_{1}, L_{1}\right)\) and \(\left(\phi_{2}, L_{2}\right)\) and want to calculate the distance \(s\) between them, as well as the direction from one to the other. We'll use the constants defining the Earth's ellipsoid as before:

• Earth ellipsoid semi-major axis (i.e. equatorial radius): \(a=6378137.0\) meters.
• Earth flattening factor \(f=1 / 298.257223563\). This is defined as the difference between semi-major and semi-minor axes, divided by the semi-major axis: \(f=(a-b) / a\).

In performing the following calculations, work with all angles in radians. We begin by calculating

\[
\begin{align}
U_{1} & =\tan ^{-1}\left[(1-f) \tan \phi_{1}\right] \label{1} \\ \notag\\
U_{2} & =\tan ^{-1}\left[(1-f) \tan \phi_{2}\right] \label{2} \\ \notag\\
L & =L_{2}-L_{1} \label{3} \\ \notag\\
b & =(1-f) a \label{4}
\end{align}
\]

Now set an initial value \(\lambda=L\). Then iterate on Eqs. (53.28) through (53.35) until \(\lambda\) converges:

\[
\begin{align}
& \sin \sigma=\sqrt{\left(\cos U_{2} \sin \lambda\right)^{2}+\left(\cos U_{1} \sin U_{2}-\sin U_{1} \cos U_{2} \cos \lambda\right)^{2}} \\ \notag\\
& \cos \sigma=\sin U_{1} \sin U_{2}+\cos U_{1} \cos U_{2} \cos \lambda \\ \notag\\
& \sigma=\arctan \frac{\sin \sigma}{\cos \sigma} \\ \notag\\
& \sin \alpha=\frac{\cos U_{1} \cos U_{2} \sin \lambda}{\sin \sigma} \\ \notag\\
& \cos ^{2} \alpha=1-\sin ^{2} \alpha \\ \notag\\
& \cos ^{2}\left(2 \sigma_{m}\right)=\cos \sigma-\frac{2 \sin U_{1} \sin U_{2}}{\cos ^{2} \alpha} \\ \notag\\
& C=\frac{f}{16} \cos ^{2} \alpha\left[4+f\left(4-3 \cos ^{2} \alpha\right)\right] \\ \notag\\
& \lambda=L+(1-C) f \sin \alpha\left\{\sigma+C \sin \sigma\left[\cos \left(2 \sigma_{m}\right)+C \cos \sigma\left(-1+2 \cos ^{2}\left(2 \sigma_{m}\right)\right)\right]\right\}
\end{align}
\]

When \(\lambda\) has converged to the desired degree of accuracy, continue calculating:

\[
\begin{align}
& u^{2}=\left(\cos ^{2} \alpha\right)\left(\frac{a^{2}-b^{2}}{b^{2}}\right) \\ \notag\\
& A=1+\frac{u^{2}}{16384}\left\{4096+u^{2}\left[-768+u^{2}\left(320-175 u^{2}\right)\right]\right\} \\ \notag\\
& B=\frac{u^{2}}{1024}\left\{256+u^{2}\left[-128+u^{2}\left(74-47 u^{2}\right)\right]\right\} \\ \notag\\
& \Delta \sigma=B \sin \sigma\left\{\cos \left(2 \sigma_{m}\right)+\frac{1}{4} B\left[\cos \sigma\left(-1+2 \cos ^{2}\left(2 \sigma_{m}\right)\right)-\frac{1}{6} B \cos \left(2 \sigma_{m}\right)\left(-3+4 \sin ^{2} \sigma\right)\left(-3+4 \cos ^{2}\left(2 \sigma_{m}\right)\right)\right]\right\} \\ \notag\\
& s=b A(\sigma-\Delta \sigma) \\ \notag\\
& \alpha_{1}=\arctan \left(\frac{\cos U_{2} \sin \lambda}{\cos U_{1} \sin U_{2}-\sin U_{1} \cos U_{2} \cos \lambda}\right)\\ \notag\\
& \alpha_{2}=\arctan \left(\frac{\cos U_{1} \sin \lambda}{-\sin U_{1} \cos U_{2}+\cos U_{1} \sin U_{2} \cos \lambda}\right)\\ \notag\\
\end{align}\]

Then \(s\) is the distance between the two points.

Example \(\PageIndex{1}\)

Find the distance between the sounding rocket in Chesapeake Hall at Prince George's Community College ( \(38^{\circ} 53^{\prime} 16.87^{\prime \prime} \mathrm{N}, 76^{\circ} 49^{\prime} 23.14^{\prime \prime} \mathrm{W}\) ) and the top (apex) of the Great Pyramid of Giza in Egypt \(\left(29^{\circ} 58^{\prime} 45.03^{\prime \prime} \mathrm{N}, 31^{\circ} 08^{\prime} 03.69^{\prime \prime} \mathrm{E}\right)\).

Solution

The given parameters are the coordinates \(\phi_{1}=38.888019^{\circ}, L_{1}=-76.823094^{\circ}, \phi_{2}=\) \(29.979175^{\circ}, L_{2}=+31.134358^{\circ}\). Employing Vincenty’s formulæ (inverse method), we find:

\[
\begin{aligned}
U_{1} & =38.794230^{\circ} \\
U_{2} & =29.895958^{\circ} \\
L & =339.15856744^{\circ} \\
b & =6356752.3 \text { meters } \\
\lambda & =108.139490^{\circ} \\
u^{2} & =0.00393162979 \\
A & =1.00098218405082 \\
B & =9.809796134747123 \times 10^{-4} \\
\Delta \sigma & =0.054160886^{\circ} \\
s & =9351378.858 \text { meters } \\
\alpha_{1} & =55.910048^{\circ} \\
\alpha_{2} & =131.801775^{\circ}
\end{aligned}
\]

So the distance \(s=9351.378858 \mathrm{~km}\) (5280 miles, 3576 feet, 10 inches), in the direction \(55.910048^{\circ}\) (10.91 south of northeast).