56.5: Vincenty’s Formulæ- Direct Problem

In the direct problem, we're given the latitude \(\phi_{1}\) (north positive) and longitude \(L_{1}\) (east positive) of one point on the Earth's surface; a distance \(s\); and a direction \(\alpha_{1}\) (measured clockwise from north). The goal of the direct problem is to find the latitude \(\phi_{2}\) and longitude \(L_{2}\) of the point you would reach by starting at ( \(\phi_{1}, L_{1}\) ) and traveling a distance \(s\) in the direction \(\alpha_{1}\).

We're also given the following constants that define the size and shape of the Earth ellipsoid:\({ }^{1}\)

• 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: \(f=(a-b) / a\).

We begin by finding the semi-minor axis \(b\) of the Earth's ellipsoid:

\[b=(1-f) a .\]

Then calculate the following, step by step, working with all angles in radians:

\[
\begin{align}
\tan U_{1}=(1-f) \tan \phi_{1} \\[8pt]
U_{1}=\tan ^{-1}\left(\tan U_{1}\right) \\[8pt]
\sigma_{1}=\arctan \left(\frac{\tan U_{1}}{\cos \alpha_{1}}\right) \\[8pt]
\sin \alpha=\cos U_{1} \sin \alpha_{1} \\[8pt]
\cos ^{2} \alpha=(1-\sin \alpha)(1+\sin \alpha) \\[8pt]
u^{2}=\left(\cos ^{2} \alpha\right)\left(\frac{a^{2}-b^{2}}{b^{2}}\right)
\end{align}
\]

\[
\begin{align}
A=1+\frac{u^2}{16384}\left\{4096+u^2\left[-768+u^2\left(320-175 u^2\right)\right]\right\} \\[8pt]
B=\frac{u^2}{1024}\left\{256+u^2\left[-128+u^2\left(74-47 u^2\right)\right]\right\}
\end{align}
\]

Then, using an initial value \(\sigma=s / b A\), iterate Eqs. \(\PageIndex{10}\) through \(\PageIndex{12}\) until there is no significant change in \(\sigma\) :

\[2 \sigma_{m}=2 \sigma_{1}+\sigma\]

\[\Delta \sigma=B \sin \sigma\left\{\cos \left(2 \sigma_{m}\right)+\frac{1}{4} B\left[\cos \sigma\left(-1+2 \cos ^{2}\left(2 \sigma_{m}\right)\right)-\frac{1}{6} B \cos \left(2 \sigma_{m}\right)\left(-3+4 \sin ^{2} \sigma\right)\left(-3+4 \cos ^{2}\left(2 \sigma_{m}\right)\right)\right]\right\}\]

\[\sigma=\frac{s}{b A}+\Delta \sigma\]

Once \(\sigma\) is obtained to sufficient accuracy, calculate:

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

Then \(\left(\phi_{2}, L_{2}\right)\) are the latitude and longitude of the ending point (in radians).

Example \(\PageIndex{1}\)

If you travel exactly 1000 miles northwest of the sounding rocket in Chesapeake Hall at Prince George's Community College ( \(38^{\circ} 53^{\prime} 17.62^{\prime \prime} \mathrm{N}, 76^{\circ} 49^{\prime} 23.40^{\prime \prime} \mathrm{W}\) ), where do you end up? (Give the answer as latitude, longitude, and describe the location.)

Solution

The coordinates of Chesapeake Hall are: \(\phi_{1}=+38.888228^{\circ}, L_{1}=-76.823167^{\circ}\). The given distance is 1000 miles \(=1609.344 \mathrm{~km}\), and the given azimuth \(\alpha_{1}=315^{\circ}\). Employing Vincenty's formulæ (direct method), we find:

\(b =6356752.3 \text { meters } \)
\(U_{1} =38.794230^{\circ} \)
\(\sigma_{1} =48.663693^{\circ} \)
\(\cos ^{2} \alpha =0.696266995365 \)
\(u^{2} =0.0046924891470 \)
\(A =1.0011720921377 \)
\(B =0.0011703772996 \)
\(\sigma =14.482402^{\circ} \)
\(\phi_{2} =48.206878^{\circ} \)
\(\lambda =-15.357896^{\circ} \)
\(C =5.84547783404 \times 10^{-4} \)
\(L =-15.331156^{\circ} \)
\(L_{2} =-92.154324^{\circ}\)

Hence the ending point is at latitude \(48^{\circ} 12^{\prime} 24.76^{\prime \prime} \mathrm{N}\), longitude \(92^{\circ} 09^{\circ} 15.56^{\prime \prime} \mathrm{W}\). This is in northern Minnesota (St. Louis county), within Superior National Forest, just a few miles south of the Canadian border.

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\({ }^{1}\) These are the values used for the WGS-84 ellipsoid, used by GPS receivers.