Shortest flight paths

If we want to fly from Urbana, USA to Delhi, India, which path should we follow? The locations of the two cities are:

	Latitude	Longitude
Urbana	$\phi = 40^\circ 6' 35'' \rm N$	$\lambda = 88^\circ 12' 15'' \rm W$
Delhi	$\phi = 28^\circ 36' 36'' \rm N$	$\lambda = 77^\circ 13' 48'' \rm E$

Geographic coordinates are typically given in spherical coordinates, but without the radius and with the angles given in degrees, minutes, and seconds, with the latitude first, and the direction given as North/South or East/West rather than positive/negative. Also note that latitude is the elevation angle up from the equator, whereas spherical coordinates often use the inclination angle down from the polar axis.

Plotting the city locations on a map of the Earth gives the figure below. Such a map can give a very misleading idea of the shortest route, however.

Shortest paths on the sphere

Finding the shortest distance between two points on the sphere is not a simple calculation given their latitude and longitude. As proved below, the shortest path on the sphere is always a great circle, which is the intersection of the sphere with a plane through the origin. The shortest distance calculation thus reduces to finding the angle between the vectors $\vec{OA}$ and $\vec{OB}$, which can be easily done by finding their dot product after changing them to rectangular coordinates.

Proof that great circles are shortest paths

Given any two points $A$ and $B$ on the surface of the Earth, we choose coordinate axes and use spherical coordinates so that $A$ is at $(r = r_0$, $\theta = 0$, $\phi = \pi - \phi_0)$ and $B$ is at $(r = r_0$, $\theta = 0$, $\phi = \pi + \phi_0)$. Here $r_0$ is the radius of the Earth and $0 \le \phi_0 \le \pi/2$. The total angle between $\vec{OA}$ and $\vec{OB}$ is $\Delta\phi = 2\phi_0$, so the great-circle distance between $A$ and $B$ is $r_0 \Delta\phi$. We will now prove that this is the smallest possible distance.

Consider any path from $A$ to $B$ on the Earth's surface. Suppose we travel along this path, starting at $A$ at time $t = 0$ and ending at $B$ at time $t = 1$. Then the position vector at time $t$ is given by $\vec{r}(t)$ with spherical coordinates $r_0$, $\theta(t)$, $\phi(t)$ and rectangular coordinates $x(t), y(t), z(t)$.

The length $L$ of our path can be computed by integrating the speed from the start time to the end time:

\[ L = \int_0^1 \| \dot{\vec{r}} \| \, dt = \int_0^1 \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{z}^2} \, dt. \]

To switch from rectangular to spherical coordinates we use the conversion:

\[\begin{aligned} x &= r \cos\theta \sin\phi \\ y &= r \sin\theta \sin\phi \\ z &= r \cos\phi. \end{aligned}\]

Differentiating these expressions and substituting (or using the spherical velocity expression) gives the path length expression:

\[ L = \int_0^1 r_0 \sqrt{\dot{\phi}^2 + \dot{\theta}^2 \sin^2\phi} \, dt \ge \int_0^1 r_0 |\dot{\phi}| \, dt \ge r_0 \Delta \phi. \]

Here we have used the fact that $\dot{\theta}^2 \sin^2\phi \ge 0$ to see that $L$ must be at least as big as the great-circle distance $r_0 \Delta \phi$. Thus the great-circle distance is the minimum path length between $A$ and $B$ on the surface of the sphere.