# 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.

Show:

Map of the Earth in an equirectangular projection, showing two possible flight paths from Urbana, USA to Delhi, India.

To get an idea of what it looks like to fly along a great circle path, we can view the crew movies taken from the ISS. The movies taken at night are especially evocative, such as this montage by Michael König.

Technically the ISS is not following a great circle path around the Earth, because the Earth is rotating underneath it. However, because the speed of the ISS is much higher than the Earth's rotational velocity (90 min versus 24 h period), these paths are close to being great circles.

## 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.

Click and drag to rotate. | ||

Show: | Earth transparency: % | |

City A: | Latitude: | $\phi_1 = $ ° |

Longitude: | $\lambda_1 = $ ° | |

City B: | Latitude: | $\phi_1 = $ ° |

Longitude: | $\lambda_1 = $ ° |

There are many different map projections that draw the surface of the spherical Earth on a two-dimensional map. Because the sphere is curved, a map projection cannot preserve both areas and angles simultaneously, leading to different trade-offs in choosing an appropriate projection for a given purpose. Some common projections are shown below.

Equirectangular. \[ x = \lambda, \quad y = \phi \]

Mercator: cylindrical, conformal (angle-preserving). \[ x = \lambda, \quad y = \log\left(\tan\Big(\frac{\pi}{4} + \frac{\phi}{2}\Big)\right) \]

Hobo-Dyer: cylindrical, area-preserving. \[ x = \lambda, \quad y = \frac{\sin\phi}{\cos^2(37.5^\circ)} \]

Winkel tripel: non-cylindrical, compromise. \[\begin{aligned}\small x &= \frac{1}{2} \left(\lambda \cos\phi_1 + \frac{2\cos\phi\sin\frac{\lambda}{2}}{{\rm sinc}\,\alpha}\right) \\ y &= \frac{1}{2} \left(\phi + \frac{\sin\phi}{{\rm sinc}\,\alpha}\right) \\ \alpha &= \arccos\Big(\cos\phi \cos\frac{\lambda}{2}\Big) \\ \phi_1 &= \arccos\frac{2}{\pi} \end{aligned}\]

You might also be interested in what your choice of map projection says about your personality.

## 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.

All the calculations on this page have assumed a spherical model of the Earth, given by \[x^2 + y^2 + z^2 = R^2,\] where the radius of the Earth is $R = 6371\rm\ km$.

Because it spins with a period of about 24 hours, the Earth bulges outwards so the center is further from the equator than from the poles. A shape that better approximates the true Earth is thus an ellipsoid, given by: \[ \left(\frac{x}{R_{\rm xy}}\right)^2 + \left(\frac{y}{R_{\rm xy}}\right)^2 + \left(\frac{z}{R_{\rm z}}\right)^2 = 1, \] where $R_{\rm xy} = 6378\rm\ km$ and $R_{\rm z} = 6357\rm\ km$.

Computing shortest path distances on ellipsoids turns out to be much harder than shortest paths on spheres. One set of equations for doing this are known as Vincenty's formulae.

There is no single best ellipsoid that fits the Earth. Instead there are many different ellipsoids in use, each of which fit the Earth better near different countries (e.g., NAD 83 for North America and WGS 84 for GPS world-wide). When even more accuracy is needed, such as for satellite orbit modeling, then true geoid models are used, with spherical harmonic expansions determined from satellite gravity measurements.