navigation4

Navigational Mathematics


4. Great circle sailing

The calculation of the great circle track between two points A and B with given latitude and longitude is an exercise in spherical trigonometry.

The points A and B form a spherical triangle with the North Pole C. Each side of this triangle is an arc of a circle centered at the center of the earth, i.e. a great circle. The length of a great-circle arc can be read off immediately from the corresponding central angle: the measurement of the central angle in minutes of arc gives the length of the arc in nautical miles. If we call a, b, c the sides opposite vertices A, B, C, then in this triangle we know a, b, and C. Side a has length 90o minus the latitude of vertex B, and vice-versa. The angle C is the difference between the longitudes of A and B. This is enough information to solve for all the elements of the triangle, in particular side c (the great-circle distance) and angle A (the initial course).


Starting point and destination, together with the North Pole, form a spherical triangle.

Here is the problem from Dutton:

  • Compute the distance and initial course by great circle sailing from a point in Lat. 37o-42' N., Long. 123o-04'W., near Farallon Island Lighthouse, to a point Lat. 34o-50' N., Long. 139o-53' E., near the entrance to the Bay of Tokio.

In this case we compute a = 90o - 37o42' = 52o18', b = 90o - 34o50' = 55o10' and C = 360o - 123o04' - 139o53' = 97o03'.

We may solve for c using the spherical law of cosines:

cos c = cos a cos b + sin a sin b cos C.

This gives c = 74.36o or 4461.6 nautical miles.

Once c is known, A can be calculated using the spherical law of sines:

sin A / sin a = sin B / sin b = sin C /sin c.

Using the known values for a, c, and C this gives A = 57.77o, or 57o46'19".