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 90^{o} 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. 37
^{o}-42' N., Long. 123^{o}-04'W., near Farallon Island Lighthouse, to a point Lat. 34^{o}-50' N., Long. 139^{o}-53' E., near the entrance to the Bay of Tokio.

In this case we compute `a` = 90^{o} - 37^{o}42' = 52^{o}18', `b` = 90^{o} - 34^{o}50' = 55^{o}10' and `C` = 360^{o} - 123^{o}04' - 139^{o}53' = 97^{o}03'.

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` = 74.36^{o} 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` = 57.77^{o}, or 57^{o}46'19".