Plane sailing is based on the assumption that the meridian through the point of departure, the parallel through the destination, and the course line form a plane right triangle, 2205_the_plane_sailing_triangle._copy

From this: \cos C = \frac{1}{D}; \cos C = \frac{p}{D}; \tan C = \frac{p}{l};

From this: 1=D \cos C, D=1 sec C, and p=D sin C .

From this, given course and distance (C and D), the difference of latitude (l) and departure (p) can be found, and given the latter, the former can be found, using simple trigonometry.

Traverse sailing combines plane sailings with two or more courses, computing course and distance along a series of rhumb lines.

Parallel sailing consists of interconverting departure and difference of longitude. Refer to Figure beside.

DLo = p sec L p = DLo \cos L

Mid-latitude sailing combines plane and parallel sailing, with certain assumptions. The mean latitude (Lm) is half of the arithmetical sum of the latitudes of two places on the same side of the equator. For places on opposite sides of the equator, the N and S portions are solved separately. In mid-latitude sailing:

DLo = p sec Lm p = DLo \cos Lm

Mercator Sailing problems are solved graphically on a Mercator chart. For mathematical Mercator solutions the formulas are:

\tan C = \frac{DLo}{m} or  DLo = m\tan C

Following solution of the course angle by Mercator sailing, the distance is by the plane sailing formula: D = l secC

Great circle solutions for distance and initial course angle can be calculated from the formulas:

D = \cos^{-1}\left(\sin L_1\sin L_2 + \cos L_1 \cos L_2 \cos {DLo} \right)

C = \tan^{-1}\left(\frac{\sin DLo}{\cos L_1 \tan L_2 - \sin L_1 \cos DLo} \right)

where D is the great circle distance, C is the initial great circle course angle, L_1 is the latitude of the point of departure, L_2 is the latitude of the destination, and DLo is the difference of longitude of the points of departure and destination. If the name of the latitude of the destination is contrary to that of the point of departure, it is treated as a negative quantity.

The latitude of the vertex, L_v, is always numerically equal to or greater than L_1 or L_2. If the initial course angle C is less than 90°, the vertex is toward L_2, but if C is greater than 90°, the nearer vertex is in the opposite direction. The vertex nearer L_1 has the same name as L_1. The latitude of the vertex can be calculated from the formula:

L_v = \cos^{-1}\left(\cos L_1 \sin C \right)

The difference of longitude of the vertex and the point of departure (DLo_v) can be calculated fromthe formula:

Dlo_v = \sin^{-1}\left(\frac{\cos C}{\sin L_v} \right)

The distance from the point of departure to the vertex (D_v) can be calculated from the formula:

D_v = \sin^{-1}\left(\cos L_1 \sin {DLo_v} \right)

The latitudes of points on the great circle track can be determined for equal DLo intervals each side of the vertex (DLo_{vx}) using the formula:

L_x = \tan^{-1}\left(\cos DLo_{vx} \tan L_v \right)

The DLo_v and D_v of the nearer vertex are never greater than 90°. However, when L_1 and L_2 are of contrary name, the other vertex, 180° away, may be the better one to use in the solution for points on the great circle track if it is nearer the mid point of the track.

The method of selecting the longitude (or DLo_{vx}), and determining the latitude at which the great circle crosses the selected meridian, provides shorter legs in higher latitudes and longer legs in lower latitudes. Points at desired distances or desired equal intervals of distance on the great circle from the vertex (D_{vx}) can be calculated using the formulas:

L_x = \sin^{-1}\left(\sin L_v \cos D_{vx} \right)

DLo_{vx} = \sin ^{-1}\left(\frac{\sin D_{vx}}{\cos L_x} \right)

A calculator which converts rectangular to polar coordinates provides easy solutions to plane sailings. However, the user must know whether the difference of latitude corresponds to the calculator’s X coordinate or to the Y coordinate.