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D3 - - - CALCULATED GREAT-CIRCLE SAILING - - 10/03/2004

When the navigation triangle is solved and the initial course angle at the point of departure is found it is possible to calculate the sequence of points along the great-circle path, but if it is done directly it involves the repeated solution of long and complex formulas. To simplify the calculations the parameters of the great-circle vertex are first found, and then all the other calculations involve the solution of right triangles. The vertex is that point on the great-circle path that is nearest the North or South poles of the earth and where the direction is exactly East and West.

Lv    = latitude of the vertex
L1    = latitude of the point of departure (point 1)
C1    = course angle at point 1
DLOv1 = difference in longitude from the vertex to point 1
Dv1   = distance from the vertex to point 1  (degrees)

      cos Lv    = cos L1 * sin C1
      sin DLOv1 = cos C1 * csc Lv
or    tan DLOv1 = ctn C1 * csc L1
      sin Dv1   = cos L1 * sin DLOv1
or    tan Dv1   = cos C1 * ctn L1

If a difference in longitude from the vertex is assumed (DLOvx),
then the parameters for a point x (located anywhere along the 
great-circle path) can be determined.

Lx    = latitude at point x
DLOvx = difference in longitude from the vertex to point x
Cx    = course angle at point x
Dvx   = distance from the vertex to point x  (degrees)

      tan Lx  = cos DLOvx * tan Lv
      cos Cx  = sin DLOvx * sin Lv   (usually not required)
      tan Dvx = tan DLOvx * cos Lv   (optional) 

When the above formulas are used, the distance between points will be a variable. To get equal distances the following method can be used. The incremental distance used is usually a multiple of the ship's cruising speed, so a very fast ship (super tanker) may pass a reference point every four hours. A slower ship would use six or more hours.

      sin Lx    = sin Lv * cos Dvx
      tan DLOvx = sec Lv * tan Dvx
      tan Cx    = ctn Lv * csc Dvx   (usually not required)

When sailing from San Francisco to Hilo Hawaii, the vertex will be near Lake Michigan. When sailing from San Francisco to Japan, the vertex will be just south of the Aleutian islands. This can be taken advantage of because the great-circle path will be symmetrical around the vertex and if the vertex longitude is used as a reference fewer calculations will be required. For example, 900 miles east or west of the vertex (along the great-circle path) will be at the identical latitude and the DLO from the vertex will be the same.

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