A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: David Pike
Date: 2019 Jan 18, 05:21 -0800
Sean you wrote:
Here are two formulae you can use to find the vertex:
cos(Latv) = cos(Lat1) ∙ sin(C)
sin(dLon) = cos(C) / sin(Latv)
"Latv" is the latitude of the vertex; "Lat1" is a latitude point along the great circle; "C" is the course you would travel along the great circle from "Lat1"; "dLon" is the difference in longitude from "Lat1" to the vertex.
Solving lat & long of the vertex of a great circle (GC) relies on the fact that the spherical triangles produced by the start point, the vertex and the pole, or the end point, the vertex, and the pole are right spherical triangles, which simplifies the working considerably and means you can use Napier’s Rules. Cos(Latv) = cos(Lat1) ∙ sin(C) is a variation of Rule 2, and sin(dLon) = cos(C) / sin(Latv) is a variation of Rule 8 in the order defined by Todhunter.
However, can you use Rule 8 to calculate the longitude of the vertex without first using Rule 2 to calculate its latitude, and can you solve Rule 2 without first knowing the starting course? To calculate this, won’t you need an end point and a bit more magic. DaveP