NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: David Fleming
Date: 2016 Jan 22, 12:39 -0800
Mark,
Not sure if I understand your question, but let me try to relieve the anxiety I'm hearing.
You mentioned that you are comfortable with Law of Cosine solution for great circle distance. That is a problem that allows an infinity of different values for LHA, Dec and Lat. When using a calculator that is no problem, you can enter any value for LHA before you take its cosine.
When using a tabular solution for solving the navigational triangle, you can't have an infinite size table. You must restrict the angles to say integer values and then compute corrections for the error introduced by that restriction. A further restriction is to work only with right spherical triangles, thereby reducing the table size. This decomposition of a general spherical triangle into two right triangles is always possible.
But I do not need know how to construct the right triangles. I just assign variables to designate the new parts of the problem. They are solved for using the tables as intermediate values which in the end I don't care about.
Dave