A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Bob Goethe
Date: 2016 Jan 22, 12:51 -0800
>>Its apparently so easy...<<
Mark, nothing about solving a triangle on a sphere is easy. If it WAS easy, there would be one generally accepted solution - which would probably have been derived in 250 BC by some Greek or another.
Pythagoras comes up with a2 + b2 = c2, and nobody ever needs to work that hard to solve a right-triangle again.
In contrast, people have been casting about for centuries for a good solution to the navigational triangle. Accuracy and ease-of-use are both values that drive continued exploration by mathematicians.
I can USE Ageton's equations (or Bygrave's for that matter, where one of the intermediate values, what Gary LaPook calls "W", equates to the value Q in the Ageton equations) without having the first foggy clue of how Ageton went about deriving those equations.
Hmmm...as I stare at one problem on my desk that I solved by both Ageton and Bygrave methods, I see not only that Ageton's Q = Bygrave's W, but that one of the intermediate values used by Ageton = the co-latitude of Gary's "X" intermediate value. Looks to me like Bygrave is also splitting the navigational triangle into two right-triangles. Maybe if I stare at this long enough, and keep in mind the relationship between the various trig functions, I will be able to sort out the extent to which Ageton and Bygrave are different solutions to the same problem vs. the extent to which they are the SAME solution framed in terms of different trig functions.
Perhaps before I die I can derive yet another solution to the navigational triangle, and get my name mentioned in the year 2046 edition of Bowditch. Napier, St. Hilaire, Goethe....I like the sound of that.