NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Trammell H
Date: 2026 Jan 7, 11:29 -0800
I think I did all the trigonometry almost right for deriving why the stereographic projection works (modulo a final sign error that I'm just going to ignore for now since it only means that the markings have to go the other way the plate). It's a little unexpected that mulitplying the projection by the rotation matrix produces the correct result to compute the bearing and height, but I'm glad there's a happy coincidence. The circular slide rule "does the math" with the markings and rotating the pointer, which feels a little magical that it just works.
The short summary is that projecting (Declination,LHA) to [Cos(Dec)*Cos(LHA), Sin(Dec)] / (1 + Cos(Dec)*Sin(LHA)) and then rotating by (90 - Lat) just happens to be the same as projecting Hc and Z with the same function. There are lots more equations here: https://trmm.net/astrolabe/#math
