NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Greg Rudzinski
Date: 2011 Nov 25, 08:17 -0800
Jaap,
(1) x+y = (Txma-Txmb)
(2) dHo = a/60•(x^2-y^2) (without reasoning why in the case at hand y < x use dHo = |a/60•(x^2-y^2)|
From (1) -> (3) y^2 = (Txma-Txmb)^2-2x(Txma-Txmb)+x^2
Substituting (3) in (2) -> (4) dHo = a/60•(x^2-(Txma-Txmb)^2+2x(Txma-Txmb)-x^2)
(4) can be re-writen as
(5) x = (Txma-Txmb)/2+30•dHo/(a•(Txma-Txmb))
Your above formula for solving time to meridional passage for before and after meridional passage ex-meridians is most outstanding !! I can't wait to try it out on some real observations. Thanks :)
Greg Rudzinski
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