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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Precomputed lunar distances
From: Frank Reed
Date: 2005 Apr 18, 00:16 EDT
From: Frank Reed
Date: 2005 Apr 18, 00:16 EDT
Bill you wrote: "Two sides of a spherical triangle meeting at the zenith. Both start off almost perpendicular to the horizon and progressively arc in to the zenith. So from the *observers* frame of reference, refraction is acting up and in, less up and more in the higher the bodies." I have a hunch that you're picturing a sort of "perspective drawing" of a spherical triangle in which it might appear as if the refraction is acting "in" or sideways. It's important to picture a spherical triangle from the perspective of the observer. The arcs (sides) of a spherical triangle are not curved. If you draw a spherical triangle with corners at the zenith and two arbitrary stars, all three sides look *exactly* straight as seen by the observer. Gou outside tonight and point at Spica. Now trace the side of the spherical triangle that connects Spica to the zenith. You finger should trace a line across the sky that looks (to you) exactly straight and exactly vertical. Next point at Antares. Trace the side of the triangle from Antares to the zenith. Your arm should rise straight and vertical. Finally trace the side from Antares to Spica. Your finger should move straight across the sky. How does refraction affect these triangle sides?? It lifts each star entirely vertically and so entirely within the two sides that lead to the zenith. There's no component perpendicular to those sides. Make sense?? And of course, the distance between the stars is reduced by refraction even through the refraction is completely in the vertical direction. -FER http://www.HistoricalAtlas.com/lunars