# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Precomputed lunar distances
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

```
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