# NavList:

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

**Re: dip, dip short, distance off with buildings, etc.**

**From:**Bill B

**Date:**2006 Jan 12, 16:38 -0500

Frank Wrote: > Now it's time to derive the equation > from scratch. There are several approaches to this. Let's start with the > refraction-free version and then apply the substitution on the radius of the > Earth. > > Draw yourself an arc of a circle representing the curved surface of the > Earth. Mark a point A at height h above the surface for the observer's > position, > mark point B at height H for the object's position at some distance from A (H > and h should be roughly comparable heights and both should be much less than > the radius of the Earth, and mark C for the center of the Earth at the > center of the arc. Now draw a triangle connecting all three points. The > lengths of > the two long sides are R+h and R+H. The length of the other side doesn't > matter. We want the length of the arc along the Earth's surface connecting > the > base of h to the base of H. This length is proportional to the angle in the > triangle at the Earth's center, C. Let's call that angle phi. So far crystal clear. Then things get foggy for me. > With a sextant we > are measuring the altitude of point B from point A so that means that we > know the angle in the triangle at point A, let's call that gamma. Which triangle? The small oblique or large oblique? Let's call the intersection of the object related to B with the horizon "HH." This will form small oblique triangle A B HH A, and we have measured the angle with our sextant from point A, so angle B A HH. So far all that is known about the small oblique triangle is one angle, gamma, measured from A. We also have large oblique triangle, A B C A, and we know 2 sides R+h and R+H. > Finally the > angle at the third corner, B, is related to the other two --it's just 180 > degrees-(gamma+phi). OK so far? Totally lost. Suspect there is an assumption(s) here I have missed. Unless A B HH A is treated as a right triangle, I do no have enough information to determine angle A B HH or A B C (angle at corner B). If I did know angle A B C, then I could find angle C A B using the law of sines, and of course phi (angle A C B). > Now apply the law of sines to the big > triangle. > And expand using the rules for sums of angles. I do not understand "expand using the rules for sums of angles." I apologize for being thick as a brick, and appreciate your efforts. Thanks Bill