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

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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

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