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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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