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

From: Bill (no email)
Date: Thu Jan 12 2006 - 16:38:52 EST

  • Next message: Bill: "Re: dip, dip short, distance off with buildings, etc."

    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


  • Next message: Bill: "Re: dip, dip short, distance off with buildings, etc."



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