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    Re: An All-Haversine Azimuth (from Lat, Dec and dLon alone)?
    From: Lars Bergman
    Date: 2021 Feb 22, 03:29 -0800

    Tony,

    What makes the formula look messy is that altitude, or rather zenith distance, is "hidden" behind the archav function. If you allow zenith distance to be a separate variable, z, it will look better:

    hav a = sec φ · csc z · sqrt{ hav[ 180° - (φ+δ+z) ] · hav(φ-δ+z) }
    where
    hav z = hav(φ-δ) + [ 1 - hav(φ-δ) - hav(φ+δ) ] · hav t

    Then I utilized the fact that hav(180°-x) = 1 - hav x, which adds to the mess.

    Further, 
    sec φ = 1 / cos φ = 1 / (1-2·hav φ)
    and
    csc z = 1 / sin z = 1 / sqrt[ hav(2·z) ]

    But, as you wrote, "no matter how long or complicated" ...

    If you try to avoid using z at all, then you'll end up with tan and cot functions which do not, as far as I am aware, easily translate to haversines. And finding their inverses ...

    The simplest method is, I believe, 

    hav a = [ hav p - hav(φ-h) ] / [ 1 - hav(φ-h) - hav(φ+h) ]
    where
    p is polar distance, p = 90° - δ, and h is altitude, h = 90° - z.

    Lars

       
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