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    Re: Smart and Shearme Position Line Tables
    From: Lars Bergman
    Date: 2020 May 5, 11:05 -0700

    DaveP,

    To me x is just an auxiliary variable, but it could very likely be explained as an arc or angle in a diagram. The spherical trig four-parts-formula gives

    cos(colatitude)·cos(azimuth) = sin(colatitude)·cot(zenith distance) - sin(azimuth)·cot(hour angle)

    which is the same as   sin(latitude)·cos(azimuth) = cos(latitude)·tan(altitude) - sin(azimuth)·cot(hour angle).

    After some trivial manipulation you get   sin(latitude)·[cos(azimuth) + sin(azimuth)·csc(latitude)·cot(hour angle)] = cos(latitude)·tan(altitude)

    Now, let   csc(latitude)·cot(hour angle) = tan(x), so you get

    sin(latitude)·sec(x)·[cos(azimuth)·cos(x) + sin(azimuth)·sin(x)] = cos(latitude)·tan(altitude),  and noting that [...] equals cos(azimuth - x) you finally get

    cos(azimuth - x) = cot(latitude)·tan(altitude)·cos(x)

    Solving for azimuth you get azimuth = x + arccos[cot(latitude)·tan(altitude)·cos(x)],  which indicate that x actually is one of the angles obtained when you let a perpendicular to the declination circle pass through the zenith.

    Your observation that there generally are two solutions seems plausible, but I haven't thought much about it. Maybe my initial statement that the triangle wasn't unique was correct in a way. I will ponder this.

    Lars

       
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