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    Re: POL and arctan2
    From: Lars Bergman
    Date: 2005 Nov 14, 12:58 +0100

    George and Bill have asked whether the resulting R value provide any
    useful information when doing a rectangular to polar conversion with x =
    -sin LHA y = tan DEC cos LAT - cos LHA sin LAT
    Out of the conversion comes two values, one of them being the azimuth
    AZ. Consider a plane right angled triangle where one of the acute angles
    is AZ. Let x be the opposite side and y the adjacent side. The
    hypotenuse R = sqrt(x sq + y sq) and tan AZ = x/y.
    We find that sin AZ = x/R, thus R = x/sin AZ. As R by definition is
    positive and x has to be positive (the length of a triangle's side),
    then sin AZ also has to be positive in this way of looking at things.
    For simplicity we say that sin LHA is positive and we thus have
    R = sin LHA/sin AZ
    The spherical trig sine-formula tells us that we can replace sin LHA/sin
    AZ with sin ZD/sin PD, where, in the pole-zenith-body nav triangle
    zenith distance ZD and polar distance PD are the sides (or arcs)
    opposite to the angles LHA and AZ respectively. Thus
    R = sin ZD/sin PD = cos ALT/cos DEC
    and we can solve for altitude
    ALT = arccos(R cos DEC)
    The altitude thus found is always positive, an altitude that shall be
    negative is displayed as positive; this is one drawback of using this
    method. For "ordinary" sights this is no problem. In a similar way R
    could be used for calculating great circle distance
    DGC = arcsin(R cos LAT2) where LAT2 is the latitude of destination.
    Lars Bergman 59N 18E

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