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    Re: POL and arctan2
    From: Bill B
    Date: 2005 Nov 10, 18:53 -0500

    > In the polar-rectangular conversion formula, we know that, by definition, R
    > squared = x squared + y squared.and that...
    I do not have a final solution yet, but I think I have my finger on the
    Executive summary:  Your gray matter is far superior to any computing
    device.  You can think outside of the box.  A computing device, even one
    with an expert system, can only do what it has been programmed to do (albeit
    quickly).  It cannot think outside of the box (with the possible exception
    of using an inference engine etc. for artificial intelligence).  The key
    being the rectangular-to-polar (R-to-P) and polar-to-rectangular routines
    are designed as two-dimensional (2D) routines.  While your clever equation
    manipulates it into doing your bidding in three dimensions for finding
    azimuth (you are thinking outside the box) it is still limited to thinking
    inside the 2D box. (Which I will get to later).
    In using the method you provided with inputs of Lat 40N, LHA's of 30, 60,
    70, and 330, and declinations of 10 and -10, the first number returned was
    (with one rounding error exception) less than 1.0.  Which caused me to
    wonder if it was a trig function, hence my curiosity.
    When I calculated Hc for the above inputs, it was consistently close to, but
    greater than, the cosine-1 of the first number returned.  A pattern was
    emerging.  It follows the sine-1 of the first number returned was close to
    but larger than the compliment of Hc.
    I wondered why?  Is it possible the R-to-P was "thinking" in 2D and I needed
    to dumb-down a bit and operate on its level? (Not too hard for me )  I
    reasoned a great circle on a sphere was a just a circle in two dimensions.
    What could be close to the same but not? Could the chord of an arc come into
    As a sanity check (and I state up front it is a circular argument) I used
    the compliment of Hc to determine distance from position to GP.  I then
    divided that distance by 21600 (the circumference of a circle with 60 nm per
    d) and multiplied by 360 to put it back into degrees.  As one would expect,
    the resulting angle was equal to the compliment of the Hc I started with.
    Then I calculated the chord subtending the distance arc using the complement
    of the Hc angle, a radius of 21600/2pi, and the law of cosines.  Next I
    divided the length of the chord by 21600 and multiplied by 360 to put it
    back in to degrees.  And guess what?  The sine of that angle matches the
    sine of the first figure the R-to-P routine spits out.
    I hope I have the words for the concept straight , but you see the idea
    nonetheless.  The next question is what to do with that information ;-)

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