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    Re: POL and arctan2
    From: Bill B
    Date: 2005 Nov 10, 20:04 -0500

    >> In the polar-rectangular conversion formula, we know that, by definition, R
    >> squared = x squared + y squared.and that...
    > George
    > I do not have a final solution yet, but I think I have my finger on the
    > problem.
    > 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
    > play?
    > 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 ;-)
    > Bill

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