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>> 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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