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    Re: Position from altitude and azimuth.
    From: Bill Lionheart
    Date: 2020 May 20, 19:50 +0100
    Isoazimuthal curves are not really spirals in the general meaning of the word. But the curvature and torsion do vary.  From the equation we have seen you can derive a 4th order polynomial in x y and z. This includes some strange extra stuff with azimuths 180 degrees different, due to losing signs when you square. The isoazimuthal  curve is the intersection of a quartic with a sphere. we dont know if the quartic factorizes. If it does it is the product of two generalized cylinders. This seems to be why they look a bit like the intersection of a cylinder with a sphere.

    On Wed, 20 May 2020, 19:32 David Pike, <NoReply_DavidPike@fer3.com> wrote:

    I wrote:  E.g. consider Dec 30N and azimuth 060.   Z can flip at GHA + 180  such that Z is still 60, but azimuth is now 360 – 60, so you’ll get a ring which might or might not be a perfect circle.  If you stick with ISO A°zimuth.  Your saying the direction of the star is always at 60 degrees to the meridian, so in this particular case you’d get a spiral, 

    Sorry, it was already an hour after bedtime.  Please read the above as: E.g. consider Dec 30°N and azimuth 060°.   Z could flip at LHA =180°  such that Z is still 60°, but azimuth is now 360 – 60°, so you’ll get a ring which might or might not be a perfect circle.  If you stick with ISO Azimuth, you're saying the direction of the star is always at 60 degrees to the meridian, so in this particular case you’d get a spiral. DaveP

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