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On 5/8/2014 9:51 AM, Greg Rudzinski wrote:
> The way I see this is that Hc is based on an assumed position not the
> actual position so that the azimuths will differ slightly. Not an issue
> for the LOP plot. But if an accurate fiber optic gyro were to be checked
> to it's precision then I would use Ho (not Hc or Hs)to calculate azimuth
> to the tenth of a degree.

Sean and Greg

Thanks for your input. I do see your points, but still feel like a puppy
chasing its own tail.

If I understand correctly, in the following formula I would still be
working from an assumed position as the LHA is derived from the GHA and
AP longitude, so your caveats are still valid:

Tan Z = sin LHA / ((cos LHA sin lat) - (cos lat tan dec))
If tan Z is negative, add 180 to Z
If hour angle is less than 180, add another 180 to Z

If the scenario was similar to Greg's post from the archives, and I was
working from about a known position (Ho and Hc theoretically equal),
then the azimuth derived from the Hc or Ho would yield the correct azimuth.

Have I caught up to my tail yet :-)

   

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