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From: Peter Fogg (no email)
Date: Thu Jul 07 2005 - 03:42:56 EDT
> -----Original Message-----
> From: Navigation Mailing List [mailto:]
> On Behalf Of George Huxtable
> Sent: Saturday, 19 February 2005 8:00 PM
> To:
> Subject: Re: Z tangent formula
>
> Bill asked, a few days ago-
>
> >Regarding a formula you gave for finding azimuth:
> >
> >"This formula is-
> >
> >Tan Z = sin (hour-angle) / (cos (hour-angle) sin lat - cos lat tan dec)
> >
> >and the rules for putting Z into the right quadrant, 0 to 360, clockwise
> >fron North, are-
> >
> >If tan Z was negative, add 180 deg to Z. If hour-angle was less than 180
> >deg, add another 180 deg to Z."
> >
> >By hour-angle I assume you mean local hour angle?
> >
> >I was also having trouble sporting out the parenthesis. Should the
> formula
> >look like the following, if I let ha= (hour-angle)?
> >
> >Tan Z = sine ha
> > _________________________________________
> >
> > (cos ha * sin lat) - (cos lat * tan dec)
>
> ==================
>
> Reply from George.
>
> Yes, and yes. You've got it.
>
> Remember that for this purpose ha is measured Westwards, so is always
> increasing. (To my mind, this is sensible, though some authorities define
> it the opposite way, always decreasing). Before local meridian passage, ha
> is negative and increasing toward zero, or else a positive angle
> increasing
> toward 360, which amounts to exactly the same thing.
>
> And lat and dec are both North-positive, South-negative
>
> And then Z will be an angle measured 0 to 360 degrees, clockwise from
> North.
>
> By the way, in that earlier posting I missed out on something, that's so
> obvious it hardly needed saying. But if those additions of 180 degrees
> happen to take the end result over 360 degrees, then (of course) you
> should
> subtract 360 degrees to bring it back into range, if that matters.
>
> George.
*************************************************************************
I've quoted this in full since it goes back a while. Its taken until now to
find the time to play with this.
I've taken 3 sights at random to contrast a few different (?) methods of
azimuth determination. The first comes from the problem Mike Burkes posed,
that I worked out and posted here on the 2 July.
Sun UL Azimuth - in decimal form, and to 3 decimal places if available
101 'Bennett book' azimuth tables
100.9 Sharp nav. calculator
100.982 NavPac
100.985 Z tan formula
The other two come from the first other worked sights to hand. Details
available if anyone's interested.
Miaplacidus Azimuth
183 'Bennett book' azimuth tables
182.6 Sharp nav. calculator
182.637 NavPac
182.665 Z tan formula
Dubhe Azimuth
010 'Bennett book' azimuth tables
010.2 Sharp nav. calculator
010.238 NavPac
010.214 Z tan formula
So I'm convinced, pending further tests, that the Z tan formula is a valid
and accurate method for determination of azimuth. The small differences seem
insignificant. It is not significantly more tedious or time consuming to
calculate, given a good scientific calculator, than the other methods. This
is because the calculator and NavPac require quite a lot of data to be
entered to give a result.
However, I do wonder whether one method or another does give a more
precisely accurate result, although the answer is probably irrelevant for
most practical purposes. Am hampered, admittedly, by not knowing what
methods my calculator or NavPac use. Anyone have other formulas for
comparison purposes? Or ideas on this?
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