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    Re: Z tangent formula
    From: Peter Fogg
    Date: 2005 Jul 7, 17:42 +1000

    > -----Original Message-----
    > From: Navigation Mailing List [mailto:NAVIGATION-L@LISTSERV.WEBKAHUNA.COM]
    > On Behalf Of George Huxtable
    > Sent: Saturday, 19 February 2005 8:00 PM
    > 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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