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    Re: Sight Reduction Formula Question
    From: George Huxtable
    Date: 2005 Jan 9, 16:45 +0000

    Paul Hirose pointed out that it's rather easy to plot out the azimuth onto
    a chart, without really having to think too hard about putting it in the
    right sector, and I agree. But it's if you want to feed the information
    into a computer, to make (for example) a least-squares analysis of a set of
    observations, that the azimuths need to be defined in a formal way that the
    computer will understand. And that means 0 deg to 360 deg, increasing
    clockwise from North.
    
    Note that astronomers, such as Meeus, like to define their azimuths as
    clockwise from due South, and beware.
    
    =====================
    
    Trevor Kenchington asked-
    
    >Could someone either explain how the expressions for the sine, cosine
    >and tangent of Z (as discussed on this thread) are derived or else point
    >me to a published explanation? I have been trying to equate the three
    >expressions but the trig I was taught as a 16 year-old is far too rusty
    >for me to make sense of them. Nor can I find an explanation in any of
    >the texts I have to hand.
    
    Yes. My 2-volume edition of Bowditch (vol.2 is 1981) states and derives all
    three expressions. In vol 2 it's in "part 3, Navigational calculations", in
    chapter 7, "Calculations of Celestial Navigation", paras 707-710.
    
    Let's see if I can pass on the gist of it.
    
    Start by drawing a spherical triangle with vertices A,B,and C in clockwise
    order, for which the sides opposite those vertices are a, b, c.
    
    Bowditch then quotes one case of the "five parts formula" as
    
    sin a cos B = cos b sin c- sin b cos c cos A
    
    and from the law of sines-
    
    sin a / sin A = sin b / sin B
    
    from which he gets sin a = sin A sin b / sin B
    
    and substitutes that in the first equation to get
    
    sin A cot B = sin c cot b - cos c cos A.
    
    Now he relates that triangle to the PZX triangle in which A corresponds to
    the pole P, B corresponds to the observer's zenith Z, and X is the
    direction of the body. In that case LHA is the angle at P, Z is the azimuth
    angle at the observer Z, the side opposite Z is (90 - dec), and the side
    opposite X is (90 - lat).
    
    So we can write-
    
    sin LHA cot Z = cos lat tan dec - sin lat cos LHA
    
    from which tan Z = sin LHA / (cos lat tan dec - sin lat cos LHA)
    
    ================
    
    Now for getting the azimuth from its cosine.
    
    Bowditch, using the same ABC triangle, writes "the law of cosines for sides" as-
    
    cos b = cos c cos a + sin c sin a cos B.
    
    This is equivalent to the same PZX triangle as before, but this time we
    will also need the side opposite to the pole P, which is (90 - alt).
    
    substituting these quantities for a, b, c, and B, we get-
    
    cos (90 - dec) = cos (90 - lat) cos (90 - alt) + sin (90 - lat) sin (90 -
    alt) cos Z
    which gives cos Z = (sin dec - sin lat sin alt) / cos lat cos alt
    
    ===============
    
    What seems at first sight to be the simplest method, getting  Z from its
    sine, is-
    
    take the sine formula sin b / sin B = sin a / sin A
    
    Make the same substitutions as before to give=
    
    sin (90-d) / sin Z = sin (90 - alt) / sin LHA
    
    or sin Z = sin LHA cos dec / cos alt.
    
    but this is a lot more complex than it seems, because of the difficulty of
    deciding, just knowing the sine of Z (when it's near due East or West)
    whether it's a bit North, or a bit South, of due East or West. Bowditch
    provides another trig expression to test which, but by the time you have
    added that test, it's no longer the simplest method.
    
    ===============
    
    Note that Bowditch refers to his azimuths as being so many degrees East or
    West of North or South, in the quadrantal notation you used to find on old
    compasses. To correspond with a modern compass, or to pass data into a
    computer program, a conversion is required to 0deg to 360 deg, clockwise
    from North.
    
    
    Trevor refers to "the complications of "ATAN2" calculator buttons or other
    ways of applying George's tangent approach.". However, there's no need to
    have an "ATAN2" or "POL" routine available, though it does simplify
    matters; the ordinary inverse tan will do the job perfectly well. And the
    arc tan approach applies in every case, with just one formula to learn, and
    no ambiguities. It's the one formula that you will find for azimuth in
    Meeus, though as mentioned earlier, Meeus' formula differs slightly from
    what we navigators need because he defines azimuths from the South.
    
    George.
    
    ================================================================
    contact George Huxtable by email at george---.u-net.com, by phone at
    01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
    Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ================================================================
    
    
    

       
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