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    Re: Slide Rule Azimuth
    From: George Huxtable
    Date: 2009 May 30, 20:16 +0100

    Greg Rudzinski wrote, in [8443]-
    An interesting azimuth formula presented by H.H. Shufeldt in his book
    Azimuth = INV SIN of  COS declination Sin meridian angle  divided by
    COS altitude ( Ho or Hc )
    Shufeldt states that Ho or Hc altitudes can be used. The INV SIN
    result  is added or subtracted from 360 or 180 degrees depending on
    An alternate arrangement for the formula:
    Azimuth = INV SIN of  SEC altitude COS declination SIN meridian angle
    I like the expediency of this formula but it does suffer from
    inadequate  slide rule scale resolution for azimuths approaching 270
    or 90 degrees. A trick to by-pass this problem for a sun observation
    would be to directly observe a corrected bearing of the sun (which
    should be low in the sky) for use as an altitude intercept azimuth.
    and Gary LaPook responded-
    That is the formula that I have used for years for calculating azimuth.
    You can find it in Bowditch. George has pointed out that it gets
    ambiguous near east and west but it is not a problem in real life and is
    quick and easy to do on a calculator or slide rule. For those rare cases
    near east or west another formula could be use. The Az calculated with
    this formula is between zero and ninety degrees so you have to figure
    what quadrant you are in and convert to Zn but this is also not a
    problem in real life since you know the approximate direction when you
    pointed your sextant.   See:
    This question has been around this list, and its predecessor, more than 
    once, but it might as well get another airing.
    Gary has pointed out the ambiguity, for azimuths near East and West, which 
    is the serious drawback to this method of working (more serious, in its way, 
    that the poor precision at these angles, which Greg did recognise). But he 
    pointed it out, only to dismiss it, as "not a problem in real life". I 
    suggest he should think again. The fact that it may be "quick and easy to do 
    on a calculator or slide rule" does not overcome those difficulties
    He refers to those "rare cases" when the object is near East or West. Not so 
    rare, however. In the tropics, there are two periods of the year when the 
    Sun is either nearly-East or nearly-West, the whole day through. Elsewhere, 
    it's always near East-West twice a day, in Summer, just the best time for 
    determining longitude.
    The difficulty is that it's impossible to distinguish, by this method, 
    between azimuths greater than 90�, and azimuths correspondingly less than 
    90�, such as between azimuths of 80� and 100�, as their sines are exactly 
    the same. As long as those azimuths differ sufficiently from 90�, there's no 
    problem; it's obvious which is the right value. Perhaps Gary is confident of 
    his ability to distinguish between azimuths of 80� and 100�, but could he do 
    so, in rough weather, for a high sky-object that might be 85�, or might be 
    95�? If he got that choice wrong, the resulting 10� of error could upset a 
    position calculation, unless the intercept happened to be a short one.
    Gary suggests that in such cases, a navigator could use a different formula, 
    as indeed he could. But that means he would have to keep two different 
    procedures in his mental locker, and know when to apply each one. How much 
    simpler, then, to use instead a formula that always preserves its accuracy 
    over all azimuths, and is free from ambiguity. This is the formula that 
    derives azimuth from its tan, rather than sin or cos, as follows-
    Azimuth from North = arc tan ( sin (MA) / (cos lat tan dec -cos (MA) sin 
    If the result is negative, add 180 degrees to make it positive. This is how 
    it works if, like many navigators, you always think of your meridian angle 
    as a positive quantity, whether it's East ot West. That result would be the 
    azimuth of a body if it's East of you. If the body is to your West, the 
    angle from North would be the same, but measured from North the other way, 
    in the Western hemisphere, so you have to subtract that result from 360�.
    Personally, I prefer to think of meridian angles (and longitudes) as 
    increasing Westerly, just as Hour angles do (and against the current 
    conventions), in whch case the rules for getting the angle in the right 
    quadrant are a bit different.
    Although this method may take a few more keystrokes on a calculator, it has 
    the advantage that it doesn't depend on the result of any previous 
    calculation, for altitude.
    contact George Huxtable, at  george@hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    Navigation List archive: www.fer3.com/arc
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