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    Re: Slide Rule Azimuth
    From: Hewitt Schlereth
    Date: 2009 May 31, 06:40 -0400

    There is also the formula for computing altitude on the prime
    vertical: Sin H = sind/cosL. I don't have Weems' The Secant Time Sight
    book in front of me, but I believe this is the same formula he gives
    there, though in secants and cosecants.  -Hewitt
    
    On 5/31/09, Gary LaPook  wrote:
    >
    >  The formula is very convenient for use with a slide rule in conjunction
    >  with the sine-cosine method and is the one I used to illustrate this
    >  computation.
    >
    >  A while ago I posted the Rust diagram for finding azimuth which is in
    >  Weem's LOP Book. The Rust diagram was computed with this formula so it
    >  suffers from the same ambiguity. To deal with this Rust provides an
    >  auxiliary diagram that allows you to determine if the body is north or
    >  south of east. Here is a link to the Rust diagrams,
    >  http://fer3.com/arc/img/103383.rust%20diagram.pdf
    >
    >  So, print out a copy and carry it with your calculator or slide rule.
    >
    >
    >  gl
    >
    >
    >
    >
    >  George Huxtable wrote:
    >  > Greg Rudzinski wrote, in [8443]-
    >  >
    >  > An interesting azimuth formula presented by H.H. Shufeldt in his book
    >  > SLIDE RULE FOR THE MARINER (pg. 77)
    >  >
    >  > 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
    >  > orientation.
    >  >
    >  > 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:
    >  > |
    >  > |
    >  > 
    http://groups.google.com/group/NavList/browse_thread/thread/af4f15cde5075f8f/058fe8755eeaca37?hl=en&lnk=gst&q=lapook+cosine#058fe8755eeaca37
    >  > |
    >  > |
    >  > 
    http://groups.google.com/group/NavList/browse_thread/thread/529edc05997d59d7/e002865149e31596?hl=en&lnk=gst&q=lapook+cosine#e002865149e31596
    >  >
    >  > ================================
    >  >
    >  > 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
    >  > lat))
    >  >
    >  > 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.
    >  >
    >  > George.
    >  >
    >  > 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.
    >  >
    >  >
    >  >
    >  >
    >  >
    >  >
    >  >
    >  >
    >  > >
    >  >
    >  >
    >
    >
    >  >
    >
    
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