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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:
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http://groups.google.com/group/NavList/browse_thread/thread/af4f15cde5075f8f/058fe8755eeaca37?hl=en&lnk=gst&q=lapook+cosine#058fe8755eeaca37
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http://groups.google.com/group/NavList/browse_thread/thread/529edc05997d59d7/e002865149e31596?hl=en&lnk=gst&q=lapook+cosine#e002865149e31596

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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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