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    Re: Slide Rule Azimuth
    From: Greg Rudzinski
    Date: 2009 May 31, 04:11 -0700

    Hewitt,
    
    I think you mean Sin H = sin d / sin L. This works great on a slide
    rule.
    
    On May 31, 3:40�am, Hewitt Schlereth  wrote:
    > 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/af4f15cde...
    > > �> |
    > > �> |
    > > �>http://groups.google.com/group/NavList/browse_thread/thread/529edc059...
    >
    > > �> ================================
    >
    > > �> 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 �geo...@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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