NavList:

   

Reply

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.
--~--~---------~--~----~------------~-------~--~----~
Navigation List archive: www.fer3.com/arc
To post, email NavList@fer3.com
To , email NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---

   

Reply