# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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