# NavList:

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

Message:αβγ
Message:abc
 Add Images & Files Posting Code: Name: Email:
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:
>  > |
>  > |
>  >
>  > |
>  > |
>  >
>  >
>  > ================================
>  >
>  > 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.
>  >
>  >
>  >
>  >
>  >
>  >
>  >
>  >
>  > >
>  >
>  >
>
>
>  >
>

--~--~---------~--~----~------------~-------~--~----~
To post, email NavList@fer3.com
To unsubscribe, email NavList-unsubscribe@fer3.com
-~----------~----~----~----~------~----~------~--~---
```
Browse Files

Drop Files

### Join NavList

 Name: (please, no nicknames or handles) Email:
 Do you want to receive all group messages by email? Yes No
You can also join by posting. Your first on-topic post automatically makes you a member.

### Posting Code

Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.
 Email:

### Email Settings

 Posting Code:

### Custom Index

 Subject: Author: Start date: (yyyymm dd) End date: (yyyymm dd)