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Re: Bowditch 1995 Table 18
From: Bill B
Date: 2005 Feb 2, 22:14 -0500

```> I did a rough sketch which suggests this solution is about right. I have not
> had time to sit down and draw a proper plot of the solution to check the
> result, but plan to do so.
>
> Bowditch 2002 is as sparse on further explanation as your 1995 version.  You
> might be interested in a short paper on the trigonometry of a similar
> situation, posted by a colleague:

Jim

Thanks for the reference.  I was playing with table 18 this AM, and found it
easier to solve as oblique triangle using the Law of Sines with my TI-30Xa.
15 seconds and voila.  Derived the following formula which seems to work.
Give it a test drive if you would be so kind.

Let AOB1 = Angle on the bow, first shot.
Let AOB2= Angle on the bow, second shot
Let D = Distance in nm from object at the time of the AOB2

As noted in an earlier posting, the prudent navigator may wish to correct
the angle on the bow (course steered) by the difference between in and
estimated course over ground.

D = (distance run * sin AOB1)/(sin (AOB2-AOB1))

Solving for distance from the object when abeam (and it would only actually
have been abeam during the run in 1 of 3 scenarios) is a simple matter of
constructing and solving the resulting right triangle.

Given the number of definitions for "relative bearing," perhaps the editors
of Bowditch were wise in being ambiguous.

Bill

```
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