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Re: Silicon Sea : Leg 69
From: Sam Chan
Date: 2001 Apr 10, 00:13 EDT

```----- Original Message -----
> >
> > > 3)  What is the DR position?
> > > --  ------------------------
> >
> > S = 10.7 C = 288.9 T,
> >
> >   20/05/2000 12:13:00 ZT = 20/05/2000 16:13:00 UT
> >                           -18/05/2000 22:20:00 UT
> >                          ========================
> >                             1 day     17:53:00
> > or 41.88 hrs
> >
> > Dist = 41.88 * 10.7 = 448.1 nmi
> >
> > D.Lat = 448.1 * cos 288.7 = 145.2 = 2d 25.2'
> >
> > DR Lat = 9d 50.5' + 2d 25.2' = 12d 15.7'
> >  dmp = mp(12d 15.7') - mp(9d 50.5')
> >      = 147.0
> > D.Lo = 147.0 * tan(288.9)
> >      = -429.4
> >      = -7d 09.4
> > DR Lo = -53d  00.0 - 7d 09.4'
> >       = -60d 09.4
> >
> > DR  12d 15.7'N    60d 09.4'W
> >
> > Actually this is too far off to make any sense.
> > We seem to be well past Barbados.
>
> It is off. If you are calculating use a Traverse Table interpolation may
> be accumlating an error. A better method would be to use one of the
> Sailings methods for DR. For sight reduction at least using a Law of
> 4mulas.zip from the web site.
>
Dan,
I find no problems with the details and results from Mike Wescott. The
calculation he shows is the Mercator sailing as shown in Cotter, the
Elements of Navigation and Nautical Astronomy, Chapter 13. I got the same
result doing the calculations by hand. I entered the same problem into the
Starpath Star Pilot and got 12d 15.3N, 60d 09.6W.
For question 7, I got an answer of CTS 338 C with a SMG of 10.9 Kt. The Star
Pilot gave an answer of CTS 324.4T and SMG of 10.9. Applying the Var 15dW
(add), and a Dev 1dE (subtract), I arrive at a CTS of 338.4 C. I double
checked the Star Pilot with a plot on the Mo Board and also arrived at a CTS
of 324T.
Regards,
Sam
```
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