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Re: Set and Drift
From: Eddie C. Dost
Date: 2006 Jun 7, 03:50 +0200
From: Eddie C. Dost
Date: 2006 Jun 7, 03:50 +0200
Hi Bill,
I get (almost) the same result as the solution requested:
True course is 45d.
COG is 51.9d.
Distance travelled is 7.9nm.
So far so good.
I am used to extending the distance to speed, so I get
7.9nm * 60min/h / 52min = 9.1kts SOG
Using Vector math and solving (polar (r, phi) vectors):
(9.1kts @ 52d) - (9.2kts @ 45d) = (1.12kts @ 143.6d)
Varying the SOG/COG values just a little yields very different
results. Taking such values of a plotting paper or a chart is
mostly "guess work". I use a rough estimate to call an answer
to such a question as in your example correct when drift is within
+/- 0.2 nm and set is withing +/- 10 degrees when I have to correct
such exams.
Something seems to be off with your triangle calculations, as
your results up to that point are the same as I have.
Hope this helps,
Eddie
On Tue, Jun 06, 2006 at 08:56:40PM -0400, Bill wrote:
> > I'll forward your "problem" to where I work, where I can look at the chart and
> > see where they came up with their answer. Sorry I can't help sooner. I'll
> > try to have an answer to you around 00h00 (UTC) 8 Jun.
>
> Thanks Pete
>
> I initially worked it without a chart using rectangular to polar conversion:
>
> dLat 4.9
> dlon 8.3
> Mean Lat 41d 13' 27"
> Conversion factor, lon to nm
> = mean lat cosine = .752137015
> .752137015 * 8.3' lon = 6.242737222 nm
>
> After R to P conversion:
> Distance = 7.936105344 nm
> True = 051d 52' 16.3"
>
> C (psc) 056
> D +04 E
> M 060
> V -15W
> T 045
>
> One angle of the oblique triangle
> = 051d 52' 16.3" - 045d = 006d 52' 16.3"
>
> One adjacent leg = 7.936105344 nm
> The other = time * speed = 52 min * 9.2 = 7.9733333 nm
>
> Using the law of cosines the drift leg = .954121933 nm
> Using the law of sines to derive the other angles and doing a bit of
> geometry, I come up with set of 136d 00.4'
>
> Plotting it graphically on the chart, on a plotting sheet, and in a computer
> drawing program, my results agree within +/- .05 nm and +/- 1d, so I am at a
> loss.
>
> Bill
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