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From: Bill (no email)
Date: Tue Jun 06 2006 - 23:21:02 EDT
Eddie
Immense help.
I did find one *big* blunder on my part. When I derived one of the angles
in the oblique triangle: sine angle = (sin 6d 53' 16"/.954) * 7.973333 =
88.8d I labeled the wrong angle! Therefore my geometry was off by approx 4d
(too low). Now I get 140.7 d, so close enough to 141.
It hit me like a brick while reading your response. My .954 nm is spot
on--for a 52 minute run. For an hour, more like 1.1 nm drift. Duh.
On a small craft I generally prefer to do things graphically, and extend it
out to a whole hour and find a scale that will fill an 8 1/2" x 11" sheet of
paper. Then I can check the results out with math if I need to be spot on.
The graphics work as a sanity check for math blunders. By starting out
mathematically I failed to get my head out of the boat, and kept playing
with results from a 52 minute run and resultant triangle.
Regarding your point on graphic solutions being an educated guess, I agree.
Given a professional drafting board and quality instruments, I am use to
doing much better. Given a chart with distortions by default (depending to
some degree on projection and scale), and relatively crude drafting
instruments and conditions, pulling set off a 1" line and transferring it to
a compass rose, or using a plastic one-arm protractor or other device, is at
best problematic. One of these days I need to acquire a fine (Russian?)
three arm protractor like Alex's. Not too much I can do about projecting a
sphere onto a flat surface and the resultant problems.
As for the Coast Guard tests, they are assuming a large commercial craft, so
they don't give you +/- 0.2 nm and set within +/- 10 degrees. More like .1
nm and 1d or less.
Thanks so much for getting me out of the rut I was stuck in. A lesson well
learned.
Bill
> 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
>
> --
> __________________________________________________brainaid______________
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> software Monheimsallee 45 phone +49 241 5151 138
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