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    Re: Set and Drift
    From: Bill B
    Date: 2006 Jun 6, 23:21 -0400

    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______________
    > Christian Dost          brainaid GbR              Bluecher & Dost
    > software                Monheimsallee 45          phone +49 241 5151 138
    > D-52062 Aachen            fax   +49 241 5151 139
    > ecd@brainaid.de         Germany                   cell  +49 172 9312808
    
    
    

       
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