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    Re: That darned old cocked hat
    From: Hewitt Schlereth
    Date: 2010 Dec 10, 20:06 -0400

    My method in practice was to imagine where the cocked hat would
    balance on a pin and take that point as my fix. This I think is
    essentially Gary' first and easiest way.
    
    What I was more curious about was George's explication to the effect
    that the fix had a 25% chance of being inside the hat versus a 75%
    chance of being outside it - i.e., being three times more likely to be
    outside the hat than in it. Where would that outside point be? How
    would you figure that out?
    
    Hewitt
    
    On 12/10/10, Herbert Prinz <666@poorherbert.org> wrote:
    > This post reminds me of what happened long after the Board of Longitude
    > had paid out their price: There was no end to the submission of "better"
    > solutions.
    >
    > Herbert Prinz
    >
    > On 2010-12-10 22:22, Gary LaPook wrote:
    >>
    >> So we now have four geometric constructions (plus visual estimation
    >> making a total of 5 ways) to plot the fix inside the cocked hat. John
    >> Karl's probability diagram shows the probability of each of these
    >> points to be essentially equal although the Symmedian point my be ever
    >> so slightly more probable. So, what method should you use? I think the
    >> decision should be based on ease of construction. Obviously the
    >> easiest way is by eyeball and is the method I recommend. The next
    >> easiest construction is my method of determining the centroid by the
    >> "median" method. You only have to use dividers to halve one of the
    >> LOPs, lay a straight edge from the opposite corner to this point, and
    >> then use the dividers to mark the 2/3rds point along the straight
    >> edge. The standard way to determine the centroid is the text easiest,
    >> halve two of the LOPs and draw the two lines from the opposite angles
    >> to those points. More difficult is measuring the angles of two of the
    >> three corners, then dividing them in half, and then finally plotting
    >> those lines. ( You can also find the bisectors of the azimuths and
    >> plot them, you get the same point, if the spread of azimuths exceed
    >> 180 degrees.) The most difficult point to plot is the Symmedian point
    >> which requires that you first plot both the medians and bisectors,
    >> measure the angle between each of the lines in each set, and then draw
    >> in the additional lines shifted by the angle between the lines in each
    >> set to the opposite side of the bisector. Plotting the Symmedian point
    >> takes a lot more work with no significant in probability that it
    >> represents the actual position of the vessel.
    >>
    >> ( See my diagrams on the "three body fix" thread.)
    >>
    >> gl
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