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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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>
>
>
>
>

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