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If that is indeed the question, most of the argument has been off the
point. I thought George was maintaining that no cocked hat that could
arise from 3 LOPs with whatever error distribution could enclose more
than 25% of the locations that could give rise to those LOPs. Even one
counterexample defeats that claim, so maybe he is claiming something
else.

I can give lots of counterexamples. Another one herewith:

An equilateral triangle of equivalent readings, with each vertex at the
.1 distance (.1 probability of the reading being at that or greater
distance) produces a probability of the location falling inside of 0.9
* 0.9 * 0.9 =0.73, so 73% of locations will be inside. the triangles
adjacent on each side will be inside 2 of the .9 bands, so will have a
probability less than 0.9 * 0.9 * 0.1 <= .081 so less than 8% in those
areas. The next triangles have 0.9 * 0.1 * 0.1 <= 0.009 or less than
1%.

The probability of that set of readings actually occurring is very low,
but not zero.

On Thu, 2 May 2002 08:59:04 +0100, Dr. Geoffrey Kolbe wrote:

>
>The question (I think) we are asking is, "Given this distribution of n
>cocked hats, where is the most probable position and what is the error in
>its position?" This is a much more tractable problem.
>
>Better yet to plot means for the multiple observations on each bearing and
>then use the standard deviations to say something about the errors on each
>mean and so to the error on the MPP. This is the most efficient way to
>proceed.
>


Rodney Myrvaagnes                                  J36 Gjo/a


"Curse thee, thou quadrant. No longer will I guide my earthly way by thee."  Capt. Ahab

   

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