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Subject: Re: Still on LOP's
From: Rodney Myrvaagnes (rodneym@XXX.XXX)
Date: Wed May 08 2002 - 00:12:52 EDT
Maybe Michael will be kind enough to tell me where this goes astray:
OI am going back to the triangle, since it seems to me the tight rope
is equivalent to zeroing out the error in one of the LOPs.
Assume that three LOPs are measured with symmetrical error distribution
about the platonic lines that would intersect at the true position.
Each LOP can be either side of the true position with equal
probability. A navigator who has measured three LOPs has no way of
knowing which side each line is on.
Thus we consider 6 LOPs, with each pair straddling the true position.
If we choose a cocked hat that includes the true position, then
switching any of the LOPs for its mirror will produce a cocked hat that
does not include the true position.
If we take the mirror of all three however, we get another cocked hat
that includes the true center. If I draw this out I get two triangles
that include the true position, and six that do not. This would appear
to get 1/3 inside, rather than the 1/4 everyone else gets.
If I collapse one pair of LOPs to its center, I get two pairs that
surround the center, and two that do not, for a 1/2 chance.
On Tue, 7 May 2002 11:48:29 -0400, Michael Wescott wrote:
>
>Why in that order?
>
>You've just placed POP#1 and POP#2 in relation to each other, not to
>mention the tight rope walker. And our tight rope walker's postion
>with respect to POP#2 is no longer independent of his relationship to
>POP#1.
>
>> It is just as likely that the tight rope walker is to the left or
Rodney Myrvaagnes J36 Gjo/a
"Anything really worth doing is worth doing badly."
(I know who said it, but he can't defend himself now.)
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