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    Re: Still on LOP's
    From: Rodney Myrvaagnes
    Date: 2002 May 7, 23:12 -0500

    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
    >> 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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