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> The circled combination, A1T2A3, does not seem to be
> possible in this configuration.  And there is only one position
> inside the triangle, T1A2T3.  I'm not sure what's going on and where
> A1T2A3 is hiding.  I believe George discussed this previously.
>
> Fred Hebard

If the sight had yielded A1T2A3, then the triangle would look different,
with the horizontal line moved a few miles to the north.  In that case,
A1T2A3 would be inside the triangle, most of the others would stay about
where they are now, and T1A2T3 would become impossible.

One of the hard things about thinking about this stuff is choosing an
appropriate starting point for the attack.  Once you lay out a particular
triangle, you've already made some assumptions about the observations.
I suspect some of the apparent pardoxes with the probabilities are related
to the concept of "conditional probabilities", which have some counter-
intuitive properties.  I think George has it right when he suggests starting
with a position, and making (or rolling dice to simulate) observations with
particular random error distributions, and anlyzing the collection of triangles
that result.  The navigator at sea is faced with a different problem, looking
at one particular triangle, but the statistics can only be understood from
George's approach.

I have a few comments on things that have been said recently; forgive
me if I don't remember by whom.  Paraphrasing, "Even if we know the fix
is likely to be outside the triangle, where would we put it."  This is
confusing the FIX (a navigational construct) with the ship's ACTUAL
POSITION (which is to some degree unknowable).  Of course we plot
the FIX "in the middle of the triangle" but it's important to remember that
the ACTUAL POSITION, while likely somewhere in the vicinity of the fix,
is surely not in that exact spot.  It's in this context that it's important to
understand how likely it is that the ACTUAL POSITION is inside the
triangle, as the FIX always is.

Gary (I think) says that if your triangle is so large that the sides
are 3*sigma away from the center, then surely your position is inside
the triangle.  The problem I have with this is knowing what sigma is.
The fact that my triangle is very large might suggest that sigma for
this observation is much larger than I thought.  Or perhaps it's more
likely that I've made some kind of blunder, or a systematic error.
If I assume the triangle is due entirely to systematic error, with equal
but unknown effect on each sight, the effects
can be removed by a graphical procedure (which I think is how the
current incarnation of this discussion got started).  If the sights were
not all in one half-plane, removing the systematic error yields a position
inside the triangle; if the sights WERE all in one half-plane, removing
the systematic error yields a position that is OUTSIDE the triangle.
It's easy to demonstrate this ... Draw a position and three perfect
LOP's, and tag each with the azimuth of the celestial body.  Move
each an equal distance toward its body; the resulting triangle will
enclose the position only if the azimuths were well-distributed.

Wishing I had more time to participate...
    -- Bill N.

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