NavList:

   

Reply

I've also been thinking about George's assertion that the
cocked hat contains the tru position just 25% of the time,
and, much to my surprise, I've convinced myself it's true.

The key question, in my mind, was whether the observations
could properly be considered as independent.

Let's start with the case of two LOP's. I think if they cross
at right angles, then we can all agree the observations are
independent, and the true position is equally likely to lie
in any of the four quadrants so defined.

Now suppose, as Trevor has suggested, that the azimuths are
nearly equal, so that the two LOP's define two skinny regions
and two very wide regions.  Intuitively, it seems that the
true position ought to be more likely to be in a wide region
than in a skinny one.  But I don't think that's true.  For
concreteness, let's say both LOP's run nearly north-south.
The first one is equally likely to be east of our true position,
or west of it.  Let's say it is east of our true position.
Now we lay down the second LOP.  It, too, is equally likely
to be east of our position or west of it, since it has no
"knowledge" of the previous observation.  (However, it is
likely to be west of *the other LOP* near out latitude, since
we are assuming the other LOP is too far east.  In this sense
the relative positions of one LOP to the other is not
independent of the error in one LOP.  But the direction from
one LOP to our true position *is* necessarily independent of
the direction from the other LOP to our true position.)
Therefore, the surprising fact is that our true position is
just as likely to be in a skinny region as in a wide one.
I can rationalize this by noting that the intersection
skitters away off to the north or south as the two LOP's move
east and west, so in a sense the area in the skinny region
grows lengthwise to make up for its lack of width.  It's still
counterintuitive.

Let's go back to the case that's easier to accept, with two
LOP's crossing at right angles.  Suppose one runs north-south
and one runs east-west, so they define four quadrants: NE, NW,
SW, and SE, and our true position is equally likely to appear
in any quadrant.  Now let's add a third LOP, running SE-to-NW.
Like the others, it is equally likely to be on either side of
our true position.  Depending on where it falls with respect
to the intersection of the other LOP's, it will define a
cocked hat of some size in either the NE quadrant or the SW
quadrant.  If our position was in the SE quadrant or the NW
quadrant, then the true position is clearly outside the cocked
hat -- this accounts for 50% of the cases.  If our position
was in the NE quadrant, and the new LOP is equally likely to
fall NE of us or SW of us, then half the time we'll be inside
a cocked hat in the NE quadrant, and half the time we'll be
outside the cocked hat (which may be in the NE quadrant or
in the SW quadrant).  This contributes 12.5% of cases inside
the cocked hat, and 12.5% more outside.  The same argument
holds if we are in the SW quadrant, giving a total of 25% of
cases where our true position was inside the cocked hat, and
75% where it was outside.  This argument depends on the fact
that whether the new LOP is NE or SW of *our true position*
is independent of which quadrant we are in.  There are some
other things we might see that are *not* independent.  For
example, if we are in the NE quadrant, the new LOP is likely
to fall NE of the intersection of the other LOP's.  Equivalently,
if we are in the NE quadrant, it is more likely that the cocked
hat falls in the NE quadrant than that it falls in the SW
quadrant.  Renaming George's 8 possible outcomes, our true
position could be
  N,E,NE; N,E,SW; N,W,NE; N,W,SW; S,E,NE; S,E,SW; S,W,NE; S,W,SW
of the three LOP's.  If it's N,E,SW or S,W,NE then our true
position falls inside the cocked hat -- but one of these doesn't
exist!  Does that reduce the probability to 1/7 (or even 1/8)?
No, because which one exists is not independent of our true
position.  The only thing that *is* independent is whether the
new LOP is NE of us or SW of us.

   

Reply