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Thinking about yesterday's messages, Martin seems to have found the key
to the whole problem -- or at least parts of it.

As he says, only 7 of George's 8 combinations of error directions can
actually occur and only one of them causes a true position to fall into
the cocked hat. The same is true of the 8 that I suggested (which were
differently labelled to George's 8 but were functionally the same once
drawn out as Martin suggested). So we have resolved one part: Using the
logic that both George and I employed, the chance of the true position
falling into the cocked hat isn't 12% or 12.5% but 1-in-7, which is to
say a bit less than 14.3%.

Martin also wrote:

>I have encountered this line of reasoning before, and I think
>there's something amiss with it, though I can't exactly quantify what.
>
>My worry is that I don't think the three bearings are actually independent.
>It's not as if we were dropping three long straws at random on a chart -
>we're in some sense measuring the same thing all three times.
>

which brought me back to the point I made on the 16th:

> I would suggest that the intersection of any two LOPs gives us an
> estimate of position. If a third LOP fails to pass through that
> intersection and passes, for example, further east, then it is already
> likely (though not, of course, certain) that the third LOP lies east of
> the true position.


To illustrate what I am groping towards:

First off, following George, I assume that LOPs are lines in the
mathematical sense of infinitely thin boundaries between areas. Being
infinitely thin, there is, as best as I can understand, zero chance of a
position lying on one, though a high probability that the true position
lies very close. (Rodney: If you can explain how Sir Isaac and his
calculus lead to a different conclusion, I'd be glad to hear it.)

Given just one LOP, corrected for all known errors and assumed not
subject to any asymmetrical errors, we then know that our true position
lies on one side of that line or the other, with a 0.5 (50%) probability
in each case, but we do not know where along the line it lies.

Add a second LOP roughly perpendicular to the first and we expect our
true position to be somewhere near the intersection, though it could lie
in any of the four sectors delimited by the two LOPs. Each sector is on
one or the other side of each LOP so, following George, we would expect
that the probability of the true position lying in any particular sector
is 0.5 x 0.5 = 0.25.

But consider what happens if the two LOPs cut at a shallower angle, say
30�: George's logic suggests that there is still a 0.25 probability of
the true position falling into any one sector, yet two of the sectors
are much narrower than the other two. Thus, each square metre of sea in
the narrow sectors would have a higher chance of being our true position
than a square metre (at the same distance from the point of
intersection) in one of the wide sectors. That doesn't have to be wrong
though: Two LOPs cutting at a shallow angle don't do a good job of
telling us where we are but they do confirm that we are somewhere near
being along a line that bisects the angle between those two LOPs. Thus,
perhaps we should have a higher probability per square metre of being in
one of the narrow segments. (That links back to the notion of elliptical
confidence limits, with the long axis of the ellipse aligned to bisect
the intersection of the LOPs.)

But then consider the theoretical case in which the two LOPs lie exactly
over the top of each other. The width of the narrow sector between them
is now zero and there can only be zero chance of our true position lying
there, whereas the wide sectors have expanded to 180� each and there is
a probability of 0.5 of our being in each one of them. So, if George's
logic were correct, as the two LOPs came nearer and nearer to cutting at
an angle of 0� (i.e. lying on top of each other), the probability of the
true position lying between them would stay steady right up to the
moment that the angle dropped to 0�, at which point the probability
would immediately drop to zero while that in the wide sectors would jump
from 0.25 to 0.5. Such discontinuities can occur in mathematics but I
very much doubt that this one is real. Thus, I suggest that George's
logic is wrong. Looking at any one LOP in isolation (and assuming
symmetry, as above), there is an equal chance of the true position lying
on either side but, with two LOPs together, that is no longer true.

The reason, as best as I can make it out, goes back to Martin's:

>It's not as if we were dropping three long straws at random on a chart -
>we're in some sense measuring the same thing all three times
>

The two LOPs are each estimates of the same thing -- a locus along which
the same true position lies. The first one we plot has an equal chance
of having the true position to its left or its right. But once we (with
omniscient knowledge of where the true position lies) determine that the
first LOP is too far to the right, then we know in advance that a second
LOP which almost overlies the first is also highly likely to be too far
to the right. [It is analogous to rolling double six with two dice.
Before you roll, there is a 1-in-36 chance of getting double six. But if
you roll one die and get a six, there is then a 1-in-6 chance that the
second die will give you the double.] So, as the angle between two LOPs
drops towards zero, so the chance of the true position lying in one of
the narrow sectors drops from 0.25 (when the LOPs are perpendicular) to
zero (when they overlie one another). I have no idea of the shape of
that drop but it won't be linear and it might follow a cosine curve.

The more complicated (and more interesting) question is what happens
when a third LOP is added. Consider the case in which the three LOPs cut
at 60�.  With the first two LOPs plotted, the chance of the true
position being in each of the sectors that could produce the cocked hat
would be a bit less than 0.25 each (since the 60� sectors are narrower
than the 120� ones). But the probability that it is in either of those
sectors (adding the two together) would still be reasonably high: maybe
0.4. Now, if that true position lies to the right of the intersection of
the first two LOPs, it is more likely that the third LOP will lie of the
right of the first intersection than to its left -- how much more likely
depending on the precisions of the various LOPs and how far off the
first intersection the true position lies. So the probability that the
cocked hat would be formed on the "correct" side of the first
intersection is above 50%, though I cannot say how much higher.

Of course, that leaves the question of whether the true position is in
the cocked hat so formed versus lying even further from the first
intersection than the third LOP does. Since the first intersection
exists on the plot, it is more probable that the true position lies
close to it than that it is far away, so again this is not a 50/50
chance, though I don't pretend to be able to quantify it.

IF this logic is correct (and I would stress that I am groping here),
then it suggests that the probability of the true position falling
within the cocked hat is a good bit higher than 14.3% but still a lot
lower than 50%. It also suggests that there is no fixed probability. As
I have noted before, if you chance to get a cocked hat that is small
compared to the level of precision of your LOPs, then you are likely not
inside it. (In the extreme, if all three LOPs pass through the same dot,
you should NOT assume that that is a highly accurate fix unless you have
others were unusually precise. Rather, you likely lie outside, though
close to, the intersection of your lines.)


Can anyone refute or advance any of this?


Trevor Kenchington

   

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