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"Trevor J. Kenchington" wrote:

> Given a single LOP, the true position could lie to either side of it,
> with the probability of it being some particular distance away taking
> the form of some probability density function -- hopefully a Gaussian
> one but most likely not.

> To simplify for the purposes of illustration,
> we could pretend that the true position is equally likely to lie
> anywhere within a distance "x" on either side of the LOP.

Unfortunately, you can't make this simplification. Doing this, you are implying
rectangular error distribution (such as you get from limited resolution of your
sextant) and this is an entirely different problem.

> Given a second LOP (assumed, for present purposes, to have a precision equal to
> that of
> the first LOP), the true position must lie within x of each LOP (still one either
> side). Thus, it must lie within a rhombus symmetrically arranged around the
> intersection of the two LOPs, the sides of the
> rhombus being longer than x (unless the two LOPs cut at 90�).

For rectangular error, this is correct. But then the true position cannot possibly
lie outside. For the consequences, see below. But not so for normal error. In this
case, the contours of equal probability density are ellipses.

> If a third LOP is added (still with the same precision), the true
> position must lie within the previous rhombus _and_ within x of the
> third LOP (on either side of it).

For rectangular error, yes. The band of the third LOP _must_ intersect the rhombus.
If it didn't, you would not just have a case where the true position lies outside the
cocked hat. You would, in fact, have a logical contradiction, and that is to say,
invalid data.

For normal error, however, the true position may be anywhere, with varying
probability. In particular, it may be outside any given confidence ellipse, and
certainly outside the previously obtained rhombus.

> The question must then be: How much of the overlap between the rhombus around the
> intersection of two LOPs and the zone of uncertainty (with width x) around the
> third LOP falls within the cocked hat formed by the three LOPs?

To evaluate the probability of the true position being in a given area, you have to
find the integral over the probability density over this area. In case of rectangular
error, the probability density is constant. Therefore you get away with comparing the
sizes of areas. But in case of Gaussian error, the probability density is not uniform
within the cocked hat. Things are becoming more tedious. It would actually find it
very interesting to see whether, by evaluating this integral  for the cocked hat, we
can confirm George's assertion that it comes out to 25% probability.

Until we find a volunteer to do this, the following thought experiment is much
easier: From your true position "shoot" 10000 rounds of 3 sights in three directions.
Assume _nothing_ about error distribution other than symmetry around 0. For each
azimuth we expect 5000 LOPs towards and 5000 away. Draw, therefore, any 10000 LOPs
satisfying this criterion. Examining all the 10000 cocked hats resulting from these
"observations", you will find that they enclose the true position in exactly those
2500 cases where all 3 LOPs in the series are on the same side (i.e. either all
towards or all away). (There are really only 8 cases to distinguish. I say 10000 to
make it statistically valid.)

Herbert Prinz

   

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