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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: That darned old cocked hat
From: George Huxtable
Date: 2010 Dec 10, 19:04 -0000

```John Karl writes-

"Next we acquire three LOPs.  They do not intersect at a common point.  We
don’t know where the true fix is."

Let's say we do. Take an anchored vessel at a precisely known location, say
by GPS.

Measuring the altitude of a star does not alter that location. The
probability of the observation being towards or away depends only on the
random scatter of the observation itself, being precisely 50:50. Instead of
actually measuring it, me might just as well simulate the observation by
tossing a coin. And we could simulate the observations of the other two
stars in exactly the same way.

In which case the probability of the three observations being in the same
direction, TTT or AAA, is exactly the same as the probability of getting
three coins the same (all-heads or all-tails), which is 1 in 4.

George.

contact George Huxtable, at  george@hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
----- Original Message -----
From: "John Karl"
To:
Sent: Friday, December 10, 2010 6:21 PM
Subject: [NavList] Re: That darned old cocked hat

George Huxtable wrote:

Three separate observations are made, one of each star. By our prior
condition, the  probability of each being toward or away is 50:50. There is
no correlation between them. They could have been made by three separate
observers, each putting his observation into a sealed envelope, to be
opened together after the event. Then the probability of any combination,
of the 8 possible, is one in 8. How can it be different?
--------------
Now I think we’ll back up a little.  We’ll take the same approach that I
used in my discussion of the “long-run” running fix – the Bayes (and E.T.
Jaynes) logic of inference, used in the face of incomplete knowledge:  Now
say we’ve made a lot of measurements with the device that locates our LOPs
and have found that random errors occur with a normal distribution of
standard deviation, sigma.

Next we acquire three LOPs.  They do not intersect at a common point.  We
don’t know where the true fix is.  We do know that each LOP represents a
sample from a normal distribution, each with the same sigma.  Lacking any
other information, we are forced to assume that each LOP locates the center
of a linear normal distribution in the x-y plane (see the first attachment
below).  This is a very poor assumption, but it’s the best we can do.  Are
we to assume the LOPs are located somewhere less likely in their
distributions?  (Here we see a major difference with another logic that
might start by considering the three probability distributions of LOPs
intersecting at precisely the same point, the true fix location.  But this
contradicts the Bayesian logic because, in fact, we don’t know the true fix
location.  This might be Georges logic.)

Using our logic, the three distributions are uncorrelated (in fact they are
in disagreement).  And the probability of point in the x-y plane being a
fix from all three normal distributions is the product of the three
distributions centered on the three LOPs.  This gives a plot of the
probability per unit area P(x,y) that the fix is at a given x-y point, as
shown in the second attachment.  We see from this figure that, because of
the shape of the normal distribution, there is one single point of MP
(maximum probability).  The probability that the fix is inside any given
area is the integral of P(x,y) over that area.  (I’ve numerically
calculated some of these probabilities, such as the 84% and 94% that I’ve
posted earlier.)  We see that areas of different shapes, different
locations, and different sizes can have very different probabilities that
the fix is inside them.  This explains George’s question above, how
different probabilities arise.  George seems to be counting types of
regions, rather then calculating the different probabilities that the fix
is inside them.

Since P(x,y) is a probability density – a probability per unit area -- we
don’t talk about the probability that the fix is at a specific point.  That
would be zero since a point encloses zero area.  Rather we do calculate and
plot the value of P(x,y) -- the density.  Those are the values on the
contours I’ve posted. For example a value of P = 0.013 means that the
probability that the fix is inside a unit area surrounding P(x,y) is 1.3%.
(The 100x100 plot, representing 10x10 nm had 100x100 data points.  So the
unit area is 0.1x0.1 nm.)  The values of P(x,y) are small because the total
area is so large, 10,000 data points.

It is true that as the cocked hat becomes smaller the MP inside it
increases, but it never can exceed one,  So the probability that the fix is
inside the hat decreases to zero as the hat’s area decreases to zero.  Thus
the navigator’s “desired” result of three LOPs intersecting at one point
means, that in all probability, the true fix is not there.  But of course,
the probability that the fix is somewhere in the x-y plane is always 100%.
It follows that the probability that the fix is outside the hat is 100%
minus the probability the fix is inside.  Since the probability that the
fix is inside goes to zero with the hat size going to zero, the probability
that the fix is outside goes to 100%.  There can be no constant 75% of the
fix being outside.

Finally from considering the form of P(x,y), as I’ve point out, the
location of the MP minimizes the sum of the squared distances to the three
sides of the cocked hat.  Now Herbert Prince (and Frank) tells us that this
point is called the symmedian of the cocked hat, as explained by Villarceau
in 1877.   Have we navigators been lost all this time??

JK
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