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John Karl writes "I think where George’s argument goes wrong is not 
thinking about integrating a variable probability/area over some specified 
area (like inside the hat)."

That is correct. I am instead applying reasoning that depends on pure 
logic. If Karl wishes to knock it down, he will have to do better than 
integrating a variable probability. He will have to show a flaw in the 
logic.

The argument applies to any combination of azimuths of the three bodies, 
but it's easy to think of in a particular case when they are separated by 
120º. It does not depend on the nature of the distribution, so it applies 
to the gaussian as well as to other shapes, All that it requires is that 
there is no systematic bias, but only random error, so that the deviation 
of any position line from the true position of the vessel is just as likely 
to be towards the body as it is away. And to keep things simple, we 
consider that the probability of exactly zero deviation is sufficiently 
small as to be negligible.

Then, if we put the three bodies in order, it's clear that there are 8 
possible combinations, where T is toward, and A is away-
TTT
TTA
TAT
TAA
AAA
AAT
ATA
ATT

There is no reason for any one of these combinations to be favoured above 
the others, so the probability of each is 12.5%
But of those combinations, only two, AAA and TTT, create a triangle that 
embraces the body. In the other six cases, 75% of the total, the body is 
outside the triangle. QED.

It makes no difference whether the operator is skilled or unskilled, or if 
the weather is rough or smooth. Those situations will produce cocked-hats 
of various sizes, but the probability of being inside or outside that 
triangle remains exactly the same.

Years ago, I checked it out, to remove all doubt in my mind, by setting up 
a computer simulation. It was not of exactly this same problem, but of the 
analogous problem of compass bearings to three landmarks, perturbed by the 
sea-state, in which the likelihood of the compass bearing being in error to 
the lefy, and to the right, was exactly the same.

I collected sufficient statistics to show that the probability of being 
outside the triangle was in the range 74% to 76%, which was convincing 
enough for me.

If John Karl's conclusion differs from 75%, I will suspect a flaw in his 
mathematics (or in his definitions) unless be can demolish that logical 
argument.

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: Thursday, December 09, 2010 5:53 PM
Subject: [NavList] Re: That darned old cocked hat


Yes, George once had me convinced that the probability of being outside the 
hat is 75% for all cases.  But after giving it more thought, I don’t 
believe it.  My logic goes like this:

We’re considering a normal distribution of random errors (no other type of 
errors) that is known a priori from many tests.  The probability of the 
true fix being at any given point is the product of these distributions 
(see the attached PDF file).  This Probability P(x,y) is the probability 
per unit area.  And it can never exceed 1.0.  The probability of the 
correct fix being inside any given area is the integral of P(x,y) over that 
area.  As the area of the cocked hat gets smaller, and tends to zero, 
P(x,y) does not get arbitrarily large.  In fact it can never exceed 1.0. 
Therefore as the hat’s area get smaller, the probability that the true fix 
is inside the hat goes to zero.  If you get a tiny cocked hat, in all 
likelihood you’re not there.

The probabilities quoted in the second attachment below are approximate 
because, for the integration, I approximated the hat triangle with the 
closest contour (inside and outside the triangle to get a good estimate). 
Whatever the accuracy of these numerical integrations are, they don’t 
change the above argument.

I think where George’s argument goes wrong is not thinking about 
integrating a variable probability/area over some specified area (like 
inside the hat).

The attached file also shows that all the best estimates that I have seen 
(Fermat's point, the center of gravity, the bisecting angle point, and the 
Steiner point) can not be the most probable location of the true fix.  (An 
observation that's of no practical use.)

JK
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