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George,

You wrote:

>Martin said-
>
>>When I drew this, RRR and LLL lay outside my triangle.  I was able to
>>
>>>redraw it so that RRR lay inside, but then there is no area LLL.
>>>
>
>This puzzles me. Are we perhaps at cross-purposes about the words Left and
>Right and how they relate to the diagram of the cocked hat? If we used the
>words Clockwise and Anticlockwise instead, for the changes in bearing,
>would that remove some ambiguity? It's just that it seems to me that on any
>picture of a cocked hat, RRR or LLL must always represent the triangle at
>its centre (but never both at the same time).
>

I think the last is the important point. For any one cocked hat and for
any one convention of what we mean by left and right (i.e. which end are
we looking from when designating "left" and "right"?) there can only be
one combination of Ls and Rs that denotes the area inside the cocked
hat. There are only 7 areas in the diagram and, if each is equally
likely, then each has a probability of 0.143 -- to the nearest 3
significant figures. That fact that some other cocked hat or some other
naming convention changes the label inside the triangle doesn't alter
the probability.

>=============================
>
>Perhaps it might help this discussion if, to give another view, I quote JED
>Williams' explanation of this matter, in "From Sails to Satellites", which
>is somewhat different to mine, and perhaps more elegant.
>
>On page 273 he provides figure 14.6, entitled "The probability of being
>within the cocked hat is 1/4". This shows a cocked hat, showing the value
>"1/4" in the central triangle, "1/6" in the areas adjacent to the sides,
>and "1/12" in the areas outside each corner.
>

>Williams says-
>
>"Fig 14.6 is the cocked hat formed by three position lines. If no
>information is available on the distribution of error the probability of
>being in each of the seven zones defined by the cocked hat is as shown in
>the figure. The proof is so simple that an outline of it is given in note
>13; but the writer is unaware of any text likely to be read by mariners or
>airmen of the period which even suggests that the true position is probably
>outside the cocked hat...."
>
>Note 13 to that chapter, on page 282, reads as follows-
>
>"If there is no information on the distribution of probable error, a
>position line of any shape simply divides the Earth's surface into two
>domains, and the position is as likely to be in one as in the other. We may
>arbitrarily denominate them the + domain and the - domain, and we choose to
>do so by nominating the enclosed area of the cocked hat the 3+ zone.
>

That returns us to the terminology I tried when I waded into the
discussion, labelling one side of each line the "cocked-hat side" and
the opposite one the "other side", but "+" and "-" look a lot more
mathematical!

>Every
>zone bounded by a side of the triangle is a 2+,1- zone, and each zone at
>the vertex of the triangle is a 1+,2- zone. The 3- zone is cospicuous by
>its absence. If the true position were in the the minus zone of each and
>every position line as they have severally been labelled, the cocked hat
>would necessarily be shaped in the form of its mirror image, and the
>position would be within it.
>

Nonsense. He has chosen to define the side towards the cocked hat as "+"
and so there simply is no "3-" zone. (Unless maybe it exists in a space
where the axes are measured in imaginary numbers or something of the
sort. No "3-" zone can exist in the dimensions we inhabit once "3+" is
defined as the cocked-hat area.)

> The fact that the cocked hat has taken the
>form it has tells us that the position is in the + domain of at least one
>position line. Given that we are in the + domain of one position line and
>may equally be in either the + domain or the - domain of two other position
>lines,out of four there is one chance that the domain is in the 3+ zone,
>two chances that it is in a 2+, 1- zone, and one chance that it is in a
>1+,2- zone. The probability of being in each zone is therefore as shown in
>fig 14.6."
>
>(end of quotation from Williams)
>

That is an interesting proof that is nowhere near as simple as he
claimed. As I understand it, he is admitting that the "3-" combination
is impossible and hence at least one of the three scores must be a "+".
Adding two more "+" or two "-" would involve a probability of 0.5 x 0.5
= 0.25, whereas adding either "+ -" or "- +" would end up with "2+ 1-"
and hence such combinations have a probability of (0.5 x 0.5) + (0.5 x
0.5) = 0.25 + 0.25 = 0.5. Dividing the "2+ 1-" and the "1+ 2-" amidst
the three sides and the three corners gives the probabilities that you
extracted from his figure.

It, of course, disagrees with the proof of the 0.25 probability which
you offered -- though I'm not about to say which (if either) of you is
right!

One thing that troubles me with his proof (aside from the assumption
that the errors around each LOP are independent of each other) is that
it not only fixes the probability of the true position being within the
cocked hat, regardless of the shape and size of that triangle, but it
sets the probability of the position being off any side (or any corner)
as the same as off any other side (or corner). While I realize that
common sense is a poor guide in these matters, that seems
counter-intuitive and leads me to be sceptical about the proof.

>The above is offered on an "if it helps" basis, and I am not intending to
>defend or explain its logic.
>
>===========================
>
>My thanks to all who have given a stir to this interesting pot, which is
>not yet fully cooked. Sorry if any list-members find it all boring.
>
>Meanwhile, I will continue to chew over the food for thought that Martin
>and Trevor have provided, and would welcome any further contributions to
>this discussion.
>
>Perhaps an answer might arise from a "Monte Carlo" type simulation devised
>by one of our computer experts, though its parameters would need to be very
>carefully specified beforehand, to ensure it was answering the right
>question.
>

I wondered about attempting not a Monte Carlo but a numerical solution,
much as Bill Murdock has now suggested. However, it would need to invoke
not simply a standard deviation but the probability density function of
the normal distribution, oriented perpendicularly to each LOP. I dare
say that that could be done in Excel but it would be a bitch of a
problem -- certainly not something I could run off in a few minutes.
Besides, numerical solutions only give case-specific answers. So I'll
hold out for some statistician, interested in navigation problems and
gifted with the ability to explain complex issues to benighted mortals,
who is willing to model a correct solution. I dare say I'll be waiting a
long time!


One final point for everyone who is monitoring this exchange (if anyone
is): Even if we accept that the probability of the true position being
inside the cocked hat is less than 50/50, the single most likely point
for the true position will remain near the centre of that triangle (but
drawn towards the "flat end" if the triangle is long and thin). If you
are well out in open water, the centre of the cocked hat remains the
best place to mark your fix. But if there is a danger nearby, do not
assume that you are save just because the danger is outside your cocked
hat. Indeed, even George has got us off on a tangent and the chance of
being inside the triangle is 0.8 or 0.9, it most certainly isn't 1.0 and
so nobody should be trusting to the idea that they must certainly be
somewhere within the cocked hat.

Or does anyone want to dispute those basic principles too?


Trevor Kenchington




Trevor Kenchington

   

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