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I sent the following message, in error, to Rodney Myrvaagnes directly,
rather than to the list. So here it is copied again to all. Rodney has
replied directly to me, and I hope he will now copy the relevant parts of
his reply to the list.

================================

Rodney Myrvaagnes said-

>George, I think you need to do better. The probability of the true
>position lying on a measured LOP is higher than it is on either side,
>not zero (Newton and Leibnitz took care of that when they invented
>calculus)
>
>If you want to simplify a normal distribution, a triangle would be a
>better approximation than two squares. And, to avoid cluttering the
>problem with intercepts, assume LOPs from bearings, rather than sextant
>sights.
>
>The width of the distribution, whether gaussian or triangle, must
>overlap those from the other sights, and, if the three sights are
>similar in difficulty, the width of the distribution should be the same
>for all of them.
>
>Even if you use your rectangular distributions, they overlap only at a
>point inside the cocked hat at the narrowest possible size. Since the
>cocked hat is itself your only measure of the error width, that is as
>far as you need to go. Ergo, the way we have been doing it is correct.

====================

George Responds-

It's good to see some thinking going on about this cocked hat question.

However, Rodney has got me quite wrong.

I have not suggested a square or triangular or any other distribution. I
think that suggestion came from Trevor J Kenchington on 14 April, not from
me. My argument does not depend at all on the shape of the distribution.
All it depends on is that there is some best-fit line line of position that
can be drawn through an observed landmark, and about which the true
position is equally likely to lie to its left as to its right.

Rodney added-

>And, to avoid cluttering the
>problem with intercepts, assume LOPs from bearings, rather than sextant
>sights.

To keep the problem simpler to understand, I WAS considering the case of
lines of position from terrestrial bearings, just as Rodney Myrvaagnes
suggests.  If he looks at my response to Trevor Kenchington, dated 17 April
(but under a different thread title, "Timing Noon") he can see that I
wrote-

"Well, let's say we are determining our position by bearings on three
distant landmarks, 1, 2, and 3.

There is an equal chance that, due to errors in taking the bearing from
landmark 1, that bearing will lie to the left of the true position as to
the right of it." And so on...

However, the situation with celestial intercepts is exactly analogous, so I
have seen no need to restrict the discussion to one situation rather than
another.

I said earlier that we could assume the probability of the true position
lying exactly on the observed bearing to be zero. Rodney questions this
assumption. It's simply a matter of geometry, in which a point is
infinitely small, and a line is infinitely thin, so the chance of one
falling on the other becomes zero. If Rodney disagrees, let him state what
he is going to assume for the size of the point defining his true position,
and the width of a line that defines a bearing, and what is the resulting
probability of the true position lying on that bearing.

His references to Newton and Liebnitz are quite bogus. In one case we are
considering the probability of a point lying within a defined area. In the
other, the probability of a point lying on a line. The first is infinitely
greater than the second.

George Huxtable.


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george@huxtable.u-net.com
George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
Tel. 01865 820222 or (int.) +44 1865 820222.
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