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    Re: Cocked hats, again.
    From: Gary LaPook
    Date: 2007 Mar 14, 01:57 -0700

    Gary LaPook wrote:
    
    Well at first blush is would seem then that from 3 LOPs with equal
    chances of being on one side or the other of each LOP that you would
    have 8 combinations (2^3) so one specific case would occur only 1 out
    of 8 times for a ratio of 7 to 1 not three to 1. That is my first
    thought and I haven't made any drawings yet.
    
    But take the case of just two LOPs crossing at a 90� angle. You would
    have a 1 in 4 chance of being in any one of the specified quadrants so
    the 3 to 1 ratio makes sense in the two LOP case. I will have to give
    the three LOP situation some more thought.
    
    One other problem I have with that idea is that it doesn't take into
    account the size of the triangle. With a very small triangle it
    approaches certainty that the observer is outside the triangle. At the
    other extreme, with a very large triangle it becomes less and less
    likely that the observer is outside the triangle. When the
    displacement of the LOPs equals 3.3 sigma the only place the observer
    can be is in the exact center of the triangle since this is the only
    location that satisfies the requirement that the observer be within
    3.3 sigmas of each LOP. ( I know that 3.3 sigma only equals 99.9% of
    the cases but this leaves only 1 in one thousand observation with a
    larger error. You could use 4 sigma,. 1 in 10,000 or 5 sigma, 1 in a
    million if you like.)  But is is a virtual certainty that with a
    maximum sized triangle the only place the observer could be is in the
    center of the triangle.
    
    
    On Mar 13, 6:10 pm, Fred Hebard  wrote:
    > Gary,
    >
    > I don't know why George got bogged down in that statistical
    > argument.  He has previously, several times, presented a much more
    > elegant one based on probability.  For each position line, with
    > random error, you have an equal chance of being on one side or
    > another of the line.  So the probability of being on one side is
    > 1/2.  Now count up all the different combinations of one side or
    > another for three position lines, and you end up with the 3 out of 4
    > proportion.
    >
    > Fred
    >
    > On Mar 13, 2007, at 8:03 PM, Gary LaPook wrote:
    >
    >
    >
    > > Gary LaPook wrote:
    >
    > > Well good, we seem to agree on most things. And I agree with you that
    > > many times the position of the observer will be outside of the
    > > triangle,
    > > the only reservation I have is with the strict 3 out of four
    > > proportion,
    > > although I haven't done any rigorous testing on this point. I had
    > > reviewed the series of posts in December (number 1908) and saw the
    > > diagram with eight triangles, six triangles showing the position of
    > > the
    > > observer outside and only two showing it inside. I have a question
    > > about  what these actually show. I also noted that the two
    > > triangles in
    > > which the position of the observer is inside are much larger than the
    > > six triangles showing the position is outside. In fact these two
    > > triangles appear to have a total area of 18 times larger than each of
    > > the six other triangles for a ratio of 18 to 6 or 3 to one but I don't
    > > know if the relative areas signify anything.
    >
    > > But, I am curious and will investigate further.
    >
    > > George Huxtable wrote:
    >
    > >> I hardly ever find anything to disagree with in Gary LaPook's
    > >> contributions. They are usually full of sense.
    >
    > >> That applies, too to most of his 6th March posting, Navlist 2236,
    > >> labelled "resolution of systematic error". Though actually, what he
    > >> discusses there is the situation of entirely random error, where all
    > >> systematic error has been corrected out.
    >
    > >> But Gary concludes with this-
    >
    > >> "Again, no one
    > >> is suggesting that the position of the observer is at the center of
    > >> the triangle but this represents the center of possible positions of
    > >> the observer.  In fact, the position of the observer will be outside
    > >> of the triangle often but I don't agree with the three out of four
    > >> allegation. Counter intuitively, the smaller the triangle the more
    > >> likely that the position of the observer is outside the triangle! If
    > >> you think about it, this should be obvious. Using   reducio ad
    > >> absurdum, think about a triangle only one inch in size, it would be
    > >> impossible for the observer to be within the triangle. At the other
    > >> extreme, a very large triangle with all of the displacements of the
    > >> LOPs from the center of the triangle equal to 3.3 NM (3.3 sigma's,
    > >> linear sigma's are slightly different than circular sigma's, see
    > >> Bowditch), the only place that the position of the observer could be
    > >> is at the fix in the center of the triangle!"
    >
    > >> Here, Gary is wrong. The statement, that "with entirely random
    > >> errors,
    > >> three times out of four the triangle will not embrace the position of
    > >> the observer", is precisely correct, and based on irrefutable
    > >> statistical arguments. If Gary doesn't accept this, unlikely
    > >> though it
    > >> may seem to him at first sight, he should scan through previous
    > >> discussion of cocked hats on this list, as Geoffrey Kolbe suggested.
    > >> If he remains unconvinced, we can go through the arguments once
    > >> again,
    > >> until one way or another, either he accepts it, or else he convinces
    > >> us otherwise.
    >
    > >> The best way to consider the matter is not to take a particular
    > >> triangle, and then consider "where can the true position be?", but to
    > >> take a true, known position of an observer, and three celestial
    > >> bodies. Now plot in his vicinity a set of three position lines that
    > >> have been displaced by Gaussian amounts, toward or away from the GPs
    > >> of the bodies in question. (Indeed, the argument doesn't rely on a
    > >> Gaussian distribution, just equal numbers toward and away). And then,
    > >> with the same observer position, and the same three bodies, plot
    > >> another triangle, with different displacements, and another, until
    > >> you
    > >> are tired of it. The resulting triangles will be different each time;
    > >> some will be large, some will be tiny, some long and thin, others
    > >> nearly equilateral. And in the end, if you check enough triangles,
    > >> you
    > >> find that 25% of those triangles will embrace the true position that
    > >> you started with.
    >
    > >> It doesn't matter how skilled or unskilled the observer is; it
    > >> remains
    > >> true. For a skilled navigator, of course, the triangles will
    > >> indeed be
    > >> smaller on average, which is where his skill shows itself, but still,
    > >> only 25% of those smaller triangles will include the true position.
    >
    > >> George.
    >
    > >> contact George Huxtable at geo...---.u-net.com
    > >> or at +44 1865 820222 (from UK, 01865 820222)
    > >> or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    
    
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