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Gary LaPook wrote:

Well George, I always was partial to "T's" and "A's."

I follow you argument and can close my eyes and visualize what you are
saying.

How's this for another description of your point.

Draw three LOPs surrounding the position of the observer. If there had
been no random errors, the three LOPs would have plotted at a point at
the location of the observer. Now indicate on the LOPs with an arrow
the azimuth of the body, lets say all pointing outward so this
represents the "TTT" case. Now if you flip one of the LOPs and make it
an "A" it will now plot closer to the interesection of the two
remaining "T" LOPs forming a smaller triangle over in that corner of
the original "TTT" triangle and the position of the observer wil not
be within the new small triangle. Then you can conceptionalize this to
the other possible combinations and come up with the 1 in 4 ratio.

Do I understand you point?

On Mar 14, 12:50 pm, "George Huxtable" 
wrote:
> 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.
>
> ==================
>
> Reply from George-
>
> Not so, Gary. Consider the case of an observer, at a known position,
> taking altitudes of three stars, which are not contained within a
> 180-degree arc. (If they were within 180 degrees, the argument would
> differ in detail, but not in principle, and reaches the same
> conclusion). Take, as a simple example, the simplest case of three
> stars whose bearings differ by 120 degrees, though the argument
> applies to other such angles just as well.
>
> Let's denote an intercept being Towards a star by the letter T, and
> Away by A. No doubt we will agree that in the absence of systematic
> error, T and A are equally likely, with 50 % probability. (We presume
> that the likelihood of an error of exactly zero is negligible).
>
> Then for the three stars in order, there are 8 combinations as
> follows-
>
> AAA, AAT, ATA, ATT, TAA, TAT, TTA, TTT
>
> Each of those combinations is equally likely; a probability of 1/8.
>
> Only for the two combinations AAA and TTT will the resulting triangle
> embrace the observer's position. So the probability of that happening
> is exactly 25%.
>
> I should add that Gary's disbelief is characteristic of the reaction
> of many list members when they meet that argument for the first time,
> but most, if not all, seem to have come round to acceptance in the
> end, if somewhat reluctantly.
>
> I have tried it out as a computer simulation and it checks out.
>
> It applies just as well to the similar cocked-hat derived from 3
> compass bearings of distant landmarks.
>
> It has some interesting side-effects. Say that there was a systematic
> error; for example, say all altitudes were too great, because of an
> error in the index correction. Then, (in the case of all azimuths
> spanning more than 180) the number of TTT observations would be
> enhanced. If that error was large enough to overwhelm the natural
> scatter, then EVERY observation would be a TTT one, and then the
> triangle would embrace the true position in EVERY case, not just 1 in
> 4. Some observers may take that to show a satisfactory state of
> affairs, but not at all; it points to a major error!
>
> It will be interesting to discover whether Gary can be convinced. Or,
> alternatively, whether he can convince us that we're wrong.
>
> George.
>
> contact George Huxtable at geo...@huxtable.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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