NavList:

   

Reply

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


--~--~---------~--~----~------------~-------~--~----~
To post to this group, send email to NavList@fer3.com
To , send email to NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---

   

Reply