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There was an interesting posting from Geoffrey Kolbe-

| On the subject of "COCKED HATS".
|
| In the past, I have always agreed with George that where there is no
| systematic error in the measured altitudes, the probability of the actual
| position being outside the cocked hat is three times higher than being
| inside. But my position has shifted slightly on this one. I wonder if I
can
| induce George to come along with me. (In the following analysis, assume
| that there is no systematic error of any kind).
|
| First, consider a measured altitude on Polaris. On reducing the sight, I
| will end up with an LOP which will run East-West across the chart. What
can
| I say about my actual position relative to this line? I can say that the
| probability of my position being on or near one point of the line is equal
| to that for any other point on the line. (In other words, I have no
| information regarding my longitude). I can say that the probability of my
| actual position being North of the line is 0.5, exactly the same as the
| probability of being South of the line.
|
| Now, suppose I take a sight on the star Canopus when it is on the meridian
| to the South. I reduce the sight and draw another LOP on the chart. This
| LOP also runs East-West across the chart, but I find that it is about 2
| nautical miles (say) to the South of the Polaris LOP.
|
| What can I say about my position relative to this new Canopus LOP? Since
it
| runs parallel to the Polaris line, I still have no new information about
my
| longitude. The chance of me being on or near one point of this line is
just
| the same as for any other point. What can I say about my position North or
| South of the line? Is the probability of my actual position being to the
| North of the Canopus LOP still 0.5, exactly the same as it being South of
| the line? I would now say no.
|
| Because the Polaris LOP sits to the North of the Canopus LOP, I have an
| independent piece of data which indicates that my position is actually
more
| likely to be North of the Canopus LOP than South of it. Statistically
| speaking, this is a so-called "Bayesian" approach to statistical analysis.
| I have some a priori information regarding my actual position, which can
| inform my estimation of my position with respect to the Canopus LOP.
|
| Actually, there is no reason why I should have taken the Polaris sighting
| first. I could just have easily have taken the Canopus sighting first.
That
| being the case, I can say by symmetry that the Canopus LOP can similarly
| inform, regarding my position relative to the Polaris LOP. Now, I can say
| that the probability of my position being South of the Polaris LOP is
| higher than the probability of being North of it.
|
| I will stop there as I am sure George will see where I am headed with this
| argument when applied to the "cocked hat" problem. And since the essence
of
| the argument is already laid out, I need go no further to persuade George
| if he is willing to come with me this far.

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

I'm not certain how far Geoffrey intends to take me, with that invitation...

It's an interesting philosophical question, but I reckon the answer is a
simple one.

We have two separate measurements; call them lat1 and lat2 (about 2' further
South). If we had just the first, all we could say is that in the absence of
any systematic error, we are just as likely to be North of lat1 as South of
it. And if we had just the second, all we could say is that we are just as
likely to be North of lat2 as South of it. And there's no contradiction
here, but it tells us that any random scatter (which we haven't been
informed about) must be at least a significant fraction of that 2-minute
difference, or else they couldn't both be true.

But we have two independent measurements, which differ somewhat. As we are
invited to assume that there are no systematic errors (such as scale
calibration of the bubble sextant), then it doesn't matter, at all, that the
second measurement was taken on a different day, of a different star, at a
different altitude. It could just as well have been a second measurement of
Polaris, taken immediately after the first one.

So now we have two measurements of what's effectively the same thing, that
differ by 2', so the next thing to do is to combine them by averaging. And
the end result is that we can now say that we are just as likely to be North
of the new averaged latitude as South of it, and by that averaging process,
we have reduced the initial scatter by root-2, from what it was for each
separate observation.

I suspect that I may have bypassed the question Geoffrey intended to ask,
instead of giving him a straight answer, but as he hasn't yet got round to
actually formulating that question, it's the best I can do at this stage.

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.


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