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On 14 October, I posted a message questioning Frank Reed's claim to an
accuracy of 6 miles for his proposed method to determine a position
without using a horizon.

I pointed out that if celestial positions were taken from the Nautical
Almanac, that would involve altogether 14 lookups, of quantiies that
were tabulated only to the nearest 0.1', a possible random error of
+/- 0.05 arc minutes in each one (a typo, stating that as +/- 0.5' ,
was later corrected).

and I added-

| Depending on the details of the geometry, it's possible for any one
of
| those 14 terms to affect the result by up to .05', one way or
another,
| simply because of the way they are tabulated in the Nautical
Almanac.
| Of course, most are KNOWN to a much higher accuracy, but not from
that
| Almanac.
|
| It is, of course, true that all 14 are highly unlikely to all add up
| in the same direction, to their full extent. They are also highly
| unlikely to cancel out to zero. Without making a full statistical
| analysis, it seems reasonable, to me, for the standard deviation of
| the resulting scatter to be taken as root-14 x the spread of each
| component, or 3.7 x .05', or 0.19'. If anyone can suggest a fairer
way
| to combine those errors, I hope they will.

Nobody has volunteered to do so, yet, but that approach worried me as
being rather unscientific, and I have had a closer look at the problem
of combining such errors, to see what scatter results in the answer.

No doubt, it can be solved analytically from probability theory, but I
tackled it in a brute-force way by a simulation, using what, in my
earlier trade, we would describe as a "Monte Carlo" method. Simply
adding together 14 numbers, each varying randomly in the range
between -.05' and +.05', to see how the final answer scatters about
its book-value, by doing the same thing again and again..

And my conclusion is that in two-thirds of the cases, the end result
is within 0.1 arc-minutes of what it would have been if those
quantities had been stated precisely, rather than approximated to the
nearest 0.01'. Only in one case out of three, will those 14
approximation-errors combine to displace the result by more than an
arc- minute.

So I was a bit unfair on Frank's proposal in suggesting that a spread
of 0.19' was an appropriate measure of the resulting scatter. All but
about 5% of observations will be affected by less than that amount.

The figure I gave of +/- 11 miles, for error in the resulting
position, with a high Moon, due to those Almanac approximations,
corresponds, then,  to a 95% confidence level, or thereabouts, and in
most cases the error will be considerably less.

As Geoffrey Kolbe has correctly pointed out, the back pages of the
Almanac warn of possible error in Moon GHA predictions of up to 0.3',
though I suspect part of that possible worst-case error may be
subsumed in the Almanac approximations, considered above. Somehow,
those errors have to be combined.

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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