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George Huxtable  wrote:
>   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...

>   | 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...

>   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.

The standard deviation of a uniformly distributed number is
(interval)/sqrt(12), so the standard deviation of the sum of the 14
numbers should be .1/sqrt(12)*sqrt(14) = .108, which agrees with
George's Monte Carlo analysis, and significantly less than his
original estimate .1/2*sqrt(14) = .187.

Note also that for refraction at least, one can interpolate in the
table to improve the accuracy of the reduction.

               - Jim Van Zandt

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