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    Re: Latitude by Lunar Distance
    From: James R. Van Zandt
    Date: 2006 Oct 17, 22:15 -0400

    
    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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