# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Latitude by Lunar Distance**

**From:**James R. Van Zandt

**Date:**2006 Oct 17, 22:15 -0400

George Huxtablewrote: > 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 --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To unsubscribe, send email to NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---