# NavList:

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

**Re: Still on LOP's**

**From:**George Huxtable

**Date:**2002 Apr 21, 21:44 +0100

I would go along with Chuck Taylor's proposed computer simulation. To my mind, it doesn't matter much to the argument whether or not the assumed distribution of bearings is Gaussian. But the result may be more convincing if we take the distribution to be Gaussian, or very nearly so. I wonder if this will help. A surprisingly simple way of simulating a Gaussian distribution can be found by summing 12 random numbers, each between 0 and 1. The average value of each random number will be 0.5, so the sum of 12 such numbers will be a value somewhere near 6. Then each simulated member of the Gaussian-distributed set is obtained from- bearing error = (sum - 6)*sigma, where "sum" has been calculated as above. "sigma" would be the standard deviation of the simulated bearing, a measure of the scatter, for which Chuck has proposed a value of 1 degree. The bearing error will be above zero as often as it is below zero, and its average value will be zero, as one would expect. This is an easy way of replicating, on a computer, Chuck's proposal- >For each bearing line, generate a random number from a Normal distribution >with mean 0 and standard deviation 1. (This is the distribution associated with >the classic bell-shaped curve; about half the time this number will be negative >and about half the time positive.) Treat these numbers as the bearing errors in >degrees. =============================== I have now convinced myself that, with the problem defined in the way Chuck has done, (that is, with the true position known and defined from the start, together with the positions of the landmarks), each set of observations of a particular landmark is indeed completely independent of the observations of any other landmark. I no longer have any reservations about that matter, so I will stick my neck out and predict that if someone implements the simulation according to Chuck's proposal, then he will indeed find the probability of the cocked hat embracing the true position will be 0.25. And if not, I will eat my (cocked) hat. George Huxtable. ------------------------------ george---.u-net.com George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. Tel. 01865 820222 or (int.) +44 1865 820222. ------------------------------