# NavList:

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

**Re: Still on LOP's**

**From:**Dov Kruger

**Date:**2002 May 1, 11:50 -0400

I consulted with a noted statistician friend, Roger Pinkham, because this is a complex problem and making guesses didn't appeal to me. He is thinking about it, but immediately made a few points that will shed some light. 1. You can't assume the errors are normally distributed. There will be numerous instances where the error will be asymmetrical, including for example some consistent instrument error that you don't catch, and observations of phenomena where your only possible error is one-sided (I don't think that second one is too relevant for navigation). But assuming that you are spot on except for random, normally distributed errors in each reading.... You further have to assume equal errors on all readings for the moment. 2. Making a statement like p(inside the triangle) = 0.25 assumes that each reading is independent. They are not. Going outside the triangle can put you further out on the tail on each of your readings, so the probability of being outside the triangle is much lower than that. So, while he is thinking about the problem, in the meantime, he mentions a related unsolved problem just to show how "unsimple" this is: Consider a two-variable bivariate gaussian, (shooting at a target). For n shots, what is the distribution of the size of the smallest circle that can enclose all the points? This is an unsolved problem. It isn't trivial, and he therefore doubts that the navigation problem is any easier to come up with an analytical answer. My own take on this problem: Estimating the error: You can't talk about the probability of being outside the triangle without knowing the variance of your measurements. I think the probability of being outside the triangle has nothing to do with the size of the triangle, but has everything to do with the consistency of your readings. The reason a small triangle is an indication that you've hit the jackpot, is that the three points will only coincide if a) they just happen to do so (a small probability), or b) they really are correct. So what you really want are a couple of more readings of the same bodies to give you some indication of how consistent your first readings were. I am short on time at the moment, but it seems to me the best way to calculate the probabilities of being in the triangle are: 1. select three points and the observer's location 2. compute the bearing to each point 3. add normally distributed errors to each bearing (as George says, this one is easy) 4. Determine if the position is within the resulting triangle repeat step 3-4 as long as you want. alternatively, just walk through the range of possible values for each observation, ie do the triple integration, but then you have to factor in the probability of each reading. This procedure, however, only gives the distribution for one set, and the problem is that different configurations appear from those diagrams to yield radically different answers. Still, by performing this computation on different configurations, you can at least see the range of probabilites. I suspect that the probability of being within the triangle is at least 0.5. That's just a guestimate, but I am looking forward to the first simulation to see if that's right. I will also guess that the probability of being within the triangle goes up at least quadratically if the triangle is increased in size by projecting lines parallel to the sides outside it. So if p(outside) = 0.5, with a similar triangle containing the original one that is bigger by 100%, I will guess that the probability of being outside the augmented triangle goes down to 0.125 Dov