# NavList:

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

**Re: Timing Noon**

**From:**Chuck Taylor

**Date:**2002 Apr 17, 20:51 -0700

Thanks to George Huxtable for his thought-provoking comments. My first reaction was to reject what he said out of hand, but I found it worthwhile to work through the details before coming to a conclusion. I see where he got the figure of 0.25 for the probability that your true position lies within the cocked hat. What I'm not sure about is whether that is the right question to ask. I would propose that the real question ought to be, "Given that three bearings converge in a triangle, which point (inside or outside the triangle) is the best point to take as your most probable position?" I am assuming that any systematic bias has been eliminated, that the remaining errors are random, and that one has no information on the relative accuracy of the three bearing lines. Consider George's example: George Huxtable wrote: > Well, let's say we are determining our position by bearings on three > distant landmarks, 1, 2, and 3. > > There is an equal chance that, due to errors in taking the bearing from > landmark 1, that bearing will lie to the left of the true position as to > the right of it. If we take the possibility that the bearing can be exactly > on the line of the true position to be zero, the probability of it being on > the left (L) is 0.5, the same as it being on the right (R). We can say the > same about landmarks 2 and 3. It's a bit mind-boggling to consider that all three bearings can be uncertain all at the same time, so let's simplify the problem. First, consider that the bearings on landmarks 1 and 3 are absolutely accurate, with no error whatsoever. The true position is at the intersection of these two bearing lines. Now consider what the "cocked hat" would look like first if the bearing on landmark 2 were off a couple of degrees to the right, and second if the bearing on landmark 2 were off a couple of degrees to the left. This will be easier to visualize if you draw it out on a piece of paper. What you will see is two triangles which meet at a common vertex. The true position is at the common vertex. It is in the interior of neither triangle. If you knew a priori that two of your bearings were exact and the third was not, but you didn't know which was which, what would be the logical thing to do? You would know that the true position is at one of the vertices of your triangle, but you wouldn't know which one. I would submit that the logical thing to do is to go with conventional practice and choose the center of your triangle, that point which is equidistant from the 3 vertices. Second, consider the case that the bearings from landmarks 1 and 3 are uncertain, but the bearing from landmark 2 is exactly known (perhaps it lies along a range line). If you draw this out, you will see that there are many more possibilities. In some cases the true position lies at a vertex, but again you do not know which one. In other cases the true position lies exterior to the triangle, but is closer to one of the vertices than to the others. Again, you don't know which one. Once again I would submit that the rational position to plot is the center of the cocked hat. I'll leave it to someone else to tackle the case where all three bearing lines are uncertain. Still, I would maintain that, in the absence of knowledge as to the relative accuracy of the various bearing lines, the rational action is to plot your position in the center of the cocked hat. Cheers, Chuck Taylor Everett, WA, USA > > There are eight possible combinations, if we list the three bearings, taken > in the order 1, 2, 3, as follows- > > LLL, LLR, LRL, LRR, RLL, RLR, RRL, RRR. > > For each such combination, because it combines 3 terms each with a > probability of 0.5, its probability is (0.5) cubed, or 0.125. There are 8 > such combinations, each with a probability if 0.125, so that looks right, > doesn't it? > > However, of those 8 combinations, there are only two which place the true > position inside the cocked hat. These are LLL and RRR. This can be seen if > a drawing is made showing all the 8 options. The other combinations put the > true position outside a side or outside a corner. So the probability of the > true position being inside the cocked hat is exactly 0.125 x 2, or 0.25, > which is what we set out to show. > > What seems at first so surprising is that this result is quite independent > of the skill of the navigator. The reason for this is that the better > navigator will produce, on average, a smaller cocked hat, But the > probability of it embracing the true position will remain at 1 in 4. > > George Huxtable. > > ------------------------------ > > george@huxtable.u-net.com > George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. > Tel. 01865 820222 or (int.) +44 1865 820222. > ------------------------------ > > >