# NavList:

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

**Re: Cocked hats, again.**

**From:**Bill Noyce

**Date:**2007 Mar 15, 21:25 -0400

> The circled combination, A1T2A3, does not seem to be > possible in this configuration. And there is only one position > inside the triangle, T1A2T3. I'm not sure what's going on and where > A1T2A3 is hiding. I believe George discussed this previously. > > Fred Hebard If the sight had yielded A1T2A3, then the triangle would look different, with the horizontal line moved a few miles to the north. In that case, A1T2A3 would be inside the triangle, most of the others would stay about where they are now, and T1A2T3 would become impossible. One of the hard things about thinking about this stuff is choosing an appropriate starting point for the attack. Once you lay out a particular triangle, you've already made some assumptions about the observations. I suspect some of the apparent pardoxes with the probabilities are related to the concept of "conditional probabilities", which have some counter- intuitive properties. I think George has it right when he suggests starting with a position, and making (or rolling dice to simulate) observations with particular random error distributions, and anlyzing the collection of triangles that result. The navigator at sea is faced with a different problem, looking at one particular triangle, but the statistics can only be understood from George's approach. I have a few comments on things that have been said recently; forgive me if I don't remember by whom. Paraphrasing, "Even if we know the fix is likely to be outside the triangle, where would we put it." This is confusing the FIX (a navigational construct) with the ship's ACTUAL POSITION (which is to some degree unknowable). Of course we plot the FIX "in the middle of the triangle" but it's important to remember that the ACTUAL POSITION, while likely somewhere in the vicinity of the fix, is surely not in that exact spot. It's in this context that it's important to understand how likely it is that the ACTUAL POSITION is inside the triangle, as the FIX always is. Gary (I think) says that if your triangle is so large that the sides are 3*sigma away from the center, then surely your position is inside the triangle. The problem I have with this is knowing what sigma is. The fact that my triangle is very large might suggest that sigma for this observation is much larger than I thought. Or perhaps it's more likely that I've made some kind of blunder, or a systematic error. If I assume the triangle is due entirely to systematic error, with equal but unknown effect on each sight, the effects can be removed by a graphical procedure (which I think is how the current incarnation of this discussion got started). If the sights were not all in one half-plane, removing the systematic error yields a position inside the triangle; if the sights WERE all in one half-plane, removing the systematic error yields a position that is OUTSIDE the triangle. It's easy to demonstrate this ... Draw a position and three perfect LOP's, and tag each with the azimuth of the celestial body. Move each an equal distance toward its body; the resulting triangle will enclose the position only if the azimuths were well-distributed. Wishing I had more time to participate... -- Bill N. --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To unsubscribe, send email to NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---