NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Where are you most likely to be in the triangle?
From: Bill Lionheart
Date: 2017 Jan 25, 12:56 +0000
From: Bill Lionheart
Date: 2017 Jan 25, 12:56 +0000
I was wrong about the Fermat point. That minimizes the sum of distances from the vertices. Minimizing the distances from the sides is easy. It turns out that is just the vertex opposite the longest side. So it your errors were exponentially distributed it is easy to find the ML estimate! Here is a neat proof on Strack Exchange http://math.stackexchange.com/questions/463698/point-which-minimises-the-sum-of-the-distances-from-the-sides-in-a-triangle Bill On 22 January 2017 at 10:23, Bill Lionheartwrote: > Thanks for everyone's reply. It is delightful to wake up on Sunday morning > and find it is "well known to those who know" and they are on NavList of > course. I was also pleased with myself for working it out (and then > eventually finding what it was called). I agree completely with Frank that > we need to train to understand where the symmedian point is for bad (ie high > aspect ratio) traingles. But also we should understand the limitations of > the "Maximum Likelihood" estimate. There is a probability distributions on > the position and as a training looking at plots of contours of probability > is good to get a feeling for it - the whole "hill" not just the summit. > > Just as another mathematical nuance the errors are approximately normal only > after averaging a lot. For example truncation error (reading only to so many > digits and neglecting the next one) is approximately uniformly distributed > If the errors were exponentally distributed (ie like exp(-|x|) rather than > exp(-x^2) you would want to minimize the sum of the distances to the line. > (Fermat point?) > > Now what was keeping me awake, and it sounds like someone here will know, is > the same problem but for a fix made using three bearings, the cocked hat of > the traditional coastal navigator). Assuming the error in the bearings is > normally distributed we want to find a point that minimizes the sum of > squares of the angular deviations from the position lines. The angles here > are taken at the objects we are taking bearings on of course rather than the > vertices of the cocked hat. > > Bill Lionheart > > -- Professor of Applied Mathematics http://www.maths.manchester.ac.uk/bl