NavList:

Thanks for the problem, Frank. I spent a "fun" evening revisiting some basic geometry to find the solution, and then enjoyed reading the other interesting responses. Here's my story...

Not having as much navigational background I approached this as a simple geometry problem, namely what is the most probable intersection of three lines that have equal measurement errors associated with them in the special case where two of the lines are perpendicular. This is a simple case of a common problem in particle physics (my field) where you want to find the most probable "event vertex" when you have measured the tracks made by particles radiating from the event.

My intuitive response to the problem was that the desired point was the "center of gravity" of the triangle formed by the three lines. First appealing to simplicity, the center of gravity is the easiest single point to define which characterizes the triangle. Second it is in fact the average of three two line intersections, the vertices of the triangle.

Now all I had to do was test my intuition by solving for the point which minimizes the "sum of the squared residuals" or the chi square. Here the residuals are the perpendicular distances of each of the lines from the point. This turned out as Frank suggested to be a fairly messy algebraic undertaking (see below), and I needed to go through it a couple of times to be sure of my answer. The answer when properly simplified was quite a surprise - contrary to my hallowed intuition the point in question was halfway along a line drawn perpendicularly from the skewed side (hypotenuse) to the opposite vertex! And it didn't depend on the angle of this line!

To salvage some of my intuitive pride I thought of a special case that made it clear that this was the right answer and not the center of gravity. For a very narrow right triangle where the hypotenuse and one other side are almost parallel, the center of gravity is one third of the way down the triangle (away from the right angle). But this is clearly not a very good choice for the most probable intersection, since it is quite a distance from the short side making chi square unnecessarily large. The correct answer is a point that is very near the short side as one would expect for a minimum chi square. So here is a case where simple averaging of results (and use of intuition) does not give the best answer.

For anyone interested in the derivation of the most probable intersection here is a summary of my calculation. I defined a coordinate system with its origin at the right angle of the triangle with two sides along x and y. The y side has length a, the x side has length b, the hypotenuse has length c, and the internal angle at the end of x side is theta. The sides a, b, and c can all be trivially expressed in terms of theta and d, the perpendicular distance (per Frank) from the hypotenuse to the opposite vertex. I wrote out an expression for chi square in terms of a, b, c, and the coordinates of the intersection point, x' and y'. Chi square is then minimized for variations of x' and y'. The initially surprising answer I got was x' = (a^2 b) / (2 c^2), y' = (a b^2) / (2 c^2). However, this answer can be rewritten as x' = d sintheta /2, y' = d costheta/2. which is just the point halfway up the line d! x' and y' do depend on theta as they must, but when the intersection is described as being along d, then the theta dependence disappears.

So I'll have to be more careful about relying on my intuition...
George B

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