# NavList:

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

**Re: That darned old cocked hat**

**From:**Frank Reed

**Date:**2010 Dec 11, 21:30 -0800

George Brandenburg, you wrote:

"Sorry, but now I think you've got it wrong because you're playing John Karl's game of actually measuring three specific LOPs. In such a case you have to calculate the probability contours, and you will not get exactly 25% probability contained within the triangle. If the lines almost intersect the probability will approach zero even though this appears to be a good outcome. If the lines miss by a wide margin the probability of including the location will be large.

Your 25% probability of inclusion applies to an average over all possible LOP triplets. In this game you measure N LOP triplets where N is a very large number. The larger the number N of measurements taken, the closer you will get to the ideal 25% figure.

This is what I meant about both of you having the right answer to different problems."

I think I've got a line on one other aspect of this. There are two ways to set the standard deviation of the observations when calculating error ellipses: (1) from known observation history or reasonable expectation, or (2) from the current set of observations themselves and no other information.

If the current set of observations consists of a dozen of more sights, then it's probably better to get the s.d. from the sights since they will reflect current conditions. But if you have four or five observations, it's probably better to set the s.d. to some reasonable value based on previous sights or maybe some sort of weighted average based on previous experience and the current set of sights. Now, what happens if I choose to set the standard deviation from the current observations when there are only THREE sights like this triangle case? Then, clearly, a small triangle implies a low standard deviation while a large triangle implies a correspondingly larger standard deviation. Since the error ellipses are scaled by the standard deviation, this means that the error ellipses scale with the size of the triangle. I don't think that it's appropriate to get the standard deviation from the current set of observations when they are so few in number, but if we do, it creates a sort of scale invariance: a small triangle implies a small error ellipse. And NOTE: many navigational software packages DO get the s.d. from the sights even when there are only three LOPs (they're following the rules in an old book from HMNAO a bit too slavishly). I haven't worked this out, but it seems to me that the integrated probability within any triangle will remain the same when this choice is made. If instead we treat the standard deviation of the observations as an INPUT to the problem, which I think is by far the better choice with a small number of LOPs, then the probability ellipses will have a FIXED size (different shapes and orientations, of course) with various different sizes of triangles. And in that case, the integrated probability over the triangle --the probability of being inside and not outside will vary from one triangle to another. But in the long run, the probability of being inside ANY triangle formed from three LOPs will average out to 25%.

-FER

PS: I am linking a simple simulator (Windows only, I can compile for Mac upon request) that generates endless three-body fixes and then checks to see whether the original position is inside the triangle or outside. It's a simple thing. You can speed it up or slow it down with the little arrow controls. The statistical probability will trend towards 25% but the value after tens of thousands of trials is slightly off from 25% (you might see 24.8%) primarily because the in/out test is done graphically: flood-filling around the LOPs in grey leaves the interior of the triangle white. The azimuths of the LOPS are chosen at random but required to be separated by a minimum of ten degrees. The intercepts are chosen from a Gaussian distribution with a standard deviation of 25 pixels. Neither of these choices affect the outcome. Other random distributions for the azimuths or intercepts produce the same results.

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