# NavList:

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

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Re: The Darn Old Cocked Hat - the sequel 1
From: John Karl
Date: 2013 Mar 13, 14:36 -0700

Geoffrey said:
Actually, the Goudsmit result can be achieved by averaging over the entire grand canonical ensemble of possible cocked hats within the probability distribution you describe, as was done by somebody using Monte Carlo methods (I forget who) the last time we went through this. The two approaches are equivalent.

No, they’re not equivalent in any sense. The symmedian problem is a problem in estimation theory, getting the best answer from incomplete data. The Goudsmit problem is one of probability theory, finding the probability of a given event from the known probabilities of the factors effecting that event.

The symmedian problem:
1. does not know the ship’s true position.
2. centers the given distributions on the given three LOPs, not on the ship’s true position.
3. does not average sights.
4. gives the symmedian for every hat.

The Goudsmit problem:
1. starts with ship’s true position known (answer to the navigation problem).
2. centers the distributions on this true position.
3. averages many sights.
4. gives the ship inside the infinity of hats 25% of the time.

The symmedian problem is what the navigator faces at sea, getting the best estimate of his position from a single hat from an unknown location. The Goudsmit problem starts out with the navigational answer, and determines the sight’s navigational errors.

And yes, it may be surprising that no matter how good your sights are, if you plot many, many, of them, three-at-a-time as hats, your known position will fall inside each hat only 25% of the time. Interesting, but not useful. Has any NavList member done that??

And incidentally, the Goudsmit statement is easy to prove, no need for Gaussian distributions or Monte Carlo methods: Only assume a symmetric altitude error distribution about the known position. Then draw eight diagrams with three LOPs. For each LOP draw two hats: first with that LOP spaced on one side of the ship’s position, the second with the same LOP spaced the same amount on the other side. Doing that for each LOP gives eight diagrams with the ship’s position falling inside a hat just twice, 25% of the hats.

Happy Hat,
JK

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