# NavList:

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

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Re: The old cocked hat- a synthesis
From: Tom Sult
Date: 2010 Dec 18, 22:42 -0600
To apply the industrial standard of excellence we really need to think of our error in terms of 6 sigma. Suddenly a dot seems pretty silly.  Back to my instructor "looks like we are around here someplace". Not said to discount rigor but as a reminder of the error.

Thomas A. Sult, MD
Sent from iPhone

On Dec 18, 2010, at 21:12, Gary LaPook <glapook@pacbell.net> wrote:

 We have been having a very interesting conversation about cocked hats this month, which I have really appreciated. What I take from both John Karl's and George Huxtable's postings is that a high percentage of the time the actual position of the vessel will be outside the cocked hat but, even so, any point within the triangle is a more probable location for your vessel any point outside the triangle. (John Karl's diagrams also show, that for some shapes, that a point slightly outside the triangle may be slightly better than some of the points inside the triangle.) But we have to remember that although a point within the triangle may be the "most probable position" (and I think we have been convinced that it is the Symmedian point) it is still not very probable. Although all the points within the triangle may have an aggregate probability of 25% (or something slightly different) any particular "point" you choose inside the triangle will have a very low probability of being the actual position of the vessel. As an analogy, for example, your ship is 100 feet long with a beam of 20 feet and the triangle is a right equilateral triangle with legs one NM long. The area within the triangle is approximately 18,000,000 square feet and the area occupied by your ship is only 2,000 square feet. This means that there could be 9,000 ships of your size inside the triangle so (if all points were equally probable) the odds are 8,999 to 1 that you have chosen the wrong point for the fix. Even using the Symmedian point and assuming that that point is twice as probable as any other point within the triangle (which is a gross exaggeration as shown by John Karl's curves) then the odds are still 4,499 to one that the vessel is actually at that point. John's curves show very slight variation in probability for points within the triangle. So, as I said before, pick any point within the triangle you like, by eyeball or by any construction you like, and use it for planning the next leg of your voyage since there is very little likelihood that any other point is any better than the point you have chosen, to represent the actual position of your ship. gl
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