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    Re: The Darn Old Cocked Hat - the sequel 1
    From: Gary LaPook
    Date: 2013 Mar 13, 23:33 -0700
    I don't see how you get the hole in the center of your graphic. Look at the table of probabilities versus standard deviation.

    http://fer3.com/arc/img/114399.extended%20table%20q7d.pdf

    Using the numbers from that table I calculated the areas of each annulus and each ring has a greater area than the next smaller ring. The second attached graphic shows the areas of each annulus in square standard deviation units. Since each ring has a 10% probability (except for the outer ring which has only 9.9%) each ring will contain 10% of the fixes or 10 if using a total of 100 fixes. So dividing 10 by the area of each ring we can find the density of fixes per unit area. Since the center circle has a smaller area than any of the other rings the density of fixes in the center is greater than further out so no hole in the center. The area of the central 10% ring is 0.66 so the density is 15.1 per square standard deviation unit. The 20% ring has an area of 0.74 so the density is down to 13.5 and the density gets less and less until the last 9.9% has a density of only 0.3.

    gl

    --- On Tue, 3/12/13, Hanno Ix <hannoix---.net> wrote:

    From: Hanno Ix <hannoix---.net>
    Subject: [NavList] Re: The Darn Old Cocked Hat - the sequel 1
    To: garylapook---.net
    Date: Tuesday, March 12, 2013, 1:14 PM


                   "Sorry,  I can speak correctly or  understandably - not both" .
                    Heisenberg.
    ________________________________________________________________________

    John:


    This is a big peace of work that you sent, and I am seriously oblidged!

    In responding, let me pick just a couple of issues.
    To get started, please answer the 2 questions I posed re: CelNav cruise on the
    US carrier "Barak Obama"  in San Diego waters, will you? :) happy

    Next, please read the attached part 2 of my memo  "The Darn Old..."
    I hope it will show my goals of these 2 memos.

    Further, more questions that might be helpful to clear up things:

    You are trying to find a zero, say a zero of a standard parable, and will use Newton's
    iteration method, well, just because ... 

    My questions are:

      * Will it always converge, if it does, to the same point no matter your personal approximate choice at the start?
      * Does the distribution of the step errors close to the zero depend on this choice of yours?
      * Are we in a comparable situation when using St. Hilaire?
       * What is the difference to using an algebraic solution that does not start with a guess??
       * What is the difference to ignoring the LOPs and their error distributions and using only the corner points?.

    Would you please consider them?

    As you can see, I seem to be missing something in your argument.

    Best Regards, John

    h

    ___________________________________________________________________________________



    PS:
    1.    http://fer3.com/arc/imgx/f1-Cocked-Hat-V2.pdf;  is not available on the navlist.
    2.   I cannot locate any paper by Goudsmith or even their titles.

    Attached File: 122810.celnav error part2.pdf (no preview available)

    View and reply to this message: http://fer3.com/arc/m2.aspx?i=122810

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