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    Re: Watches as chronometers (correction)
    From: Gary LaPook
    Date: 2013 May 31, 00:42 -0700
    Trying one more time.

    Should be:
    A = 306/266/+40 fast:  B = 528/536/ -8 slow:  
    C = 886/906/-20 slow

    Typo corrected, no change in conclusions.

    gl


    --- On Thu, 5/30/13, Gary LaPook <garylapook@pacbell.net> wrote:

    From: Gary LaPook <garylapook@pacbell.net>
    Subject: [NavList] Re: Watches as chronometers (correction)
    To: garylapook@pacbell.net
    Date: Thursday, May 30, 2013, 10:18 PM


    Should be:
    A = 306/266/+40 fast:  B = 528/536/ -8 slow:  
    C = 906/886/-20 slow

    gl


    --- On Thu, 5/30/13, Gary LaPook <glapook{at}pacbell.net> wrote:

    From: Gary LaPook <glapook{at}pacbell.net>
    Subject: [NavList] Watches as chronometers
    To: garylapook{at}pacbell.net
    Date: Thursday, May 30, 2013, 10:11 PM


    This is another update on my experiment using three cheap, $17.00,
    watches as a chronometer.

    I obtained another set of data today, May 31, 2010 Z at
    0108 Z, 1351 days since the start, more than three years
    and eight months. See attached photo taken at 01:08:00Z,
    (WWV radio signal)
    watch "A" is 5 minutes and 6 seconds fast, etc. (The seconds
    are in the top windows.)
    I have not had to replace any of the batteries.
    At the beginning of this experiment I determined the daily rate of
    each watch from a 99 day sample. Watch "A" has a rate
    of 0.1919 seconds fast per day; watch "B", 0.3737 fast; and watch
    "C" is 0.6263 fast. Using these values I calculated what the
    predicted corrections would be, the values that would be
    used by the navigator to correct the chronometer time,
     and compared with the measured values.
    The differences show how much of a navigational error would have
     resulted from the navigator using the predicted times.

    In the format for A, B, and C in seconds: actual error; 
    predicted error; difference.


    A = 306/266/+40 fast: B = 528/536/ -8 slow: C = 906/846/-20 slow

    Averaging these differences equals 4 seconds fast.

    Using only one watch with the largest error, 40 seconds, would
    resulted in a 10 minute of longitude error. Using the average of
    all three watches would have produced an error of only one minute
    of longitude, not bad for fifty-one dollars worth of watches
    after more than three and a half years.

    See my prior reports at:

    http://fer3.com/arc/m2.aspx/Watches-chronometers-LaPook-oct-2011-g17175


    gl

    Attached File:


    : http://fer3.com/arc/m2.aspx?i=124211

    : http://fer3.com/arc/m2.aspx?i=124212

       
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