Oops! I am so embarrassed because I did not allow for the one leap second inserted during the test period on June 30, 2012. I can account for it now by either subtracting one second from the measured error or by adding one second to the calculated error and I will do it the second way.
Trying one more time.
Should be:
A = 306/267/+39 fast: B = 528/537/ -9 slow:
C = 886/907/-21 slow
Averaging these differences equals 3 seconds fast.
Using only one watch with the largest error, 39 seconds, would
resulted in less than a 10 minute of
longitude error. Using the average of
all three watches would have produced an error of only 3/4 of one
minute of longitude,a maximum of 4,557 feet, 1,389 meters at the equator.
Sorry.
gl
gl
--- On Fri, 5/31/13, Bill B <billyrem42@earthlink.net> wrote:
From: Bill B <billyrem42@earthlink.net>
Subject: [NavList] Re: Watches as chronometers
To: garylapook@pacbell.net
Date: Friday, May 31, 2013, 11:42 AM
On 5/31/2013 1:11 AM, Gary LaPook wrote:
> 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.
Being it is you I am almost embarrassed to ask, but did you factor leap
seconds into your computations?
Bill B
: http://fer3.com/arc/m2.aspx?i=124231
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