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Dear Brian,/snip/
I'm glad we agree!
You might find this interesting to ponder:
In 2009, October clock change, rather than putting it back one hour it was advanced by 23 hours. This meant that the free decay of the pendulum was even slower, because the gravity arms were lifted off the pendulum. But the interesting question is the restart. The exponential curve that shows the swing amplitude getting back to normal surely tells us something about the drive?
<http://trin-hosts.trin.cam.ac.uk/clock/?menu_option=�; data&from=� 25/10/2009&channel=�amp&channel2=�0&to= �25/10/2009&xmax=�2&skip=� 0&ymax=�54&ymin=�37>
Food for thought.
Best wishes
Hugh
On 18 May 2015 4:57:49 am brian whatcott <betwys1@sbcglobal.net> wrote:
Thank you for responding, Hugh. Your working clarified that the cause of
Bernard's rather high value for the Q of the Trinity Pendulum was his
use of the escapement energy contribution at each cycle as about 0.0029
j in lieu of the 0.0038 j which would be needed to maintain a Q value as
low (!!) as your free decay Q value. This escapement energy
contribution value was evidently derived from a Trinity Clock site
mention of the Denison gravity arms weighing 50 grams each and
descending 3 mm.
Your note was also a useful reminder of Denison's use of an
air-brake ('fly') with a lossy friction clutch to cope with the varying
clock drive needed during the year while maintaining a sensibly constant
impulse to the pendulum, and as important, eliminating the escapement
skip not uncommon in tower clocks prior to 1860.
Thanks again
Brian Whatcott
On 5/17/2015 12:21 PM, Hugh Hunt wrote:
> Dear Bernard, Doug, Brian
>
> I'm very interested in your calculations. I am Keeper of the Clock at
> Trinity and Rick Lupton, who wrote the report, was my student.
>
> The pendulum free decay (without drive) can be see here t clock change
> <http://trin-hosts.trin.cam.ac.uk/clock/?menu_option=data&from=26/10/2008&to=26/10/2008&width=700&channel=amp&&xmax=1&skip=0&ymin=32&ymax=53.6>
>
>
> Note that the decay is nicely exponential. The width of the plot is
> one hour, so the decay is from 53.6mrad to 29.8mrad in about 3050 seconds
> Using the usual definition of Q from where it derived (Quality Factor
> for resonant circuits) then A2/A1 = exp(- zeta w T )
> where A1 and A2 are two amplitudes separated by time T for a damped
> oscillator with resonance frequency w (rad/s)
> The Damping Factor zeta = 1/(2Q)
>
> So A1=53.6 A2=29.8 T=3050 s w = 2*pi/3 for our 3s pendulum
> (half period = 1.5s)
> gives zeta = 0.0000919 or
>
> Q = 5,441