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Re: [Phys-L] Q for band width is not valid. Was: Re: finding Q from band width?




On 2013, Oct 21, , at 07:18, brian whatcott <betwys1@sbcglobal.net> wrote:

You now appear to accept that losses vary in a non-linear way with amplitude.


I accepted this and published my models and experiments showing this for both quadratic and constant (da Vinci friction) dissipation in a clock pendulum over a year ago.

HSN 2012-3 (July)

http://www.cleyet.org/Pendula,%20Horological%20and%20Otherwise/HSN/HSN%20published%20articles/PDFs/Q%20of%20a%20%22grand%20mother%22%20clock's%20pendulum.pdf

And had noticed that Q varied in the free decay of a Synchronome pendulum about five years before that. Published in HSN 2007-4 (August)

http://www.cleyet.org/Pendula,%20Horological%20and%20Otherwise/HSN/HSN%20published%20articles/PDFs/*%20*Q%20from%20t%20@%20BDC%20(submitted%20v.).pdf


Though a math solution to an experimental difficulty is forlorn (in my view) I can sketch an experimental approach which may be helpful.

I must ask a maths. friend, evidently.

It is wise to recall that the drag loss is not the only loss, and that there may be other losses acting in a non-linear way, in addition to the flexure loss which is also amplitude dependent.

It's generally agreed among the horological community to be very minor except in the most accurate pendula. That is one reason why some better clocks use silica pendula in addition to reduced thermal sensitivity.


If one applies a narrow magnetic impulse to the pendulum at the bottom dead center position to find the frequency for maximal amplitude for constant power,

Both because of circular error and dissipation frequency sensitivity, the frequency will change depending on the value of the constant power -- One reason why I eschew this method. Incidentally? I suspect this sensitivity will vary depending on the dissipation regime, which is another reason for solving the driven nonlinear dissipation regimes' ODE.

then the variation in losses may then be mapped for varied energy values and pendulum amplitudes at the SAME driven frequency.


I did this in two trials, but using a displaced solenoid and driven sinusoidally. The force tho is v. sensitive to the amplitude, as it's, what?, fourth power of the separation.


Given an experimental protocol which can then relate pendulum amplitude to total losses; varying the drive frequency to establish the two frequencies which result in the experimentally derived amplitude for half input power will provide you with an improved procedure for estimating the Q at a specific amplitude.


The bandwidth Q formula is defined for a linear damping, not, AFAIB**, for any other dissipation regime. I think one reason for the lack of agreement between the energy and bandwidth methods is because the dissipation was largely quadratic.

I've lost my experimental pendulum. Gate Keeper Seese insisted on its return to the stairwell landing controlling its Westminster chime and hour gong.

I've now reinstalled my original experimental pendulum. [Vernier RMSensor with LabQuest-LabPro acquisition system, and the Mumford supplied crude? E-M clock drive] I'll use the same solenoid and Pasco sig. gen. to explore driving the pendulum while being run by the E-M "escapement". This should be very interesting, as I know not of this done before.

bc, from a dining room physicist to a coffee table one.

** As Far As I Believe

Sincerely

Brian Whatcott Altus OK

On 10/20/2013 5:19 PM, Bernard Cleyet wrote:
On 2013, Oct 07, , at 20:12, Bernard Cleyet <bernard@cleyet.org> wrote:

I'm attempting to find the Q of a clock pendulum while free from the escapement. (crutch and crown removed, so only the spring suspension in use) using the bandwidth method, i.e. Q = "natural" or resonance frequency / FWHPower, no?

I very stupidly thought the simple formula [ frequency / half power bandwidth] was valid in the case of a quadratically damped oscillator. Obviously not as the Q is not constant, but is a function of the amplitude. But can one not derive a valid one from the particular amplitude solution of the appropriate ODE?

I've found many sources for the complimentary solution of the driven ODE, but not the particular one.

Any maths mavens out there who can show me?

bc maths declined, so won't even try a trial solution.

I'm driving it E-magnetically using a PM waxed the the edge of the bob (lenticular) and a coil waxed to the case opposite the PM. Not only is this asymmetrical, but also the force is a strong (4th power?) function of the separation (near field). Therefore, I vary the drive to maintain the same amplitude, which is measured by the blocking time of a photogate at ~ BDC, as I manually change the drive frequency in 0.001Hz steps. I assume the drive power is the square of the applied EMF. The undriven frequency agrees w/ the minE drive frequency (~ 5%) [0.565Hz]


So if the min drive E is 0.3V^2 then the diff between the two frequencies at 0.6V^2 is the FWHP, yes??

The result is ~ 1k, which agrees ~ with the ring down method.

bc
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