Re: [Phys-l] question on averaging.

[Bernard Cleyet <bernardcleyet@redshift.com>]

John!

I'll interleave
On 2009, Feb 19, , at 02:56, John Denker wrote:

> On 02/18/2009 10:21 PM, Bernard Cleyet wrote:
> > I'm analyzing the free decay of a pendulum à la Siegel [AJP 76 (10)
> > 956.]
> >
> > However, the pendulum has such a large Q (few hundred => ~2k) that I
> > must average over several data points.  Since the necessary values
> > are the speed squared there are two orders.  One to average the
> > squared speeds and tother to square the averaged.  The difference is
> > not large as over most of the decay the loss is rather small. [Which
> > is why I must average!]  Also the factor of interest is the
> > difference between the squared points (speeds).  So the question,
> > which is the correct  order, if there is one?
> >
> > bc  thinks this is more a meta-physics question
>
> Metaphysics is not required.  Physics suffices.
>
>  -- Energy is (strictly) conserved.  Speed is not.

System not closed; dissipation found.  constant (Coulomb due to ball  
bearing suspension),  linear (viscous drag), and quadratic (turbulent  
drag).

>  -- Energy is almost constant from cycle to cycle in this system.

Yes, the result is significant proportion is noise requiring averaging.


>   Speed is not.

No it's very slowly changing, but very noisy

>  -- Q is conveniently defined in terms of energy.

def.:  2Pi*speed squared / delta(speed squared)

Yes, but not nearly as revealing as a* [ = ~ delta(speed squared /  
(speed* period)]  {Siegel}

>  -- KE goes like speed squared (at corresponding points in
>   the cycle).
>
> Therefore, unless there are special additional requirements, as-yet
> unmentioned, speed squared (or, better, energy) is the way to go.
> A big part of physics consists of connecting experiment to theory,
> and in this system the connection is easiest in terms of energy.
>
> Also note that "averaging" is a lame substitute for curve-fitting.


The first time I plotted Q (bc's continuous method) using a ~ >10K Q  
pendulum (observatory type) Q values varied as much as Q = +/- 10^10!!!

The decay is so little that noise is dominant.  Fitting to Q is a  
solution, but very impractical, because the Q is not well behaved.  A  
fit requires many terms w/ difficult physical interpretation.  The v.  
hi. Q pendulum is so poorly behaved that even fitting to a* is not  
practical.   It acts as if two pendula with a smooth transition.

I interpret John has answered my question,   Square the speeds then  
massage.

[I thought the question was more interpretive than "cut and dried",  
hence my meta-physics.]

bc thanks JD

p.s. I will be pleased to send plots to requesters.

[refs.   *Siegel/sync 6/a*;  *sync5;  3mm with a*]