Richard and list,
It's funny you should say something about metric stuff. I've worked metric
for 10 years and not until the other night did I really have to stop and
think about how much force to start with for the string tension (maybe I
should have ask Newton) to get 440Hz. Somehow I couldn't come up with a
first guess on how many Newtons though I figured it was around 120 lbs.
You had a good thought. The first case should have been a string fixed only
at one end. The ones you described later where a 420mm wire is looped is
more difficult (it is a large deflection problem and is non-linear case).
The ones where the string is stretched by a known amount can easily be done,
but the stress levels would also need to be closely followed to make sure
the string wouldn't break (or go to plastic deformation). Since I do a lot
more frequency prediction than stress analysis, I'll stick mostly with
prestressed linear modal analysis for this phase.
Case #3.
Length: 420mm (16.5354")
Diameter: 0.8mm (.0315" sorry, but it was an even metric number. Tell me
what diameter and length you want!)
Constrained on one end only.
No forces or gravity effects.
FREQUENCY
FUNDAMENTAL 3.2205
2nd Mode 20.184
3rd Mode 56.516
4th Mode 110.75
5th Mode 183.08
Case #4.
Length: 420mm (16.5354")
Diameter: 0.8mm (.0315")
Constrained on one end.
The other end is stretched 1mm (.0394")
No other added forces or gravity effects.
FREQUENCY
FUNDAMENTAL 14.124
2nd Mode 45.772
3rd Mode 95.503
4th Mode 163.32
5th Mode 249.24
Realize that Case 4 is different than Case 2. In case 4 the string and the
termination point is stretching so the "speaking" length has gotten 1mm
longer (which will affect the frequency). In case 2 a force was applied to
put tension on the string so it stretched, but in the analysis the speaking
length was still 420mm (these are the games that are easy to play in
modeling...) Maybe tomorrow I will find out how far to stretch the string
to equal the force that was applied in case 2 to show the difference in the
frequencies.
That's all for today's boring lecture.
doug richards
San Jose, CA
> -----Original Message-----
> From: Richard Moody [SMTP:remoody@easnet.net]
> Sent: Friday, August 28, 1998 10:26 PM
> To: pianotech@ptg.org
> Subject: Re: Predicting string behavior
>
> Imagine the string clamped on one end with vice grips and dangling
> vertically. That would be one case of no tension.
> If the string is held horizontaly clamped at each end with no tension, the
> degree of slackness should still affect freq. I imagine. Imagine the vice
> grips are 20mm apart but the wire is 420 mm long. What freq? Now the
> vice grips are 42mm apart, what freq now? Imagine the vice grips 420 mm
> apart, what is the tension? what is the freq? Suppose the vice grips are
> 421 mm apart, what is the tension, and the freq? and so on for the vice
> grips 430mm apart what are the measurements there?.
>
> `How does this model compare with your model? Regarding your model, how
> did you come up with your numbers for no tension?
>
> Please for the "metric illiterate" could you include American lengths and
> wire guages?
>
> you wrote
> > If the fundamental is A1 and the second partial is A2,
> > Cents deviation =A2-2*A1
> > It that correct?
>
> That is the diff of freqs. Cents has to be figured from the diff of freqs.
> The theoretical diff between A1 and a second partial A2 is one octave, or
> 1200 cents. The fundamental is usually expressed in cps, (cycles per
> second) The 2nd partial is usually measured at close to 2*A1 but in cps
> not cents. If the freq of A1 is known and the freq of A2 is measured then
> the diff in cents can be figured by.this formula from Niklas Eliasson
>
> >>>>> 1200*(log(F/f)/log(2)) = cents
> >
> > F = first frequency
> > f = second frequency.
> >
> > Next how would one solve for f or F for any value of cents
>
> Just put it
>
> F=f*2^(cents/1200)
>
> or
>
> f=F*2^(-cents/1200)
>
> (note the minus-sign in the second formula!)
>
> It all goes back to the fact that an octave is divided into 12 equal
> parts and each of those parts (semitones) is divided into 100 parts.
>
> Conclusion - an octave is divided into 1200 equal parts, and each
> part is called a "cent".
>
>
> Be in touch if you get into any problems!!! I am always glad to help!
>
>
> Niklas E,
> pianotech, Linkoping
> > From: Niklas Eliasson <e96nikel@isy.liu.se><<<<<<<
>
> Doug wrote
> > If anyone is following this, next time I could simulate bending
> the
> wire
> > over a bridge and look at how good and poor bridge termination affects
> > string frequencies.
>
> Yes I am intersted esp in what is considered a good or poor bridge
> termination.
>
>
> Richard Moody
>
> ----------
> > From: Doug Richards <Doug.Richards@quantum.com>
> > To: 'pianotech' <pianotech@ptg.org>
> > Subject: Predicting string behavior
> > Date: Friday, August 28, 1998 4:00 AM
> >
> > Hi all,
> >
> > I couldn't sleep tonight so thought I would put in my two cents for the
> year
> > (I'm a long time lurker).
> >
> > With all the discussion on string inharmonicity, I finally decided to
> finish
> > a little study using the modeling tools I use to predict structural
> > properties of disk drives (my day job).
> >
> > I decided to start very simple.
> > Steel string, 420mm long and 0.8mm diameter.
> > The string ends are constrained like a cantilever beam (it would be
> > something like somehow welding both ends into huge steel blocks)
> >
> > Case #1 is with no tension on the string.
> > I included this case to show that the upper bending modes of the string
> do
> > not relate to an even number multiple of the fundamental.
> > FREQUENCY CENTS DEVIATION
> > FUNDAMENTAL 14.12378 (0)
> > 2nd PARTIAL 45.77155 (17.52)
> > 3rd PARTIAL 95.50265 (67.26)
> > 4th PARTIAL 163.32415 (135.08)
> > 5th PARTIAL 249.24185 (220.99)
> >
> >
> > Case #2 applies enough tension to "tune" the string to very near 440 Hz.
> > FREQUENCY CENTS DEVIATION
> > FUNDAMENTAL 440.00518 (0)
> > 2nd PARTIAL 880.55096 (0.54)
> > 3rd PARTIAL 1322.17697 (2.16)
> > 4th PARTIAL 1765.42088 (5.40)
> > 5th PARTIAL 2210.81749 (10.79)
> > If you curve fit the CENTS DEVIATION, the best fit is a 3rd order
> polynomial
> > equation.
> > y = 0.0896x^3 + 0.0023x^2 - 0.0924x
> >
> >
> > One thing I wanted to verify is that the Cents deviation is calculated
> > correctly.
> > If the fundamental is A1 and the second partial is A2,
> > Cents deviation =A2-2*A1
> > It that correct?
> >
> > I have been able to make avi files of the animated mode shapes, but they
> are
> > ~5Mb for each frequency.
> > If anyone is interested in what they look like .......
> >
> > If anyone is following this, next time I could simulate bending the wire
> > over a bridge and look at how good and poor bridge termination affects
> > string frequencies.
> >
> > Comments?????????
> >
> >
> > C. Douglas Richards
> Quantum
> > Corporation
> > Mechanical Engineer
> > 408.894.4592
> > Finite Element Specialist
> > doug.richards@qntm.com
> > DPSG Mechanical Design Group 500 McCarthy Blvd,
> Milpitas
> > CA
> >
> > Oh yeah, also PTG member in the Santa Clara Chapter (in case it
> matters).
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