Piano Tuning Pin Rotation: A Physics Problem

inkling

Hello everyone, I am new here and so I apologize in advance
for the many forum faux pas I am likely to commit.

I am not particularly well educated in the field of physics,
but I am a piano tuner by trade, and we work with issues every
day that are basically problems of physics and engineering, and
some us tend to be interested in the theoretical side of things.
A discussion on our trade specific mailing list has hit a dead
end, and so I find myself here at your collective mercy.

Here is my question:

Imagine a length of music wire affixed at one point to a unmoving
anchor point, and to the other coiled around a tuning pin secured
in a unmoving pinblock.

When the tuning pin is rotated by a tuning "hammer" (wrench)
The termination points remain the same, but the tension increases,
causing the pitch to rise (obviously)

Disregarding (for the moment) all factors of flex of the anchor
points, and also disregarding any friction points, how would I
calculate relationship between the rotation of the tuning pin and
the pitch of the string?

The known factors are as follows:

String length: 972mm (38.267")
String diameter: 1.09mm ( .043")
String density: 7900 kg/m3 [is that right for high grade steel?]
String tension: 86.583kg (190lbs)
Tuning pin diameter: 7.01mm (.276")
Note Pitch: 174.61hz (Note F-33)

(I cannot guarantee the accuracy of these figures,
so feel free to point out any errors.)

So, I rotate the pin say, 0.5 degrees-
How does that translate into pitch change?

How would I calculate this on my own?

Thanks so much in advance for your time.


Kurt

Danger

Welcome to PF, Kurt.
Wow, what a question to bring up! I'm not the one to answer it, but I certainly look forward to seeing the responses from the guys who know what they're talking about.
I suspect that the specific composition of the wire would be a major factor, as well as the ambient temperature. The number of turns of the wire around the peg would probably make a difference as well, since that would in essence change the properties of the peg.

inkling

On a piano tuning pin, there may be 4 coils, but they lie along side each other- the wire
never overlaps itself. The termination point will always be flush against the side of the
tuning pin.

The extra coils may indeed cause the pin to act stiffer, but we are ignoring flex or twist
at the moment.



[kurt]

pam

You have to know the elasticity of the wire, given by what is called "Young's modulus", and make a calculation. But the accuracy with which you can measure the angle of turn is so much less than you could measure with beats from a tuning fork (or even by ear) , that the formula would be useless.

inkling

Agreed. I guess what I am looking for is some idea of how tiny the pin movements need to
be to get a really perfect note.

inkling

I found this:

(here: http://www.mmdigest.com/Archives/Dig....04.02.08.html)

I don't need precision, I am trying to get ANY idea of the size of motion we are
talking about.


[kurt]

Q_Goest

Hi Kurt,
Do you really need to know how the rotation of the pin translates to pitch change? If so, you need to determine the change in tension due to the stretching of the string, because the pitch is a function of the string tension. Here's a couple references to string pitch based on tension:
http://www.cs.helsinki.fi/u/wikla/mu.../wwwscalc.html
http://hyperphysics.phy-astr.gsu.edu...es/string.html

To determine the change in tension due to rotation of the pin, you need to determine how much the string stretches. As pam points out, you need the modulus of the string. Steel is roughly 30,000,000 psi (change to metric) which is generally represented by E. So you have:
S = F/A
or: F = S A
Where:
S = Stress in the string
F = Force (tension) on the string in Newtons
A = cross sectional area of the string

And

S = e E
Where:
e = string stretch (mm/mm)
E = Young’s Modulus

Combining:
F = S A = e E A

F is the tension in the string. Use that in the calculator to determine pitch. Play with this a bit and with the calculator (see attached above) to see if you can get what you’re looking for.