whatkindapianoisitanyway?

[McNeilTom@aol.com]

Hi, Michael -

Here's a stab at an 'intuitive' answer to your question:

If you leave everything else the same (i.e., length and tension), increasing
the diameter of wire will also increase the mass.  This will lower the
vibrating frequency.  You can compensate for the lowered frequency by
increasing the tension on the larger diameter wire - BANG!

Both the vibrating frequency and the breaking strength are related to the
cross-sectional area of the wire.  Increasing the diameter gives increased
strength, but lower frequency.  It's a trade-off that you cannot win if the
speaking length is too long - which it is in your example.  Since the piano
presumably worked at one time, but not now after rebuilding, I'd suspect that
the relationship between the bridge and plate has changed.  In other words,
the strings are slightly longer now than originally.

I like Horace's hypothesis about the magic in 19th c. wire.  That's certainly
well documented as regards 17th c. harpsichord wire.  However, I think that in
both old piano wire and old harps wire the charm lies in the warmth of tone,
not in superior strength.  I think modern music wire is probably the strongest
yet.

   -     Tom McNeil, RPT     -
Vermont Piano Restorations


In a message dated 98-06-04 13:56:38 EDT, you write:

<< ts a shomer grand.
 
 You should ignore the column that has the gauge of the wire listed.  My
 point is well taken with Mr. Greely's contribution, gauge has nothing to do
 with the breaking point of the wire even though it seems it should have
 everything to do with it.
 
 Equation for tension is:
 
 Tension = (frequency^2 + Length^2 + diameter^2)/434
 
 Equation for Break pt.:
 
 % break point = tension/(2528 * diameter^2)
 
 So the diameter^2 in the denominator of the "% brake point" equation
 cancels out the diameter^2 in the tension equation.  So % brake point is
 independent of the diameter of the string. This means you can try all kinds
 of gauge sizes and it will never alleviate the string brakeage problem.
 Now, this seems counterintuitive to me but I know its right. Can someone
 explain it?

Michael Wathen