[CAUT] String stiffness questions

RicB ricb at pianostemmer.no
Mon Jan 8 16:33:57 MST 2007


Hi Ken...

This seems like a good clue. But I'm not sure its the same thing they
are using in their respective spreadsheets.  In anycase refering to the
relevant paragraph in McFerrin there is the Inharmonicity Coefficient
and it does indeed vary with the density of the wire.  It is designated
as "B"

Curiously there is a simplification that I think yields an odd way of
arriving at the Youngs Modulus (Q) if one first knows the density (D) of
the material.

Page 43 of McFerrin.... gives two equations for B, one the result of
putting in the values of Q and D, the other with these unknown.  These
two variations then form an equivalence if I am not mistaken... so

B = (1731 * pi^2 *d^2 *Q)/(2 * 64 * f^2 * L^4 * D)  =  (3.4 * 10^13 *
d^2)/( f^2 * L^4)

If you simplify these equivalence you end up with

Q = (2 * 64 * 3,4 * 10 ^13 * D) / (1731* pi^2)

which works out to

Q = 2,54737025541 * 10^11 * D

He doesn't specify what units of volume D are figured  in as far as I
can see.

Anyways...  it would seem you can figure the Youngs modulus directly
from the density of the the string material using the constant
2,54737025541 * 10 ^11

That is unless I'm off my bonkers again...:)

Cheers
RicB





    Hey, Alan & Ric,

    I believe Ric is asking about the inharmonicity constant.  I don¹t
    know if
    that is the name used by engineers, but that¹s what I have heard it
    called.
    It relates length, diameter, and tension to predict the
    inharmonicity of a
    particular string/note.  PScale used it to generate an inharmonicity
    number
    for each note, and ETDs use it to build their stretch curves.  I
    have only
    seen one value referring to music wire in general, and I have wondered
    whether it might be different for Mapes vs Roslau.  I would
    certainly expect
    Pure Sound to be different.  I can¹t lay my hands on a precise value,
    though.  Ron? Del? I¹m getting in over my head here!  Bail me out!

    Regards,
    Ken Z.




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