Hi Folks
Well, I used a little time thinking about this latest post of Ron Ns on
the subject, and droped back a couple weeks on the previous related
thread to find a relevant claim in more specifics then in his last. I
ran into this quote, from
http://ptg.org/pipermail/pianotech.php/2007-March/203148.html
"A swelling bridge cap moving the string up the pins 0.2mm
(0.0079", which is less than I have observed with controlled
tests) will change the length around 0.025mm, or ten times as
much as soundboard rise."
An interesting tidbit for back engineering to see what he did and how he
came up with it. First, let me say that its not the speaking length we
are talking about here. The significant increase in the overall string
length this causes is on the surface of the bridge itself. The speaking
length would increase, if ever so slightly on long strings... and the
back length a bit more until they get close to the shorter speaking
lengths.. but we'll put aside any additions to those lengths for now and
perhaps deal with them later.
By my figuring, you end up with somewhat different results then Rons.
Heres what I come up with for the following assumptions.
- Bridge width 20 mm
- Angle of string displacment 10 ¤
- pins slanted at 10 degrees.
To begin with, a 0.2 mm rise up a 10 ¤ slanted pin will be displaced
sideways by
Sin 10 * (0.2 / Cos 10).
Since there are 2 pins we have to take this times to resulting in a
figure of 0.07053079228 mm.
2 * Sin 10 * (0.2 / Cos 10)
To find out how much this lengthens the string across the bridge surface
you need to add this added sideways displacement to the existing
displacement. Then simply square the result, add it to the square of 20
and take the root of the sum (finds the new string length), finnally
subtracting the origional length (given by (20 / Cos 10)
The string length before the added displacment is given by
20 / Cos 10
The existing sideways displacement is this figure time Sin 10
Sin 10 * 20 / Cos 10
Adding the additional sideways displacement to this you have
2 * Sin 10 * (0.2 / Cos 10) + Sin 10 * 20 / Cos 10
Finding the new string length using the old a^2 + b^2 = c^2 you have
SQRT( (2 * Sin 10 * 0.2 / Cos 10 + Sin 10 * 20 / Cos 10)^2 + 20^2)
The origional string length is simply
20 / Cos 10
Which is subtracted from the new length. The whole operation is then.
SQRT( (2 * Sin 10 * 0.2 / Cos 10 + Sin 10 * 20 / Cos 10)^2 +
20^2) - (20 / Cos 10)
And the result of this is about half what Ron comes up with. Presumably
because he figured on a 20 degree pin slant. A 0.2 mm rise in this
example only causes a 0.012366 mm increase in string length. In
addition however will be a very slight increase in the speaking length
and likewise for back lengths. In long strings the increase to the
speaking length is nearly nothing. For the 406 mm long string in Rons
example using my 10 degree pin slant, (see his whole post at the address
above) with a 100 mm back length and a 200 mm length from the front
termination to the tuning pin thrown in... and assuming a 0.85 mm Ø
would result in a roughly 5 cent change in pitch. A 50 mm long string
otherwise similiar with a 0.8 mm Ø would see a 22 cent change, and a
1000 mm long string with a 0.925 mm Ø would only see around a 3 cent change.
It should be noted that this increase in string length will affect the
pitch of a string just like an increase in vertical deflection would.
And as such a 0.124 mm increase will affect shorter strings much more
then longer strings. Assuming then such a 0.2 mm rise in the bridge
surface is consistant over the entire bridge... then with nothing else
working on the system the effect of this would be to cause an increased
pitch rise inversely proportional in some sense to the length of the
strings. I.e. the shorter the string the a larger the pitch change.
Another interesting point is that this increased side displacment will
leave its footprint in the indentation... causing it to be wider then it
would be had side displacement remained constant. In Rons example this
would be an extra 0.15 mm width.... quite visable to the eye. In my
example 0.07 mm.
Seems to me like the consequences of this idea dont really fit well with
what we see in pianos. This kind of an effect would dominant greatly the
pitch picture in the high treble, and there are some questions about
what must compensate then for the abrubt change in pitch change that is
seen between the tenor / treble break and again at the treble /
diskant. And then why doesnt the very top just fly off the chart ?
More thoughts for troubled minds... :)
Cheers
RicB
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