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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