Bridge Seating / food for thought

Ric Brekne ricbrek at broadpark.no
Sun Sep 10 07:28:51 MDT 2006


Hi Steve.  My reply is below... sorry about the length.

    Ric,
    I'm curious how you got these numbers.  Seems backwards to me. 
    Please help
    me understand how an effective speaking length change of 0.003 mm would
    require that much deflection of the bridge pin.
    Thanks,
    Steve Fujan

    Ric Brekne <ricbrek at broadpark.no> wrote:
    "To get a  rougly 4 beat per second false beat this string needs to
    change in length by  around 0.003 mm.   That would require the pin to
    deflect  nearly 1 mm !! "

First look at it from an intuitive perspective.  The string is 
<<anchored>> for all practical purposes in this context at the front 
termination. If you loosen the string grasping it exactly at the point 
it contacts the front bridge pin and move it left or right then it will 
follow an arc yes ??  no length difference at all there yes ?  Ok.. so 
to get an increase in speaking length the path of the string moving in 
any sideways motion has to be outside that arc to get a change in 
distance. How much increase in length of this string do you imagine will 
occur with a 0.1 mm sideways movement from the pin ?   Not much... not 
much at all.

For simplicities sake and in keeping with the wobbly pin theory.... lets 
say the pins path of movement is the normal to the strings path. ie. 90 
degrees.  You have then a simple triangle to solve for.  The first 
length is going to be the long leg, and the second the hypotenuse. 
Figure the short leg which will be the distance the pin would have to 
give sideways to allow for the change in length.

If then... a string is 100 mm at rest, and when excited pin wobble 
allows it to achieve a 0.1 mm increase in length for the horizontal 
component... the effectively you have  100.1^2 - 100^2 = (Pin 
deflection)^2.  The old a^2 + b^2 = c^2.  Here the deflection distance 
is a, the length of the string at rest is b and the increased length is 
c.  Solving for a you have a^2 =  c^2 - b^2.   In this example, the 
needed sideways deflection to get that 0.1 mm increase is 4.47 mm !  
There is just no way the sideways pin wobble can possibly effect a large 
enough change in string length to cause any significant false beat.  
Its important to remember that any string deflection that actually does 
occur will spread this tension out over the entire length of the 
string.. not just the speaking length. This in effect reduces the net 
change.

On the other hand... the idea that the pin terminates the vertical 
component and a regressed notch of about 0.1 mm terminating the 
horizontal component makes a whole lot more sense. You have two distinct 
lengths that fit the bill, and you dont need a bridge pin to be for all 
practical purposes in two places at the same time, let alone achieve 
distances that are impossible.  Tho, as I said in my reply to Frank I 
would still like to know why there are so many exceptions to the  rule.

You could argue that it would be more accurate to figure the change in 
string length by using the two bridge pins and the front termination as 
the triangle to figure since any sideways movement in the front pin also 
lowers tension on the bridge face string segment. I didnt here to make 
my reply as simple as I can. Doing even with large angles between the 
two pins still doesnt get you anywhere near the amount of needed 
sideways deflection from the front bridge pin to account for any 
hearable beat.

Cheers
RicB





More information about the Pianotech mailing list

This PTG archive page provided courtesy of Moy Piano Service, LLC