[CAUT] 1098 slight redesign

Ric Brekne ricbrek@broadpark.no
Thu, 16 Feb 2006 23:38:33 +0100


Hi Jim... others

Just did what I think is the correct way of figuring the change in the 
angle of the downwards force vector from the string to the bridge. Since 
you said it took about 3 mm to get 0 bearing, I figured it as the string 
being a straight line if you raised the backlength 3 mm.  So if you had 
84 mm back length, 1068 mm speaking length, then before raising the 
backlength 3 mm you have about 177.95 degrees between the two lengths.  
Raising the back length then so that the string is straight angles the 
downwards force vector backwards then about 1 degree if I have it 
right.  That would be splitting the 177.95 in two and subtracting the 
difference from the 180 degrees angle thats created when you raise the 
back length 3mm.

Grin... I havent got a clue as to whether a 1 degree change in the 
direction force is applied to the bridge makes a heap of difference 
tho... proabbbly not to much considering at the same time you are 
lessening the amount of downward force to next to nothing.

If someone sees an error in the way I did this... I'd like to get 
straightened out.  Fun and games :)

Cheers
RicB

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