Thanks guys... dont know why on earth I got the idea that I had to split the 160 pounds in half... grin.. tension remains the same regardless. Then on top that thinking I had to only take one side of the <<triangle>>... sheesh. Anyways, thanks for straighting that out for me. A question came to mind however in reading Rons post when he mentioned the following. "When looking at a given piano, I suggest that you set up a spreadsheet to calculate the downbearing force you are planning to set up per rib." Right under a rib I suppose the support is strongest... and gets a bit weaker as you move away from the rib. How do you figure the distance needed between two ribs then ? Thanks RicB Steve Fujan wrote: String Tension x Sin(downbearing angle) =3D Downbearing Force i.e. 160 x Sin(2 degrees) =3D 5.58 pounds (3.49% of string tension) Ron Overs wrote: The downbearing (vector) force on the sound board is equal to the SIN of the angle of deflection times the string tension.
This PTG archive page provided courtesy of Moy Piano Service, LLC