Phil Ford wrote: > >>The formula for center of percussion is: > >CP = I / Mr > >Where I would be inertia of hammer and shank assembly about the axis of >rotation >M is mass of the hammer and shank assembly >r is distance to center of mass of the assembly > >Taking a couple of real examples: > >Say the hammers are hung at 5 inches from the hammer center. >Assume the shank weighs 4 grams. For the sake of simplicity assume it is a >straight cylinder with its CG halfway out, at 2.5 inches. > >To bracket this a bit: >For a bass hammer, assume hammer mass is 11 grams (a pretty heavy bass >hammer). >For a treble hammer, assume hammer mass is 4 grams (a pretty light treble >hammer). > >Do the math and you get: > >For the bass hammer CP = 4.74 inches from hammer center >For the treble hammer CP = 4.44 inches from hammer center<<<<<< >______________ Dean May wrote: >I fear you are making the same mistake I was making earlier. You are >calculating the Center of Gravity, not the Center of Percussion. Your >formula for CP appears to be correct, but you fail to calculate the value >for "I" which is not the inertia, but the Moment of Inertia..... >Now for Center of Gravity: >CG=(4*2.5+11*5)/15 >CG=4.33 in (I'm not sure why I got a different number than you did) Because I was calculating Center of Percussion. This number is the Center of Gravity. >Going back to your formula for Center of Percussion: > >CP=I / mr >CP = 308 / (15*4.33) >CP= 4.74 in Which is exactly the number that I got. See above. Is this some sort of trick question? Whatever. The result is still that the Center of Percussion is not at the hammer. Phil Ford >... >You are correct in that there is another "sweet spot", that is related to >the natural frequency of the bat. I had never before heard of it having an >effect on transferring energy to the ball. But it has been too long since I >was a real engineer. > > >Dean, P.E.
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