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Old 09-20-2012, 09:54 AM   #60
stoneage
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Join Date: Apr 2010
Location: Stockholm, Sweden
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Quote:
Originally Posted by A.Reader View Post
@stoneage Thanks for your many technical posts. This is an interesting attempt to develop a new way of looking at the weight parameters of racquets.

However, I believe there is a mistake in the formula in your effective_mass.xlsx spreadsheet. There is an extra factor of 2 in the denominator there. When you take that out, the plots become monotonic with 15/r and you get much larger effective masses on the right side.
I hate to admit it, but you are absolutely right! The 2 shouldn't be there. I have updated the Excel sheet. The change alters the curves as you have noticed. I don't think that there is a much difference in the relation between the racquets, but since the curve rises faster it is more difficult to see the difference. It is therefore even more motivated to use the normalized diagram instead (which never had the irritating 2). Thanks a lot for pointing it out even if it was embarrassing

Quote:
In the eqv_mass_norm.xlsx spreadsheet, I don't understand the expression for the reference racquet that you use in the denominator there. Can you explain how Ip for the reference racquet equals 320 (1033 + 50*r + r*r) ?

Using I[end] = (M*L^2)/3 for a uniform rod, I get
Ip = 320 (70*70/3 + r*r) for the reference racquet. This makes the curves look quite different.
Not quite. r is defined as the distance to the point 10 cm up the handle where the force is applied. So starting with the MOI around the midpoint, Ip for the rod is:
Ip=M*L^2/12+M*(c+r)^2

and c for an evenly balanced rod is L/2-10 = 25 so I get
Ip=M*70*70/12+M*(25+r)^2= M*(1033+50*r+r*r)

Also remember that the parallel axis theorem can only be used in relation to the MOI around the center of mass, so you can't calculate Ip the way you did, even if r would have been to the endpoint of the racquet.
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