At 11:53 PM 1/20/03 -0800, Alan Brooks wrote:
Are you sure about those numbers? A 40-ft/sec change in velocity in the half millisecond I have read the ball is in contact with the club is an average acceleration of 80,000 ft/sec/sec (a delta-V of 40 divided by a delta-t of 0.0005). Seems a bit high. A 90-mph club head is about 132-ft/sec (60-mph is 88-ft/sec). A change in velocity due to impact of 40-ft/sec is about a 30% velocity change and I thought it was much smaller than that (about 10%).When you figure the momentum transfer, the clubhead loses just about 24% of its velocity during impact. (Math below. I'm sure Alan is interested; I don't know who else is.)
Bernie has sent me a copy of the polar velocity plots, and I don't see a full 40 ft/sec difference, but nowhere near as small as Alan's guesstimate. Here are the differences for the two plots:
Wood: 140 ft/sec to 110, or 30 ft/sec drop (23%)
2-iron: 130 ft/sec to 95, or 35 ft/sec drop (27%)
Given the rough eyeball precision of the numbers I took from the graph, these are pretty consistent with the math of momentum transfer that I figured.
Is there any reason a strong 'hitter' with a soft shaft could not have the shaft bent back at impact?Alan, let me discuss this with you off-line. I've thought about it quite a bit, and would like to try some ideas on you.
Cheers!
DaveT
MOMENTUM TRANSFER EQUATIONS:
Working the problem for the driver; other clubs would have slightly different numbers.
(1) Clubhead speed before (H0) and after (H1) impact, and ball speed (B1) after impact are governed by:
200*H0 = 200*H1 + 45*B1
(Typically 200g driver head weight and 45g ball weight)
(2) The ratio of ball speed after to clubhead speed before is given in the appendix of Cochran (or derivable without much difficulty from equations for conservation of momentum and energy):
B1/H0 = (1 + COR) / (1 + 45/200)
If we use a COR of .7 (what it probably was with the clubs and balls of Bobby Jones' day, when the plot was made), we get
B1/H0 = 1.39.
(3) Plug (2) back into (1)
200*H0 = 200*H1 + 45*1.39*H0
Solving this for H1/H0, we get;
H1/H0 = 200 / (200 + 1.39*45) = .76
This means the clubhead retains 76% (or loses 24%) of its speed during impact.
