Check Out this picture. Probably not your best player but it can happen with a late loading of the shaft.
llhack ----- Original Message ----- From: "Alan Brooks" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Tuesday, January 21, 2003 3:36 PM Subject: Re: ShopTalk: 20 min talk > Thanks Dave, > > This morning I dug into the Appendices in Jorgensen's book and he > calculates the energy fraction in the club head after impact at 0.531, > which indicates a velocity loss of about 30%. I was in error. That > definitely changes my thinking about what happens after impact. > > I'm home at lunch to let the wallpaper guy in, so I'll take another look at > all this this evening. > > Thanks again, > > Alan > > At 09:28 AM 1/21/03 -0500, you wrote: > >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. > > >
<<clip_image002.jpg>>
