Al I agree with your assessment of the situation. Toe bob is one of the reasons I have got into the habit of aligning the stiff plane in the 12-6 plane. You refer to the TT research. I've not seen this research although I have looked for it and know of it. Do you have a source?
Cheers Graham ----- Original Message ----- From: Al Taylor <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Tuesday, December 31, 2002 5:00 AM Subject: Re: ShopTalk: shaft flex v.s. frequency > Alan, > We went through the 1/2 cycle discussion several years ago. You might want > to look at True Tempers research that they did with their Shaft Lab. It is > amazing what the head does in the last few milliseconds prior to > impact. Toe Bob, I think, is what they call it. They found that pros had > much less to nil amount of toe bob, compared to amateurs. They figured it > was their ability to maintain a load on the shaft, due to continuous > acceleration, into impact. The concept of spine aligning, which was later > indicated by tests done by Golfsmith, tended to minimize this bobbing > affect. The results were more "on center" hits. This, in effect, made for > longer hits, by way of reducing distance loss due to off center hits. It's > a maddening game some of us play, trying to fit golfers. > > Al > > At 10:48 AM 12/29/2002, you wrote: > >Hi Al, > > > >Not from my perspective, I think that is exactly correct. Also, keep in > >mind that a shaft only undergoes a half cycle of oscillation during a golf > >swing (at least the part of the swing we care most about). Things that > >happen after many cycles of oscillation don't really effect the way the > >club hits the ball. > > > >Regards, > > > >Alan > > > >At 03:20 AM 12/29/02 -0500, you wrote: > >>Well Alan, > >>We may be getting there, but will let you be the judge. I understood > >>your example and how the formula applies, but couldn't equate/apply the > >>example to a golf shaft. Instead, I pictured two parallel springs (or a > >>thick bar and a skinny one), one stiff, one weak, both tied together at > >>both ends. I can intuitively now see how the force required to bend this > >>in either direction would be the same and also how that would apply to a > >>golf shaft. I just can't see how the formula applies in my case. > >>(probably just as well too!) > >> > >>Anyway, assuming I am getting close, let me make a proposition. I would > >>say that once you have found this point in a shafts axial rotation that > >>allows it to oscillate in a flat plane, when twanged as we do it, and > >>that since the force to bend it a certain amount in either direction > >>would be the same, then the shaft will respond the same from equal but > >>opposite direction pulls off neutral. That being the case (hahah I > >>hope), it would then seem to me that it would make no difference which > >>side of this planer oscillation was placed towards the target. > >> > >>Did I get too brave? > >> > >>Al > >> > >> > >> > >>11:41 PM 12/28/2002, you wrote: > >>>The simple answer is that the two legs have the same load on them and > >>>the paper leg is weaker than the steel one. But that really isn't a > >>>good analogy to what goes on in a bending mode. Let's try this > >>>one. You have two bars of equal length with the ends tied together with > >>>springs of equal length so that, with the springs just pulled tight the > >>>two bars are parallel, except that one of the springs is twice as stiff > >>>as the other one. You grab the bars in the middle and pull them apart, > >>>stretching the springs. The two springs have the same load applied to > >>>them, but the less stiff one stretches further than the stiff one, so > >>>the ends of the bars with the less stiff spring separate further apart > >>>than the ends with the stiffer spring. The more you pull the greater > >>>the difference becomes, or the two bars are rotating away from each > >>>other the more you pull. Basically, in order to balance the load on the > >>>ends of the bar so that they don't spin out of your hands the less stiff > >>>spring stretches further so the pull on the end of the bar is the same > >>>as that from the stiffer spring and the bars are stationary in your > >>>hands but rotated away from each other. From a mechanics standpoint the > >>>'moment' (force times distance) about the pivot point, the point you are > >>>pulling the bars apart from, has to balance. Because the distance is > >>>the same (you are grabbing the bar in the middle), the forces applied to > >>>the ends of the bar by the springs also have to be equal, but this means > >>>that the less stiff spring has to be stretched further than the stiff > >>>spring so the bars separate further on one end than on the other. This > >>>is kind of your steel and paper table legs, same load but different deflection. > >>> > >>>Now lets move the pull point on the bars 1/3 of the length of the bar > >>>from the stiff spring end, hence two thirds of the length from the less > >>>stiff spring end, so that the distance from the pull point is twice as > >>>far from the less stiff spring end than the stiff spring end. Now pull > >>>the bars apart. The bars will remain parallel because the moment about > >>>the pivot point from the stiff spring is the same as the moment about > >>>the pivot from the less stiff spring, and the bars do not rotate > >>>apart. The deflection in the less stiff spring, which is the same as > >>>the deflection in the stiff spring, applies the same moment to the bar > >>>(although half the force) because the distance to the pivot (the point > >>>you are pulling on) is twice as large (half the force times twice the > >>>distance). This new pull (or pivot) point is the 'neutral axis'. If > >>>you could push on the bars they would still remain parallel because the > >>>'stiffness' is the same in both directions and the forces on the end of > >>>the bar are still producing the same moment about the pivot point. This > >>>is roughly analogous to what goes on in a beam with an applied bending load. > >>> > >>>See if this helps. If not, I'll be glad to try again. > >>> > >>>Regards, > >>> > >>>Alan > >>> > >>> > >>> > >>> > >>> > >>>At 10:05 PM 12/28/02 -0500, you wrote: > >>>>Alan, > >>>>Thanks for the explanation. It seems to make more sense that > >>>>way. Question: You have a table or statue or any other object that > >>>>stands on legs. One leg is made of steel the other is made of > >>>>paper. It falls over into the leg of paper. Why the paper leg and not > >>>>the steel leg? I am smart enough to know this is too simple to apply > >>>>to shafts, so await the dressing down. Is this not similar to the two > >>>>sides of a shaft? As Dave referred to, it seems intuitively that it > >>>>would bend in one direction more easily than the other, though in > >>>>reality it doesn't. I may get out of 9th grade physics yet. > >>>> > >>>>Al > >>>> > >>>>At 05:18 PM 12/28/2002, you wrote: > >>>>>In an attempt to understand where the 'weak and strong' sides of a > >>>>>shaft concept came from it occurs to me that one of the problems with > >>>>>understanding this concept is that it is easy to visualize a shaft > >>>>>that is stronger on one side than the other; a thicker wall on one > >>>>>side in a steel shaft, or more fibers on one side in a composite > >>>>>shaft. This will, indeed, result in a shaft that is 'stronger' on > >>>>>that side - to a tensile (or compressive) load applied parallel to the > >>>>>axis of the shaft! Assuming the shaft remains straight the strain and > >>>>>stresses in the shaft material will be the same through a cross > >>>>>section but, because of the greater cross sectional area on the > >>>>>'thick' side of the shaft more of the reaction force to the axial > >>>>>shaft load will be carried on that side of the shaft, so, in a sense, > >>>>>it is 'stronger'. In a bending situation, however, because of the > >>>>>redistribution of stresses that occurs in the shaft to balance the > >>>>>forces on either side of the neutral axis, this does not result in a > >>>>>shaft being stiffer in one direction than in the opposite > >>>>>direction. In a given bending plane the shaft has the same stiffness > >>>>>in both directions. The 'neutral axis' is defined, by the way, as the > >>>>>line of zero stress through the cross section of a shaft under bending > >>>>>load and is not always at the geometric center of the shaft. The > >>>>>stresses in the material on one side of the neutral axis are > >>>>>compressive and tensile on the other for bending in one direction and > >>>>>then reverse for bending in the other, but the neutral axis remains in > >>>>>the same location, hence the resistance to bending (stiffness) is the > >>>>>same. I hope this helps. > >>>>> > >>>>>Regards, > >>>>> > >>>>>Alan > >>>>> > >>>>> > >>>>> > >>>>>At 04:58 PM 12/26/02 -0500, you wrote: > >>>>>>At 04:32 PM 12/26/02 -0500, Al Taylor wrote: > >>>>>>>I'm impressed. I have no clue if you answered my question, but I > >>>>>>>was impressed. John, you still there? ;-) > >>>>>>>Al > >>>>>> > >>>>>>OK, John and Alan and I all said, counterintuitive as it may seem, > >>>>>>yes it bends exactly the same TOWARD and AWAY FROM the spine, as long > >>>>>>as it's in the same plane. > >>>>>> > >>>>>>You could weld a small steel rod to the shaft, to give it as stiff a > >>>>>>spine as you want. It will still have exactly the same stiffness in > >>>>>>BOTH DIRECTIONS in the same plane. > >>>>>> > >>>>>>Twirling it in a spine finder might or might not say that. But > >>>>>>measuring the REAL stiffness will. I have posted here how to measure > >>>>>>true stiffness, several times over the past week. > >>>>>> > >>>>>>Hope this answers it. > >>>>>>DaveT > > >
