Alan, The GS spine studies are discussed in a GS Clubmaker article in the September/October 1999 issue on page 18, authored by john a. meng who has included many comments by Tom Wishon. Good article.
Bernie Writeto: [EMAIL PROTECTED] ----- Original Message ----- From: "Alan Brooks" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Monday, December 30, 2002 6:11 PM Subject: Re: ShopTalk: shaft flex v.s. frequency > Thanks for that info Al, do you have any idea where I could find > information on what was done? Either by TT and/or Golfsmith? > > Alan > > At 04:00 PM 12/30/02 -0500, you wrote: > >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 > > > > >
