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? ;-)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.
Al
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
