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
> >
>
>
>
