I disagree strongly with your depiction of events. During the release swivel phase and HH action phase of the swing, the different club parts are not rotating about the imaginary axis between the grip and the CG of the clubhead. What is happening is that the imaginary axis is moving in space due to movement of the flat left wrist/hand in space. The movement of the flat left wrist/hand occurs along a circular arc (relative to an imaginary center of the circle situated "somewhere" within space and all those parts (toe of the club, sweetspot of the club, hosel of the club) have the same angular velocity as the rotating flat left wrist/hand relative to the center of the imaginary circle. In other words, any point on that imaginary axis between the grip and the CG of the clubhead has the same angular velocity as the hosel and toe of the club - with respect to the instantaneous center of the imaginary circle of the flat left wrist/hand's rotational arc. The key point is "instantaneous" - if the flat left wrist/hand suddenly rotates along a tighter curve, then there must instantaneously be a different imaginary circle of rotation for all the rotating units (left hand, hosel, sweetspot, toe of the clubhead) which always have the same angular velocity relative to the center of that instantaeous imaginary circle of rotation. The only time that this system will breakdown is if the left wrist suddenly bends and doesn't remain flat.
Jeff.
Nothing imaginary about it.
Sorry Jeff but...
(Now that I think I understand what is being argued here ...) I have to agree with Yoda and (probably Yodaslurk earlier) position(s). The physics cause the CG (call it sweet spot if you must) to remain stationary as the club twists about the Z axis (in MY diagram above). The CG is on the line of action of the centripetal acceleration.
Centripetal acceleration is a fact of life.... Its line of action (through the c.g. on the (instantaneous) plane of the golf swing, is the axis of rotation the club parts move about as a result of the golfer's hand torque.
Sometimes thing are best proven to one's self by experiment.
-Take a carpenter's square or some such thing.
-Tie a string to one end and the other end of the string to a door knob or something fixed.
-Take a bungee cord and attach one end of it to the middle (CG... sweet spot etc.) of the other leg of the carpenter's square.
-Have your assistant pull back on the bungee to simulate centripetal loading.
Now you give the thing a little twist. Observe ... what axis is it rotating about? That's right... the line of action of the bungee.. ie line of action of centripetal acceleration (less any adverse effects on the experiment due to gravity).
The club operates identical to this example in 3D.
Last edited by no_mind_golfer : 01-25-2009 at 02:14 PM.
This means there are 6 degrees of freedom (DOF) which fully describe the motion of any three dimensional object in it. They are 3 translations and three rotations. We're (me, myself and I) talking about only one of those 6 DOF's. We're talking about ROTATION about the Z axis in our arbitrarily defined golf swing coordinate system.
That Z axis is defined by a line through instaneous center of the instantaneous plane of rotation of the golf club. It just so happens that the COG of the Clubhead (by the very definition of centripetal acceleration resides (ALWAYS) on that instantaneous plane. Any club part that are not co-linear with said "line of action" will, when subject to a torque about said "Z" axis, move OFF OF the plane of the swing. Ergo... when the golfer's hands provide the torque which closes the club face, the hossel moves away from the target and the toe of the club moves towards the target OFF OF THE INSTANTANEOUS PLANE of the golf swing.
Jeff ... what you are saying is correct ONLY when the butt end of the grip where the golfer's hands are located, is not co-linear with the line of action of the instantaneous swing plane. During the release, for all practical purposes they share the same line of action.
Confusing as H*** isn't it.... The real difficulty is getting us all on the same page via type-written word.
Last edited by no_mind_golfer : 01-25-2009 at 02:41 PM.
You wrote-: "We're (me, myself and I) talking about only one of those 6 DOF's. We're talking about ROTATION about the Z axis in our arbitrarily defined golf swing coordinate system."
You might be writing about rotation about the Z axis. I am not - I am writing about how the left hand rotates in space during the golf swing and drags the Z axis with it so that both the left hand and the Z axis rotate as an unit around the instantaneous axis of rotation of the flat left wrist/hand. I do not believe that the left hand is applying a torque force around the Z axis when the flat left wrist/hand moves in space. I think that a torque force around the Z axis only develops if the body stops rotating post-impact and the arms keep moving uncontrollably around the stalled body. The key to a good golf swing is keeping the arms rotating in synchrony with the pivoting/rotating torso so that the flat left wrist/hand rotates slowly in space. If a golfer follows AJ Bonar's recommendation in the next photo, then he may be applying a torque force around the Z axis.
YodasLuke got a little impatient with all the theoretical mumbo jumbo and decided to get out the 'big club' and ball we keep around for demonstration purposes. They come in handy for communicating the finer points of TGM (such as Hinge Action). Hard to miss that Clubface!
So, does the Clubshaft rotate onto the Sweetspot's Plane or vice versa? Ted decided to put this argument to bed once and for all. Click here and see for yourself . . .
Yoda - I applaud the effort that Yodas Luke has made to resolve this issue with a personal demonstration using the big club. Unfortunately, he couldn't keep the club on-plane post-impact so it will not convince the sceptics.
Consider my analysis of Yodas Luke's swing.
Image 1 - Address, The clubshaft is on a plane and I will arbitrarily label that plane the elbow plane. You have drawn another plane line through the sweetspot, and I will call that sweetspot plane 2. I am not calling it sweetspot plane 1, because sweetspot plane 1 apparently goes from the sweetspot to PP#3. So, I will call it sweetspot plane 2.
Image 2
That's a perfect on-plane takeaway. Note how the hosel and clubshaft remain on the elbow plane - despite a >90 degree startup swivel action. The hosel didn't leave the elbow plane and rotate to sweetspot plane 2 despite a >90 degree clubface swivel action.
Image 3
He starts the downswing with a slight OTT move that moves the hands slightly forward off the elbow plane. That causes the hosel to momentarily leave the elbow plane.
Image 4
Later in the downswing, he pulls the hands back a bit which causes the clubshaft to stay on-plane better during the release swivel phase of the downswing. Note that the hosel and clubshaft are on the elbow plane during the release swivel phase (as they were during the start up swivel phase).
Image 5
At impact, the clubshaft and hosel is on-plane, and still on the elbow plane.
In other words, Yodas Luke kept the clubshaft and hosel on the elbow plane during the start up swivel phase and also during release swivel phase of the swing, and the hosel never left the elbow plane (except momentarily when his hands went forward in a minor OTT move at the start of the downswing action).
Image 6
The clubshaft is now off-plane. It looks like he is performing a HH action, but he has allowed the clubshaft to leave the inclined plane (elbow plane). According to 10-10-D, the clubshaft should remain on the inclined plane during a HH action. Now, if anybody thinks that the hosel is leaving the elbow plane in order to rotate to the sweetspot plane 2 (as if the hosel is rotating around the sweetspot), then that interpretation cannot be correct because that should only happen by the 4th parallel. The clubface is only partially closed at this early post-impact time point, and the hosel is already on sweetspot plane 2. This is "steering" - due to a failure to keep the clubshaft on-plane post-impact. I think that Yodas Luke would have to pivot more leftwards post-impact so that he can keep the clubshaft on the inclined plane (elbow plane) during the followthrough. The inclined plane during the followthrough phase of the swing must be the elbow plane (and not sweetspot plane 2) because it has to be the same inclined plane post impact as it was pre-impact (during the release swivel phase). One cannot shift planes between the 3rd parallel and the 4th parallel and call that a symmetrical on-plane swing.
Not too shabby an effort, though. Especially considering that the club weighs about 15 pounds, its grip is about 1 1/2" in diameter, and the sweetspot is about 9" from the hosel.
Hybrid Mode
