An uncompensated swing - The most efficient application of force, based on the laws of physics, geometry and anatomy.
Good answer EdZ!
Also I believe with an "Uncompensated swing" the alignments allow for maximum efficiency of power applied going into the ball plus the least amount of stress on the body parts.
Here are six alignments within the "Address" that I have found allow for maximum use of centrifugal force and least need for timing to hit the ball to the target.
1.The Left Wrist to the Club Shaft.
2.The left Wrist to the clubface.
3.The Club Shaft to the body center of gravity.
4.The Elbows to the hips.
5.The Spine to the ground.
6.The Clubface and Body to the ball’s target.
With these alignments in place and with a correct "pushing" action on the "pressure points" in the hands, all other alignments that are involved in the "Uncompensated Swing" are possible'
Hope this helps.
"An "On Plane" Right Shoulder from Start Down to Follow Through"
Impossible....
Think of it this way - the left arm and clubshaft are inline at followthrough.... thus the left arm is onplane and if this straight line is onplane this means the left shoulder is onplane.... Ok now think if the right shoulder is onplane at followthrough which by definition is both arms straight... both shoulders must be onplane, now if both shoulders are onplane and lets say for simplification purposes, you are using the between the shoulders pivot center point which you are a fan of anyways...
Now with this information imagine the stroke in reverse from follow through backwards. Both shoulders are onplane - the point between the shoulders is onplane - now as the right shoulder goes back to the top ... how can the left shoulder ever leave the inclined plane - hint hint it can't....
The onplane right shoulder movement occurs during the initial startdown to throw the primary lever assembly via creating a pressure at pp4 to drive it into impact..... it cannot stay onplane till followthrough....
"An "On Plane" Right Shoulder from Start Down to Follow Through"
Impossible....
Think of it this way - the left arm and clubshaft are inline at followthrough.... thus the left arm is onplane and if this straight line is onplane this means the left shoulder is onplane.... Ok now think if the right shoulder is onplane at followthrough which by definition is both arms straight... both shoulders must be onplane, now if both shoulders are onplane and lets say for simplification purposes, you are using the between the shoulders pivot center point which you are a fan of anyways...
Now with this information imagine the stroke in reverse from follow through backwards. Both shoulders are onplane - the point between the shoulders is onplane - now as the right shoulder goes back to the top ... how can the left shoulder ever leave the inclined plane - hint hint it can't....
The onplane right shoulder movement occurs during the initial startdown to throw the primary lever assembly via creating a pressure at pp4 to drive it into impact..... it cannot stay onplane till followthrough....
Disagree - does anyone believe the downswing is a simple pivot around the spine? Isn't this necessary for your argument to hold?
Disagree - does anyone believe the downswing is a simple pivot around the spine? Isn't this necessary for your argument to hold?
Chris
This is a power package issue and you if try to draw an onplane right shoulder in 3d with the primary lever assembly being inline at followthrough, you'll see exactly what I mean.
De-fogging: Flying Wedges, Hinge Action and Basic Stroke Elbow Location
Originally Posted by Thom
Nice list Tongzillium
Sorry if I'm threadjacking but....
I've just had an AHA-moment reading the third sentence.
The perpendicular relationship of the Flying Wedges: the Flat Left Wrist and Level Right Wrist. Isn't it right that if you keep this relationship the right elbow position will follow the hingeaction:
-so if your're swinging with horizontal hinging, you'll turn the flat left wrist to the plane, that'll automatically put the right elbow in pitch position.
-and if you're hitting with angled hinging, you'll turn the flat left wrist less, and the perpendicular level right wrist will automatically put the right elbow in punch/push position.
Am I right?
I think you got a few concepts mixed up here: The Flying Wedges, Hinge Action, and the Basic Stroke Elbow Locations at Release.
The relationship of The Flying Wedges with itself (the Left Arm Wedge and Right Forearm Wedge) are the same whether you’re Hitting or Swinging. In other words, you should always have a Strong Single Action Grip (10-2-B) with a Flat Left Wrist and Level Right Wrist from Start Up to Follow Through. However, the relationship of The Flying Wedges (as a whole) to the Inclined Plane may differ between Hitting and Swinging.
All differences between Hitting and Swinging (including Hinge Action and Elbow Location at Release) essentially stems from their different Acceleration methods -- Longitudinal (Pull) for Swinging vs Radial (Push) for Hitting. And it is because Swingers need to Drag Load (10-19-C) the Club that they need to use Standard Wrist Action (10-18-A) which gives the Pitch Basic Stroke Elbow.
So, I’d say your concept that the Left Wrist is “turned less” with Hitting than Swinging is right, but you are confused about why such differences occur.
You are almost there, I hope this post will clear some fog.
This is a power package issue and you if try to draw an onplane right shoulder in 3d with the primary lever assembly being inline at followthrough, you'll see exactly what I mean.
I'm still unconvinced...
Say you did keep the right shoulder onplane to followthrough, and the primary lever assembly ended up inline, in what way would you be off plane (e.tg. over/under etc), or where would the left shoulder be?
Say you did keep the right shoulder onplane to followthrough, and the primary lever assembly ended up inline, in what way would you be off plane (e.tg. over/under etc), or where would the left shoulder be?
Chris
You could draw that one picture in a position - but just tell me how your going to get there. If the right and left shoulder are on a plane any line between that point is going to be onplane too. So if you use a point between the shoulders center for simplicity - that will be onplane too(not quite true but close enough for our purposes).
Now if the stationary point stays onplane - the right shoulder stays onplane - how can the left shoulder leave that plane. Now try to visualise the shoulder motions and ask yourself, how can you have a top of the backstroke where this could happen....it can't....
You could draw that one picture in a position - but just tell me how your going to get there. If the right and left shoulder are on a plane any line between that point is going to be onplane too. So if you use a point between the shoulders center for simplicity - that will be onplane too(not quite true but close enough for our purposes).
Now if the stationary point stays onplane - the right shoulder stays onplane - how can the left shoulder leave that plane. Now try to visualise the shoulder motions and ask yourself, how can you have a top of the backstroke where this could happen....it can't....
Independent movement of the shoulders? They are not a single T-bar. In fact this independent movement need not even occur during the downswin. If both shoulders are forward at the top then they curve forward from the spine. I agree any point between the shoulders would need to be on plane, just not that the spine would need to be between the shoulders - in a linear fashion that is!
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