Release – 2

[jake2]

To me this is not that complicated in principle. I do not know how well I can explain my thinking and if it's exactly correct, but I'll give it a try.

Here we go:

In order to maintain circular path of an object that has a mass, a force toward the center of rotation is required (centripetal). The required force is a function of mass, velocity and radius. Mass stays the same in this case.
- As velocity increases, the force required to hold the radius increases.
- The smaller the radius, the higher the required force.

A club(head) that's not released travels around at shorter radius than a club that has released.

Force to maintain the shorter radius can only be exerted to the grip end by the hands. The force needs to generate torque to the club that maintains the (wrist) angle (can we call it negative torque).
- Accelerating (not constant velocity) the grip end provides negative torque
- Significant amount of negative torque can also be exerted by pulling with the right hand (keeping right elbow bent).

As the velocity increases in the down swing, at some point this negative torque is not enough anymore to keep the radius and that causes the release.

We do not need a force to release the club, we need a force to keep it from releasing.

Of course, positive torque can be also exerted by actively straightening the right arm, but that is not necessary for the release to happen.

When the club is fully released (shaft inline with the center of rotation of the hands), the shaft provides the required force and no torque is needed.

 

[tobell]

Momentum transfer

Jake2,

I think I follow your post. In a broader sense regarding Mandrin's multi-segment single joint model, wouldn't the simplest explanation be momentum transfer from the inner segment to the distal outer segment of equal or lesser mass?

Thanks.

 

[jake2]

Jake2,
I think I follow your post. In a broader sense regarding Mandrin's multi-segment single joint model, wouldn't the simplest explanation be momentum transfer from the inner segment to the distal outer segment of equal or lesser mass?
Thanks.

I do not see how relative mass and momentum transfer would have anything to do with the release actually happening or not. As the release happens, yes there is momentum transfer. Also, as I understand it, momentum transfer happens also from a lesser mass to a larger mass, it just doesn't generate the same speed.

However, momentum transfer might have a lot to do with how efficiently club head speed is being generated in a real life golf swing.

 

[LiftOff]

Radius

Jake2,

What specifically is your 'radius'? From what to what?

When it's said that straightening the left leg just before impact shortens the radius, how is that accomplished?

Thanks.

 

[hcw]

Spot On

To me this is not that complicated in principle. I do not know how well I can explain my thinking and if it's exactly correct, but I'll give it a try.

Here we go:

In order to maintain circular path of an object that has a mass, a force toward the center of rotation is required (centripetal). The required force is a function of mass, velocity and radius. Mass stays the same in this case.
- As velocity increases, the force required to hold the radius increases.
- The smaller the radius, the higher the required force.

A club(head) that's not released travels around at shorter radius than a club that has released.

Force to maintain the shorter radius can only be exerted to the grip end by the hands. The force needs to generate torque to the club that maintains the (wrist) angle (can we call it negative torque).
- Accelerating (not constant velocity) the grip end provides negative torque
- Significant amount of negative torque can also be exerted by pulling with the right hand (keeping right elbow bent).

As the velocity increases in the down swing, at some point this negative torque is not enough anymore to keep the radius and that causes the release.

We do not need a force to release the club, we need a force to keep it from releasing.

Of course, positive torque can be also exerted by actively straightening the right arm, but that is not necessary for the release to happen.

When the club is fully released (shaft inline with the center of rotation of the hands), the shaft provides the required force and no torque is needed.

j2,
this is essentially what i was describing in post #31 of this thread...your hands hold on to the club in order to move it and your trail wrist bends to allow the clubhead to move further and gain the mechanical advantage of going from moving in a smaller arc to a larger arc after release...this allows for a large increase in clubhead speed (ie clubhead accleration) coming into impact...

-hcw

 

[jake2]

j2,
this is essentially what i was describing in #31of this thread..
-hcw

No argument there.
To me Mandrin's simplified examples show it clearly that no magical force etc. is needed for the release to happen. Also, the hands do not have to slow down for it to happen.
In my opinion simple dynamics principles (F=ma) can explain this.

 

[jake2]

Jake2,
What specifically is your 'radius'? From what to what?

What makes things a bit complicated in a real life golf swing is that there is no constant radius anywhere. It does not follow a circle.
When the radius keeps changing, you have to look at it individually at each point. The way I think about it in simple terms (and I only try to think in simple terms) is that if you draw a circle using the momentary curvature, you find the center and the radius. Center also is always at a line that is perpendicular to the momentary tangent.

When it's said that straightening the left leg just before impact shortens the radius, how is that accomplished?
Thanks.

Any time you increase the curvature of the motion, radius gets shorter. The trick here is that the center moves.
You could start with a straight line horizontal motion. If you pull up, it makes the path curve and the radius becomes smaller (was infinite with the straight line).

 

[jake2]

j2,
this is essentially what i was describing in #31of this thread..
-hcw

No argument there.
To me Mandrin's simplified examples show it clearly that no magical force etc. is needed for the release to happen. Also, the hands do not have to slow down for it to happen.
In my opinion simple dynamics principles (F=ma) can explain this.

 

[LiftOff]

Jake2

Yep, I follow all that. Makes sense. Rather simple when explained like that.

Many Thanks

 

[Bronco Billy]

That Sucker Sounds Good......

To me this is not that complicated in principle. I do not know how well I can explain my thinking and if it's exactly correct, but I'll give it a try.

Here we go:

In order to maintain circular path of an object that has a mass, a force toward the center of rotation is required (centripetal). The required force is a function of mass, velocity and radius. Mass stays the same in this case.
- As velocity increases, the force required to hold the radius increases.
- The smaller the radius, the higher the required force.

A club(head) that's not released travels around at shorter radius than a club that has released.

Force to maintain the shorter radius can only be exerted to the grip end by the hands. The force needs to generate torque to the club that maintains the (wrist) angle (can we call it negative torque).
- Accelerating (not constant velocity) the grip end provides negative torque
- Significant amount of negative torque can also be exerted by pulling with the right hand (keeping right elbow bent).

As the velocity increases in the down swing, at some point this negative torque is not enough anymore to keep the radius and that causes the release.

We do not need a force to release the club, we need a force to keep it from releasing.

Of course, positive torque can be also exerted by actively straightening the right arm, but that is not necessary for the release to happen.

When the club is fully released (shaft inline with the center of rotation of the hands), the shaft provides the required force and no torque is needed.

Man That Sucker Sounds Good..... Doesn't It...... I'd Like to See nm Rip the BGesus Out of it..... Mandrin Kinda Got a Little Wimpy in His Old Age..... Maybe He'll use His Third Language Excuse and still Bury it Diplomatically....If Neither One of them Can Bury it then We Got a Winner..... Ball Game Over..... Have a Great Day.....

 

[golfspike]

To me this is not that complicated in principle. I do not know how well I can explain my thinking and if it's exactly correct, but I'll give it a try.

Here we go:

In order to maintain circular path of an object that has a mass, a force toward the center of rotation is required (centripetal). The required force is a function of mass, velocity and radius. Mass stays the same in this case.
- As velocity increases, the force required to hold the radius increases.
- The smaller the radius, the higher the required force.

A club(head) that's not released travels around at shorter radius than a club that has released.

Force to maintain the shorter radius can only be exerted to the grip end by the hands. The force needs to generate torque to the club that maintains the (wrist) angle (can we call it negative torque).
- Accelerating (not constant velocity) the grip end provides negative torque
- Significant amount of negative torque can also be exerted by pulling with the right hand (keeping right elbow bent).

As the velocity increases in the down swing, at some point this negative torque is not enough anymore to keep the radius and that causes the release.

We do not need a force to release the club, we need a force to keep it from releasing.

Of course, positive torque can be also exerted by actively straightening the right arm, but that is not necessary for the release to happen.

When the club is fully released (shaft inline with the center of rotation of the hands), the shaft provides the required force and no torque is needed.

Sounds right that you need a force to stop releasing the club to early
Jake your examples of negative torque can you explain them more and would arching the left hand ( for a right handed player) be considered negative torque? and how about braking the pivot is that positive torque?

 

[jake2]

Jake your examples of negative torque can you explain them more and would arching the left hand ( for a right handed player) be considered negative torque? and how about braking the pivot is that positive torque?

If braking the pivot causes slowing down the hands, that would cause positive torque. Simply here the positive torque tries to un-cock the left wrist and negative would keep it cocked.

The arching part .. I think it's more of a result of properly executed pivot (than a cause for something). It gets more complicated, because we get into other axis of rotation that closes the club face (and there's separate forces and positive/negative torque rotating the shaft around it's axis).
Left wrist is un-cocking, forearms are rotating, so visualizing how the forces act is not that easy in my opinion. In this case the hands may be slowing down, but that's because release causes a momentum transfer to the club. Good pivot keeps the left arm pulling and that probably has something to do with the arched left wrist.

 

[golfspike]

If braking the pivot causes slowing down the hands, that would cause positive torque. Simply here the positive torque tries to un-cock the left wrist and negative would keep it cocked.

The arching part .. I think it's more of a result of properly executed pivot (than a cause for something). It gets more complicated, because we get into other axis of rotation that closes the club face (and there's separate forces and positive/negative torque rotating the shaft around it's axis).
Left wrist is un-cocking, forearms are rotating, so visualizing how the forces act is not that easy in my opinion. In this case the hands may be slowing down, but that's because release causes a momentum transfer to the club. Good pivot keeps the left arm pulling and that probably has something to do with the arched left wrist.

Thanks Jake so a good pivot causes an arched left hand/wrist would Ben Hogan be a good example of this or not? I have read a lot of conflicting opinions on the whole supination deal and even his golf swing from tgm that he became a 4-b hitter and others say hands manipulated swinger. You may possibly be able to clear up this matter although where and when a player exerts pressure in the hands/grip into and at impact is individual

 

[jake2]

Thanks Jake so a good pivot causes an arched left hand/wrist would Ben Hogan be a good example of this or not? I have read a lot of conflicting opinions on the whole supination deal and even his golf swing from tgm that he became a 4-b hitter and others say hands manipulated swinger. You may possibly be able to clear up this matter although where and when a player exerts pressure in the hands/grip into and at impact is individual

Golfspike, you are asking a lot. I'm not sure I fully understand what someone means by 4-barrel hitter. I do not think he was pure anything but then again, as Brian says, there are no pure hitters (did it apply for pure swingers too?). I guess 4-b in the classification implies that is not really pure, so it may not be easy to argue that it is not correct.

I do not know what sort of hand manipulation Hogan supposedly did. When I hear hand manipulation, I think of using forearm muscles to actively move wrists and hands coming to impact. I cannot believe that actively moving hands that way is going to help consistency. Other thing then may be if one uses the same muscles to maintain certain wrist alignments, that might be different. Whatever Hogan did, it was consistent.

 

[mandrin]

Our 'linear golfer' is learning and progressing.

 

[golfspike]

Our 'linear golfer' is learning and progressing.

Mandrin your the expert in physics give us a break down of Hogan's swing since he had the best Release ever

 

[tongzilla]

Mandrin your the expert in physics give us a break down of Hogan's swing since he had the best Release ever

why is his release the best?

 

[dbl]

golfspike, that is a tall order and something for a different thread. Hogan played magnificently, but modelling the dynamics of his 24 parts (or any golfer) would be quite a task. As you may read, Mandrin has been working on basic models (like the nunchuck) which have quite a few simpler parts and motion. As to judging the the quality of Hogan release, I cannot at the outset say that is "more efficient" than Purtzer's, Trevino's, CHIII's etc. It would take quite a bit to say what is a "best" release when each golfer is trying to achieve differend kinds of golf strokes. What makes a random release better than a snap release (or vice versa) would vary golfer by golfer. BTW, a golfer could also zero out several accumulators and then vastly simplify the release, and that somehow might fit within a notion of best.

From Mandrin's linear model, you might see that a reduced amount of effort, or effort applied at a certain time, can produce almost equal results for a golfer's clubhead. So maybe that is something Hogan learned from digging it out of the dirt, but perhaps WE can find ways to maximise our own effectiveness through this basic modelling.

 

[pecky987]

golfspike, that is a tall order and something for a different thread. Hogan played magnificently, but modelling the dynamics of his 24 parts (or any golfer) would be quite a task. As you may read, Mandrin has been working on basic models (like the nunchuck) which have quite a few simpler parts and motion. As to judging the the quality of Hogan release, I cannot at the outset say that is "more efficient" than Purtzer's, Trevino's, CHIII's etc. It would take quite a bit to say what is a "best" release when each golfer is trying to achieve differend kinds of golf strokes. What makes a random release better than a snap release (or vice versa) would vary golfer by golfer. BTW, a golfer could also zero out several accumulators and then vastly simplify the release, and that somehow might fit within a notion of best.

From Mandrin's linear model, you might see that a reduced amount of effort, or effort applied at a certain time, can produce almost equal results for a golfer's clubhead. So maybe that is something Hogan learned from digging it out of the dirt, but WE can perhaps find ways to maximise our own effectiveness through this basic modelling.

great post dbl
btw, i havnt looked at this thread alot, can we hav a sit-rep of what we now know??

 

[Dariusz J.]

why is his release the best?

Hogan's type of release (push release) is the best because it minimizes timing factor = making the swing more repetable, more consistent and more automatic.
It's not the best release type if someone wants to be a Re-Max champ, though. But this is of secondary importance.

Cheers