Release – 2

[mandrin]

Back to the drawing board

Mandrin,
I meant that if the velocity of inner stick is constant or accelerating (the distances between red dots (chain) are constant or getting bigger) the outer stick will not release - the angle between sticks will remain constant - because the chain is much looser than a human wrist.

Dariusz,

I am afraid that you have to do a bit of re-thinking on release.

Figure shows release for constant angular velocity of inner stick.

Notice that there remains a nice release occurring nevertheless.

 

[Dariusz J.]

Mandrin,

Imagine a machine holding the inner stick and turning with a constant velocity, imitating a swing motion on a constant plane. The only one difference would be that the machine is able to turn 360 degress over and over. Would you suggest that the outer stick will release in some point ?
IMHO, the angle between sticks (say, lag of outer stick) will remain more or less the same. I wrote more or less, since I am not sure if gravity and the weight of the outer stick may be changing the angle a bit in given stages of the circular motion. I cannot imagine a nunchakoo acting like on the above diagram of yours (provided the velocity and plane is constant). Please enlight me where I am wrong in my thinking - I think it may be a crucial info for laics like me to follow your work further with understanding. Thanks.

Cheers

 

[dbl]

Mandrin, is that nunchuck example in post 21 in a vertical plane and so gravity has a significant affect? I would have thought a simplified versin of Darius' example was that the inner stick was on a horizontal plane; at rest the outer stick droops vertically and then with a constant angular velocity would trail and also rise some. But I don't see the action of the outer stick passing the inner stick for that case. P.s. One can tape two pens together to make a small model. So we need a bit more information on what plane things are happening and what forces you are having in play, if you don't mind.

 

[Biffer]

No passing zone

The outer stick will "release" and become inline with the inner stick and remain in that position until the inner stick slows.

 

[Dariusz J.]

One more thing that I am sure is important. What is the velocity of the inner stick ? Too slow and the gravity will overcome and the outer stick will release; fast enough and the gravity will be overcomed and the outer stick does not release.
I remember a trick of water that cannot go out of the small bucket that is tied to a string and one turns it around ona vertical plane; the speed must be high enough not to let water pour out of the bucket.

Cheers

 

[Bronco Billy]

The Gravity Release....

One more thing that I am sure is important. What is the velocity of the inner stick ? Too slow and the gravity will overcome and the outer stick will release; fast enough and the gravity will be overcomed and the outer stick does not release.
I remember a trick of water that cannot go out of the small bucket that is tied to a string and one turns it around ona vertical plane; the speed must be high enough not to let water pour out of the bucket.

Cheers

Damn...... This Means I Can't Play Golf in a Weightless Environment.....

 

[mandrin]

Mandrin,
Imagine a machine holding the inner stick and turning with a constant velocity, imitating a swing motion on a constant plane. The only one difference would be that the machine is able to turn 360 degress over and over. Would you suggest that the outer stick will release in some point ?
IMHO, the angle between sticks (say, lag of outer stick) will remain more or less the same. I wrote more or less, since I am not sure if gravity and the weight of the outer stick may be changing the angle a bit in given stages of the circular motion. I cannot imagine a nunchakoo acting like on the above diagram of yours (provided the velocity and plane is constant). Please enlight me where I am wrong in my thinking - I think it may be a crucial info for laics like me to follow your work further with understanding. Thanks.

One more thing that I am sure is important. What is the velocity of the inner stick ? Too slow and the gravity will overcome and the outer stick will release; fast enough and the gravity will be overcomed and the outer stick does not release.
I remember a trick of water that cannot go out of the small bucket that is tied to a string and one turns it around ona vertical plane; the speed must be high enough not to let water pour out of the bucket. Cheers

Dariusz,

Gravity only plays a minor role in a real golf swing hence don’t complicate things with gravity. Simply think in terms of slender rods resting on a horizontal table surface, with no friction, and rotating around a fixed center. I am trying to simplify things to the utmost without loosing the essential features.

Translating your idea of imposing constant angular velocity on the motion of the inner rod, than, mathematically derived, there is a release as I have shown. However, even when imposing your suggestion of angular acceleration for the inner segment there still will be a release as shown in figure above.

 

[mandrin]

Mandrin, is that nunchuck example in post 21 in a vertical plane and so gravity has a significant affect? I would have thought a simplified versin of Darius' example was that the inner stick was on a horizontal plane; at rest the outer stick droops vertically and then with a constant angular velocity would trail and also rise some. But I don't see the action of the outer stick passing the inner stick for that case. P.s. One can tape two pens together to make a small model. So we need a bit more information on what plane things are happening and what forces you are having in play, if you don't mind.

dbl,

Imagine simply the two rods to be resting on a horizontal surface presenting no friction and rotating around a fixed inner center. Let’s keep it as simple as possible and hence eliminate gravity as it only plays a minor role in developing clubhead speed in a real golf swing.

What I have done simply is to translate Dariusz’ suggestions for velocity and acceleration for the motion of the inner rod into mathematics. In real life this is not necessarily easy, even for an Iron Byron, but in mathematics it poses no problem whatsoever.

 

[mandrin]

The outer stick will "release" and become inline with the inner stick and remain in that position until the inner stick slows.

Biffer,

The figures I have produced in posts #21 and #27 show respectively the inner segment to have either constant velocity or increasing velocity at and beyond the inline situation. Yet releases occurs quiet readily in both cases. Just hang on and I might be able to make it all clear slowly and progressively as we go along. Well, let's hope so, otherwise I have created only confusion.

 

[Dariusz J.]

Dariusz,

Gravity only plays a minor role in a real golf swing hence don’t complicate things with gravity. Simply think in terms of slender rods resting on a horizontal table surface, with no friction, and rotating around a fixed center. I am trying to simplify things to the utmost without loosing the essential features.

Translating your idea of imposing constant angular velocity on the motion of the inner rod, than, mathematically derived, there is a release as I have shown. However, even when imposing your suggestion of angular acceleration for the inner segment there still will be a release as shown in figure above.

Mandrin,

OK, let's think about a horizontal system, as you propose to get rid of the minor role of gravity.
One more question: if you make all red dots form a perfect circle (360 degrees) and apply a constant velocity of the inner rod turning over and over (even no acceleration) - why should the outer rod release ? and if it will - where should it release if the circle is perfect ? Please explain.

The only thing that comes to my mind is that the wrist (chain, red dot) does not move on an ideal circle (e.g. on a elipse) and any change of the shape of the path influences the ability of outer rod to release in a specific point/time even if the velocity is maintained or even there is acceleration.

Cheers

 

[hcw]

Experiment with a Golf Club

Take a club, set up and go the the top of your backswing and then make your pivot move down in as "slow motion" as possible and focus on keeping your trail wrist bent. Going slowly, you can hold off release for a very long time. In fact you can even get the clubhead back to the ball, just quite a bit above it!. Now if you repeat the exercise over and over, pivoting a little faster each time you will see that it gets harder and harder (indeed at some point impossible) to hold off release and then the clubhead will drop to the ground. Actually it will get a little closer and closer as you speed up the pivot and eventually you just cannot keep the clubhead from fully releasing. Where/when that point is depends on how much inertia/momentum/force your pivot has imparted to the clubhead versus how strong you are. You probably could also add to this natural release via activating the appropriate muscles, but that takes timing/coordination! Hopefully this real world experiment hasn't offended my buddy mandy's preference for "in silico" discussions and I'm sure he will have an impressively intelligent and witty response (punctuated appropriately with emoticons) telling me how/why I'm wrong about at least something.

-hcw

 

[mandrin]

Mandrin,

OK, let's think about a horizontal system, as you propose to get rid of the minor role of gravity.
One more question: if you make all red dots form a perfect circle (360 degrees) and apply a constant velocity of the inner rod turning over and over (even no acceleration) - why should the outer rod release ? and if it will - where should it release if the circle is perfect ? Please explain.

The only thing that comes to my mind is that the wrist (chain, red dot) does not move on an ideal circle (e.g. on a elipse) and any change of the shape of the path influences the ability of outer rod to release in a specific point/time even if the velocity is maintained or even there is acceleration.

Cheers

Dariusz,

Have a closer look and check it out - the red dots in the figures ( P#19, 21, 27) are all situated on a perfect circle.

Let me pose you a question - why do you think that the outer rod should not release when there is a constant velocity imposed on the inner segment?

 

[Dariusz J.]

Dariusz,

Let me pose you a question - why do you think that the outer rod should not release when there is a constant velocity imposed on the inner segment?

Mandrin,

I try to answer you (forgive me for using not proper words): because I think that there is a force missing in the system that could cause the outer rod release. If the velocity is constant, the outer rod should always be in the same position in relation to the inner rod. I simply cannot see why the position of outer rod should change in any moment of the circular motion of the inner rod in such a stable system.
Perhaps I am missing a very simple and very important thing - but it's my common sense talking now.

Cheers

 

[mandrin]

Mandrin,

I try to answer you (forgive me for using not proper words): because I think that there is a force missing in the system that could cause the outer rod release. If the velocity is constant, the outer rod should always be in the same position in relation to the inner rod. I simply cannot see why the position of outer rod should change in any moment of the circular motion of the inner rod in such a stable system.
Perhaps I am missing a very simple and very important thing - but it's my common sense talking now.

Cheers

Dariusz,

We are discussing issues related to golf down stroke. This is a motion limited in time and space. It starts form zero initial conditions and attains a certain motion and energy. At no point in time or space is there any equilibrium situation or position.

You however are referring to a very different situation comprising a continuous motion where steady state conditions have been reached. No wonder that we are not quite on the same wavelength. But it is not appropriate to be used as some reference for a golf down stroke.

Don’t worry about English, it is my third language and I fully sympathize with any problem of expressing ideas clearly in another language as I do spend a lot of time doing my best in this regard.

 

[mandrin]

Lots of POSITIVE TORQUE involved.

Brian,

It would definitely be interesting to get a little more flesh on the bones.

 

[Dariusz J.]

Dariusz,

We are discussing issues related to golf down stroke. This is a motion limited in time and space. It starts form zero initial conditions and attains a certain motion and energy. At no point in time or space is there any equilibrium situation or position.

You however are referring to a very different situation comprising a continuous motion where steady state conditions have been reached. No wonder that we are not quite on the same wavelength. But it is not appropriate to be used as some reference for a golf down stroke.

Don’t worry about English, it is my third language and I fully sympathize with any problem of expressing ideas clearly in another language as I do spend a lot of time doing my best in this regard.

Mandrin,

If English is your 3rd language I do admire your linguistic talents

Ad rem: yes, I did refer to a hypothetical completely different model than a human golf swing in order to clear all aspects from the beginning.

Therefore, as I understand, you confirm that in a model I have described the outer rod will not release provided the inner rod is constantly turning with a stable angular velocity.
If the situation is different in reality, I can see only one reason why - because the downswing has its start and has its end. Correct ?

Cheers

 

[miki111]

Mandrin,
so at the end you can say, that golfclub release because hands (or arms or shoulders or something closer to axis of rotation then club and the clubhead) will slow down prior or during release?

 

[golfspike]

Mandrin,

If English is your 3rd language I do admire your linguistic talents

Ad rem: yes, I did refer to a hypothetical completely different model than a human golf swing in order to clear all aspects from the beginning.

Therefore, as I understand, you confirm that in a model I have described the outer rod will not release provided the inner rod is constantly turning with a stable angular velocity.
If the situation is different in reality, I can see only one reason why - because the downswing has its start and has its end. Correct ?

Cheers

Dariusz
I get what you are saying but i think the red dot ( the wrist) in the golf swing has bending cocking hingeing or whatever its called so even if the inner rod is moving at a stable speed there will still be a release factor and clearly because the ball needs to be hit. Now if there is no ball there that can be a totally different story but that's not golf
Cheers

 

[Dariusz J.]

Golfspike,

No. As far as I am concerned, Mandrin proved that no matter if there is a chain instead human wrists, the outer rod will release this way or another.
If you go back to the thread titled "Release-1" you will notice that the question of intention to hit the ball or not has already been discussed and Mandrin said that it does not matter.

Cheers

 

[golfspike]

Golfspike,

No. As far as I am concerned, Mandrin proved that no matter if there is a chain instead human wrists, the outer rod will release this way or another
If you go back to the thread titled "Release-1" you will notice that the question of intention to hit the ball or not has already been discussed and Mandrin said that it does not matter

Cheers

i would have thought that if it were a chain instead of a wrist it would release more and the intention to strike the ball does not matter!! really hmmm can scientists prove this???