Acceleration.....does it matter?

[Golf nerd]

"If a truck hits me at 100 mph, will it matter if it was accelerating or hitting the brakes? I will get squished the same amount won't I?"

Elk ragout, hehehe.

[Krt22]

[quote name='Golf nerd' timestamp='1412703742' post='10250069']
[quote name='larrybud' timestamp='1412686838' post='10248467']
[quote name='Golf nerd' timestamp='1412620426' post='10244127']
[quote name='larrybud' timestamp='1412615693' post='10243675']
[quote name='Golf nerd' timestamp='1412614207' post='10243535']
Acceleration is vital to keep the clubhead behind the hands (lag) in the downswing. As soon as acceleration of the arms and hands stops, the release takes place (P6).
Usually acceleration of body parts stops around P6
[/quote]

You have cause and effect backwards. The body slows because the release takes place. Conservation of angular momentum. Just like an ice skater in a twirl slows when they expand their arms outward from their body... they don't slow their body to expand their arms, their body slows BECAUSE they expand their arms.
[/quote]

Don't think[b] I[/b] have it backwards.
Study Dave Tutleman and the effect of the double pendulum.
What causes the release is as I stated, the centrifugal force or call it angular momentum, which forces the lag angle to open, when the inertia of the club no longer stays behind the accelerating hands and arms.
[/quote]

That's not what you said. You said release is caused when the acceleration of the hands and arms stop. That's not true. The hands and arms slow down mostly because of other parts which speed up, i.e. the clubhead. (Note, I say "mostly" because the body has certain range limitations and the muscles can contract only so much).

The hand and arms slow down because the radius of the arc is increasing as the player's wrists unhinge, just as the spin rate of an ice skater slows as the arms are extended (the radius of the arm is increased).

Due to conservation of angular momentum, the arms slow down because the momentum of the arm/club segment remains constant, that is, energy is conserved. If you actively slow down muscle segment of your body, total momentum of the unit is reduced.
[/quote]

Maybe we are talking about different things. I said the release starts when the acceleration of hands and arms stops. You say that is not true. Then you say the clubhead is causing the release, because the clubhead is speeding up. I ask you: what is speeding up the club?

My answer:
As long as you accelerate the acceleration keeps the clubhead behind. That's called moment of inertia.
[/quote]

That is incorrect, the moment of inertia of a body is a fixed parameter based on its mass and geometry

[larrybud]

[quote name='Golf nerd' timestamp='1412703742' post='10250069']
[quote name='larrybud' timestamp='1412686838' post='10248467']
[quote name='Golf nerd' timestamp='1412620426' post='10244127']
[quote name='larrybud' timestamp='1412615693' post='10243675']
[quote name='Golf nerd' timestamp='1412614207' post='10243535']
Acceleration is vital to keep the clubhead behind the hands (lag) in the downswing. As soon as acceleration of the arms and hands stops, the release takes place (P6).
Usually acceleration of body parts stops around P6
[/quote]

You have cause and effect backwards. The body slows because the release takes place. Conservation of angular momentum. Just like an ice skater in a twirl slows when they expand their arms outward from their body... they don't slow their body to expand their arms, their body slows BECAUSE they expand their arms.
[/quote]

Don't think[b] I[/b] have it backwards.
Study Dave Tutleman and the effect of the double pendulum.
What causes the release is as I stated, the centrifugal force or call it angular momentum, which forces the lag angle to open, when the inertia of the club no longer stays behind the accelerating hands and arms.
[/quote]

That's not what you said. You said release is caused when the acceleration of the hands and arms stop. That's not true. The hands and arms slow down mostly because of other parts which speed up, i.e. the clubhead. (Note, I say "mostly" because the body has certain range limitations and the muscles can contract only so much).

The hand and arms slow down because the radius of the arc is increasing as the player's wrists unhinge, just as the spin rate of an ice skater slows as the arms are extended (the radius of the arm is increased).

Due to conservation of angular momentum, the arms slow down because the momentum of the arm/club segment remains constant, that is, energy is conserved. If you actively slow down muscle segment of your body, total momentum of the unit is reduced.
[/quote]

Maybe we are talking about different things. I said the release starts when the acceleration of hands and arms stops. You say that is not true. Then you say the clubhead is causing the release, because the clubhead is speeding up. I ask you: what is speeding up the club?

My answer:
As long as you accelerate the acceleration keeps the clubhead behind. That's called moment of inertia. As soon as you acceleration is over, the angle is released and the clubhead is now getting faster than the hands and is accelerating, while hands and arms slow down! The same happens in a car for example. While accelerating your body is pushed back into the seat, because the mass of your body has a high moment of inertia. As soon as your acceleration stops, your body is not pushed into the seat anymore. If you brake, your body goes forward. The example is not perfect, because it is linear. Take a curve and feel the centrifugal force. The tighter you curve the more are you forced to the outside, although you are not driving faster.
P6 to impact is a tight curve.

Like RBImGuy mentioned, pros accelerate during chip and pitch shots that the clubhead lags behind (thru inertia). If you decelerate into impact, the clubhead speeds up and pass the hands before or during impact.
[/quote]

The wrists unhinging (unc0ck) is what speeds up the club head because it increases the arc length. The length of the arc increases, and conservation of angular momentum slows the hands/arms. I will go back to the ice skater. When she is spinning, and her hands are close to her body (her hands being like the clubhead in this case) the body is spinning fast. When the arms are extended, the hands are now going faster compared to the speed they were going when they were close to her body. The radius of the arc of her hands has increased, and her body has slowed. She didn't slow her body to speed up her hands, her body slowed BECAUSE she extended her arms, which sped up her hands.

I'm sorry, but you don't understand your own analogy: When you brake a car, your body does NOT accelerate relative to the ground. Your body is going faster than the car because your body has momentum and the car is decelerating. You certainly don't brake the car to speed up your body relative to the ground. If that was the case your body would never slow down!

I'll give you another analogy: Are you familiar with a medieval weapon call a trebuchet? The arm of the trebuchet slows as the sling expands outward from the arm due to conservation of angular momentum. There's no mechanism (ie. muscles) to have the arm of the trebuchet slow down. It slows because the radius of the arc of the sling increases.

Just to drive the point: If slowing your arms and hands sped up the clubhead as you state, then the goal in golf would be to stop your hands from moving completely before impact.

[Lefthook]

[quote name='iteachgolf' timestamp='1412648974' post='10246961']
No the only thing that matters is club head speed at separation. Obviously assuming loft, AOA, and all those variables are constant. Nothing "packs more punch". Speed when the ball leaves the face is what matters.
[/quote]

Yes,

Ch speed at separation.

But you have argued that ch speed at first contact is all that matters and I would like to see the evidence.

[iteachgolf]

[quote name='Lefthook' timestamp='1412710789' post='10250663']
[quote name='iteachgolf' timestamp='1412648974' post='10246961']
No the only thing that matters is club head speed at separation. Obviously assuming loft, AOA, and all those variables are constant. Nothing "packs more punch". Speed when the ball leaves the face is what matters.
[/quote]

Yes,

Ch speed at separation.

But you have argued that ch speed at first contact is all that matters and I would like to see the evidence.
[/quote]

Where did I argue that first contact is all that matters. I used separation and not contact in every post on this thread. Do you instantly try to discredit every post I make before actually reading all the post in the thread

[Lefthook]

I teach,

I was also referring to this:

[quote name='iteachgolf' timestamp='1412641341' post='10246071']
If speed, contact, and AoA are the same assuming same face angle both horizontally and vertically the ball will fly the same. What makes you think that keeping all variables constant would change the results?
[/quote]

Your responses are ambiguous. You have been arguing against those who believe that what goes on in the impact interval makes a difference. Yet, you switch from speed to ch speed on separation in between posts here. The last is typically Tgm stuff and homer k would never emphasize this unless he thought impact made a difference... And the only significant difference in this regard is work during the impact interval. And the only way to add work is to add force while the club is in contact with the ball. Leverage ....

Look, I have high regards of your skills and knowledge. Very high actually. But quite often you appear to be too sure about topics that hasn't been properly researched imo. And you come out pretty hard on people who ask smart questions. And impact is one of those areas that is not properly understood.

[iteachgolf]

And where did I say first contact? I didn't switch. Every comment where I specified a point in time referred to separation. How is it ambiguous? I don't think I could have been clearer. There was no switching. I first asked why a person believed keeping those variables constant would result in different flights. And then added clarity by stating if all else is the same at separation then flights are the same regardless of what happened before. I then explained how the differences in flight would be from those variables not being the same as humans aren't machines and different swings produce different alignments. What does TGM or Homer Kelly have to do with anything? Where am I arguing against people who say the impact interval makes a difference? I simply made statements of fact, and they are facts. The OP didn't say anything about impact interval just that if clubhead speed was constant when hitting the ball (I read that as separation) would there be a difference if one was slowing down PRIOR TO IMPACT.

Where am I wrong and where am I not clear on what I stated? I never even discussed what happens from impact to separation. What I said was simply if all else is equal at separation than what happened before makes no difference. You are the one trying to turn this into something it isn't and simply trying to call me out every chance you get.

[iteachgolf]

[color=#ff0000]Here are all the post I made in this thread so we can take a look at how ambiguous I was and how hard I was on people who asked "smart" questions. [/color]


[quote name='iteachgolf' timestamp='1412641341' post='10246071']
If speed, contact, and AoA are the same assuming same face angle both horizontally and vertically the ball will fly the same. What makes you think that keeping all variables constant would change the results?
[/quote]

[color=#ff0000]So here all I said is that if you change no variables the results won't change. I know so ambiguous [/color]

[quote name='Albatross85' timestamp='1412643232' post='10246265']
[quote name='iteachgolf' timestamp='1412641341' post='10246071']It just seems there has to be something that differentiates flights and distances from one to the next. Where does leverage come into the equation? And lag? You can't tell me all 95 mph swings are the same. Human beings are not machines so there has to be some other factors in play here
[/quote]

They aren't. Contact, effective loft, and aoa all vary between players. What is leverage? More shaft lean results in a lower effective loft and changes contact point on face which results in very different flights. The consistency between swings varies as well individual to individual. But that is not what you asked. You asked if those variables are the same would the flight be the same and that answer is yes
[/quote]

[color=#ff0000]And here I stated if the flights are different than all the variables aren't the same. I know ground breaking[/color]

[quote name='iteachgolf' timestamp='1412644054' post='10246369']
No this was proven invalid a long time ago. "resisting deceleration" has no effect. Same club head speed at separation will equal same distance all else being equal. Regardless of what the speed reached prior to separation
[/quote]

[quote name='iteachgolf' timestamp='1412648974' post='10246961']
No the only thing that matters is club head speed at separation. Obviously assuming loft, AOA, and all those variables are constant. Nothing "packs more punch". Speed when the ball leaves the face is what matters.
[/quote]

[color=#ff0000]And the only times I actually get specific about impact I use the word separating. And simply made statements of fact that seem pretty clear to me. Never said one work about impact interval or initial contact. I was pretty consistent about using the word separation and only the word separation. Don't think I was hard on anyone or said anything that is the least bit controversial that "[/color][color="#ff0000"]hasn't been researched enough"[/color]

[DeNinny]

OP, I am an engineer and here is almost as far* as I can take the physics and math (*I stop short of detailed potential energy equations). In short, I think acceleration matters mainly because it is the only way that the shaft can stay loaded until impact. In order for the shaft to stay bent, a constant force (or torque to be exact) needs to be applied on the shaft. By accelerating the swing, the clubhead mass will also accelerate and thus impart the proper force at the end of the shaft in order to bend and load it. At impact, this unloading of the shaft is additional energy imparted on the ball in addition to the velocity based (i.e. non-accelerated) kinetic energy of the clubhead. I think this is an important part of the physics and it is evidenced in the real world with golfers getting fitted with the proper shaft flex, in conjunction with proper swing weight, for one’s tempo and swing speed.

I also want to be clear that I still think the clubhead velocity is the most important aspect, but I myself cannot deny my own experience being slightly longer using regular flex shafts vs stiff or x-flex. I have a 90 mph driver SS, so regular flex fits me (along with my tempo) best.

Ok that was my short answer. Warning and my pre-apology for the length of the rest of this as my detailed “proof”. In the below diagram is what is happening at impact in three phases, which I call “contact”, “compress”, and “detach”:

[attachment=2449683:20140925_192048.jpg]

During the “contact” phase, the clubhead is just impacting the ball. At the point of impact, time t0 = 0, the clubhead with mass, Mc, and velocity, Vc, contacts the ball with mass, Mb. Assuming no acceleration, the total energy of impact is simply the kinetic energy, KEc, of the clubhead. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- assumes no acceleration)

where, by definition, kinetic energy is one-half times the mass times the velocity squared. Again, I want to emphasize I have neglected acceleration at this point in order to keep the equation simple.

During the “compress” phase, the clubhead and ball are traveling together as one object. Together they travel a total distance Dc for a time t1. The kinetic energy is transferring from the clubhead to the ball at this time as well. Since the ball and clubface are not rigid objects, they both undergo a compression. Both are feeling the energy of impact from the clubhead’s total energy. The middle diagram shows this with an oblong ball and curved clubface. Because of the compression, the molecules of the ball end up rubbing together and generating heat from the friction, which I have noted as Hc. This is a loss of some (although I doubt it is much) energy from the energy of impact. I want to note also that this is the actual time that forces are imparted on the ball. (Again, I’m still ignoring forces from clubhead acceleration for now.) I also want to emphasize that during the latter part of the this time, t1, the ball is also decompressing and returning to normal shape, progressively tending toward this just prior to detaching from the clubface at the end of this phase. Lastly, I believe others have mentioned that there is deceleration at impact, and while I agree that this is likely happening initially in this phase, I believe there is also some acceleration by the ball decompressing (just like a spring unloading) and other reasons, like the shaft unloading, during the end of this brief phase.

During the “detach” phase, the ball with mass, Mb, then leaves the clubface at a velocity, Vb, based on the remaining energy of the system. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- equation from before)

= Hc + ½ x Mb x Vb x Vb (<- heat loss from the ball’s compression and the remaining energy is transferred to the ball)

Now let’s incorporate the energy from loading the shaft, PEs, and also just the pure acceleration from the golfer, PEg. Both of these energies are considered potential energies because they are both accelerative forces that have yet to enact this force onto the ball. There is acceleration (or the potential for it), but at this point there is no energy transfer because there is no object for which to apply the force. This is why it is a potential energy. Now the total energy equation becomes

Total Energy = KEc + PEs + PEg (<- kinetic energy of clubhead plus the potential energies of the loaded shaft and from the golfer’s acceleration)

I have defined KEc by the basic kinetic energy equation, so now I’ll explain the shaft potential energy, PEs. The only way to have stored energy from the shaft, you have to keep it in a bent position. The only way to keep the shaft in a bent position is to apply a constant force to it. The only way to have a constant force is to have an acceleration on the clubhead. By swinging with acceleration, the clubhead weight on the other end of the shaft essentially creates a lag force (or rather, torque) that bends the shaft. (Remember that Force = Mass x Acceleration.) As soon as acceleration is stopped and the clubhead is traveling at constant velocity, the shaft will unbend or unload. Therefore, [i][b]acceleration is needed in order for the shaft to stay bent (i.e. loaded) until impact[/b][/i].

I have not defined the PEg, the potential energy from the golfer accelerating throughout the “compress” phase. I can tell you it will have its own set of variables. My gut feel is that the kinetic energy of the clubhead velocity and the potential energy from the shaft unloading play the greater role in overall shot distance than this potential energy. In fact, the shaft staying loaded is already taking (or storing) a fair portion of the golfer’s acceleration energy, so not much is “left over”.

To close this out, here is the final total energy equation with everything I've explained:

Total Energy = KEc + PEs + PEg = ½ x Mc x Vc x Vc + PEs + PEg

= Hc + ½ x Mb x Vb x Vb (<- assumes the ball is leaving the clubface at constant velocity)

Ultimately you could solve this equation for Vb, the velocity of the ball. You can see that having PEs and PEg as part of this equation will increase the resulting Vb, ultimately resulting in more distance. Thus in my personal opinion,

KEc > PEs >> PEg

The kinetic energy of the clubhead matters the most, but the potential energy of the bent shaft plays a significant role too. The loaded shaft is stored potential energy up until impact where it is able to unload this stored energy into the ball. Again, this is supported by the fact that you should get fitted for the proper shaft flex in order to maximize your potential distance. If shaft flex [i]weren’t [/i]important, then we could play with the stiffest and lightest shafts possible and still be able hit it as far as we do now.

I’m going to stop here unless you want more equations about the shaft bending and such. I hope this gives you an idea of the physics and math involved to truly answer your question.

[Lefthook]

[quote name='iteachgolf' timestamp='1412641341' post='10246071']
If speed, contact, and AoA are the same assuming same face angle both horizontally and vertically the ball will fly the same. What makes you think that keeping all variables constant would change the results?
[/quote]

[color=#ff0000]Iteach: So here all I said is that if you change no variables the results won't change. I know so ambiguous [/color]

At least your sarcasm is crystal clear....

The question in this thread deals with whether acceleration profile matters. You imply that it doesn't... If it doesn't you may as well speak about ch speed at first contact, since the relationship between first contact and separation will be a constant.

[quote name='iteachgolf' timestamp='1412641341' post='10246071'][quote name='Albatross85' timestamp='1412643232' post='10246265']
It just seems there has to be something that differentiates flights and distances from one to the next. Where does leverage come into the equation? And lag? You can't tell me all 95 mph swings are the same. Human beings are not machines so there has to be some other factors in play here
[/quote]

They aren't. Contact, effective loft, and aoa all vary between players. What is leverage? More shaft lean results in a lower effective loft and changes contact point on face which results in very different flights. The consistency between swings varies as well individual to individual. But that is not what you asked. You asked if those variables are the same would the flight be the same and that answer is yes
[/quote]

[color=#ff0000]Iteach: And here I stated if the flights are different than all the variables aren't the same. I know ground breaking[/color]

Here you are neglecting the essence of OP's question: Does the acceleration profile make a difference to the ball flight? As far as I can see, you imply that the answer is "no".

[quote name='iteachgolf' timestamp='1412644054' post='10246369']
No this was proven invalid a long time ago. "resisting deceleration" has no effect. Same club head speed at separation will equal same distance all else being equal. Regardless of what the speed reached prior to separation
[/quote]

And here you are explicit of a related topic: Resisting deceleration. I have seen some "evidence" too. But it isn't convincing. It has been dismissing axial forces up the shaft. There may be other variables too that hasn't been accounted for.

The extreme case of not resisting decel is shaft breakage with a wedge right at impact. The club head will deflect towards the ground and the ball flight will be lower. The opposite would be (relatively) fast moving hands that resists this deflection.

There are indications that the golfer makes a difference through the ball. KM has posted some very interesting videos about impact that puts a big question mark beside the massively simplified mainstream way of interpreting impact. Side spin in the wrong direction (gear effect overriden by what?). Different spin rates and ball flights due to minute differences during the impact interval between flop wedges, punches and those low spinning two bounce & stop approaches. He has also detected different durations of the impact interval between those shots.

[quote name='iteachgolf' timestamp='1412648974' post='10246961']
No the only thing that matters is club head speed at separation. Obviously assuming loft, AOA, and all those variables are constant. Nothing "packs more punch". Speed when the ball leaves the face is what matters.
[/quote]

How do you know that? Either you have evidence that I haven't seen or you have evidence that doesn't make the cut. It would be nice to see it in any case.

[color=#ff0000]Iteach: And the only times I actually get specific about impact I use the word separating. And simply made statements of fact that seem pretty clear to me. Never said one work about impact interval or initial contact. I was pretty consistent about using the word separation and only the word separation. Don't think I was hard on anyone or said anything that is the least bit controversial that "[/color][color=#ff0000]hasn't been researched enough"[/color]


You did not bring up separation in your first response, so the consistency is only so-so. But how can you respond to the topic without being specific about impact? Being ambiguous helps a lot: As in giving a clear answer that isn't really clear.

And of course you dismissed that resisting decel makes a difference too. If that isn't being specific about impact I don't know what is.

I don't know what you really think of impact after this. To me it looks like a lot of D plane stuff and very simplified, but I may be wrong because you have a habit of not showing your hand.

The more I look into impact, and the more others do the same, it seems clearer and clearer that the impact interval is unchartered territory.

[Golf nerd]

[quote name='larrybud' timestamp='1412707697' post='10250369']


The wrists unhinging (unc0ck) is what speeds up the club head because it increases the arc length. The length of the arc increases, and conservation of angular momentum slows the hands/arms. I will go back to the ice skater. When she is spinning, and her hands are close to her body (her hands being like the clubhead in this case) the body is spinning fast. When the arms are extended, the hands are now going faster compared to the speed they were going when they were close to her body. The radius of the arc of her hands has increased, and her body has slowed. She didn't slow her body to speed up her hands, her body slowed BECAUSE she extended her arms, which sped up her hands.

I'm sorry, but you don't understand your own analogy: When you brake a car, your body does NOT accelerate relative to the ground. Your body is going faster than the car because your body has momentum and the car is decelerating. You certainly don't brake the car to speed up your body relative to the ground. If that was the case your body would never slow down!

I'll give you another analogy: Are you familiar with a medieval weapon call a trebuchet? The arm of the trebuchet slows as the sling expands outward from the arm due to conservation of angular momentum. There's no mechanism (ie. muscles) to have the arm of the trebuchet slow down. It slows because the radius of the arc of the sling increases.

Just to drive the point: If slowing your arms and hands sped up the clubhead as you state, then the goal in golf would be to stop your hands from moving completely before impact.
[/quote]

Buddy, lets end this conversation, because you don't understand what I am saying. I understand my analogies very well, but maybe it wasn't the best. Of course we don't put the brakes on to speed up something, but it should give you an idea about inertia. The trebuchet analogy is in line with my understanding of the arm/clubhead movement in the golf swing. The arm is accelerating up to a certain point. Then the the acceleration max is achieved and the mass of the object is passing the arm. Trebuchet like golf swing. If you can not follow my description thats okay, but further discussion will not help to clarify that.

[Albatross85]

[quote name='DeNinny' timestamp='1412742733' post='10253563']
OP, I am an engineer and here is almost as far* as I can take the physics and math (*I stop short of detailed potential energy equations). In short, I think acceleration matters mainly because it is the only way that the shaft can stay loaded until impact. In order for the shaft to stay bent, a constant force (or torque to be exact) needs to be applied on the shaft. By accelerating the swing, the clubhead mass will also accelerate and thus impart the proper force at the end of the shaft in order to bend and load it. At impact, this unloading of the shaft is additional energy imparted on the ball in addition to the velocity based (i.e. non-accelerated) kinetic energy of the clubhead. I think this is an important part of the physics and it is evidenced in the real world with golfers getting fitted with the proper shaft flex, in conjunction with proper swing weight, for one’s tempo and swing speed.

I also want to be clear that I still think the clubhead velocity is the most important aspect, but I myself cannot deny my own experience being slightly longer using regular flex shafts vs stiff or x-flex. I have a 90 mph driver SS, so regular flex fits me (along with my tempo) best.

Ok that was my short answer. Warning and my pre-apology for the length of the rest of this as my detailed “proof”. In the below diagram is what is happening at impact in three phases, which I call “contact”, “compress”, and “detach”:

[attachment=2449683:20140925_192048.jpg]

During the “contact” phase, the clubhead is just impacting the ball. At the point of impact, time t0 = 0, the clubhead with mass, Mc, and velocity, Vc, contacts the ball with mass, Mb. Assuming no acceleration, the total energy of impact is simply the kinetic energy, KEc, of the clubhead. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- assumes no acceleration)

where, by definition, kinetic energy is one-half times the mass times the velocity squared. Again, I want to emphasize I have neglected acceleration at this point in order to keep the equation simple.

During the “compress” phase, the clubhead and ball are traveling together as one object. Together they travel a total distance Dc for a time t1. The kinetic energy is transferring from the clubhead to the ball at this time as well. Since the ball and clubface are not rigid objects, they both undergo a compression. Both are feeling the energy of impact from the clubhead’s total energy. The middle diagram shows this with an oblong ball and curved clubface. Because of the compression, the molecules of the ball end up rubbing together and generating heat from the friction, which I have noted as Hc. This is a loss of some (although I doubt it is much) energy from the energy of impact. I want to note also that this is the actual time that forces are imparted on the ball. (Again, I’m still ignoring forces from clubhead acceleration for now.) I also want to emphasize that during the latter part of the this time, t1, the ball is also decompressing and returning to normal shape, progressively tending toward this just prior to detaching from the clubface at the end of this phase. Lastly, I believe others have mentioned that there is deceleration at impact, and while I agree that this is likely happening initially in this phase, I believe there is also some acceleration by the ball decompressing (just like a spring unloading) and other reasons, like the shaft unloading, during the end of this brief phase.

During the “detach” phase, the ball with mass, Mb, then leaves the clubface at a velocity, Vb, based on the remaining energy of the system. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- equation from before)

= Hc + ½ x Mb x Vb x Vb (<- heat loss from the ball’s compression and the remaining energy is transferred to the ball)

Now let’s incorporate the energy from loading the shaft, PEs, and also just the pure acceleration from the golfer, PEg. Both of these energies are considered potential energies because they are both accelerative forces that have yet to enact this force onto the ball. There is acceleration (or the potential for it), but at this point there is no energy transfer because there is no object for which to apply the force. This is why it is a potential energy. Now the total energy equation becomes

Total Energy = KEc + PEs + PEg (<- kinetic energy of clubhead plus the potential energies of the loaded shaft and from the golfer’s acceleration)

I have defined KEc by the basic kinetic energy equation, so now I’ll explain the shaft potential energy, PEs. The only way to have stored energy from the shaft, you have to keep it in a bent position. The only way to keep the shaft in a bent position is to apply a constant force to it. The only way to have a constant force is to have an acceleration on the clubhead. By swinging with acceleration, the clubhead weight on the other end of the shaft essentially creates a lag force (or rather, torque) that bends the shaft. (Remember that Force = Mass x Acceleration.) As soon as acceleration is stopped and the clubhead is traveling at constant velocity, the shaft will unbend or unload. Therefore, [i][b]acceleration is needed in order for the shaft to stay bent (i.e. loaded) until impact[/b][/i].

I have not defined the PEg, the potential energy from the golfer accelerating throughout the “compress” phase. I can tell you it will have its own set of variables. My gut feel is that the kinetic energy of the clubhead velocity and the potential energy from the shaft unloading play the greater role in overall shot distance than this potential energy. In fact, the shaft staying loaded is already taking (or storing) a fair portion of the golfer’s acceleration energy, so not much is “left over”.

To close this out, here is the final total energy equation with everything I've explained:

Total Energy = KEc + PEs + PEg = ½ x Mc x Vc x Vc + PEs + PEg

= Hc + ½ x Mb x Vb x Vb (<- assumes the ball is leaving the clubface at constant velocity)

Ultimately you could solve this equation for Vb, the velocity of the ball. You can see that having PEs and PEg as part of this equation will increase the resulting Vb, ultimately resulting in more distance. Thus in my personal opinion,

KEc > PEs >> PEg

The kinetic energy of the clubhead matters the most, but the potential energy of the bent shaft plays a significant role too. The loaded shaft is stored potential energy up until impact where it is able to unload this stored energy into the ball. Again, this is supported by the fact that you should get fitted for the proper shaft flex in order to maximize your potential distance. If shaft flex [i]weren’t [/i]important, then we could play with the stiffest and lightest shafts possible and still be able hit it as far as we do now.

I’m going to stop here unless you want more equations about the shaft bending and such. I hope this gives you an idea of the physics and math involved to truly answer your question.
[/quote]DeNinny you sir are the man! This is exactly what i was looking for. I had completely forgotten about the shaft loading aspect. I think this could also be what the term efficiency means in the golf swing, ones ability to properly load the shaft with minimal effort. This would also go to explain in part why some people are longer without swinging harder.

All responses and opinions are welcome in this thread but it does seem that some people are very quick to think they know the answer and there is no gray area. I believe DeNinny just took us all to school with some cold hard facts. There is no need for a pissing contest. I know Iteach knows a tremendous amount about the golf swing as do others but we can always keep learning so lets keep an open mind in this discussion.

[larrybud]

[quote name='Golf nerd' timestamp='1412765034' post='10253991']
[quote name='larrybud' timestamp='1412707697' post='10250369']
The wrists unhinging (unc0ck) is what speeds up the club head because it increases the arc length. The length of the arc increases, and conservation of angular momentum slows the hands/arms. I will go back to the ice skater. When she is spinning, and her hands are close to her body (her hands being like the clubhead in this case) the body is spinning fast. When the arms are extended, the hands are now going faster compared to the speed they were going when they were close to her body. The radius of the arc of her hands has increased, and her body has slowed. She didn't slow her body to speed up her hands, her body slowed BECAUSE she extended her arms, which sped up her hands.

I'm sorry, but you don't understand your own analogy: When you brake a car, your body does NOT accelerate relative to the ground. Your body is going faster than the car because your body has momentum and the car is decelerating. You certainly don't brake the car to speed up your body relative to the ground. If that was the case your body would never slow down!

I'll give you another analogy: Are you familiar with a medieval weapon call a trebuchet? The arm of the trebuchet slows as the sling expands outward from the arm due to conservation of angular momentum. There's no mechanism (ie. muscles) to have the arm of the trebuchet slow down. It slows because the radius of the arc of the sling increases.

Just to drive the point: If slowing your arms and hands sped up the clubhead as you state, then the goal in golf would be to stop your hands from moving completely before impact.
[/quote]

Buddy, lets end this conversation, because you don't understand what I am saying. I understand my analogies very well, but maybe it wasn't the best. Of course we don't put the brakes on to speed up something, but it should give you an idea about inertia. The trebuchet analogy is in line with my understanding of the arm/clubhead movement in the golf swing. The arm is accelerating up to a certain point. Then the the acceleration max is achieved and the mass of the object is passing the arm. Trebuchet like golf swing. If you can not follow my description thats okay, but further discussion will not help to clarify that.
[/quote]

lol. ok. I totally understand what you're claiming, since you've said it 3 times. You claim the arms slow down, which causes the club head to speed up.

Since you understand the trebuchet, then you'll understand that one does not slow down the trebuchet arm to speed up the throw. The trebuchet arm slows BECAUSE the length of the arc increases. Thanks for playing.

[RBImGuy]

while som flex is due to the camera cant keep up with the motion its accelleration at its best.

310+ carry

[tembolo1284]

[quote name='DeNinny' timestamp='1412742733' post='10253563']
OP, I am an engineer and here is almost as far* as I can take the physics and math (*I stop short of detailed potential energy equations). In short, I think acceleration matters mainly because it is the only way that the shaft can stay loaded until impact. In order for the shaft to stay bent, a constant force (or torque to be exact) needs to be applied on the shaft. By accelerating the swing, the clubhead mass will also accelerate and thus impart the proper force at the end of the shaft in order to bend and load it. At impact, this unloading of the shaft is additional energy imparted on the ball in addition to the velocity based (i.e. non-accelerated) kinetic energy of the clubhead. I think this is an important part of the physics and it is evidenced in the real world with golfers getting fitted with the proper shaft flex, in conjunction with proper swing weight, for one’s tempo and swing speed.

I also want to be clear that I still think the clubhead velocity is the most important aspect, but I myself cannot deny my own experience being slightly longer using regular flex shafts vs stiff or x-flex. I have a 90 mph driver SS, so regular flex fits me (along with my tempo) best.

Ok that was my short answer. Warning and my pre-apology for the length of the rest of this as my detailed “proof”. In the below diagram is what is happening at impact in three phases, which I call “contact”, “compress”, and “detach”:

[attachment=2449683:20140925_192048.jpg]

During the “contact” phase, the clubhead is just impacting the ball. At the point of impact, time t0 = 0, the clubhead with mass, Mc, and velocity, Vc, contacts the ball with mass, Mb. Assuming no acceleration, the total energy of impact is simply the kinetic energy, KEc, of the clubhead. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- assumes no acceleration)

where, by definition, kinetic energy is one-half times the mass times the velocity squared. Again, I want to emphasize I have neglected acceleration at this point in order to keep the equation simple.

During the “compress” phase, the clubhead and ball are traveling together as one object. Together they travel a total distance Dc for a time t1. The kinetic energy is transferring from the clubhead to the ball at this time as well. Since the ball and clubface are not rigid objects, they both undergo a compression. Both are feeling the energy of impact from the clubhead’s total energy. The middle diagram shows this with an oblong ball and curved clubface. Because of the compression, the molecules of the ball end up rubbing together and generating heat from the friction, which I have noted as Hc. This is a loss of some (although I doubt it is much) energy from the energy of impact. I want to note also that this is the actual time that forces are imparted on the ball. (Again, I’m still ignoring forces from clubhead acceleration for now.) I also want to emphasize that during the latter part of the this time, t1, the ball is also decompressing and returning to normal shape, progressively tending toward this just prior to detaching from the clubface at the end of this phase. Lastly, I believe others have mentioned that there is deceleration at impact, and while I agree that this is likely happening initially in this phase, I believe there is also some acceleration by the ball decompressing (just like a spring unloading) and other reasons, like the shaft unloading, during the end of this brief phase.

During the “detach” phase, the ball with mass, Mb, then leaves the clubface at a velocity, Vb, based on the remaining energy of the system. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- equation from before)

= Hc + ½ x Mb x Vb x Vb (<- heat loss from the ball’s compression and the remaining energy is transferred to the ball)

Now let’s incorporate the energy from loading the shaft, PEs, and also just the pure acceleration from the golfer, PEg. Both of these energies are considered potential energies because they are both accelerative forces that have yet to enact this force onto the ball. There is acceleration (or the potential for it), but at this point there is no energy transfer because there is no object for which to apply the force. This is why it is a potential energy. Now the total energy equation becomes

Total Energy = KEc + PEs + PEg (<- kinetic energy of clubhead plus the potential energies of the loaded shaft and from the golfer’s acceleration)

I have defined KEc by the basic kinetic energy equation, so now I’ll explain the shaft potential energy, PEs. The only way to have stored energy from the shaft, you have to keep it in a bent position. The only way to keep the shaft in a bent position is to apply a constant force to it. The only way to have a constant force is to have an acceleration on the clubhead. By swinging with acceleration, the clubhead weight on the other end of the shaft essentially creates a lag force (or rather, torque) that bends the shaft. (Remember that Force = Mass x Acceleration.) As soon as acceleration is stopped and the clubhead is traveling at constant velocity, the shaft will unbend or unload. Therefore, [i][b]acceleration is needed in order for the shaft to stay bent (i.e. loaded) until impact[/b][/i].

I have not defined the PEg, the potential energy from the golfer accelerating throughout the “compress” phase. I can tell you it will have its own set of variables. My gut feel is that the kinetic energy of the clubhead velocity and the potential energy from the shaft unloading play the greater role in overall shot distance than this potential energy. In fact, the shaft staying loaded is already taking (or storing) a fair portion of the golfer’s acceleration energy, so not much is “left over”.

To close this out, here is the final total energy equation with everything I've explained:

Total Energy = KEc + PEs + PEg = ½ x Mc x Vc x Vc + PEs + PEg

= Hc + ½ x Mb x Vb x Vb (<- assumes the ball is leaving the clubface at constant velocity)

Ultimately you could solve this equation for Vb, the velocity of the ball. You can see that having PEs and PEg as part of this equation will increase the resulting Vb, ultimately resulting in more distance. Thus in my personal opinion,

KEc > PEs >> PEg

The kinetic energy of the clubhead matters the most, but the potential energy of the bent shaft plays a significant role too. The loaded shaft is stored potential energy up until impact where it is able to unload this stored energy into the ball. Again, this is supported by the fact that you should get fitted for the proper shaft flex in order to maximize your potential distance. If shaft flex [i]weren’t [/i]important, then we could play with the stiffest and lightest shafts possible and still be able hit it as far as we do now.

I’m going to stop here unless you want more equations about the shaft bending and such. I hope this gives you an idea of the physics and math involved to truly answer your question.
[/quote]

Post of the year right there.

[larrybud]

Deninny,

While acceleration may cause the shaft to load differently, and the type of shaft being fitted properly is important for a specific golfer to optimize his potential, all of those things happen *before* impact, and may cause the club head speed to be different *at impact* than someone who is decelerating. Yes, two different shafts may give you different results because two different shafts will deliver the club head with different variables into the ball. Nobody is arguing otherwise.

But that's not what the original question was about. The original question was about two different club heads with the same speed at impact, one accelerating and one decelerating.

Once impact occurs that shaft has little if no influence on the ball. as tested by Cochan/Stobbs in "Search for the Perfect Swing". The club head might as well be swinging on a rope. (I don't have the book in front of me but I'd gladly give you a reference when I get home tonight).

What you're ignoring is that the "compress" phase is so short that any accelerating/deceleration of the club head within that 3/4" is so small as to be inconsequential in the ball speed. We're talking tenths of mph.

[northgolf]

[quote name='tembolo1284' timestamp='1412771185' post='10254235']
[quote name='DeNinny' timestamp='1412742733' post='10253563']
OP, I am an engineer and here is almost as far* as I can take the physics and math (*I stop short of detailed potential energy equations). In short, I think acceleration matters mainly because it is the only way that the shaft can stay loaded until impact. In order for the shaft to stay bent, a constant force (or torque to be exact) needs to be applied on the shaft. By accelerating the swing, the clubhead mass will also accelerate and thus impart the proper force at the end of the shaft in order to bend and load it. At impact, this unloading of the shaft is additional energy imparted on the ball in addition to the velocity based (i.e. non-accelerated) kinetic energy of the clubhead. I think this is an important part of the physics and it is evidenced in the real world with golfers getting fitted with the proper shaft flex, in conjunction with proper swing weight, for one’s tempo and swing speed.

I also want to be clear that I still think the clubhead velocity is the most important aspect, but I myself cannot deny my own experience being slightly longer using regular flex shafts vs stiff or x-flex. I have a 90 mph driver SS, so regular flex fits me (along with my tempo) best.

Ok that was my short answer. Warning and my pre-apology for the length of the rest of this as my detailed “proof”. In the below diagram is what is happening at impact in three phases, which I call “contact”, “compress”, and “detach”:

[attachment=2449683:20140925_192048.jpg]

During the “contact” phase, the clubhead is just impacting the ball. At the point of impact, time t0 = 0, the clubhead with mass, Mc, and velocity, Vc, contacts the ball with mass, Mb. Assuming no acceleration, the total energy of impact is simply the kinetic energy, KEc, of the clubhead. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- assumes no acceleration)

where, by definition, kinetic energy is one-half times the mass times the velocity squared. Again, I want to emphasize I have neglected acceleration at this point in order to keep the equation simple.

During the “compress” phase, the clubhead and ball are traveling together as one object. Together they travel a total distance Dc for a time t1. The kinetic energy is transferring from the clubhead to the ball at this time as well. Since the ball and clubface are not rigid objects, they both undergo a compression. Both are feeling the energy of impact from the clubhead’s total energy. The middle diagram shows this with an oblong ball and curved clubface. Because of the compression, the molecules of the ball end up rubbing together and generating heat from the friction, which I have noted as Hc. This is a loss of some (although I doubt it is much) energy from the energy of impact. I want to note also that this is the actual time that forces are imparted on the ball. (Again, I’m still ignoring forces from clubhead acceleration for now.) I also want to emphasize that during the latter part of the this time, t1, the ball is also decompressing and returning to normal shape, progressively tending toward this just prior to detaching from the clubface at the end of this phase. Lastly, I believe others have mentioned that there is deceleration at impact, and while I agree that this is likely happening initially in this phase, I believe there is also some acceleration by the ball decompressing (just like a spring unloading) and other reasons, like the shaft unloading, during the end of this brief phase.

During the “detach” phase, the ball with mass, Mb, then leaves the clubface at a velocity, Vb, based on the remaining energy of the system. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- equation from before)

= Hc + ½ x Mb x Vb x Vb (<- heat loss from the ball’s compression and the remaining energy is transferred to the ball)

Now let’s incorporate the energy from loading the shaft, PEs, and also just the pure acceleration from the golfer, PEg. Both of these energies are considered potential energies because they are both accelerative forces that have yet to enact this force onto the ball. There is acceleration (or the potential for it), but at this point there is no energy transfer because there is no object for which to apply the force. This is why it is a potential energy. Now the total energy equation becomes

Total Energy = KEc + PEs + PEg (<- kinetic energy of clubhead plus the potential energies of the loaded shaft and from the golfer’s acceleration)

I have defined KEc by the basic kinetic energy equation, so now I’ll explain the shaft potential energy, PEs. The only way to have stored energy from the shaft, you have to keep it in a bent position. The only way to keep the shaft in a bent position is to apply a constant force to it. The only way to have a constant force is to have an acceleration on the clubhead. By swinging with acceleration, the clubhead weight on the other end of the shaft essentially creates a lag force (or rather, torque) that bends the shaft. (Remember that Force = Mass x Acceleration.) As soon as acceleration is stopped and the clubhead is traveling at constant velocity, the shaft will unbend or unload. Therefore, [i][b]acceleration is needed in order for the shaft to stay bent (i.e. loaded) until impact[/b][/i].

I have not defined the PEg, the potential energy from the golfer accelerating throughout the “compress” phase. I can tell you it will have its own set of variables. My gut feel is that the kinetic energy of the clubhead velocity and the potential energy from the shaft unloading play the greater role in overall shot distance than this potential energy. In fact, the shaft staying loaded is already taking (or storing) a fair portion of the golfer’s acceleration energy, so not much is “left over”.

To close this out, here is the final total energy equation with everything I've explained:

Total Energy = KEc + PEs + PEg = ½ x Mc x Vc x Vc + PEs + PEg

= Hc + ½ x Mb x Vb x Vb (<- assumes the ball is leaving the clubface at constant velocity)

Ultimately you could solve this equation for Vb, the velocity of the ball. You can see that having PEs and PEg as part of this equation will increase the resulting Vb, ultimately resulting in more distance. Thus in my personal opinion,

KEc > PEs >> PEg

The kinetic energy of the clubhead matters the most, but the potential energy of the bent shaft plays a significant role too. The loaded shaft is stored potential energy up until impact where it is able to unload this stored energy into the ball. Again, this is supported by the fact that you should get fitted for the proper shaft flex in order to maximize your potential distance. If shaft flex [i]weren’t [/i]important, then we could play with the stiffest and lightest shafts possible and still be able hit it as far as we do now.

I’m going to stop here unless you want more equations about the shaft bending and such. I hope this gives you an idea of the physics and math involved to truly answer your question.
[/quote]

Post of the year right there.
[/quote]

Not quite, the bolded part makes a statement that is not born out in reality. The shaft is never "loaded" at impact - hi speed global shutter cameras and strain gauges on shafts evidence that there is forward bend (and toe down bend) in the shaft at impact.

[chris87]

I realize this thread has kind of moved on from simple answers, but I thought it might be helpful to mention that the people who cite Newton's F=m*a as proof that an accelerating clubhead transfers more force to a ball are misinterpreting this equation. You cannot mix objects in the equation.....The "a" term is the net acceleration of the object being accelerated (the ball), and the "F" term is the net force acting on the object being accelerated (the ball). So the correct statement is "Net force on the ball is equal to mass of the ball multiplied by net acceleration of the ball".

You could also say that the net force on the clubhead is equal to the mass of the clubhead * the acceleration of the clubhead. This would only tell you that if the clubhead is accelerating then the golfer / gravity must be applying a force to it....which is obvious to everyone and tells you nothing about what is happening to the ball. In conclusion, you cannot mix objects in the equation, for example, by saying that the force on the [u]ball[/u] is the mass of the [u]clubhead[/u] multiplied by the acceleration of the [u]clubhead[/u].

Energy and momentum are two concepts you could use to determine what will accelerate the ball. Both are related to mass and velocity, not acceleration.

[Krt22]

[quote name='larrybud' timestamp='1412771464' post='10254263']
Deninny,

While acceleration may cause the shaft to load differently, and the type of shaft being fitted properly is important for a specific golfer to optimize his potential, all of those things happen *before* impact, and may cause the club head speed to be different *at impact* than someone who is decelerating. Yes, two different shafts may give you different results because two different shafts will deliver the club head with different variables into the ball. Nobody is arguing otherwise.

But that's not what the original question was about. The original question was about two different club heads with the same speed at impact, one accelerating and one decelerating.

Once impact occurs that shaft has little if no influence on the ball. as tested by Cochan/Stobbs in "Search for the Perfect Swing". The club head might as well be swinging on a rope. (I don't have the book in front of me but I'd gladly give you a reference when I get home tonight).

What you're ignoring is that the "compress" phase is so short that any accelerating/deceleration of the club head within that 3/4" is so small as to be inconsequential in the ball speed. We're talking tenths of mph.
[/quote]

Agreed, and I am an engineer as well. As much as it would nice to account for all of the energy storage in the shaft, head, etc..all of those variables are just contributing to the final velocity at impact. And the main crux of his argument hinges on acceleration of the club head stopping and the shaft potentially unloading before or during impact. Now the clubhead can be at constant LINEAR velocity, but still undergo angular acceleration which will keep the shaft loaded. but either way, totally negligible AT IMPACT, since again that only will contribute to its final velocity at impact. Its really a basic elastic collision with a known COR.

If (linear) acceleration during impact could increase final ball speed, what acceleration value would OEMs use? If they assume its too high, they will never react full COR with real users. if they made it too little, real users could exceed the COR.

The bigger question is, why would anyone try to decelerate into impact?

[DeNinny]

[quote name='Krt22' timestamp='1412796652' post='10256463']
[quote name='larrybud' timestamp='1412771464' post='10254263']
Deninny,

While acceleration may cause the shaft to load differently, and the type of shaft being fitted properly is important for a specific golfer to optimize his potential, all of those things happen *before* impact, and may cause the club head speed to be different *at impact* than someone who is decelerating. Yes, two different shafts may give you different results because two different shafts will deliver the club head with different variables into the ball. Nobody is arguing otherwise.

But that's not what the original question was about. The original question was about two different club heads with the same speed at impact, one accelerating and one decelerating.

Once impact occurs that shaft has little if no influence on the ball. as tested by Cochan/Stobbs in "Search for the Perfect Swing". The club head might as well be swinging on a rope. (I don't have the book in front of me but I'd gladly give you a reference when I get home tonight).

What you're ignoring is that the "compress" phase is so short that any accelerating/deceleration of the club head within that 3/4" is so small as to be inconsequential in the ball speed. We're talking tenths of mph.
[/quote]

Agreed, and I am an engineer as well. As much as it would nice to account for all of the energy storage in the shaft, head, etc..all of those variables are just contributing to the final velocity at impact. And the main crux of his argument hinges on acceleration of the club head stopping and the shaft potentially unloading before or during impact. Now the clubhead can be at constant LINEAR velocity, but still undergo angular acceleration which will keep the shaft loaded. but either way, totally negligible AT IMPACT, since again that only will contribute to its final velocity at impact. Its really a basic elastic collision with a known COR.

If (linear) acceleration during impact could increase final ball speed, what acceleration value would OEMs use? If they assume its too high, they will never react full COR with real users. if they made it too little, real users could exceed the COR.

The bigger question is, why would anyone try to decelerate into impact?
[/quote]

Larrybud, Krt22, and northgolf,

I understand your points about perhaps the shaft is unloaded by the time of impact. My main point is that acceleration is absolutely needed, at some point during the downswing, in order to load the shaft. If the unloading is occurring just prior to impact, then the ball can still benefit from this. What will happen is that the stored potential energy will convert to additional clubhead kinetic energy as the shaft unloads. In this case there is additional velocity added to the clubhead, which I will call Vu for "unloading velocity". Mathematically now we can define a new total clubhead velocity, Vc', defined as

Vc' = Vc + Vu

where Vc is the velocity of the clubhead from the golfer swinging the club (defined earlier) and Vu is the added velocity of the shaft unloading just prior to impact. The new Vc' can then be used in the standard kinetic energy equation instead of Vc.

Ultimately whether the shaft unloads during impact or just prior to impact, the unloading shaft, which is a result of acceleration, contributes to the final ball velocity. Is it much? That is up for debate and actually crunching the numbers...

[Krt22]

You even state it yourself, it can occur just prior to impact and thus contributes to final impact velocity, so Vc' is the only value that will come into play during the actual collision.

[nikeman1982]

?

[HappyGolf]

If the clubhead is accelerating and the shaft is 'loaded' then I don't see how it can contribute to impact as it is in a stable condition, only when the acceleration slows can the unloading of the shaft contribute anything, and I'm a firm believer that you want to be as close to your highest possible swing speed either AT or just past impact whether or not you are playing for a late hit with a lot of acceleration (and shaft loaded) or even (dread the thought) swinging from the top and having the shaft unloaded is of little consequence - to me. The ball only reacts to how fast the clubhead is travelling.

The words thrown around in tuition such as 'accelerate' are far better than telling someone to '[b]de[/b]celerate'.

[DeNinny]

[quote name='northgolf' timestamp='1412783949' post='10255329']
[quote name='tembolo1284' timestamp='1412771185' post='10254235']
[quote name='DeNinny' timestamp='1412742733' post='10253563']
....

I have defined KEc by the basic kinetic energy equation, so now I’ll explain the shaft potential energy, PEs. The only way to have stored energy from the shaft, you have to keep it in a bent position. The only way to keep the shaft in a bent position is to apply a constant force to it. The only way to have a constant force is to have an acceleration on the clubhead. By swinging with acceleration, the clubhead weight on the other end of the shaft essentially creates a lag force (or rather, torque) that bends the shaft. (Remember that Force = Mass x Acceleration.) As soon as acceleration is stopped and the clubhead is traveling at constant velocity, the shaft will unbend or unload. Therefore, [i][b]acceleration is needed in order for the shaft to stay bent (i.e. loaded) until impact[/b][/i].

I have not defined the PEg, the potential energy from the golfer accelerating throughout the “compress” phase. I can tell you it will have its own set of variables. My gut feel is that the kinetic energy of the clubhead velocity and the potential energy from the shaft unloading play the greater role in overall shot distance than this potential energy. In fact, the shaft staying loaded is already taking (or storing) a fair portion of the golfer’s acceleration energy, so not much is “left over”.

To close this out, here is the final total energy equation with everything I've explained:

Total Energy = KEc + PEs + PEg = ½ x Mc x Vc x Vc + PEs + PEg

= Hc + ½ x Mb x Vb x Vb (<- assumes the ball is leaving the clubface at constant velocity)

Ultimately you could solve this equation for Vb, the velocity of the ball. You can see that having PEs and PEg as part of this equation will increase the resulting Vb, ultimately resulting in more distance. Thus in my personal opinion,

KEc > PEs >> PEg

The kinetic energy of the clubhead matters the most, but the potential energy of the bent shaft plays a significant role too. The loaded shaft is stored potential energy up until impact where it is able to unload this stored energy into the ball. Again, this is supported by the fact that you should get fitted for the proper shaft flex in order to maximize your potential distance. If shaft flex [i]weren’t [/i]important, then we could play with the stiffest and lightest shafts possible and still be able hit it as far as we do now.

I’m going to stop here unless you want more equations about the shaft bending and such. I hope this gives you an idea of the physics and math involved to truly answer your question.
[/quote]

Post of the year right there.
[/quote]

Not quite, the bolded part makes a statement that is not born out in reality. The shaft is never "loaded" at impact - hi speed global shutter cameras and strain gauges on shafts evidence that there is forward bend (and toe down bend) in the shaft at impact.
[/quote]

Thanks for this. I never knew what has been proven or not at impact and was more just trying to emphasize the importance of acceleration in order to reap the benefit of a loaded shaft. As I mentioned in a separate post to also larrybud and Krt22, if the shaft unloads just before impact, then the energy of the shaft unloading will still result in added clubhead velocity.

For OP and all,
Here are more equations and diagrams on the shaft bending (or loading) and the resulting potential energy from this and also the shaft unbending (or unloading) and the resulting kinetic energy:

[attachment=2450897:202508.jpg]

The first diagram is simply the shaft and clubhead before any work or energy is done on them. Fb is the force that will be required to bend the shaft and Ls is simply the length of the shaft from the point of anchor to the point at which the force is applied. Although it is not necessary at this point, the torque, Tb, required to bend the shaft is from this standard torque equation:

Tb = Fb x Ls

The middle diagram is of the loaded shaft. Since the shaft is bent, it has stored potential energy. As soon as whatever is preventing the shaft from straightening is removed, the potential energy will be converted to kinetic energy. At this point, the total energy is the potential energy since there is zero kinetic energy. Mathematically, the energy required to bend the shaft is defined by the following equation:

Total Energy = PEs = Fb x Lb

where

PEs = potential energy of the bent shaft (mentioned earlier)

Fb = force required to bend the shaft

Lb = length over which the force, Fb, was applied

Now let’s break down Fb in more detail. As mentioned, this is typically defined as the product of the object mass and the acceleration of the mass. Mathematically,

Fb = Mc x Ac

where

Mc = mass of the clubhead

Ac = acceleration of the clubhead

Note that Ac is what is created from the golfer accelerating his swing. Ultimately, the potential energy of the loaded shaft is

PEs = Mc x Ac x Lb

I want to emphasize again that a force applied on an object over a distance is considered energy (or work to be exact). More detail can be learned here: [url="http://en.wikipedia.org/wiki/Work_(physics)"]http://en.wikipedia..../Work_(physics)[/url].

In the bottom diagram, I have shown the unloaded shaft. The point of fully unloading the stored potential energy is when the shaft returns to a straight position. At the moment it reaches this point, it will have zero potential energy and 100% kinetic energy. The velocity of the clubhead is designated as Vu. Mathematically the total energy is now kinetic energy, defined per the following equation:

Total Energy = KEu = ½ x Mc x Vu x Vu

where

KEu = kinetic energy of the unloaded shaft

Mc = mass of the clubhead

Vu = unloaded final velocity of the clubhead

Now we can combine the two Total Energy equations per the following:

Total Energy = PEs = KEu

Total Energy = Mc x Ac x Lb = ½ x Mc x Vu x Vu

The total energy here is analogous to that of a falling object starting at a given height and undergoing the force of gravity as it falls this height. The only difference is that Ac is the gravitational acceleration constant for the earth and Lb is the given height.

From the equation you can see that the more the golfer can increase Ac, his swing acceleration, the bigger Vu gets. This is ultimately why acceleration is important.

Furthermore, after the input from larrybud, northgolf, and Krt22, it is actually best if the shaft is unloading precisely at impact rather than during impact. This way the ball will receive the maximum energy from the clubhead. If it is unloading during impact (or “compress” phase as I defined earlier), the value for Lb will be equal to Dc, which I defined earlier as the distance the ball and clubhead travel in contact with each other. During this time, the unloading shaft can only transfer the energy (or work) while there is contact. Since Dc is (likely, as mentioned by larrybud) much less than Lb, the potential energy for this is lower by the proportion that Dc is smaller than Lb.

So ultimately it is best that the shaft fully unloads right at the point of impact in order to get the maximum benefit of the potential energy of the loaded shaft. This will give the maximum Vu and thus give the maximum Vc’ which ultimately results in the maximum Vb, or ball velocity.

Now if the shaft unloads well before impact, then there will be zero benefit of loading it because there is no more potential or kinetic energy to transfer to the ball. It dissipated before contact.

[LaggingBehind]

I just started a thread about pretty much the same topic... stupid me
The jist of it was about maximizing the change in acceleration at impact, known as jerk. This is from the summation of all accelerating components of the swing, and I am under the belief it has much to do with the unhinging of the wrists. The arc of the hip turn is wider than the arc of the shoulder turn, which is wider than the arc of the wrist hinge, which is wider than the arc of shaft unloading. As the energy of the accelerating components move down this curved path of decreasing radius the acceleration increases each time, resulting in jerk. [url="http://en.wikipedia.org/wiki/Jerk_%28physics%29"]http://en.wikipedia..../Jerk_(physics)[/url]

I think of it as a series of Easement curves
[url="http://en.wikipedia.org/wiki/Track_transition_curve"]http://en.wikipedia....ransition_curve[/url]

[Silky]

There are many issues that are still muddy.
1. What is the weight of the clubhead that used in the calculation of the kinetic energy above? Should one includes the hosel and or a lower portion of the club shaft?
2. In reality, near impact, the club is translating (linear motion) and also rotating. So the kinetic energy of the club (not just the club head) will include both motions. Moreover, the club can be rotating around the longitudinal axis (opening/closing of club face). This adds to the total kinetic energy too. Lastly, the club shaft can be vibrating (loading/unloading). This needs to be added too. How and how much these different kinds of kinetic energies get transferred to the golf ball?

Please help clarify just for fun, not that it will improve my golf swing at all.

[Lefthook]

3, trampoline effects etc on the club face
4. Axial force - up the golf shaft. It travels by the sound of speed in steel / graphite. The impact shock will (in theory at last) reach the hands roughly in the middle of the impact interval. It will probably be significant losses and delays in practice because of shaft flexing and absorption in the grip etc, but how the hands moves and how well they resist the impact shock could make a difference. I think it does. "Heavy hands". Not so much in terms of ball speed perhaps, but ball flight, spin etc.

DeNinny assumes in his model that the grip is fixed in his shaft unloading model. In a golf swing, this will be a variable depending on how the golfer uses the hands. Worst case is no forward force and no torque, which would simply create a vibrating shaft between two stationary hinges. The other extreme is towards how it is modeled, where the unloading moves the club head.

[Soloman1]

The shaft is not "unloaded" to a straight line at impact. It returns to neutral then bends along the arc of the swing toward the target. And yes, shaft droop is real.

Let's return to the original idea of the thread.

Trying to accelerate through impact will in all likelihood improve your swing and perhaps give you the feel of being able to maintain the clubface direction through the ridiculously short impact interval, but you will not hit it farther. One reason is that the difference is so small you can't measure it, or be sure of the cause of that unmeasurable result was your feeling of accelerating. We're talking less than 1/2 of a percent here.

So if you hit your 5 iron the standard GolfWRX 5-iron distance (240 yards carry), the difference might be less than a yard.

Decelerating intentionally nearing impact would contribute nothing positive to any golf swing, from driver through putter.

So, go ahead and feel like your body is accelerating and you will do no harm and likely have better results more often.

[mikah]

[quote name='northgolf' timestamp='1412783949' post='10255329']
[quote name='tembolo1284' timestamp='1412771185' post='10254235']
[quote name='DeNinny' timestamp='1412742733' post='10253563']
OP, I am an engineer and here is almost as far* as I can take the physics and math (*I stop short of detailed potential energy equations). In short, I think acceleration matters mainly because it is the only way that the shaft can stay loaded until impact. In order for the shaft to stay bent, a constant force (or torque to be exact) needs to be applied on the shaft. By accelerating the swing, the clubhead mass will also accelerate and thus impart the proper force at the end of the shaft in order to bend and load it. At impact, this unloading of the shaft is additional energy imparted on the ball in addition to the velocity based (i.e. non-accelerated) kinetic energy of the clubhead. I think this is an important part of the physics and it is evidenced in the real world with golfers getting fitted with the proper shaft flex, in conjunction with proper swing weight, for one’s tempo and swing speed.

I also want to be clear that I still think the clubhead velocity is the most important aspect, but I myself cannot deny my own experience being slightly longer using regular flex shafts vs stiff or x-flex. I have a 90 mph driver SS, so regular flex fits me (along with my tempo) best.

Ok that was my short answer. Warning and my pre-apology for the length of the rest of this as my detailed “proof”. In the below diagram is what is happening at impact in three phases, which I call “contact”, “compress”, and “detach”:

[attachment=2449683:20140925_192048.jpg]

During the “contact” phase, the clubhead is just impacting the ball. At the point of impact, time t0 = 0, the clubhead with mass, Mc, and velocity, Vc, contacts the ball with mass, Mb. Assuming no acceleration, the total energy of impact is simply the kinetic energy, KEc, of the clubhead. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- assumes no acceleration)

where, by definition, kinetic energy is one-half times the mass times the velocity squared. Again, I want to emphasize I have neglected acceleration at this point in order to keep the equation simple.

During the “compress” phase, the clubhead and ball are traveling together as one object. Together they travel a total distance Dc for a time t1. The kinetic energy is transferring from the clubhead to the ball at this time as well. Since the ball and clubface are not rigid objects, they both undergo a compression. Both are feeling the energy of impact from the clubhead’s total energy. The middle diagram shows this with an oblong ball and curved clubface. Because of the compression, the molecules of the ball end up rubbing together and generating heat from the friction, which I have noted as Hc. This is a loss of some (although I doubt it is much) energy from the energy of impact. I want to note also that this is the actual time that forces are imparted on the ball. (Again, I’m still ignoring forces from clubhead acceleration for now.) I also want to emphasize that during the latter part of the this time, t1, the ball is also decompressing and returning to normal shape, progressively tending toward this just prior to detaching from the clubface at the end of this phase. Lastly, I believe others have mentioned that there is deceleration at impact, and while I agree that this is likely happening initially in this phase, I believe there is also some acceleration by the ball decompressing (just like a spring unloading) and other reasons, like the shaft unloading, during the end of this brief phase.

During the “detach” phase, the ball with mass, Mb, then leaves the clubface at a velocity, Vb, based on the remaining energy of the system. Mathematically,

Total Energy = KEc = ½ x Mc x Vc x Vc (<- equation from before)

= Hc + ½ x Mb x Vb x Vb (<- heat loss from the ball’s compression and the remaining energy is transferred to the ball)

Now let’s incorporate the energy from loading the shaft, PEs, and also just the pure acceleration from the golfer, PEg. Both of these energies are considered potential energies because they are both accelerative forces that have yet to enact this force onto the ball. There is acceleration (or the potential for it), but at this point there is no energy transfer because there is no object for which to apply the force. This is why it is a potential energy. Now the total energy equation becomes

Total Energy = KEc + PEs + PEg (<- kinetic energy of clubhead plus the potential energies of the loaded shaft and from the golfer’s acceleration)

I have defined KEc by the basic kinetic energy equation, so now I’ll explain the shaft potential energy, PEs. The only way to have stored energy from the shaft, you have to keep it in a bent position. The only way to keep the shaft in a bent position is to apply a constant force to it. The only way to have a constant force is to have an acceleration on the clubhead. By swinging with acceleration, the clubhead weight on the other end of the shaft essentially creates a lag force (or rather, torque) that bends the shaft. (Remember that Force = Mass x Acceleration.) As soon as acceleration is stopped and the clubhead is traveling at constant velocity, the shaft will unbend or unload. Therefore, [i][b]acceleration is needed in order for the shaft to stay bent (i.e. loaded) until impact[/b][/i].

I have not defined the PEg, the potential energy from the golfer accelerating throughout the “compress” phase. I can tell you it will have its own set of variables. My gut feel is that the kinetic energy of the clubhead velocity and the potential energy from the shaft unloading play the greater role in overall shot distance than this potential energy. In fact, the shaft staying loaded is already taking (or storing) a fair portion of the golfer’s acceleration energy, so not much is “left over”.

To close this out, here is the final total energy equation with everything I've explained:

Total Energy = KEc + PEs + PEg = ½ x Mc x Vc x Vc + PEs + PEg

= Hc + ½ x Mb x Vb x Vb (<- assumes the ball is leaving the clubface at constant velocity)

Ultimately you could solve this equation for Vb, the velocity of the ball. You can see that having PEs and PEg as part of this equation will increase the resulting Vb, ultimately resulting in more distance. Thus in my personal opinion,

KEc > PEs >> PEg

The kinetic energy of the clubhead matters the most, but the potential energy of the bent shaft plays a significant role too. The loaded shaft is stored potential energy up until impact where it is able to unload this stored energy into the ball. Again, this is supported by the fact that you should get fitted for the proper shaft flex in order to maximize your potential distance. If shaft flex [i]weren’t [/i]important, then we could play with the stiffest and lightest shafts possible and still be able hit it as far as we do now.

I’m going to stop here unless you want more equations about the shaft bending and such. I hope this gives you an idea of the physics and math involved to truly answer your question.
[/quote]

Post of the year right there.
[/quote]

Not quite, the bolded part makes a statement that is not born out in reality. The shaft is never "loaded" at impact - hi speed global shutter cameras and strain gauges on shafts evidence that there is forward bend (and toe down bend) in the shaft at impact.
[/quote]

a flexible shaft will decelerate clubhead at impact as much as 25%, compared to 8% for a very stiff shaft.

Titleist Performance Institute has studied the science of impact as much as anyone.
Deceleration begins prior to impact. There would be no acceleration of clubhead without it.


[b]"Good Segmental Stabilization:[/b]
[color=#282828]This last fundamental is the key to generating power and speed in the golf swing. In order to pass energy from one part of your body to the next, there must be a deceleration of the previous segment so that energy can be effectively transferred to the next body part. A great analogy is the cracking of the whip. In order to create a loud snap at the end of a whip, you must rapidly accelerate the handle of the whip and then quickly stopped or decelerate the handle. It is this deceleration of the handle that allows speed to be transferred to the next part of the whip."[/color]
[url="http://www.mytpi.com/articles/swing/5_key_fundamentals"]http://www.mytpi.com...ey_fundamentals[/url][color=#282828] [/color]

[LaggingBehind]

I believe that is jerk^
The increase in acceleration over time of the club head results in an increase in torque/time. I don't know what the accepted term is for dtorque/dt though (or if there even is one).