Nick Price....

So… Nick Price… Hitter right?

Definitely:

Explains why he has always been considered one of the best iron players out there…

A hitter’s release here.

It has been written that when Price struck his irons that it sounded like a gun. This could only occur with him holding shaft flex, which means the club had to be accelerating. Formula Force is mass x acceleration.

As I understand it the force you get from that equation is the the force required to make mass accelerate, not the amount of force that is created by that mass accelerating. A golf club has mass, to accelerate that mass one must apply force. Force would equal the amount of effort it takes to make the golf club accelerate. The force that is being measured is the force the muscles had to create to accelerate the club.

Does anyone know how to apply the formula so that it shows the amount of force that is created in the clubhead. I would like to know the amount of force stored in the accelerating clubhead, not the force it takes to make the clubhead accelerate.

As we know most players hit the ball using momentum, without acceleration. Even without acceleration, a force is applied to the golf ball, since force is required to make a still object move. The golf ball will be accelerating when hit, even if the club is not accelerating. We know it will accelerate faster in relation to the force applied. Again, if Nicks clubhead speed without a ball is measured before impact at 110mph and 114 mph after impact. That proves the mass is accelerating, but the force measured is his muscular effort to move the clubhead from 110mph to 114mph.

I don’t think this can be answered because the concept of force being stored in the clubhead seems flawed. Can you rephrase your question? What is it that you want to know?

It could be that there is not an answer. I am trying to figure out the differences in the amount of force that in the traveling club head, so that you can know how much is being applied to the ball. The thing that I can’t get is that there is force being applied to the ball, and this force can occur in 3 ways depending on swing.

1. Club head accelerating at impact
2. Club head at a constant speed at impact
3. Club head decelerating at impact

All have force because it is required to move the ball.

Force = mass x acceleration, noting that force is the push or pull that is applied to an object to change its momentum

Momentum = mass x velocity

I am thinking like this. The force(club head) is the push that is applied to the object(golf ball) to change its(golf ball) momentum.

Force in simple terms is the pressure applied to the surface of an object. More pressure, more compression, more energy. The difficult part is that force causes acceleration

1. Accelerating causes stored energy in the shaft to be transferred to the ball through the club head, causing more compression of the ball. This is the commonality of all great ballstrikers.
2. Constant speed, the shaft would not bow forward to lose shaftflex. I know this is better than deceleration, but do not know the specifics of how it would make the ball fly. It would not be bad.
3. Deceleration would cause the shaft to bow forward. I know some will argue that a club head moving at a higher velocity at impact is more important to compression than a slower moving club head that is accelerating.

I want to be able to show how an accelerating club head compresses the ball better than just faster club head speed without acceleration. I know this to be fact through experience. When I hold shaftflex into impact, and I don’t mean the shaft is straight, I mean it is bowed backwards when the ball is hit, the difference is astonishing. I have done this no more than 30 shots in 2 years(the maximum I can flex the shaft), I feel that I hold flex decent, but what I am referring to has to be what Hogan did. The ball does sound like it was shot out of a gun, the speed at which it travels to the green is mind boggling, and the flight is so impressive that I have become obsessed with figuring out how to make my swing do that on a consistent basis. My ball flight is ok, very high and travels in an arc, reaches high point 60 to 70 percent of the way to the target . On the shots I hold flex the ball shoots at a more direct trajectory and rises to the same height, but the point it reaches maximum height is at least 90 percent of the way to the target. It shoots off the club face so much faster than normal that I look up and the ball is at the green, it is like the ball is compressed so much that when it rebounds from the compression it comes off expending all of its energy in a direct line with little curve, then falls in as close to vertical line to the pin as I can imagine. It is different to imagine a golf ball not arcing at its highest point, but instead to just drop.

I know some will argue, and I can’t prove that acceleration is causing this with formulas because I do not know how to apply the formulas to show the ball behaving differently when compressed due to accelerating speed instead of just a fast speed. It is made even harder because of longer face contact creating more energy transfer.

I am guessing that the rate at which the golf ball gains momentum would give some indication as to the force applied, but how can you differentiate the force being due to mass and a constant or decelerating speed of the club head instead of that force being due to mass and increasing speed of the club head. Then it is not a question of force alone because the ball will have longer contact with the face the more it is compressed, therefore allowing more energy to be transferred to the ball. You could have slower speed at impact while accelerating, but increased transfer of energy to the ball because of longer face contact. There are so many variables that it seems the measurement can’t be accurate. Would Trackman be able to measure everything about the ball flight, then derive what happened in reverse to create the flight.

To give those a simplistic idea of the difference from the 3 speeds you stated…constant, decelerating, and accelerating.

Im going to use a very simple car and barrel example/analogy. Lets take a 1 mile drag strip or airport runway and set the barrel at the 1/2 mile mark. We then get the car to get up to a speed of say 50 mph. There will be a white line 100 feet before impacting into the barrel in which we can either keep a constant speed, let go of the gas pedal (decelerating), or Floor the gas pedal (accelerating).

Which do you think can get the barrel to move the farthest?

Better yet, what if we use a Hummer compared to a Mini cooper? Then we throw more mass into this equation.

Flatlies: Thank you for rephrasing the question. I have come up with a very simple statement that I would like to run by you:

If the golfer applies force to the clubhead through impact, and if contact is perfect (centered, square, zero loft), then that same force is applied to the ball, given that the clubhead is a perfectly stiff object.

Perfectly simple, and no F=m*a is needed in the above argument. Of course it doesn’t answer what you want to know. I am sure that you would like to compare the above force to the additional force on the ball resulting from the clubhead’s mass traveling into the ball with speed. That is a bit harder to work out, but it is fun to think about…

I have come up with a way to estimate the force on the ball. An ultra-high speed camera might come in handy!

Measure the ball-speed v after impact. The ball’s kinetic energy is then 0.5mv^2 (m is the mass of the golf ball). Assume that a force F is uniformly applied to the ball over a distance d (say 1 cm). The work put into the ball is Fd, and this equals the resulting kinetic energy. Thus F = 0.5m*v^2/d. I am too lazy to run the numbers now, but I might get to it during the weekend.

The above provides an estimate for the total force put into the ball. It is some sort of average force of course, because the force is not applied uniformly. And I am sure there are other shortcomings in the above approach.

To single out the force applied to the clubhead by the backward shaft-flex, all it takes is to determine the amount of flex prior to impact, and to then recreate it by applying a directly measurable force (for example by hanging a weight from the clubhead, with the shaft horizontal, and the club held in the grip area).

Thanks for answering. I know that the car being floored would move the barrel further, trying to work out how that energy transfer can be quantified at impact. I know all of this has to do with impulse, but can’t work it out from there.

I will explain with the car example. Which would move the barrel the furthest, mass is same.

1. Car going constant 70mph.
2. Car going 60mph 100 feet before barrel, and 80mph 100 feet after barrel. Constant rate of acceleration, 70 mph at impact.
3. Car going 110 mph 100 feet before impact, 50 mph 100 feet after barrel. Constant rate of deceleration, 80 mph at impact.
4. Car going 50mph 100 feet before impact, 90mph 100 feet after barrel. Constant rate of acceleration, 70 mph at impact.
5. Car going 10 mph 100 feet before impact, 110 mph 100 feet after barrel. Constant rate of acceleration, 60 mph at impact.

I don’t care what force it would take to accelerate or decelerate the car, just how far the car would move the barrel as an indication of energy transfered from car to barrel.

This is where I am going. If the object is to move the barrel the furthest, which is more important and when. The car going from 10mph to 110mph seems that it would move it the furthest, given it is the fastest rate of acceleration by far. Does that fast acceleration make up for the 20mph less it will impact the barrel when compared to the decelerating car or the 10mph less than the constant impact speed. When does a faster constant speed at impact move the barrel further than a slower accelerating speed at impact. When does the rate of acceleration become more important than impact speed, or is it always. If there is no limit on acceleration could this be answered.

Use a fixed object.

If it was a brick wall 2 feet thick, the goal is to drive through the wall with the least resistance. Which is more important in this, I think this is much more simple. In golf you want to hit through the ball. It takes a tremendous amount of force to accelerate an object, the quicker the acceleration the more force required. Simple, all things equal it will take more horsepower to move a car from 0-60mph in 4 seconds than it would in 5 seconds. For the brick wall to stop the car it has to exert that exact amount of force back to the car, only way it will stop. Which goes through the wall with the least amount of difficulty, car at constant 100mph at impact or a car going 80 mph at impact accelerating at a rate of 20 mph every 100 feet.

Exactly fade, agree. The force applied to the golf ball from the clubhead is figured by multiplying the golf balls mass by its acceleration. But this does not tell you if the clubhead is accelerating at impact, this is what I want to figure and prove. That formula with force and distance won’t work. The ball is on the face longer when flex is held, use your 1cm, say its without flex and would be 2cm(made up to show longer face contact) with flex. All that shows is that the ball was compressed more and was on the face longer, still does not show how.

Plastic ruler against your arm, hold one side firm and pull the other away from arm. The ruler is flexed, let go and it will hurt. The reason is that energy is held in the ruler, the energy hits your arm and is transferred. The result is pain. Anyone that says energy is not stored in the shaft, and can’t be transferred knows less about golf than a newborn. Did an experiment with a guy I work with. Ball is on the tee, I hold an iron at impact postion without moving. He is behind me, pulling the clubhead backwards behind the ball. The face of the iron is touching the ball. Clubhead is not moving before touching the ball. X shaft, clubhead is 3 inches behind the grip, visible bow in shaft. The ball obviously goes toward the target. If the stored energy in the shaft could not be transfered to the the ball the the ball would remain on the tee without moving. The fact that players and teachers argue this is the same as arguing the Earth is the center of the universe or that it is flat.

I know when this happens by the flight of the ball, but this is always subjective and the reason people disagree about what creates great ballstriking. If you prove that a ball behaves differently when hit with shaftflex by an exact measurement then it is objective, can’t be argued because it is fact instead of opinion.

It is so obvious to me when I hit a ball with shaftflex at impact. It frustrates me that it is so obvious, but has not been proven by measurement. If I can’t figure this out, I am going to the local college physics department. They might try to figure as a challenge. Seems that even physics has tried to figure this out and conflicts with itself. This isn’t quantum mechanics, physics can’t claim two different results happen from the same set of criteria for objects larger than atoms. There is an exact science.

Hanging a weight wouldn’t work.

1. The pressure pulling the shaft back is static because the club is static. Force has to constantly increase to maintain acceleration, this means the force is constanlty increasing through the impact area. When force stops increasing, acceleration stops at that instant. The force applied to the club to make the shaft flex backward is much greater than what the weight would show because in has to increase, the weight pulling the shaft does not.
2. Hanging weight is in one direction, pulling the shaft backwards leaves out many other pressures. The clubhead pulling the shaft backwards is caused by the clubhead trying to catch up with the grip. The clubhead does the same as the grip, it is rotating in a circle controlled by suppination of the hands, it is trying to expand away from the body because of body rotation, it is going toward the target in a line at impact, and it is going downward from transition. All these happen at once.

Clubhead in a doorjam facing a hallway

1. Push grip toward floor to represent downward force, change in height toward the ball
2. Push grip forward down the hall to represent linear force, straight line toward target
3. Push grip to the left with body as you would during pivot, this represents centrifugal force, outward force that draws a rotating body away from the center of rotation
4. Suppinate hands, feel you are trying to turn the left palm toward the sky by trying to rotate it under the grip, this represents torque.

Do all of these at once, this is what it takes to fully stress the shaft. Hogan was doing work.

I would suggest an exploration of Tutelman’s site for basic physics and calculations … http://www.tutelman.com/golf/index.php

No. F=m*a. a=F/m. If m is constant, then a constant force produces constant acceleration.

You seem to focus on acceleration in your arguments. I prefer to think in terms of force, because I see force as the cause of acceleration. Impact with a ball will cause forces that produce a good bit of deceleration of the clubhead, and acceleration of the ball. In reality, the backward force on the clubhead (and the forward force on the ball) during momentum transfer in the collision is much larger than the force that the golfer puts into the clubhead in the impact zone, even when he/she tries. To prove that to yourself, do one of your static club-pushing experiments with your driver, and try to compress a ball as much as you see it compress in high-speed video.

Fade, your the man thanks for thinking this through with me. You are correct.

Force has to constantly increase to maintain acceleration is what I wrote, think I worded incorrectly for what I meant. Force has to constantly increase to maintain same rate of acceleration.

“Say you are starting up from a stop light in your car (for simplicity assume you have an automatic transmission). If you push the gas pedal part way down and hold it in the same position, then the car will start out at its maximum acceleration. The faster the car goes, the less the acceleration will be.”

An object can increase velocity while decreasing acceleration. Above acceleration is not maintained and constant, it is decreasing. The car is still accelerating, but at a decreased rate. The pedal will have to be pushed farther to maintain the same rate of acceleration.

Car with a single gear to make it easy goes from 10mph to 11mph, rpms go from 2,000 to 2,100 to increase force to make the mass accelerate. Stop at 2,100rpms and the car will not go from 11mph to 12 mph. Rpms must increase from 2,100rpms to 2,200 rpms to make the car accelerate from 11mph to 12mph. The rate of acceleration is maintained at 1mph, from 10 to 11 then 11 to 12. I know units are not correct. 2,200 rpms is more force required to push the car than 2,100 rpms for the same rate acceleration. The engine is working harder to create the force required to accelerate the same 1mph. Rate of acceleration is maintained, force has to constantly increase to keep the car accelerating at 1mph. The numbers are made up, it is not an equal 100 rpms for 1mph. The rpms have to multiply rapidly as velocity increases to keep acceleration constant at 1mph. This is why a car acclerates faster from 0-60 than 60-120. The engine represents the muscles moving the club.

I am not sure what you mean by rate of acceleration. Wikipedia defines acceleration as follows:

In physics, acceleration is the rate at which the velocity of a body changes with time. In general, velocity and acceleration are vector quantities, with magnitude and direction, though in many cases only magnitude is considered (sometimes with negative values for deceleration). Acceleration is accompanied by a force, as described by Newton’s Second Law; the force, as a vector, is the product of the mass of the object being accelerated and the acceleration (vector). The SI unit of acceleration is the meter per second squared (m/s^2).

Acceleration will decrease, but only because opposing forces acting on the car increase with increased speed. Assume that through a perfect automatic transmission, the force produced to move the car is constant, and in this case at maximum. With increasing speed the air-drag on the car increases, and produces an increasing force that works against the force produced by the engine. The net force on the car is therefore reduced at increasing speeds. When the car reaches the speed at which the air drag force is equal to the force produced by the engine, the net force on the car becomes zero, and the car no longer accelerates: It has reached its maximum speed. In the absense of opposing forces however, say in a vacuum , the car in the above example would keep gaining speed steadily.

I have given air-drag as an example of something that decreases the net force that the car is moved by. Do you think this is a major problem for a golf-club? If not, what do you think is pushing back the clubhead against the force you apply to it?

In a car with a single gear engine-rpm translates straightforwardly to mph via the multiplication of a constant (1/200 in the above). The speedometer and tachometer will behave the same way, and you could have just one meter labeled with both rpm and mph.

You must get the units correct. Not because I am a stickler, but because it encourages understanding the issue. Acceleration is the rate at which the velocity of a body changes with time. If velocity increases by 1 mph per second, then the acceleratioin is 1 mph/s. Note that a unit of time is included twice. A cleaner form of this: If velocity increases by 1 m/s (meter per second) per second, then the acceleration is 1 m/(s*s) or 1 m/s^2.

PS I hope we are not giving Nick Price a headache after he googles himself!

Price would have known these forces by what he felt, only headache would be the one from the sound of impact. He would be good here, does not like new equipment.