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virge666
7th July 2011, 09:45 AM
Yes, it is a golf question. Having a flashback from year 10 physics.

Force = Mass x Acceleration.

So where is velocity in all this... was there a F=MV^2 somewhere ?

Bruce
7th July 2011, 09:52 AM
Speed is acceleration x time.

v = u + at

v2 = u2 + 2as

s = ut + 1/2 at2

average velocity = (v + u)/2

virge666
7th July 2011, 09:58 AM
Speed is acceleration x time.

v = u + at

v2 = u2 + 2as

s = ut + 1/2 at2

average velocity = (v + u)/2


So what is the relationship between velocity and force ?

just
7th July 2011, 10:00 AM
The relationship occurs through acceleration, which is the rate of change of velocity over time (in its simplest terms).

virge666
7th July 2011, 12:12 PM
The relationship occurs through acceleration, which is the rate of change of velocity over time (in its simplest terms).

What if something is not accelerating.

if an object of 100kg travelling at 5m/s hits an object - there must be some force in the impact ?

Pieface
7th July 2011, 12:37 PM
Yep the rapid deceleration of the moving mass and the acceleration of the struck object means there is plenty of force in this collision.

What is it they say Virge...

It's not the speed that kills it's the sudden stop.

Hawkers2008
7th July 2011, 12:40 PM
[QUOTE=virge666;661591]What if something is not accelerating.

I buggered it up I am too rusty

Hawkers2008
7th July 2011, 12:50 PM
I should add that when you consider all vector/directional info F=ma equation for a golf ball impact would be somewhat complicated.

I think you need to be looking at impact rather than force. Impact is change in momentum.

However after impact the velocity/speed of both objects will change - so there wil be acceleration or deceleration at that point and the force could be calculated.

http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html

Bruce
7th July 2011, 01:35 PM
Are you thinking about the collision between a golf ball and a club head?

Here's some of the simpler concepts you will need.
http://www.animations.physics.unsw.edu.au/jw/momentum.html

Force is the wrong thing to think of in this. The clubhead arrives at the point of collision with a certain amount of momentum based on its' speed and mass and then an impulse occurs to transfer some of that energy to the ball resulting in the acceleration over the time of the collision. Thinking back I think you do the impulse calculation first then extrapolate back to determine what the force applied by that impulse was.

Toolish
7th July 2011, 01:47 PM
What if something is not accelerating.

if an object of 100kg travelling at 5m/s hits an object - there must be some force in the impact ?

What happens after the items touch. The moving object will slow, hence accelerate, hence there is a force.

What is the context you are looking at, a golf club and ball? At impact club slows and ball speeds up, the exact details of the collision are a nightmare to go through due to it being glancing blow, both items deforming, the way the shaft does or does not support the clubhead etc etc etc.

markTHEblake
7th July 2011, 01:54 PM
Virge. Tutelman covers the collision formula in his article on forged v cast. you can easily find this in the golf club articles then materials section( i am sure)

virge666
7th July 2011, 03:01 PM
Gotcha - thanks lad.

What I am trying to figure out is this.

At what point is mass more important than velocity.

is it better to have a lighter club travelling faster - or a heavier club travelling slower. Dont care about accuracy, ball charactistics, impact of the ball.

I think Momentum will come into it somewhere though... the radius (shaft length will also come into it)

if I make my club 1" over size - the club's heft will be heavier and the swing speed will be slower
if I make the club STD - the club's heft will be lighter and the swing speed will be faster.

then we can change shafts weights as well - Lets take S300 shafts at 126gr. and Superlites at 92gr.

So where is the formula we can plug in some variables to work it... and try and get some answers...

So far - the big point is that the club MUST be accelerating through impact - that one is a given...

Where's Gerry ?

virge666
7th July 2011, 03:03 PM
What happens after the items touch. The moving object will slow, hence accelerate, hence there is a force.

What is the context you are looking at, a golf club and ball? At impact club slows and ball speeds up, the exact details of the collision are a nightmare to go through due to it being glancing blow, both items deforming, the way the shaft does or does not support the clubhead etc etc etc.

Lets forget it is a golf ball - lets say a stick on a brick wall.

Long lite stick - vs cricket stump ?

Where is the sweetspot where lite and long is better than the cricket stump ? There must be a speed that they both inflict the same anmount of force or a formula we can punch in...

Peppas
7th July 2011, 03:20 PM
Because I am a lazy physicist

http://www.golf-simulators.com/physics.htm

Peppas
7th July 2011, 03:23 PM
And I have plenty of physics textbooks at home if you want to borrow some :)

Bruce
7th July 2011, 04:18 PM
So far - the big point is that the club MUST be accelerating through impact - that one is a given...

NOTE: Assuming Newtonian Laws and conservation of momentum. Things get messy with elastic collisions etc.

NO NO NO. delta V1 at this point doesn't add anything to the equation once the collision starts.

Go back to your original equation F=ma - mass and acceleration have a linear relationship. They are locked together. What you are looking for is a combination that gets the most speed from the most mass possible. the tradeoffs will lie in the golfers ability to accelerate that mass to the point of impact. A 1000g head arriving at 50m/s has the same moment as a 200g head travelling at 250m/s. Both will have the same effect in a collision. Lighter and Faster == Slower and Heavier. It is the same. There is no inflection point where one becomes more important - the relationship is a straight line.

You might be mixing in the conservation of angular momentum now too with the idea of a light weight on the end of a stick vs cricket stump analogy. The centre of mass and the radius of the circle are what count here. Put some energy into a rotational system - like spinning around in your office chair - and pull your arms in tight. Goes Faster. Put your arms out wide. Slows down. Do it with dumb bells in your hands and note the difference. Much faster in close and much slower out wide. In this - the angular momentum is the same.

I don't think physics equations are telling us anything we didn't already know other than get as much mass and velocity into the collision as possible, and technique and strength will allow you to do that most efficiently.

virge666
7th July 2011, 04:39 PM
I don't think physics equations are telling us anything we didn't already know other than get as much mass and velocity into the collision as possible, and technique and strength will allow you to do that most efficiently.

I am trying to take all that strength and technique out of it... it is too much of a variable. Forget golf swings and glof clubs and all that.

stick 46" long with a 200gr weight on the end of it travelling at 90mph, hits something solid that can record impact forces.
vs
stick 43" long with a 220gr weight on the end of it travelling at 95mph, hits something solid that can record impact forces.



A 1000g head arriving at 50m/s has the same moment as a 200g head
travelling at 250m/s. Both will have the same effect in a collision. Lighter and
Faster == Slower and Heavier. It is the same. There is no inflection point where
one becomes more important - the relationship is a straight line.


This means a lot though...

momentum = Mass x Velocity.

so P = .2 x 90
P = 18

or
P = .22 x 95
P = 20.9

Therefore a clubhead weighing 220gr travelling at 95mph will hit the ball further than a clubhead weighing 200gr travelling at 90mph. I can get clubhead speed as a constant . . . so then I should be able to build the clubhead to get the most amount of momentum out of a swing...

A golfer will always swing a shorter club faster than a longer club . . . so if I can match the headweight up with the shorter club - I should be able to provide more power at impact.

But now we need angular momentum . . .

that would be P = Radius x Mass x Velocity...

the plot thickens...

Hawkers2008
7th July 2011, 04:55 PM
faster arm speed yes, whether the clubhead is going faster is less certain, if its a longer shaft the head might still be going faster

Marto65
7th July 2011, 04:58 PM
JEEZ !!! .. It's golf meets Big Bang Theory .... with the comedy taken out.

virge666
7th July 2011, 05:04 PM
faster arm speed yes, whether the clubhead is going faster is less certain, if its a longer shaft the head might still be going faster

Maybe - but the weight will slow it down... I reckon there is a momentum value that I will be able to find that will suit the golf swing...

I can test a lot of really good players - and then compare to some less really good players. I am hoping for some correllation.

The plan is to see why in the old days - everyone had 43.5" shafts with 105gr. Then we all went to 46" shafts with 60gr shafts.

The latest burner is 46.5" and a 196gr head - FFS !!



JEEZ !!! .. It's golf meets Big Bang Theory .... with the comedy taken out.

Sorry mate - the plan is to dumb it right down to stick hitting ball with more force. Forget all the other shit...

Peppas
7th July 2011, 05:38 PM
Ok serious answer... it's been awhile since I've done mechanics at uni.

Have you considered in your calculation longer shaft = longer radius of curvature.

The longer the radius of curvature for a given mass, the faster the velocity. Just looked it up, average speed is = 2 x pi x radius / time

So radius is related to velocity, I don't think you can have it in as a variable in your momentum equation. Ok I probably didn't help...

mike
7th July 2011, 05:42 PM
The Big Bang Theory is a comedy? Well I never.

Dotty
7th July 2011, 07:15 PM
Maybe - but the weight will slow it down... I reckon there is a momentum value that I will be able to find that will suit the golf swing...

I can test a lot of really good players - and then compare to some less really good players. I am hoping for some correllation.

The plan is to see why in the old days - everyone had 43.5" shafts with 105gr. Then we all went to 46" shafts with 60gr shafts.

The latest burner is 46.5" and a 196gr head - FFS !!




Sorry mate - the plan is to dumb it right down to stick hitting ball with more force. Forget all the other shit...
Let me know when you are ready for a 'less really good' guinea pig.

We might have to steal the 30yo laminated timber Lady Status 3 wood from the rental clubs at Berridale. I reckon I hit that further than most clubs.

ddasey
7th July 2011, 07:16 PM
Sorry gents, couldn't resist

12533

Bruce
7th July 2011, 09:05 PM
Maybe - but the weight will slow it down...


You're almost there. The weight doesn't slow it down per se. It just makes it tougher to accelerate up to maximum speed. That's why lighter is the trend as it is easier for the average golfer to maximise the clubhead speed within the length of a golf swing. The materials of clubs in 2011 allow lighter but still stiff enough to be controllable. If you'd tried to make a 60g shaft in the age of persimmon it would have been made of aluminium foil.

How do you, of all people, not know this?

virge666
11th July 2011, 05:30 PM
How do you, of all people, not know this?

Because - it aint that easy... The transition is where is all goes wrong.

At the top of the backswing on the change of direction... the killer move is the "over the top" move where the right shoulder gets involved and compensates for the weight of the club on the change of direction.

Gravity is pulling the clubhead down towards the ground whilst your body is pulling the shaft back around.

This is why good players take it back outside and let it drop inside... whilst every bad player on the planet takes it back inside and then brings it back down outside... if you watch the TV, every single pro will bring the club down on an equal flatter plane than he took it up. KJ Choi may be an exception... but not much and he plays a gay fade.

THE MOST IMPORTANT bit about clubfitting is watching what a players body does on the transition. It is without any doubt at all the MOST important thing. Does the player SWING the club around a stable pivot - or does the player counter-balance the swing on an unstable pivot.

So what I am trying to work out is the difference in force and momentum between getting the arms swinging faster vs getting a longer lighter shaft moving.

How does that mess all sound... ??

henno
11th July 2011, 06:29 PM
This thread gives me a nerd-boner.

Bruce
12th July 2011, 11:46 PM
Gravity isn't what is causing the problem from the top, nor the weight of the club. Anyone could comfortably hold a club in a position at the top with nothing but gravity to act against. It is the acceleration action to change the angular momentum of the club that has people casting etc. Take the club away and replace it with a weight machine with pulleys etc and ask someone to move the handle from a top position around to an impact like position as fast as they can. If that is the sole goal expect to see them moving their shoulders around, their weight and all sorts of things that are not considered good moves for a golf swing.

A casting action probably gives a feel of maximising the acceleration of the club head which is one of the desirable actions in a golf swing - it just ****s up a few other components.

virge666
13th July 2011, 01:39 PM
Gravity isn't what is causing the problem from the top, nor the weight of the club. Anyone could comfortably hold a club in a position at the top with nothing but gravity to act against. It is the acceleration action to change the angular momentum of the club that has people casting etc.

I use to think that - but a lot of the Euro teachers are not finding that to be the case. The difference is this....

When you are statically holding the club at the top, your body has time to adjust to the position of the club. Your core holds it and your fingers hold the weight in the club. it is "Static" and your body is braced to hold that extra weight.

What they have found is that the less the "heft" of the club the smaller the OTT move. What they think is happening is that due to the extra weight in the head or "heft", the right shoulder and forearms braces that little bit more and stops it from moving and seperating correctly.

So a club that has the weight more evenly distributed may sometime be a better option than a lite shaft with a head further away.

I think it is the perceived change in weight that is the issue. An overall lite club... 50gr shaft and 200gr head requires more CHANGE of Angular momemtum than a 75gr shaft and 200gr head. And it is that difference that can accentuate the OTT move.

Make sense ?

Bruce
13th July 2011, 04:48 PM
The delta of momentum happens around the centre of mass. Which with the heavier shaft will be moved towards the middle.

Your third paragraph about the heft is what I am trying to explain to you. It isn't anything to do with gravity, it is about accelerating the centre of mass. Those with a fast transition probably feel like they are pouring loads of energy in, but in reality are just getting it back to zero before rotating down towards the ball. With the weight more to the middle, the centre of mass has less distance to travel hence less work required to move it around which I think is something people can feel.

If I set you the task of accelerating a club to the maximum possible speed in the impact zone, and nothing else mattered other than speed - would an OTT move be the best way to achieve that? I have my suspicions that for most people this is the case - hence its' prevalence.

virge666
13th July 2011, 05:21 PM
The delta of momentum happens around the centre of mass. Which with the heavier shaft will be moved towards the middle.


Right - this is the important bit.

Please - just forget this is a golf swing for 5 minutes - think of it as an axe or a sledgehammer - I need you distance it from a golf swing for just this bit.


Your third paragraph about the heft is what I am trying to explain to you. It isn't anything to do with gravity, it is about accelerating the centre of mass. Those with a fast transition probably feel like they are pouring loads of energy in, but in reality are just getting it back to zero before rotating down towards the ball. With the weight more to the middle, the centre of mass has less distance to travel hence less work required to move it around which I think is something people can feel.


Before the move down mate... the head has weight. it has too - your hands are keeping that golf club head in the air, it might not feel like much . . . but you are doing it. The further the head from the grip - the more effort you have to put in to keep that head in the air... as the fulcrum is moving away from you with each increase in length of the shaft.

You with me . . .

If I keep making the shaft longer, the head gets harder and harder to kick into motion . . .

counter balancing love that fulcrum back towards the grip . . . I want a formula that I can create that I can plug in to find out the best counterbalancing weight to put in a club . . ot the best place to have the fulcrum on a club . . .

You can really see this with juniors struggling to keep the clubhead in the air at the top of the swing... this will really help them and seniors.


If I set you the task of accelerating a club to the maximum possible speed in the impact zone, and nothing else mattered other than speed - would an OTT move be the best way to achieve that? I have my suspicions that for most people this is the case - hence its' prevalence.

No - because you lose the lag angle on the downswing. and secondly,it is impossible to swing OTT without flipping or hitting a pull shot 90m left.

If you flip - their is no accelleration through impact - therefore there is no force. There is only momentum.

Can you see where I am going with this ? it is not speed I want - I want force. Force = ball speed. Speed just equals clubhead speed with is frigging useless with regards to fitting.

Cheers

Greeny
13th July 2011, 05:22 PM
This thread is all torque.

virge666
13th July 2011, 05:38 PM
This thread is all torque.

Not a bad pun . . . but I have not mentioned torque . . .

I think Torque is Radius x Force ?

:) :)

Bruce
13th July 2011, 09:29 PM
Virge. Go back and read a year 10 physics book. You just aren't getting it.

Your last post is full of uninformed crap.

rubin
13th July 2011, 10:14 PM
I think I know what ur trying to achieve virge. My physics is a little rusty, but I think the formula u are looking for isn't as simple as what u may be hoping.

Another suggestion - maybe speak to the physics and/or biometrics department at a uni, there is bound to be a golf mad student who would love to take it on as a doctorate or masters project.

Peppas
13th July 2011, 10:17 PM
I think I know what ur trying to achieve virge. My physics is a little rusty, but I think the formula u are looking for isn't as simple as what u may be hoping.

Another suggestion - maybe speak to the physics and/or biometrics department at a uni, there is bound to be a golf mad student who would love to take it on as a doctorate or masters project.

x2

I actually had a browse through my old mechanics text from uni. I see what Bruce is saying regarding momentum in there. It gets complicated after that. If you want to borrow the text I'll get it upto you sometime :)

virge666
13th July 2011, 10:53 PM
I think I know what ur trying to achieve virge. My physics is a little rusty, but I think the formula u are looking for isn't as simple as what u may be hoping.

Another suggestion - maybe speak to the physics and/or biometrics department at a uni, there is bound to be a golf mad student who would love to take it on as a doctorate or masters project.

You may be right. Bruce is missing the effect that gravity has on the head of the golf club at the top of the backswing... I scored 93 in my HSC for physics so I am not a total muppet, probably just explaining myself badly.

I am coming back to momentum and force... I need a way to explain it with out all the letters and numbers... Otherwise no one will take it on board and just start to glaze over it.

It comes down to players swinging the club head vs players swinging the shaft.... One is more powerful than the other, I would just like some numbers behind it.

How about a club that is accelerating will decelerate less at impact ?

Maybe a video tomorrow . . .

Peppas
13th July 2011, 11:14 PM
Rigid body mechanics + dynamic balancing for the club at the top of the swing.

Jono
14th July 2011, 01:09 AM
Virge,

1) You're not applying the equation F=ma properly.

F = ma as it pertains to the golf ball ...

F is the force the golf club exerts on the golf ball

m is the mass of the ball

a is the acceleration of the ball (not the club)

2) You should think of the clubhead as a free moving body through impact. ie. 200g clubhead travelling at 100mph will exert same force on the ball whether it is acclerating or decelerating.

3) As the whether you should go with heavier clubhead or a lighter one, you should look at this equation.

golf ball speed after impact =mass of clubhead/(mass of clubhead + mass of golf ball) x (1 + coefficient of restitusion) x clubhead speed before impact

Or more eloquently written:

V2 = m1/(m1+m2) x (1+e) x u1

m1 is the mass of the clubhead

u1 is the velocity of the clubhead before impact

m2 is the mass of the ball

v2 is the velocity of the ball after impact

e is the COR of the club

For completeness, u2 would bethe velocity of the ball before impact and v1 would be velocity of the clubheadafter impact.

This equation assumes perfect contact.

The above equation is derrived from the conservation of momentum equation

m1u1 +m2u2 = m1v1 + m2v2 ………..(1)

and the COR equation

e = (v2-v1)/(u1-u2) ……………………(2)

In the collision between clubhead and golf ball, golf ball is at rest before impact, so u2 would be 0.

So equation (1) becomes:

m1u1 = m1v1 + m2v2 ……. (3)

And equation (2) becomes:

e = (v2-v1)/u1 ………(4)

v1 = v2 – e.u1

Substitute v1 in equation (3) with v2 – e.u1 and you get the original collision equation:

v2 = m1/(m1+m2) x (1+e) x u1

Now mass of the ball is aconstant, about 50g, and e for a legal golf club is 0.83. So the above equation becomes:

v2 = m1/(m1+50) x 1.83 x u1

Let’s say that you can swing a 200g clubhead at 100mph. So your momentum would be 20000g.mph

v2 = 200/250 x 1.83 x 100

= 80 x 1.83.

Now let’s assume that you can achieve same clubhead momentum with a 5g clubhead. So your clubhead speed would be 4000 mph.

v2 = 5/55 x 1.83 x 4000

= 363.6 x 1.83


Now what this tells you is this.

Providing you can achieve the same preimpact momentum of the clubhead, lighter clubhead is better than a heavier one.

The big problem with this of course is that many people can swing a 200g clubhead at 100mph but nobody can swing a 5g clubhead at 4000mph. While they have the same momentum, the latter has a much bigger kinetic energy than the former… 40 times more. We just cannot generate this kind of energy.

So it’s a trade off. If you are the type that can swing a lighter clubhead proportionally faster, then that will give you a higher ballspeed. However, if you don’t have enough fast twitch fibers to move light objects much faster, you are better off with a heavier head.

If you want to find out what clubhead weight is optimal for you, get a really light clubhead (eg R7 driver head minus the weight inserts) and measure the clubhead speed. Gradually add weight in small increments and measure the corresponding clubhead speeds. Plug in the numbers into the above equation and you will find what driver head weight will give you the maximum ball speed.

Caveat: above is only concerned with production of maximum ball speed ASSUMING a perfect contact. If you tend to achieve a better contact with a heavier clubhead, then that may get you the maximum ball speed.

And of course, the heavier clubhead may give you more consistency in terms of direction which is pretty important. ;)

idgolfguy
14th July 2011, 01:46 AM
What's up with the colour of the text? Can't see the text clearly enough.

But I see you reasoning! Nice work, Jono.

Goldy
14th July 2011, 07:59 AM
Is THIS the toast thread?

Leon Phelps
14th July 2011, 08:21 AM
My small brain hurts so bad ......

Hawkers2008
14th July 2011, 08:30 AM
Bravo Jono,


takes me back to HSC physics nearly 30 years ago.

virge666
14th July 2011, 08:53 AM
Jono,

Welcom back !!!

Can you change your text to black please - lights on users cannot see it.

virge666
14th July 2011, 08:55 AM
So it’s a trade off. If you are the type that can swing a lighter clubhead proportionally faster, then that will give you a higher ballspeed. However, if you don’t have enough fast twitch fibers to move light objects much faster, you are better off with a heavier head.

If you want to find out what clubhead weight is optimal for you, get a really light clubhead (eg R7 driver head minus the weight inserts) and measure the clubhead speed. Gradually add weight in small increments and measure the corresponding clubhead speeds. Plug in the numbers into the above equation and you will find what driver head weight will give you the maximum ball speed.

Caveat: above is only concerned with production of maximum ball speed ASSUMING a perfect contact. If you tend to achieve a better contact with a heavier clubhead, then that may get you the maximum ball speed.


Agree on all that and many thanks...

But back to the problem - the clubhead at the top of the swing.

I must be explaining this really badly - let me do a video today and post it.

AndyP
14th July 2011, 08:59 AM
I could have fixed it, but I've changed my mind after your last post.

Bruce
14th July 2011, 09:40 AM
If I didn't know you Virge and you didn't have a good pile of credibility I would think you were trolling. But you are known generally not to be a dickhead so I guess you just aren't explaining what the problem is you are trying to solve.

Thanks for getting all the formulas in Jono - I've been posting from either my phone or an iPad the last couple of days and so copying and pasting them in was more of a chore than I was prepared to undertake.

Virge, you started out thinking you needed Force on the ball and that clubhead speed didn't matter for some reason. I hope Jono's formula list shows that the force is a function of clubhead speed and in the collision to accelerate the ball is the MOST important factor.

Now you want to talk about the clubhead at the top? Why? This is the goalpost shifting that normally characterises trolling

virge666
14th July 2011, 09:52 AM
Now you want to talk about the clubhead at the top? Why? This is the goalpost shifting that normally characterises trolling

I assure you I am not trolling...

Let me do a quick video and post it... I am not to interested in the ball and the mechanics of hitting the ball. it is all about the abiltity to put the club in motion from the top of the swing.

I have gone back to my Kinetic Energy formulas - the force required to get an abject in motion . . . this may be where I or we need to go.

Jono
14th July 2011, 10:18 AM
I think it is the perceived change in weight that is the issue. An overall lite club... 50gr shaft and 200gr head requires more CHANGE of Angular momemtum than a 75gr shaft and 200gr head.

That is not true. If both clubs are the same length and presuming both shafts have similar balance points, the second club would have the higher moment of inertia. Therefore if they were moving at the same angular velocity, the second club would require more change in angular momentum.

Angular momentum = moment of inertia x angular velocity



counter balancing love that fulcrum back towards the grip . . . I want a formula that I can create that I can plug in to find out the best counterbalancing weight to put in a club . . ot the best place to have the fulcrum on a club . . .


Virge,

Think of the club as having 3 heft properties
1) weight - this is just the total weight of the club
2) 1st moment (mR) - this is the total weight x radius (ie distance from the balance point of the club to the grip end)
3) moment of inertia (mR^2) - this is the sum the mass of individual point along the club multiplied by the distance of that point from the grip end of the club squared.

Counterbalancing may move the balance point towards the grip end but it also increases the total weight of the club. End effect is that it does NOT decrease any of the above 3 heft properties.

If you put the counterweight right at the grip end, you increase the weight of the club without increasing the first moment or the MOI. This can make the club FEEL lighter to swing for a golfer but it is not. You'll require just as much torque to swing the club if not more.




it is not speed I want - I want force. Force = ball speed. Speed just equals clubhead speed with is frigging useless with regards to fitting.


Virge, you have to set your parameters correctly before you can start your analysis.

There's only three forces on the ball during impact.
1) the force exerted by the clubhead
2) gravity
3) tee holding the ball up against gravity

The ball doesn't feel the force the golfer exerts on the grip end of the club.

The only force we need to be really concerned about is the force exerted by the clubhead. Now that force is very much dependent on the clubhead velocity along with the clubhead mass, the quality of the contact etc.

virge666
14th July 2011, 10:31 AM
Thanks Jono - enjoying the lesson.

Back shortly.

The force I mention is important in my eye because if the clubhead has force behind it . . . then it is more resistant to decelarating at impact...

so if the hands are ahead at impact - the club will not decel as much as it would if the hands are behind . . .

in other words - hands in front - accerating through impact, hands behind = decelerating at impact.

which I am hoping will lead back to the F=ma equaution.

But all this is for later on - after the Moment of inertia conversation.

Cheers

Jono
14th July 2011, 10:52 AM
Thanks Jono - enjoying the lesson.

Back shortly.

The force I mention is important in my eye because if the clubhead has force behind it . . . then it is more resistant to decelarating at impact...

so if the hands are ahead at impact - the club will not decel as much as it would if the hands are behind . . .

in other words - hands in front - accerating through impact, hands behind = decelerating at impact.

which I am hoping will lead back to the F=ma equaution.

But all this is for later on - after the Moment of inertia conversation.

Cheers

Virge,

Most golf physicists consider clubhead at impact as a floating body. If you look through Dave Tutelman's stuff, it will tell you why.

The clubhead will decel after impact no matter what you try to do. The law of conservation of momentum demands it.

As to hands ahead vs hands behind ... generally, hands ahead is better because it sets up better impact conditions ... such as delofting the club, better delivery path etc.

Yes, having hands ahead might resist decel BEFORE impact and thus have more cluhead speed at impact. However, it will not resist decel THROUGH impact.

Jono
14th July 2011, 10:54 AM
I have gone back to my Kinetic Energy formulas.

Don't do it.

http://en.wikipedia.org/wiki/Lagrangian_mechanics

Bruce
14th July 2011, 11:01 AM
in other words - hands in front - accerating through impact, hands behind = decelerating at impact.

I the impact were perfect with the 2 centres of mass colliding in line then you could let go of the club at impact. One of the reasons we need to hang on is to deal with the less than perfect impact we are always going to generate and the other is to keep accelerating until the collision is over - you can add a bit more to the equation by doing this, but I don't think it is as much as you want it to be.

Velocity is acceleration over time. You want the time to be as long as possible so you get maximum velocity. It all keeps coming back to clubhead speed mate.

Pieface
14th July 2011, 11:32 AM
Looks like the discussion has moved on but here's my very laymans (and most probably wrong! ) interpretation of why you want to be accelerating the club at impact...

If you are accelerating the clubhead the shaft is under load. Under load the shaft will deform according to it's characteristics. A good shaft will be very uniform in the way that it deforms under load, within it's tolerances. This uniform and consistent deformation allows the player to remove some of the variables from what occurs at impact. If the shaft is not under load - constant velocity - or is decelerating the relationship of the clubhead to the shaft will be much more variable leading to inconsistent contact.

Velocity of the clubhead determines the maximum possible distance you can hit the ball
Accelerating through the ball gives you the best chance of making consistent, quality contact with the ball.

*Dons flameproof suit*

Greeny
14th July 2011, 12:39 PM
Virge,

1) You're not applying the equation F=ma properly.

F = ma as it pertains to the golf ball ...

F is the force the golf club exerts on the golf ball

m is the mass of the ball

a is the acceleration of the ball (not the club)

Spot on Jono.

virge666
14th July 2011, 08:59 PM
Don't worry boys, it can't be done.

The hands don't do anything to start the transition... The body starts the rotation.. So the calculations are mute.

Don't know why you are all still talking about impact though when it has absolutely nothing to do with the question...

Anyway many thanks to those who helped out...


Bugger

virge666
15th July 2011, 09:41 AM
I may just use swingweight.

D2 is 42 swing weight points . . . if you add an inch to the club - it becomes 48 swing weight points - so it is 14.2% harder to get the club into motion.

Pretty caveman like but it may have to do.

greeneye
22nd July 2011, 11:50 PM
Momentum = mass x velocity
Kinetic energy = 1/2 m v^2