Bryson DeChambeau

bren2001

GreeBo wrote: »

The two bold sentences would seem to contradict each other?
If the weight of the golfer doesnt contribute to the mass at impact, why is it take into account? The player doesnt hit the ball, the clubhead does.

You've no doubt seen trick shots where the shaft of the club is a rubber hose...how is the golfers weight or mass being transferred to the ball in that scenario?

What I'm trying to say in that sentence is that for the law of conservation of momentum the mass is not the mass of the club + the golfer. You would need to calculate the effective mass of both or the contributory mass of the golfer. It would be relatively complex to do.

Last time I checked, both the player and the club hit the ball. Otherwise, the golfer wouldn't feel anything. The force is transferred up and subsequently resisted by the body (mainly).

GreeBo

bren2001 wrote: »

"The club has no idea what mass is being used to swing it" - what does that even mean? The clubs an inanimate object. It's not about what the club knows, its about really basic maths. The club and the body are one system, they are connected at the hands. Regardless of how much you dispute it, it is still a fixed connection (whatever the word "fixed" means in this context). In any collision mass is important. Your body moves through impact, hence, your weight has an effect. If your body was still, it would be different.

If your body is colliding with something in your golf swing you are doing it wrong.
You still havent explained how the golfers weight is relevant to the golf ball. We have all agreed that their weight *may* impact their ability to generate clubhead, but 100mph is 100mph regardless of the weight of the person swinging it.
I thikn perhaps you are confusing force (f=ma) but in the golf swing its the mass of the clubhead, not hte mass of the person swinging the clubhead.

A fat guy and a skinny guy throw a golf ball at your head, in both scenarios the ball is travelling at 50mph when it hits you. There is zero difference between these two scenarios, even though they both had a "fixed connection" with the ball when they threw it.

Conservation of momentum doesn't come into a dynamical system where two bodies collide. Right. Care to explain that one? That's actually the formula you would use to calculate initial ball speed. How else would you calculate it? What else would you use? F=ma? Funny enough, mass is in there and the conservation of momentum is just an expansion of that.

Yeah, the mass of the club and the ball, not the golfer.

Care to draw a free body diagram and outline how that's the case? You feel the impact and your body is what keeps the club on line. The club pushes back against your hand as the ball is struck. Your body, not just your hands, resists this and pushes against the club and the ball. Hence, your weight has to be taken into account. If you generate 90 mph clubhead speed but your body stays perfectly still while your arms move, you won't hit the ball as far as someone who generates 90 mph clubhead speed but turns through the shot. Why? The golfers weight.

Sorry but thats just not correct. the same club hitting the same ball at the same speed imparts the same force on the ball, regardless of the person being 100kg or 10kg.
Again if this is not the case, explain how a rubber shafted club hits the ball? How is the mass of the player impacting the ball through a rubber hose?

If you added 250kg to yourself and, in the exact same manner, then yeah, the ball goes further. If you add it to your feet, then yes, in theory the ball will go further. However, adding weight above your centre of gravity will have a larger effect.

How? How does the ball go further?
How is my extra 250kg of mass getting to the ball? The ball gets hit at the same speed by the same mass (the mass of the clubhead) everytime, everything else is irrelevant.
Why isnt every Pro golfer lashing on a weight belt before every shot?
My centre of gravity?! Why is this suddenly relevant to the ball?
Sure, it might allow me to swing faster, but thats whats impacting the ball, not my COG or COM or what I had for breakfast.

GreeBo

bren2001 wrote: »

What I'm trying to say in that sentence is that for the law of conservation of momentum the mass is not the mass of the club + the golfer. You would need to calculate the effective mass of both or the contributory mass of the golfer. It would be relatively complex to do.

Last time I checked, both the player and the club hit the ball. Otherwise, the golfer wouldn't feel anything. The force is transferred up and subsequently resisted by the body (mainly).

No, the club hits the ball, the player doesnt.
The impact, vibration, etc are passed to the body, but the ball isnt pushing back against the body, the ball is rebounding off the face, thats why COR is so important to distance.
Again (again!) if what you are saying is correct, how does a rubber shafted club hit the ball? aA rubber shaft has no ability to resist anything, so how is this resistance passed to the golfers body?

bren2001

GreeBo wrote: »

No, the club hits the ball, the player doesnt.
The impact, vibration, etc are passed to the body, but the ball isnt pushing back against the body, the ball is rebounding off the face, thats why COR is so important to distance.
Again (again!) if what you are saying is correct, how does a rubber shafted club hit the ball? aA rubber shaft has no ability to resist anything, so how is this resistance passed to the golfers body?

If it's only the club that hits the ball, why do I feel it?

GreeBo

bren2001 wrote: »

If it's only the club that hits the ball, why do I feel it?

erm, because you are holding the things thats hitting it?

bren2001

GreeBo wrote: »

erm, because you are holding the things thats hitting it?

Exactly. One solid mass.

Apply F=ma from the club to the ball and F=ma from the shaft to my hands. The force of golfer transfer through the club to the ball i.e. the mass of golfer is a factor in determining both forces. Nothing to do with speed, all to do with free-body diagrams in statics i.e. at the point of impact everything is frozen from a mathematics perspective.

Now just put those two equations together and the golfer and club are a combined mass. Which is how you would actually do it in practice.

Explain why that's wrong.

That's the simplest way of putting it. You might want to brush up on your statics theory.

GreeBo

bren2001 wrote: »

Exactly. One solid mass.

How is it "one solid mass"?

Apply F=ma from the club to the ball and F=ma from the shaft to my hands. The force of golfer transfer through the club to the ball i.e. the mass of golfer is a factor in determining both forces. Nothing to do with speed, all to do with free-body diagrams in statics i.e. at the point of impact everything is frozen from a mathematics perspective.

All the ball knows about is f=ma of the club.

Now just put those two equations together and the golfer and club are a combined mass. Which is how you would actually do it in practice.

I have a 4 time limit, so this will be my last time. Explain how the rubber shafted golf club hits the ball?

Explain why that's wrong.

That's the simplest way of putting it. You might want to brush up on your statics theory.

Indeed.

5 axioms of Statics Theory wrote:

1. A rigid body acted upon by two forces is in a state of static equilibrium if and only if the two forces are of the same intensity, lie along the same line of action, and are oriented in opposite directions along the line.
2. If a system of two forces in equilibrium is added to or extracted from a given system of forces, the way that the system of forces acts on a rigid body undergoes no change.
3. The resultant of two forces acting at the same material point is equal to the vector sum of the two forces. The line of the resulting force's action contains the material point. This axiom obeys the principle of vector summation.
4. Two interacting bodies react on each other with two forces of equal intensity, and along the same line of action, but in opposite directions along the line. This axiom is also known as principle of action and reaction.
5. If a deformable body is in a state of static equilibrium, it would also be in static equilibrium if the body were rigid. This axiom is also known as the principle of solidification.

Definition of rigid wrote:

unable to bend or be forced out of shape; not flexible.

Now remind me which bit of a golfer swinging a club is a rigid body?

The clubhead imparts a force onto the golfball at impact and, due to acxiom 4) (& Mr Newton's 3rd law) the ball imparts a force on the club. However these are not balanced forces, they do not cancel each other out, so the ball moves.
The shaft bends due to the difference in force applied to the head and the grip, the golfer holding the grip end. This bend is storing energy which is released into the ball.
The golfers grip resists the force from the ball (transferred via the shaft, that feeling you asked about earlier) but also the non-rigid shaft absorbs this impact, resulting in vibrations, again the feeling you feel.

The club slows down and the ball gets out of the way.
If the ball weighed 50kg then it still wouldnt matter what the golfer weighs as its not a rigid system, try hitting an impact bag, your wrists will break down, no matter how much you weigh.


So, unless your argument is regarding the scenario where the golfer weighs less than the ball, the golfers weight, when all other factors are the same, is irrelevant. (and even in that scenario its not clear cut)

Golf is my Game

Yeh, there no serious debate about this one. Brysons hitting it out there long because he has putt on more muscles and can turn faster an so get more club head speed. His weight in itself has nothing to do with in any scientic discussion but golf is not known for scientific sound discussion, its usually non science, psudoscien e or just nonsense. If Brayson had just milled the doughnuts instead of pumping iron and put o the same weight he'd be hitting it no further than he was last year. The ball is hit with the club heat through sever elastic and yielding stages, like the flexy shaft itself, he's wrists, elbows, shoulders, torso, hips, knees, andles, and he isn't bolted to the ground either. So he isn't a rigid body. Muscle is what does it.

bren2001

GreeBo wrote: »

How is it "one solid mass"?

All the ball knows about is f=ma of the club.

I have a 4 time limit, so this will be my last time. Explain how the rubber shafted golf club hits the ball?



Indeed.




Now remind me which bit of a golfer swinging a club is a rigid body?

The clubhead imparts a force onto the golfball at impact and, due to acxiom 4) (& Mr Newton's 3rd law) the ball imparts a force on the club. However these are not balanced forces, they do not cancel each other out, so the ball moves.
The shaft bends due to the difference in force applied to the head and the grip, the golfer holding the grip end. This bend is storing energy which is released into the ball.
The golfers grip resists the force from the ball (transferred via the shaft, that feeling you asked about earlier) but also the non-rigid shaft absorbs this impact, resulting in vibrations, again the feeling you feel.

The club slows down and the ball gets out of the way.
If the ball weighed 50kg then it still wouldnt matter what the golfer weighs as its not a rigid system, try hitting an impact bag, your wrists will break down, no matter how much you weigh.


So, unless your argument is regarding the scenario where the golfer weighs less than the ball, the golfers weight, when all other factors are the same, is irrelevant. (and even in that scenario its not clear cut)

You're a 4-time limit? You've not clarified a single point on your position mathematically nor have you answered mine.

By that definition, nothing is rigid. The rigid part of the golf swing is the golf club combined with the person, at the moment of impact they are rigid bodies. That's what those rules refer to.

Can I ask what qualifications you have to speak on statics theory? I'm an engineer. You keep telling me I'm wrong yet I'm fully qualified in the area.

The rubber shaft and golf ball analogy doesn't explain your position at all, the weight of the golfer does transfer through the rubber but it's highly elastic so cannot transfer the energy well. Instead, it absorbs the impact of the golf ball and elastically deforms. If you want to get into finite element analysis, let go. I'll start discussing that no bother.

ckeng

bren2001 wrote: »

You're a 4-time limit? You've not clarified a single point on your position mathematically nor have you answered mine.

By that definition, nothing is rigid. The rigid part of the golf swing is the golf club combined with the person, at the moment of impact they are rigid bodies. That's what those rules refer to.

Can I ask what qualifications you have to speak on statics theory? I'm an engineer. You keep telling me I'm wrong yet I'm fully qualified in the area.

The rubber shaft and golf ball analogy doesn't explain your position at all, the weight of the golfer does transfer through the rubber but it's highly elastic so cannot transfer the energy well. Instead, it absorbs the impact of the golf ball of elastically deforms. If you want to get into finite element analysis, let go. I'll start discussing that no bother.

For what it's worth I'm an engineer too and the golf swing isn't static. You should be looking at angular momentum here. The lever arm between the centre of mass of the golfer and the centre of rotation of the swing is tiny, so the contribution to the angular momentum of the club head is correspondingly tiny - more than corresponding really since lever arm to angular momentum is a squared relationship.

bren2001

ckeng wrote: »

For what it's worth I'm an engineer too and the golf swing isn't static. You should be looking at angular momentum here. The lever arm between the centre of mass of the golfer and the centre of rotation of the swing is tiny, so the contribution to the angular momentum of the club head is correspondingly tiny - more than corresponding really since lever arm to angular momentum is a squared relationship.

No, its not but you can treat it as a static problem with a few simplifications. But thank you, you agree the weight of the golfer does contribute it's just negligible. I wasnt arguing the later but I don't agree with it.

ckeng

bren2001 wrote: »

No, its not but you can treat it as a static problem with a few simplifications. But thank you, you agree the weight of the golfer does contribute it's just negligible. I never disputed the later but I don't agree with it.

What are the simplification's out of curiosity?

thecomedian

Weight doesn’t matter. Just ask Jamie Sadlowski

GreeBo

bren2001 wrote: »

No, its not but you can treat it as a static problem with a few simplifications. But thank you, you agree the weight of the golfer does contribute it's just negligible. I wasnt arguing the later but I don't agree with it.

As an engineer, can you give me your definition of negligible?
And would also be great if you can explain the simplifications that turn a dynamic, deformable system into a rigid system and still talk about negligible forces...

bren2001

GreeBo wrote: »

As an engineer, can you give me your definition of negligible?
And would also be great if you can explain the simplifications that turn a dynamic, deformable system into a rigid system and still talk about negligible forces...

There is no definition of negligible, it is entirely dependent on the application. If I were to design a dishwasher and said the water was 60C for the cycle but the machine actually heated the water 62, who cares? If I worked in a pharmaceutical company and heated a chemical to 62 instead of 60. That may not be ok. I've also stated several times, I'm happy for you or anyone to say the mass of the golfer is negligible but I don't think it is. At that point, it's an argument over what negligible means. I'm not going to argue semantics.

However, the point stated by you and Golf_is_my_Game is that the mass of the golfer has no effect. You didn't say it was negligible. You said "The club has no idea what mass is being used to swing it". Do you accept that the sentiment of this statement is incorrect? The mass of the golfer does have an effect even if it's small.

How do you turn a dynamical problem into a statics problem? Apply a force to stop the moving object and you can do your analysis from there. It's one of the most common approaches used. It's no different than dropping a box on a cantilever beam and studying the system as a static system or the "stationary observer" philosophy of relative motion. If you want to argue none of them are static problems, go ahead but I don't see how it's relevant.

You seem hell-bent on this idea of rigid as if its a smoking gun. Both static theory and dynamical theory assume a rigid body. Technically, the theory of dynamics in mathematics doesn't apply either. However, I think a good assumption is to model the golf club as a rigid body at the moment of impact. I don't think the deformation of the club caused by the ball when it is struck has a significant impact on distance. I think the effect is negligible. The shape of the club just before impact and just after impact are so similar, I'd model them as the same and hence can call the system rigid. Even though it's not but no system is strictly rigid. The club may deform greatly during your downswing but that isn't relevant at the moment of impact. Put in a different material for the shaft and it's a different story.

If you want to argue that my model for the system is poor, I won't argue back. It makes a rake of assumptions that would need to be taken into account. However, when modelling something, it depends what you end goal is. If I construct a model to study a system in the frequency domain, time domain analysis will just throw up garbage. Do I care? No. In this case, my model is solely to demonstrate that the mass of a golfer has an effect on distance. My model demonstrates that. ckeng, a fellow engineer, confirms that mass has an effect. Are both me and ckeng, qualified engineers, wrong?

Henry Ford III

PokeHerKing wrote: »

You can't figure out why his irons don't go the same distance and you're calling him dim:pac:

Yes.

Please explain why equal length say 4 iron and 7 iron go different distances.

Mobile32

There is definitely science to all what Bryson is doing. Unfortunately it is not the good kind. Genie out of the box now. Golf will have to take drug testing seriously and clamp down. Even most of top players are openly referring to Bryson Steroid use.

Henry Ford III

Henry Ford III wrote: »

Yes.

Please explain why equal length say 4 iron and 7 iron go different distances.

Loft, launch and spin.

Henry Ford III

ronnoco13 wrote: »

Loft, launch and spin.

That's a list of terms not an explanation.

Let's start with this:-

On an otherwise identical swing a longer club will have greater clubhead speed, and will therefore hit it further.

GreeBo

bren2001 wrote: »

. I don't think the deformation of the club caused by the ball when it is struck has a significant impact on distance. I think the effect is negligible. The shape of the club just before impact and just after impact are so similar, I'd model them as the same and hence can call the system rigid.

I was going to reply to each of your points until I read the above. So your backup example is to say that COR is negligible? And also that the shaft bending during the downswing means nothing at impact?

You should go tell club and shaft manufacturers that they are wasting their time.
I'm certainly no longer going to waste mine talking to you about it!

GreeBo

Henry Ford III wrote: »

That's a list of terms not an explanation.

Let's start with this:-

On an otherwise identical swing a longer club will have greater clubhead speed, and will therefore hit it further.

If the loft, launch and spin are the same then it might, otherwise it may not.
Otherwise why do we have different lofts?

Golf is my Game

Henry Ford III wrote: »

That's a list of terms not an explanation.

Let's start with this:-

On an otherwise identical swing a longer club will have greater clubhead speed, and will therefore hit it further.

Thats true. Which is why Bryson has to have different lofts on his clubs to ones used by regular sets that also vary the distance they his be having different lengths.
The change in loft changes the proportions of club head speed transferred to the ball in the horizontal and vertical vectors. You could hit a ball with a club having 89° on it (if you hade the precision to hit the ball) but most of your effort would go into making the ball rise almost vertically in the air and then drop a meter in front of you. It like Mickelson does with his trick shots when he opens 60deg and takes a full lash at it but only send the ball 20 yards. The club is still the same length. But sending a ball up agasint gravity uses energy which is wasted as far as getting the ball forward goes. So thats why.
So his 4 iron is shorter than normal. He has to have less loft on it to make up for the slower head speed but still get a useful 4 iron distance. His Wedge is longer than normal, so he has higher loft so the ball doesnt go as far as it would for being extra long.

For regular golfers, the varying of club shaft length as well as loft is a good compromise to get a spread of distances. Because they just can get as much club speed as the top lads. They have no problem with clubspeed so can launch it no bother even with lower loft to get 4 iron distance. So if you can get a clean strike it doesnt matter which proportion of the two thing you have. Bryson is going on the idea that theirs less variables in the swing if he just uses one length. But really theres so many variables that it doesnt matter. Its a mental thing so in that way helps him but really its an illusion and hed play just as well with a regular set but he got caught up in a hole physics science thing and thats his thing now. Like other get all into fitness or mental psychology or whatever. Its just his schtick like they say. Its no resolution in golf thinking though or breakthrough in performance or useful to anyone else. Its as scientific as Gary Player always wearing black because he read black absorbed the suns energy more than bright colours so he would have more energy if he wore black. Or a lad tried to explain to me once who was always washing his car that it was so he would get more miles per gallon. On about friction and dirt and that the car was more slippery through the air if it was clean and had wax on it. Which is true I guess. But spending his time washing it with fancy polishes and the like to save money on petrol was nonsence. It made him happy anways which is partly what contributes to Bryson doing well so its good from that angle. But for the rest of us to copy or to think is more than just mumbo jumbo is just silly and thats a fact.

Matmania

Ok for all saying hes not doing anything different here go's. He's using the same size clubs who has done that. His loft on the driver is 5.5 i think average on the tour i think is 10.5. Also his swing is a tad and i mean a tad different. I know we going to get here to argue but he is doing things different why are people talking about it.

Weather it works or not who knows. 25/1 was a great price for the masters 3 weeks ago gone know. I had lucas glover 175/1 ernie 50/1 etc i know my golf. Play off 7. This guy is trying something different and if you cant see it good luck to you because you arguing with yourself. He is targeting the masters. Bubba won the masters by destroying the par fives. Taking the corner out on o think the 13. I f he drives like he can he will have 8 max into the par fives at augusta massive advantage and the fairways are wide. We shall see but anyone saying hes noting doing anything different hasn't got a clue. That's a given.

Matmania

waiting for the oh hes not doing anything different. because everyone on the tour is using the same size clubs and a 5 degree driver. anyone name anyone who has done that with his swing. because i cant. the longest drivers comp i thing use 4 degree. can anyone name one golfer doing what he is doing. probably not. because he is doing something quite different. like like or not he is. also since he is doing it not bad going.

Golf is my Game

Oh hes not doing anything differrent with his swing. Its pretty normal and there much more lads with bigger devaiations from what Butch and Faldo and whoever would say is the ideal swing. On a normal curve, he is a tad off the centreline, but only a tad and so is still in the 95%.

No one is arguing hes not doing something different with his clubs. Thats not even arguable. What me and maybe others are saying is that theirs nothing BETTER in what hes doing. And its no genius breakthrough of scientifica analysis. People have been experimenting with different length loft specs since we started playing golf on scottish wastelands. Theyve all been tryed before. I remember the 'EQL' I think it was by John Letters irons in the 80s or 90s that they tried to push with the same 'logic' that has led Bryson to them but they failed, no one copied them, and that was the end of that. Because they were just a marketing gimmick. Not a bad one for a minor golf club manufcturer that was failing and trying to have something different to offer and get a few sales. But it would have been the same trick on any lads that bought them as Bryson is playing on himself. Different sure. And with a very limited and flawed justification that kids the vulnerable of mind to think the line that goes from 'same length so less variability in your swing from iron to iron' leads to the effect of 'less variability in your swing will have you hit your irons better'. Which is the bit where the science which we say with a laugh, breaks down. But marketing is fine. Like I sayed before, If like using a lucky poker chip as your ball marker gives you comfort that you will putt better, then theres benefit in him using them (probably - its possible he is holding himself back by uising them and that the millions and millions of golfers using varied length iron shafts for the last hundred were actually doing it for the very good reason that its better than single length ones) if he believes theres something behind it even if there isnt.
Hows Bryson irons accuracy for length and dispersion compared to other top ranking pros ? Which takes putting, driving, short game, out of it ?
Its a gimmick lads. Its as useful to him as the Hogan cap.

Golf is my Game

In fact, the more I think about it, the most likely situation until there is evidence to the contrary, is that it is doing his game more harm than good.

bren2001

GreeBo wrote: »

I was going to reply to each of your points until I read the above. So your backup example is to say that COR is negligible? And also that the shaft bending during the downswing means nothing at impact?

You should go tell club and shaft manufacturers that they are wasting their time.
I'm certainly no longer going to waste mine talking to you about it!

At what point did I say any of that was negligible? I don't know if you're intentionally misunderstanding what I said about shaft bend or not but when you're doing a mathematical analysis at the point of impact, what has happened before is irrelevant. The bend is taken into account by the shape of the club at the point of impact. That's how shaft bend is taken into account in a simple model. My statement was relating to the deflection of the shaft in the moment before impact and the moment after. An infintely small amount of time. I don't believe the deflection that occurs due to the strike of the ball in this time-frame is a significant factor.

The model and mathematical analysis I've stated in this thread is solely to demonstrate that the weight of a player has an impact on distance. It is not a full model to actually calculate the distance a ball would travel. You would need a much more sophisticated model for that. Most likely a DAE.

I don't particularly want to waste my time talking to a brick wall. My only point is that the weight of a golfer impacts the distance a ball will travel. I am not saying what other factors are important.

It seems to me that you've accepted that weight does matter in a golf swing but you don't want to admit you are wrong and are now trying to twist your argument or move the goalposts. Very simply, do you now accept that the weight of a golfer does impact the distance a ball will travel? That's the only point I've been trying to convey.

bren2001

Henry Ford III wrote: »

That's a list of terms not an explanation.

Let's start with this:-

On an otherwise identical swing a longer club will have greater clubhead speed, and will therefore hit it further.

If I gave you a cannon with a basketball in it and asked you to shoot it through a hoop, how would you do it assuming the ball comes out at the same speed every time?

You'd tilt the cannon until you got it through the hoop. That's the same as loft on a golf club. You're just changing the angle the ball leaves the ground at like the cannon. That's how loft impacts distance.

Henry Ford III

bren2001 wrote: »

If I gave you a cannon with a basketball in it and asked you to shoot it through a hoop, how would you do it assuming the ball comes out at the same speed every time?

You'd tilt the cannon until you got it through the hoop. That's the same as loft on a golf club. You're just changing the angle the ball leaves the ground at like the cannon. That's how loft impacts distance.

Magnificent strawman. Simply outstanding.

Why does every other pro use different length irons?

bren2001

Henry Ford III wrote: »

Magnificent strawman. Simply outstanding.

I wasn't and am not trying to strawman you. It's how I would teach anyone on an introductory to projectiles course.

If you want a more mathematical approach, if your initial ball speed is u and the angle it leaves the ground is alpha, your horizontal speed is ucos(alpha) and vertical speed is usin(alpha) using the Pythagoras theorem.

To find the length of time the ball stays in the air, solve 0 = usin(alpha) - g*t where g is gravity or 9.81 m/s.

To find the distance the ball travels solve s = ut + 0.5*a*t^2, assuming no air resistance, a is 0 and its just s = ut where s is displacement, u is ucos(alpha) and t is from your equation above (you double it).

Test this is for all values of alpha and you'll see alpha of 45 degrees gives max distance. Loft on a golf club controls alpha and helps in deciding distance.

That's the mathematical reason but I figured you didn't particularly want an explanation like that.

GreeBo

bren2001 wrote: »

At what point did I say any of that was negligible? I don't know if you're intentionally misunderstanding what I said about shaft bend or not but when you're doing a mathematical analysis at the point of impact, what has happened before is irrelevant. The bend is taken into account by the shape of the club at the point of impact. That's how shaft bend is taken into account in a simple model. My statement was relating to the deflection of the shaft in the moment before impact and the moment after. An infintely small amount of time. I don't believe the deflection that occurs due to the strike of the ball in this time-frame is a significant factor.

The model and mathematical analysis I've stated in this thread is solely to demonstrate that the weight of a player has an impact on distance. It is not a full model to actually calculate the distance a ball would travel. You would need a much more sophisticated model for that. Most likely a DAE.

I don't particularly want to waste my time talking to a brick wall. My only point is that the weight of a golfer impacts the distance a ball will travel. I am not saying what other factors are important.

It seems to me that you've accepted that weight does matter in a golf swing but you don't want to admit you are wrong and are now trying to twist your argument or move the goalposts. Very simply, do you now accept that the weight of a golfer does impact the distance a ball will travel? That's the only point I've been trying to convey.

No I don't.
I started saying the ball didn't care about anything other than impact, you started saying the golfers weight is relevant, I'm saying is not and you haven't been able to demonstrate anything to the contrary. I asked would iron Byron hit it further if you have it heavier and you ignored it. I asked various similar questions and you ignored them.

The ball is hit by the club and distance depends on the force, which is the mass times acceleration of the club. The same club at the same speed hits the ball the same distance.
Why is Rory longer than koepka with the same swing speed and 20kg lighter?

GreeBo

Henry Ford III wrote: »

Magnificent strawman. Simply outstanding.

Why does every other pro use different length irons?

I don't think anyone knows what point you are trying to make... Can you explain?

bren2001

GreeBo wrote: »

No I don't.
I started saying the ball didn't care about anything other than impact, you started saying the golfers weight is relevant, I'm saying is not and you haven't been able to demonstrate anything to the contrary. I asked would iron Byron hit it further if you have it heavier and you ignored it. I asked various similar questions and you ignored them.

The ball is hit by the club and distance depends on the force, which is the mass times acceleration of the club. The same club at the same speed hits the ball the same distance.
Why is Rory longer than koepka with the same swing speed and 20kg lighter?

Right, here's an article titled "Are heavier baseball players better hitters? Basically, no, says physics". The conclusion is that heavier baseball players don't hit it further because their swing changes. The same is true for golfers . I have not disputed that and specifically stated earlier that "fat players typically won't perform well but if they had the exact same swing as a much lighter golfer. I would expect the fat golfer to hit it further". The point being, two players with the same swing but different weights will hit the ball different distances. That's the only point I'm trying to make.

You're claiming "the same club hitting the same ball at the same speed imparts the same force on the ball" and that "the club has no idea what mass is being used to swing it" i.e. the weight of the golfer has no impact on distance.

The article states: "It’s all about kinetic energy, the kind of energy used in motion. It’s described by the formula k = (1/2)mv2, where m is the mass of the system—here, the weight of the hitter and the bat—and v is the velocity, or speed, which here accounts for the movement of hitter and bat. "...."in theory, if a bigger guy is just as fast or strong as another guy, he should be able to hit the ball further" note: the weight of the batter is taken into account.

Towards the end of the article (3rd last paragraph) it states "If—and only if—players can keep their speed constant while making perfect contact with the ball on a perfectly hard bat, they would be able to hit the ball harder if they’re bigger" where bigger means heavier i.e. the weight of a baseball player impacts distance. I'm making the exact same point about golfers.

Can you either explain why a baseball players weight would matter but not a golfers or explain why researchers a Penn State University and I are incorrect?

https://qz.com/801738/why-are-baseball-players-fat/

Golf is my Game

With yere maths theres also the mistake in trying to make it to a static situation because it isnt. While the time the clubs in contact with the ball is very small and so intuitively we tend to think of it as an instant in fact the time the ball is in contact is very important and is a critical element in the pros getting their distance and spin on the ball. The ball is very elastic, and the club head slightly elastic. The acceleration/deceleration of the clubhead and what your swing timing is doing to it for the period of contact is crucial to the compression of the ball. the transfer of clubhead speed to ball speed, and to the spin put on it as a result. Swing timing that maximises the time the club is in contact with the ball is what allows top players put such back spin on the ball that they can control it on the green, even with short clubs where high clubhead speed is not required they can do it. Its a dynamic impact. Low level golf simply never get such spin on the ball to have the ball spin back on the green because they arent able to sustain the club head speed as it contacts the ball (they fool them selves sometimes that they do when a ball jumps back out of its pitch mark but that werent the spin).

Golf is my Game

Henry Ford III wrote: »

Magnificent strawman. Simply outstanding.

Why does every other pro use different length irons?

Because its the best way found over the years to get best gaps between the clubs by spreading the distance gained between them between a little loft change and a little length channge. Its a balanced compromise. You could do it with same length clubs like Bryson. Or you could do it with same loft on them all and change the length much more. Either works. A mix works best though top elite players with great club head speed as a result of their swings can nudge it towards the loft being the difference. Club golfer could more likely benefit the otherway. Put more loft on the club and make it longer. The ball will get in the air easier. But theyl find they run out of length in the irons then. But us hyrids from about 6 iron down and they would do well. Seniors and lads in the 20 handicaps range and so on. In fact they sort of being doing it for years anyway with 7 woods and the lokes of the 'gentlemans persuader' type clubs for decades even before hybrids were found to be a more easily hit low twenty degree club for the less skilled.

bren2001

Golf is my Game wrote: »

With yere maths theres also the mistake in trying to make it to a static situation because it isnt. While the time the clubs in contact with the ball is very small and so intuitively we tend to think of it as an instant in fact the time the ball is in contact is very important and is a critical element in the pros getting their distance and spin on the ball. The ball is very elastic, and the club head slightly elastic. The acceleration/deceleration of the clubhead and what your swing timing is doing to it for the period of contact is crucial to the compression of the ball. the transfer of clubhead speed to ball speed, and to the spin put on it as a result. Swing timing that maximises the time the club is in contact with the ball is what allows top players put such back spin on the ball that they can control it on the green, even with short clubs where high clubhead speed is not required they can do it. Its a dynamic impact. Low level golf simply never get such spin on the ball to have the ball spin back on the green because they arent able to sustain the club head speed as it contacts the ball (they fool them selves sometimes that they do when a ball jumps back out of its pitch mark but that werent the spin).

Your point was that the weight of the golfer has no impact on the distance they can hit it. I've said if two golfers of different weight but all other variables are the same hit a golf ball at the same speed, with the same swing, at the same angle etc. the heavier golfer will hit it further. You stated this was incorrect and the only things to consider are loft and length (the later of which is incorrect).

If all variables are kept the same except for the golfers weight what you're saying wouldn't need to be taken into account to demonstrate that weight can impact distance.

I've stated that the static model was solely to demonstrate my point and nothing else. It would not be how I would estimate distance mathematically.

First Up

The ONLY factor affecting distance is the speed at which the club hits the ball. Fat golfers, thin golfers, light golfers, heavy golfers, tall golfers and short golfers will all hit the ball exactly the same distance if they hit it with the same clubhead speed.

The real question is how does a golfer's weight influence the clubhead speed they can generate. I don't think it does - but muscular strength might.

bren2001

First Up wrote: »

The ONLY factor affecting distance is the speed at which the club hits the ball. Fat golfers, thin golfers, light golfers, heavy golfers, tall golfers and short golfers will all hit the ball exactly the same distance if they hit it with the same clubhead speed.

The real question is how does a golfer's weight influence the clubhead speed they can generate. I don't think it does - but muscular strength might.

Can you explain why a golfers weight doesn't effect distance but the linked article above shows that a baseball players weight effects distance? What is the fundamental difference? Why would a golfers weight not feature in the conservation of momentum or energy laws?

First Up

bren2001 wrote:

Can you explain why a golfers weight doesn't effect distance but the linked article above shows that a baseball players weight effects distance? What is the fundamental difference? Why would a golfers weight not feature in the conservation of momentum or energy laws?

Is there evidence that baseball players (or golfers) of different weights but the same swing speeds achieve different distance?

In any case, baseball is not a good comparison; a baseball is moving before being hit which is an important variable. A golf ball is static.

bren2001

First Up wrote: »

Is there evidence that baseball players (or golfers) of different weights but the same swing speeds achieve different distance?

In any case, baseball is not a good comparison; a baseball is moving before being hit which is an important variable. A golf ball is static.

See the article here:
https://qz.com/801738/...aseball-players-fat/
Research by Penn State.

Yes there is a difference but why would the mass of the golfer be taken out in the kinetic energy calculation but not for the baseball player?

First Up

bren2001 wrote:

Yes there is a difference but why would the mass of the golfer be taken out in the kinetic energy calculation but not for the baseball player?

Weight is only a factor if it contributes to force, generating speed.

Weight on it's own is meaningless, unless it translates into strength, generating greater force (speed).

bren2001

First Up wrote: »

Weight is only a factor if it contributes to force, generating speed.

Weight on it's own is meaningless, unless it translates into strength, generating greater force (speed).

The article I linked explains that weight is a factor in its own right for generating force through kinetic energy. Why would that be the case for baseball but not for golf?

The equation for kinetic energy is 0.5mv^2 and the article clearly states the baseballers weight is taken into account.

First Up

bren2001 wrote:

The article I linked explains that weight is a factor in its own right for generating force through kinetic energy. Why would that be the case for baseball but not for golf?

Baseball and golf are different (I have played both) but I still await evidence that a golfer weighing 120 kilos generating the same clubhead speed will hit a golf ball further than a golfer weighing 90 kilos.

bren2001

First Up wrote: »

Baseball and golf are different (I have played both) but I still await evidence that a golfer weighing 120 kilos generating the same clubhead speed will hit a golf ball further than a golfer weighing 90 kilos.

They are different but at the end of the day, in both cases, you swing some form of stick/bat. If two baseball players have identical swings and clubs, the heavier one will generate more kinetic energy and hit the ball further. The article linked above explicitly states that. By logical extension, a golfer will generate more kinetic energy too. From an energy calculation perspective, there is little difference between a club and a bat. One is more stiff than the other. Can you please explain why this would not be the case? Why does the case of baseball not logically extend to golf?

In terms of a moving ball you mentioned earlier, it's irrelevant when calculating the kinetic energy of the player/bat/club before impact as it is independent of the ball.

Otherwise, can you please explain from a mathematical perspective why the weight of the golfer would not be taken into account in either the law of conservation of momentum or energy. If neither are applicable (they both are), can you please outline how you would calculate the initial ball speed from a kinematics perspective?

First Up

bren2001 wrote:

Otherwise, can you please explain from a mathematical perspective why the weight of the golfer would not be taken into account in either the law of conservation of momentum or energy. If neither are applicable (they both are), can you please outline how you would calculate the initial ball speed from a kinematics perspective?

I'd prefer to do it from a golfing perspective. There is plenty of data showing the direct relationship between clubhead speed and distance. None of it is qualified by the weight of the person generating it.

Golf is my Game

First Up wrote: »

I'd prefer to do it from a golfing perspective. There is plenty of data showing the direct relationship between clubhead speed and distance. None of it is qualified by the weight of the person generating it.

Yeh, if you go to your pro he never tells you you need to put on a few stone or has a scales in his teaching bay to see if youve got enough momentum through the ball. Not on TV. They dont comment on how players who lose weight like Bubba last year, or I though Rahm is looking slimmer this weekend, is going to lose distance and its going to cost him so he should go back on the pasta and cream cakes. Looking at the long hitters out there theres no correlation evident between the long lads and the short ones and their weight or even size. Rory is a small enough and light lad for example. Its all about speed.
How many yards would you say or would your equations back up, that a stone is worth to a tipical golfer ? Im an engineer too, but washing machines and dishwashers mainly rather than sports but, so still have a good feel for mechanics.

bren2001

First Up wrote: »

I'd prefer to do it from a golfing perspective. There is plenty of data showing the direct relationship between clubhead speed and distance. None of it is qualified by the weight of the person generating it.

The relationship between clubhead speed and distance is well established. I'm not saying that's untrue. I'm sayi g the weight of a person makes a difference, that's all. That difference might be very very small.

Ok, from a golfing perspective, the kinetic energy before impact is 0.5mv^2, the m refers to the mass of the golf club and golfer. Why is that wrong?

bren2001

Golf is my Game wrote: »

Yeh, if you go to your pro he never tells you you need to put on a few stone or has a scales in his teaching bay to see if youve got enough momentum through the ball. Not on TV. They dont comment on how players who lose weight like Bubba last year, or I though Rahm is looking slimmer this weekend, is going to lose distance and its going to cost him so he should go back on the pasta and cream cakes. Looking at the long hitters out there theres no correlation evident between the long lads and the short ones and their weight or even size. Rory is a small enough and light lad for example. Its all about speed.
How many yards would you say or would your equations back up, that a stone is worth to a tipical golfer ? Im an engineer too, but washing machines and dishwashers mainly rather than sports but, so still have a good feel for mechanics.

I never once suggested that adding weight is an advisable way to gain distance. It's clearly a silly way to do it. At no point in this did I say that.

You're original point was: "length the ball will travel is set by two characters of the club, the loft on the face and the length of it". That was incorrect, length is indirect. You've changed your position and now it's all about speed. I'm not arguing that this is the dominant characteristic.

However, you also said: "Weight, is incorrect. Speed and loft are all that matters.". This is wrong. Weight is a factor. How much does it contribute? I don't know and am not going to sit down and do the equations to give a definite answer. However, the weight of the golfer does impact distance. If it's true for baseball as outlined by Penn State researchers, it's true for golf.

How can Rory hit it so far? Very fast clubhead speed and he keeps the club square and on-line. If he weighed more and swung the club the same way, he'd hit it further. I've never stated weight is the dominant factor. It's not but it does contribute. That's all i've been saying.

You're an engineer, you've probably completed a course in mechanics or kinematics. Please explain how the weight of a golfer does not contribute to the kinetic energy at the point of impact. The formula being 0.5mv^2. Law of conservation of energy applies, the golf ball has no kinetic energy before impact. It has plenty after. Or, please explain why the findings of the following baseball study cannot be applied to golf:
https://qz.com/801738/why-are-baseball-players-fat/

Rippeditup

The thing with Bryson is the weight has been added with a very clear focus on overall strength while trying to not lose mobility, the strength/weight is being used heavily on how he loads up on the back swing and how he drives through on the left side on the downswing which is all based on coiling up his body and weight to get behind the hit..
with Rory he has insane mobility on the back swing getting his back, glutes and legs all coiled before releasing..
The swing speed is determined by the weight you load on the right side before driving the club through impact from the shift to the left side.. strength, weight, mobility and stability of the body all leads to more power...

I am no engineer but have been a long hitter of a ball since being a small kid and even today at 40 and been out for years would carry 290+ with big launches and I am no pro just understand body movement..
it’s like Olympic lifting, bigger lads will move more weight but vs body % will be nowhere near the smaller lads as the coil and power is better due to technique


Rory has exceptional technical ability which Bryson has also bit he has put in the raw power also with the weight gain to increase overall output

Golf is my Game

bren2001 wrote: »

The relationship between clubhead speed and distance is well established. I'm not saying that's untrue. I'm sayi g the weight of a person makes a difference, that's all. That difference might be very very small.

Ok, from a golfing perspective, the kinetic energy before impact is 0.5mv^2, the m refers to the mass of the golf club and golfer. Why is that wrong?

That you are counting all of the body of the golfer contributing to the mass in the 1/2mv2 equation. v is a vector with direction as well as magnitude. You are using all that v as if it were in the direction of the target. The 1/2mv2 component that resolves for the movement of the body of the golfer is neglectably zero since the v of the movement of the body as a mass is almost zero. The body itself simply isnt moving fast. And the remainder of the compnonent that is, is almost a zero sum, since mass of the body that is actually moving back (the left half for a right handed golfer) from the ball negates that which is moving toward the ball (the right side). Leaving no body momentum component that has any measureable contribution to energy transferred to the ball.