Bryson DeChambeau

bren2001 · 10-07-2020 04:30PM

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 · 10-07-2020 05:05PM

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 · 10-07-2020 05:09PM

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 · 10-07-2020 05:13PM

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 · 10-07-2020 05:15PM

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 · 10-07-2020 05:23PM

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 · 10-07-2020 06:16PM

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 · 10-07-2020 06:28PM

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 · 10-07-2020 06:33PM

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 · 10-07-2020 06:50PM

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 · 10-07-2020 06:56PM

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 · 10-07-2020 06:59PM

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 · 10-07-2020 07:06PM

Weight doesn’t matter. Just ask Jamie Sadlowski

GreeBo · 11-07-2020 06:51AM

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 · 11-07-2020 08:04AM

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 · 11-07-2020 08:10AM

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 · 11-07-2020 08:17AM

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.

11-07-2020 08:27AM

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 · 11-07-2020 09:36AM

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 · 11-07-2020 09:59AM

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 · 11-07-2020 10:02AM

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 · 11-07-2020 10:08AM

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 · 11-07-2020 10:46AM

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 · 11-07-2020 10:52AM

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 · 11-07-2020 11:07AM

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 · 11-07-2020 11:11AM

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 · 11-07-2020 12:01PM

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 · 11-07-2020 12:12PM

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 · 11-07-2020 12:26PM

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 · 11-07-2020 12:35PM

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.

Bryson DeChambeau

Comments