Golf Ball Impact: Does Face or Clubhead Direction Matter?

[DumbGolfer]

A frequent topic of debate for golfers is the initial direction that the golf ball follows when it is struck with the club face angled right or left relative to the direction of the moving clubhead (i.e., an "open" or "closed" face for those familiar with golf terminology).

The common belief is that the initial flight of the ball is primarily influenced by the clubhead's direction, rather than the direction the clubface is pointing at separation.

Can anyone on this forum please offer a definitive answer to this question? I will gladly clarify my request if necessary.

 

[kuengb]

If there were no friction between club and ball, i. e. the ball would not start to rotate, and the impact was totally elastic, the angle of the ball's flight would be the same as the club's angle at the impact. In reality the ball is rotating which decreases the angle of the flight, but still the angle must be bigger than 5/7 of the club's angle. I can explain these "5/7" if you want, just ask.
Anyway, if I didn't overlook an important factor, that's it. It won't help you for a hole-in-one though.

Best wishes, Bruno

 

[DumbGolfer]

Originally posted by kuengb
If there were no friction between club and ball, i. e. the ball would not start to rotate, and the impact was totally elastic, the angle of the ball's flight would be the same as the club's angle at the impact. In reality the ball is rotating which decreases the angle of the flight, but still the angle must be bigger than 5/7 of the club's angle. I can explain these "5/7" if you want, just ask.
Anyway, if I didn't overlook an important factor, that's it. It won't help you for a hole-in-one though.

Best wishes, Bruno

Bruno,

Thanks for the reply. If you don't mind, I am very interested in learning more about the 5/7 aspect.

 

[kuengb]

First I'll explain something else: Imagine you are Bowling and you throw your bowl without rotation. So, only a moment after it has touched the ground, it will slide, not roll. But since there's a frictional force acting, the bowl will constantly increase rotation during its slide; until it has reached the point where it revolves fast enough that it can roll on the ground, from then on there won't be any friction. But: The same frictional force that has increased the bowl's rotation has slowed down its speed during this starting period.

Now, if you do the calculations you will find that the final speed is 5/7 of the initial speed, no matter how big the friction is. The form of the bowl (a sphere, I've ignored the three little holes ) is an important factor for this number.

The connection may not be obvious, but the golf ball impact is basically the same problem: The angle of the flight is proportional to the transversal speed. And since this tr. speed is "splitted up into rotation and translation", as in the Bowling example, v_{friction}^{tr} = 5/7*v_{nofriction}^{tr}, hence the relation between the two angles is 5/7. And, as I said, the actual angle must be somwhere in between.

By the way, in what situations is this "open/closed" thing used? Laugh at me if you want but isn't it the best to simply shoot straight? Hahaha, I guess that's on the FAQ-list of "Golf For Dummies".

Bruno

 

[JohnDubYa]

By the way, some of the most egregious examples of bad physics occurs in golf talk. Read the advertisements in golf magazines sometime.

Another area of bad physics is high fidelity audio.

 

[DumbGolfer]

Originally posted by kuengb
First I'll explain something else: Imagine you are Bowling and you throw your bowl without rotation. So, only a moment after it has touched the ground, it will slide, not roll. But since there's a frictional force acting, the bowl will constantly increase rotation during its slide; until it has reached the point where it revolves fast enough that it can roll on the ground, from then on there won't be any friction. But: The same frictional force that has increased the bowl's rotation has slowed down its speed during this starting period.

Now, if you do the calculations you will find that the final speed is 5/7 of the initial speed, no matter how big the friction is. The form of the bowl (a sphere, I've ignored the three little holes ) is an important factor for this number.

The connection may not be obvious, but the golf ball impact is basically the same problem: The angle of the flight is proportional to the transversal speed. And since this tr. speed is "splitted up into rotation and translation", as in the Bowling example, v_{friction}^{tr} = 5/7*v_{nofriction}^{tr}, hence the relation between the two angles is 5/7. And, as I said, the actual angle must be somwhere in between.

By the way, in what situations is this "open/closed" thing used? Laugh at me if you want but isn't it the best to simply shoot straight? Hahaha, I guess that's on the FAQ-list of "Golf For Dummies".

Bruno

Bruno,

If it weren't for the cruel intentions of golf course architects ("Why on Earth did they put a tree there?!?") we wouldn't have to worry about such things as "fading" the ball right or "drawing" it left.

Back to our problem, we've taken friction into account, but what about the compression (and deformation) of the ball against the clubface?

Also, forgive me if I'm beating this "5/7" thing to death, but are we talking about a range of initial direction angles that would go from 0/7 (a frictionless collision which sends the ball on a path perpendicular to the face) to 5/7 (at an angle closer to the path of the clubhead than the angle of the clubface)? And is it strictly clubhead velocity that affects the angle (i.e., the harder you hit the ball, the straighter it will go)?

 

[kuengb]

I won't win a poetry contest with that but at least it's "clear text": If c is the direction of the club's movement, p the line perpenticular to the clubface, b the ball's direction and \alpha the angle between c and p, then the angle c-b is at least 5/7*\alpha and at most \alpha.

Deformation certainly has an influence on the angle, but it's hard to say anything about that influence except that it probably is very small since the deformation is not too big and it's mostly "symmetric". Therefore it cannot cause strong transversal forces during the impact, hence no big angle change. I would say it decreases the angle slightly.

The primary effect of deformation (more precisely: inner friction due to deformation) is a "loss" of speed. Without that, the ball's speed would be twice the clubhead's speed, in reality it is maybe ten, twenty percent smaller.

If my assumptions are more or less correct, the angle should not or only a bit depend on the clubhead's velocity, and the 5/7-7/7 "rule" is always valid.

Bruno

 

[DumbGolfer]

Thanks, Bruno, for all your help. You're welcome in my foursome any time.

 

[Michael D. Sewell]

Bruno,
Would you be kind enough to show me the math on the 5/7 rule and conservation of momentum as applied to the bowling ball?
Thank you,
Mike

 

[kuengb]

Ayy! My bew'ful latex...and the private messenger ate it all up...OK, so i'll try it again:

The condition on a rolling body is v=r \omega. So we have to find out how long it takes until this holds for the bowling ball. Acceleration during the slide is a=F_f / m, angular acceleration \alpha = \frac {F_f r} I with I=2/5 m r^2for the bowl.
Solve the equation

r \omega(\tau) = r \alpha \tau = v_0-a \tau

for the time tau and then

v_1 = v_0- a \tau = v_0 - (2/7*m/F_f*v_0)(F_f/m) =5/7*v_0
Of course this is no closed system therefore neither angular nor linear momentum is conserved.

Bruno

 

[Michael D. Sewell]

Bruno,
Nicely done. I appreciate your time and effort. Thank you so much for your help.
Best wishes,
Mike

P.S. I don't know what made me think that this was a closed system. I am such an idiot.