Ball Actually "Pinched" Against Ground or Not

[winders]

Is the Boo Weekley video linked in this thread?

Look at my post here:

http://thesandtrap.com/forum/showpos...9&postcount;=48 S-

[jambalaya]

As the club head contacts the ball in its downward arc to the low point of the swing just in front of the ball, the ball sticks to the club for some fraction of a second until friction is overcome and the ball travels up the face and is launched according to the loft of the club. Since the ball is already sitting on the ground it does get driven down into the ground so to speak before it is launched. I think this is what people talk about when they say to pinch the ball. Or has this already been said and I am just opening a can of worms?

[immts007]

Anyone else being reminded of the "airplane on a conveyor belt" thread?

[iacas]

Anyone else being reminded of the "airplane on a conveyor belt" thread?

Heh:

http://thesandtrap.com/forum/showpos...admill%20plane

Or has this already been said and I am just opening a can of worms?

Already been said and still wrong. I think we're done here now, right?

[Joe Mama]

I gather that after a lengthy debate the question of whether the ball makes significant contact with the ground when struck by the club was answered in the negative. Couldn't the issue have been resolved more quickly by observing divots? I don't recall ever seeing divots with craters the diameter of golf balls embedded in them. Tiger's divots are all dollar-bill shaped rectangle of nearly uniform depth.

[Joe Mama]

A few posters invoked laws of physics to justify their position, so I thought I would try my hand at applying highschool physics to the question of whether there ever could be a downward force acting on the ball during contact with the clubface that is greater than the upward force. The result I came up with involves the so-called "coefficient of kinetic friction," which I will symbolize as "k," and the club loft, which I will symbolize as @. In order that the ball be driven downward, the following condition must be satisfied: k > tan @ In the case of a 31 degree six iron, the coefficient would have to be larger than 0.6. A typical metal to polymer coefficient is 0.2-0.3, while steel on wood is 0.2-0.5. I don't know what the coefficient is for a typical golf ball coating and the metal clubface, but it is probably around 0.3--way too low for our purposes. However, it looks like to me the "physics" does, indeed, support an argument that a golf ball with a sufficiently large coefficient of kinetic friction on metal would be driven downward, contrary to the semi-rigorous arguments presented by some posters in this forum. The details of the derivation of the tangent equation above I will not provide. It is enough to say that anyone with some familiarity with force vectors and Newton's laws could reproduce for themself. Basically, I compared the vertical component of the contact force (normal force) acting on the ball to the downward component of the kinetic frictional force.

[Fourputt]

A few posters invoked laws of physics to justify their position, so I thought I would try my hand at applying highschool physics to the question of whether there ever could be a downward force acting on the ball during contact with the clubface that is greater than the upward force. The result I came up with involves the so-called "coefficient of kinetic friction," which I will symbolize as "k," and the club loft, which I will symbolize as @.

In order that the ball be driven downward, the following condition must be satisfied:

k > tan @

In the case of a 31 degree six iron, the coefficient would have to be larger than 0.6. A typical metal to polymer coefficient is 0.2-0.3, while steel on wood is 0.2-0.5. I don't know what the coefficient is for a typical golf ball coating and the metal clubface, but it is probably around 0.3--way too low for our purposes.

However, it looks like to me the "physics" does, indeed, support an argument that a golf ball with a sufficiently large coefficient of kinetic friction on metal would be driven downward, contrary to the semi-rigorous arguments presented by some posters in this forum.

The details of the derivation of the tangent equation above I will not provide. It is enough to say that anyone with some familiarity with force vectors and Newton's laws could reproduce for themself. Basically, I compared the vertical component of the contact force (normal force) acting on the ball to the downward component of the kinetic frictional force.

Cant be.  Look at this video of Tiger hitting a 2 iron.   Even as flat as the loft on that club, the ball still starts upward even though the club is still moving downward.

[bubble]

the only pinching I feel is when the contacts (club+ball) and (sole bounce + turf) are simultanious.

on short game shots I get a little extra loft and spin when this happens, the sound and fell are really nice.

[Hatchman]

A resurrected thread. To me it's a feeling of hitting down on the ball or "pinching" it like already stated. But the funny thing is I've never influenced the static ball to go downward into the turf then in a split second ricochet up from the turf swinging a dynamic load with calculable loft properties. The early tree whittlers and now modern club designers have it handled. Off a tee maybe different, just lean back and top it and see your uniform ball trench 8" long. Who hasn't done that? You won't get one in the fairway or drive your ball into the ground unless you hit above the equator. Oh, I did that too, in a hazard area and buried it in the mud

[boogielicious]

A few posters invoked laws of physics to justify their position, so I thought I would try my hand at applying highschool physics to the question of whether there ever could be a downward force acting on the ball during contact with the clubface that is greater than the upward force. The result I came up with involves the so-called "coefficient of kinetic friction," which I will symbolize as "k," and the club loft, which I will symbolize as @.

In order that the ball be driven downward, the following condition must be satisfied:

k > tan @

In the case of a 31 degree six iron, the coefficient would have to be larger than 0.6. A typical metal to polymer coefficient is 0.2-0.3, while steel on wood is 0.2-0.5. I don't know what the coefficient is for a typical golf ball coating and the metal clubface, but it is probably around 0.3--way too low for our purposes.

However, it looks like to me the "physics" does, indeed, support an argument that a golf ball with a sufficiently large coefficient of kinetic friction on metal would be driven downward, contrary to the semi-rigorous arguments presented by some posters in this forum.

The details of the derivation of the tangent equation above I will not provide. It is enough to say that anyone with some familiarity with force vectors and Newton's laws could reproduce for themself. Basically, I compared the vertical component of the contact force (normal force) acting on the ball to the downward component of the kinetic frictional force.

You are overthinking it and your physics is incorrect. In order for the ball to move downward, the face would have to be angled downward (less than 90 degrees perpendicular to the ground) at impact. It is not. Therefore the ball goes up. BTW, this is Newton's third law of motion.

[saevel25]

A few posters invoked laws of physics to justify their position, so I thought I would try my hand at applying highschool physics to the question of whether there ever could be a downward force acting on the ball during contact with the clubface that is greater than the upward force. The result I came up with involves the so-called "coefficient of kinetic friction," which I will symbolize as "k," and the club loft, which I will symbolize as @.

In order that the ball be driven downward, the following condition must be satisfied:

k > tan @

Not correct. You are not taking into account the fact that even with a downward angle of attack the forces applied on the ball are still upward pointing. See below

The movement of the club does have some say in the launch angle. 15% for driver, and 25% for the irons.

Launch angle for driver = 85% dynamic loft + 15% angle of attack.

Lets say you have a 9 degree driver and present 8 degrees of loft at impact.

0 = 85% (8 deg) + 15% (AA)

AA >= 45 degrees downward

In all other instances the ball will travel up the clubface and never hit the ground.

The only shot you ever have of hitting the ball into the ground is if you thin the ball above the equator, top the ball. If you present something like 2 degrees of loft at impact with a driver and hit down 11 degrees.

[Joe Mama]

The arrows in your drawing don't seem to account for the fact that the ball rotates counter-clockwise, causing the ball to roll up the club face. In order that this happen, a counter-clockwise torque has to be applied to the ball. This torque would be achieved by a frictional force exerted tangentially to the ball, which you don't show. You only show the contact force (also called the normal force). The three arrows you show represent a correct "resolution" of the contact force vector (the hypotenuse of the right triangle) into its horizontal and vertical components. So far, so good. But, because there is relative motion between the clubface and the ball, with the face moving downward, sliding under the ball and the ball rolling up the face, the ball is rotating counter-clockwise UP the face. How could a counter-clockwise rotation occur without a counter-clockwise torque? It couldn't. What your drawing does not show is a fourth red arrow that accounts for the kinetic friction force vector tangent to the ball. The vertical component of the missing frictional force vector would compete against your short vertical arrow (the verticall component of the contact force). As I indicated in my discussion of the tangent equation I derived, only in the case of a pair of surfaces with an extraordinarily high coefficient of kinetic friction would the vertical component of the frictional force be greater than the vertical component of your contact force vector. In sum, I believe the fault in your analysis lies in the fact that your drawing ignores friction.

[Fourputt]

The arrows in your drawing don't seem to account for the fact that the ball rotates counter-clockwise, causing the ball to roll up the club face. In order that this happen, a counter-clockwise torque has to be applied to the ball. This torque would be achieved by a frictional force exerted tangentially to the ball, which you don't show. You only show the contact force (also called the normal force).

The three arrows you show represent a correct "resolution" of the contact force vector (the hypotenuse of the right triangle) into its horizontal and vertical components. So far, so good. But, because there is relative motion between the clubface and the ball, with the face moving downward, sliding under the ball and the ball rolling up the face, the ball is rotating counter-clockwise UP the face. How could a counter-clockwise rotation occur without a counter-clockwise torque? It couldn't. What your drawing does not show is a fourth red arrow that accounts for the kinetic friction force vector tangent to the ball.

The vertical component of the missing frictional force vector would compete against your short vertical arrow (the verticall component of the contact force). As I indicated in my discussion of the tangent equation I derived, only in the case of a pair of surfaces with an extraordinarily high coefficient of kinetic friction would the vertical component of the frictional force be greater than the vertical component of your contact force vector.

In sum, I believe the fault in your analysis lies in the fact that your drawing ignores friction.

None the less, you would have to have at least a vertical clubface to come close to an actual "pinch", and even then it would be minimal.  To truly pinch a ball between the clubface and the ground the face would have to be closed past vertical.  The ball spends such a tiny instant in actual contact with the clubface that any minimal downward force is more than countered by the loft of the clubface.

[iacas]

In sum, I believe the fault in your analysis lies in the fact that your drawing ignores friction.

The flatter the face of the club the more normal the force and thus the less friction applies. That's why longer clubs with less loft spin less. Your physics are still wrong Joe. The loft presents an incredibly strong lifting force while the angle of attack is relatively shallow and cannot and does not overcome the lifting forces. In practice the face is 75-85% of the total influence on starting direction of the ball.

[Joe Mama]

None the less, you would have to have at least a vertical clubface to come close to an actual "pinch", and even then it would be minimal.  To truly pinch a ball between the clubface and the ground the face would have to be closed past vertical.  The ball spends such a tiny instant in actual contact with the clubface that any minimal downward force is more than countered by the loft of the clubface.

I think I explained in an earlier post that a net force downward (sum of all forces acting either upward or downward) is virtually unrealizable in a real golf setting using conforming golf balls. No golf ball that I know about has a coefficient of kinetic friction large enough to cause a net downward force that would result in the ball moving downward toward the ground. (Recall that at least a coefficient of kinetic friction greater than 0.6 is needed). Thus, I totally agree that not only is the "pinch" you describe not even "minimal" using conforming golf balls, it is not even possible. Having said that, I wish to make the point that the purpose of my post was to correct what I think is incorrect physics applied by iacas to justify his claim that no pinching occurs. He's right--as most all of us are--that no pinching occurs, but his opinion is biased by looking at videos that correctly show no pinch, as well as on his vector diagram, which is incomplete and therefore incorrect, for the reasons I gave earlier. The actual physics involved WILL allow a golf ball to be pinched, but it would have to have a coefficient of kinetic friction on the golf club face greater than 0.6. Again, I state that iacas physics is incorrect.

[GolfLug]

Only time ball moves downwards is if you hit it above the equator - which as far as I know you can only do it if you hit it with a leading edge or with the bounce. Terms like 'skulling' 'topping' come to mind. Technically negative dynamic loft can do it too, but that would be one hell of a forward leaning shaft.

[Joe Mama]

The flatter the face of the club the more normal the force and thus the less friction applies. That's why longer clubs with less loft spin less. Your physics are still wrong Joe. The loft presents an incredibly strong lifting force while the angle of attack is relatively shallow and cannot and does not overcome the lifting forces. In practice the face is 75-85% of the total influence on starting direction of the ball.

Please explain why the "physics" in the force diagram below is wrong, iacas. The force diagram you provided in a previous post omitted the frictional force. Why did you omit it? Is DOES exist, doesn't it?

[Joe Mama]

The drawing above shows the two forces (ignoring the gravitational force) acting on the golf ball at impact. One of the forces is the normal force (call it N) and the other (the one omitted by iacas) is the frictional force (call it f). Let @ represent the angle theta, and k the coefficient of kinetic friction. The sum of the y-component of those two forces is given below: N sin @ - kN cos @ If this sum is negative, there will be a net downward force on the ball, which means the ball will be accelerated toward the ground. In order that the sum above be negative, the coefficient must be greater than the tangent of the angle @. k > tan @ In the case of a 31 degree angle, k would have to be greater than 0.6. Rare indeed are coefficients this large, and non-existence they are for golf balls. Thus, no pinching will ever be observed, but that doesn't absolve iacas of the responsibility to offer correct physics to support his view. Iacas, if my diagram and associated mathematics is in error, please explain exactly why. (Note added: f = kN)