Pointless question

[Dhakala]

My two cents:

Aside from how much each object compresses, the "initial" point of contact would, I suppose, depend on the size and curveature of the arc of each object. In my humble theory, the initial contact point of two billiard balls is relatively small, much smaller than a carbon paper test might predict. The compression, etc happens after contact.

No, Craig. The point of contact of two spheres is an infinitesimally small point, regardless of the spheres' dimensions. Compression happens during contact. "After contact," when the spheres are no longer in contact, they cease to compress each other.

 

[Irish634]

No, Craig. The point of contact of two spheres is an infinitesimally small point, regardless of the spheres' dimensions. Compression happens during contact. "After contact," when the spheres are no longer in contact, they cease to compress each other.

Hi David, I agree. That's why I specified the word "initial" contact, and that compression actually happens after this point in time. Maybe my wording wasn't exactly how I was thinking of it, but I think we are on the same page.

Craig

 

[Jal]

The word "inelastic" when referring to collisions of objects doesn't mean "doesn't deform."

"Elastic" means all of the work of deformation is stored as potential energy that is returned as kinetic energy of the objects after collision. [Think bouncing a superball off a hard surface]

"Inelastic" means the work of deformation is not stored as potential energy that is returned. [Think trying to bounce a ball made of clay]....

At the moment of initial contact the tangent is pointing in one direction. The balls then enter each others space. Upon exit, the tangent line is pointing in a different direction. The amount of this rotation of the tangent depends on how long the balls are in contact and can be figured out directly from it. But what effect does this have on throw and spin? Do the flat spots propagate around the circumference of the balls effortlessly or is some small but perhaps significant mutual force required to move them along?

Jim

 

[RSB-Refugee]

Do the flat spots propagate around the circumference of the balls effortlessly or is some small but perhaps significant mutual force required to move them along?

Jim

I think that is what I was getting at, when I wondered, if the marks would be oblong or have a little curve to them.

Tracy

 

[RSB-Refugee]

Do the flat spots propagate around the circumference of the balls effortlessly or is some small but perhaps significant mutual force required to move them along?

Jim

How else could follow or draw be imparted on an object ball? If I had to guess, I would say, yes they do propagate.

Tracy

 

[Jal]

I think that is what I was getting at, when I wondered, if the marks would be oblong or have a little curve to them.

Tracy

Yes and yes, but probably hard to detect.

Jim

 

[Jal]

How else could follow or draw be imparted on an object ball? If I had to guess, I would say, yes they do propagate.

Tracy

Yes, I agree, they do propagate. But, although they're sort of intimatley connected by compression, it's the friction force that imparts the spin on the object ball. I was wondering that, if in addition to the friction, the compression force is skewed away from the line of centers by the necessity to push the leading edge of the flat spot down (get it out of the way as the tangent line shifts) enough to affect throw and acquired spin in some small but perhaps significant way.

It can be tested, of course, with perfectly frictionless balls. But if you see some slight spin imparted to the object ball, is this because of the skewed compression, or some residual friction?

Come to think of it, since you can perform that old hustler's trick of sending a frozen ball down the rail by wetting the contact point, I suppose the flat spot propagation must have very little effect. Maybe you see it differently?

Jim

 

[JohnPT]

what is the point to all this?

oops! didnt pay attention to the title of the thread