Pointless question
- Thread starter Slasher
- Start date
Nostroke said:
It doesn't matter. Perfectly round and perfectly inelastic objects will touch at a point - which has no size, only position.
But billiard balls are elastic - they deform upon impact and rebound to their original shape. Thus, the point of impact becomes a spot with area.
If you can find carbon paper, hold a sheet between two balls and shoot one into the other, trapping the paper in between. You'll see the spot where the surfaces of the balls touched as a carbon smudge on one ball. The harder the shot, the bigger the spot.
I saw an experiment, where paper and carbon paper were attached to an object ball. A cue ball was hit into the papers. Soft shots left smaller marks than hard shots. The two balls actually flatten at the points of contact. I'd like to see the same, where cut shots are hit with english, to see if the marks would be oblong.Slasher said:
Tracy
I just wrote about the same thing. I need to learn, to type faster.Dhakala said:
Tracy
Dhakala said:
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].
When in my business we're thinking of an idealized limit of molecules as "hard spheres" (like an idealized limit of pool balls that don't deform and have a point contact), we refer to the collisions as ELASTIC collisions of hard spheres.
Dhakala said:
The spot won't be incredibly accurate, because the paper will deform probably at least as much as the balls. The paper's contact area will be larger than the actual contact area between just the balls.
But for the OP, it's very small for very hard balls (such as billiard balls). They don't deform very much. If they didn't deform at all, the contact point would be that: a point, meaning its area is zero. Any two uniformly strictly convex objects will contact only at a single point.
Of course, if you really want to get down into the physics of materials, it could be argued that if considered on a minute enough scale, they never contact at all. Such discussion is not very relevant when talking about the observable actions and reactions of macroscopic objects, though.
-Andrew
The contact point on an object ball is define on a cut shot at a slow speed. As the cut shot speed (forward motion and spin on cue ball are both speeds) increases the contact point one aims at is ever so slightly above the define contact point. This is why most people overcut shots when they hit it real hard. The collision then becomes more of a swipe across the face of the contact point distorting the impact. To try and put it simply, as you increase your speed on the same cut shot the point of aim is above the define contact point. (this is a difficult concepte to explain on a key board but fun).
Forcing the angle is a term that fits this action. Also if the balls are clean, new and the cloth is slick this action is accentuated. This is an advanced concept, beginners are not able to feel this ball action till they have allot of years of play and increased skill.
Forcing the angle is a term that fits this action. Also if the balls are clean, new and the cloth is slick this action is accentuated. This is an advanced concept, beginners are not able to feel this ball action till they have allot of years of play and increased skill.
Andrew Manning said:
The paper's contact area with what, other than "the actual contact area between just the balls"?Bob Jewett reported the results of the carbon paper experiment in the August, 2001, issue of Billiards Digest:
"The result is that for a hard shot, the flat spot is a quarter-inch or six millimeters in diameter. How much did the surface of each ball compress during such a collision? Simple geometry says about 0.3mm or one hundredth of an inch. That's about the thickness of three sheets of typing paper."
Dhakala said:
When you cut a ball your coming across the point of contact, when you hit a straight in ball your not. This motion of hitting the side of the ball,( "swipe may be a good definition") the object ball thats being contacted is also being pushed downward, plus it turns the object ball somewhat as it (swipes) comes across the point of contact.
Island Drive said:I think ouside english would leave a longer "swipe", than inside english. It would be interesting to see the results with draw and english on the same shot. I wonder if it would leave a slightly curved "swipe".Dhakala said:
When you cut a ball your coming across the point of contact, when you hit a straight in ball your not. This motion of hitting the side of the ball,( "swipe may be a good definition") the object ball thats being contacted is also being pushed downward, plus it turns the object ball somewhat as it (swipes) comes across the point of contact.
Tracy
Dhakala said:
What I mean is that the balls will compress the paper on both sides. Picture the paper as a pillow instead of a sheet of paper. Both balls will sink into the pillow significantly and each will have a large contact area with the pillow, even if the balls don't deform at all. Now shrink the pillow down to the thickness and firmness of a sheet of paper, and this effect will still happen. The spot on the paper won't reflect exactly the contact area between the balls had the paper not been there; it reflects each ball's contact area with the paper, which depends both on ball deformation and paper deformation. With thin enough paper I suppose you could consider this effect negligible, but it will have some effect.
Sorry for the confusing wording in my first post.
-Andrew
It depends, as others have said, on how soft the balls are. The size of the flat spot can be calculated in severals ways from either dynamic or static measurements. Marlow measured the contact time. The diameter of the flat spot is about 5mm depending on ball speed.Slasher said:
Use a thin layer of wax on the balls. That eliminates most of your objection.Andrew Manning said:
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. I guess you could just place two billiard balls together (at rest & without being struck) to simulate this.
I am sure there must be a video of it here. If I can find a demonstration I'll update the post.
Thanks,
Craig
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. I guess you could just place two billiard balls together (at rest & without being struck) to simulate this.
I am sure there must be a video of it here. If I can find a demonstration I'll update the post.
Thanks,
Craig
Last edited: