Shooting Pool: Newsletter Challenge (01/17/05)

What does "scratching" mean? I've never heard it when playing pool in the UK.

Thanks

Scratching is when the cue ball goes into a pocket (a definite no-no).

However, if "scratching" is not a penalty in the UK, I would be happy to play pool with you since I seem to have more than my fair share of white ball excursions into all six pockets. ;-)

"Potting the White" is indeed a penalty here - it's just that we call it....potting the white!

Happy to have a game though!

Now back to the question....

Wow! I've only been playing pool for a little over forty years, and I just now found that I don't know much about it! Scratching happens among men (and few women) more often when they don't know what to do on their next shot, similar to what baseball players do when they're waiting for the next batter. Practice, practice, practice . . .

Well, I don't play pool much, more of a darts player, but the times I have played, I haven't been able to create a right angle with the path of the cue ball after any shot. In fact, I don't think it's physically possible for incoming and outgoing paths of a cue ball during a collision to be 90 degrees. So I assume he must be talking about a different angle, but I'm not sure what. Maybe the path of the other ball and the cue after collision? That would help my pool game a lot if it were true, never really thought about it.

Actually, the cue ball can strike another ball and deflect at any angle from near zero degrees to 180 degrees from its original course. At either extreme the cue ball simply grazes the other ball and experiences a nominal course deviation +/- of its original trajectory. A dead on collision results in a full 180 degree reversal of direction, however, most of the cue balls energy is transferred to the ball it strikes. This is particularly true for tables that have a coin slot since the cue ball is slightly larger than the rest.

For simplicity we can consider collisions between pool balls as perfectly elastic and disregard the entropy that reality holds in life. The balls are spherical, so when one ball contacts another it will transfer some or all of its energy into the ball it contacts. From a 2-D perspective consider the balls as circles with a dot representing the origin of the circle.

When balls collide their surfaces meet and an exchange of energy results. At the instance of collision there are vector forces that come into play. The contact area between balls is always directly between the origins of the balls. You can draw a line between the center of one ball and the center of the next and the contact area will exactly where that straight line intersects the circumference of the balls. The vector of the moving ball usually lies in a different direction (i.e., not a head on collision). However the force that is applied to the static ball is ALWAYS in the direction that lies from the origin of the moving ball toward the origin of the static ball at the moment of contact. The magnitude of the force applied to the static ball depends on the angle of incidence between the dynamic ball's vector and the vector created between the two balls origins. If the two angles are identical then all of the energy is transferred to the static ball. If the angles differ, then only a fraction of that energy is transferred (I think it is a sin/cosine function of those angles).

Her is an example. The cue ball will deflect at an outgoing angle that is equal to the angle of incidence of the impact. On paper, draw two circles that are in contact. Draw a line from the origin of the cue ball to the static ball's origin. Now draw a vector line from the cue ball that points just to the right (clockwise) of the static ball. Observe the angle made between the line that connects the two origins and the vector line. Let's say it is 30 degrees. Now draw a third line that starts at the origin of the cue ball and is 30 degrees clockwise from the cue ball's original vector line. This third line represents the new trajectory of the cue ball. The trajectory of the static ball will follow a vector that lies along the line that we drew from the origin of the cue ball at contact to the origin of the static ball. That vector is simply the extension of that line between origins.

The sum of the magnitudes of the resulting vectors (post collision) should be equal to the magnitude the cue ball's initial vector. This is for perfectly elastic collisions on a frictionless plane. Real world parameters have surface friction, air resistance, imperfections in the balls, and one more significant factor; English! By introducing a spin on the cue ball you can do some strange things.

Now, to the original question of what right angles have to do with pool, I haven't a clue what Chris meant. The angles of incidence is always equal to the angle of reflectance for light striking a mirror. The same holds true for perfectly elastic collisions between two spheres. I think right-angle is a misleading statement in the question.

Why can't I play pool? Even though I have knowledge of physics, I just lack the skill!

I think we are talking about different angles. For instance, you describe above "A dead on collision" which you said "results in a full 180 degree reversal of direction". I would say a dead on collision results in a 0 degree angle between the incoming path of the cue ball and the outgoing path of the cue ball. I think you're right about the 90 degree angle shot being possible though. So that means the boy must be talking about a different angle than the one created by the incoming cue ball and outgoing cue ball.

I think the boy is probably referring to the path of the "other ball" and the cue balls outgoing path must always be 90 degrees, but I'm not sure if that's true.

I would consider 0 degrees deflection as no change in the cue ball's direction. 180 degrees would be when the cue ball reversed direction and returns in the same direction it came.

Imagine throwing a handgranade and you dog grabs it and brings it back to you. That woudl be a 180 reversal of the direction of the handgranade.

The good news is it only happens once.

It's only 180 if you are measuring the angle between the projected path (without collision) and the actual path due to collision. If you are meauring the angle between the incoming path and the outgoing path, it's 0 degrees. None of this helps us answer the question though.

I think I might have used "tennis ball" rather than "hand grenade" in that example, but that's just me.

Okay, I tend to lean toward the dramatic at times. ;-)

It gets your point across, next time say a cat, in place of the dog or the hand grenade, either way works.

Ah, so it would be better if I had said: Imagine throwing a cat and you dog grabs it and brings it back to you. That would be a 180 reversal of the direction of the cat.

Well actually I was saying that "Imagine throwing a cat and your dog grabs it and brings it back to you" or "Imagine throwing a hand grenade and you cat brings it back to you" would be fine. You see, I like dogs better than cats, I wasn't considering the physics.

I like both. In fact, our German Shepherd pup "saved" our oldest cat's life last night. It is a bit of a long story, but our dog had a bit of an act of indiscretion on the hardwood floor Due to the resulting insult to the olfactory senses I opened up both front and back doors to the house (yes, it is freezing here). After a short while I heard our dog barking as if an intruder was present. To my amazement our indoor 18 year old cat (who is quite sick) decided to take an outside walk in the snow storm. If the dog hadn't alerted me, I would have never seen him go out. The cat would have been shut out and would have quickly been overcome by the weather (he has zero body fat and not much more muscle). He was one lucky cat thanks to our pup.

So complicated, yet so wrong. The exit angles are mutually-perpendicular, irrespective of the angle of incidence. Hence the reference to right angles.

In a perfecty-elastic, frictionless collision between spheres of equal mass, 100% of the momentum perpendicular to the contact plane is transferred to the target ball and 100% of the momentum parallel to the contact plane is retained by the cue ball. Result: trajectories normal to and parallel to the contact plane.

The son's advice is therefore to examine the trajectory perpendicular to that intended for the target ball. If it intersects pocket, use evasive measures such as spin or shot strenght to modify the cue ball's trajectory, or just change shot.

I don't know if the slightly smaller-diameter cue ball in pay tables is also slighty lighter: if so, the cue ball will bounce slightly behind the contact plane.

I think we're looking at the theory of collisions between perfect spheres (pool balls are pretty close).

Because the balls are spherical (ER states the bleeding obvious) the interactions are always perpendicular to the surface of the balls. This is the right angle my son is using to calculate the trajectory of the red/yellow (sorry it's spots and stripes over there isn't?!) ball after impact and the resultant velocity of the cue ball. All this from transfer of momentum calculations, ain't maths wonderful?

In a perfectly elastic collision, a head-on collision should result in all of the momentum being transferred, i.e. the cue ball will stop. This doen't happen every time, because of factors such as spin on the ball and friction from the baize. A skilled player can account for these.

Great job! I could not see what the right angle was about, but it is the tangent line between the contact point of the two balls. The resulting vector for the static ball is at a right angle to the tangent line or, as I stated in my earlier post, the line that is drawn from the cue ball's origin to the origin of the static ball.

Elastic collisions behave like light striking a mirror, with the mirror surface representing the tangent line you so well described.

Incidentally, the spin you spoke of is also known as restitution, I think.

Well put.

I don't really know all the physics behind it but I do play pool daily and this is what my son means by right angles: If you were to draw a line between the centers of the two at the moment of impact the stationary ball always travels directly along that line you drew and the cue ball will exit at a 90 degree angle to that line. The only time this is not the case is during an head on collision or spin is induced on the cue ball.

The resulting angles of trajectory from the colliding spheres is only half of the equation. I think that the right angle in reference is the cue stick on the cue ball. A cue ball struck at a right angle will glance true to the object ball however if the cue ball is struck with an angle that creates spin on the cue ball the trajectory will be changed or even stop on impact. The directional energy transfered to the object ball will remain the same. The rebounding energy is transfered to the table. The cue ball is under control. This is called putting "English" on the ball. Rack em up!

The ball being struck will always rebound according to the laws of colliding spheres. The inertia of the cue ball puts it in the pocket when the angle is correct. The spin negates the inertia.

Do you think you son said finding the right angle? What you need to do is look at the ball and find that straight line, you need to know your speed, contact and bumper bounce. If there is a light in the center of the table it sometimes works to aim the center of the cue ball towards the center of the reflection on the ball you need to pocket. The basics you will need is to practice practice practice. To many angles speeds and materials to figure out to become a good pool player. just practice!!!!!!!!

How does spin (restitution) negate inertia?

I concur that it is probably more accurate to interpret "correct angle" and not 90 degree angle. English, imparted by an off center striking of the cue ball by the cue stick imparts a spin on the cue ball. The linear momentum is dominant due to the spin encountering sliding friction. Once a component of the linear momentum is transferred to the target ball the spin can exert a more significant effect on the resulting trajectory, which becomes a vector sum of the resulting momentum of the collision (angle dependant) and the vector imparted by the traction from the spin.

The right angle he speaks of is the angle between the paths of the cue ball and the object ball after collision. Unfortunately, this rule of thumb only works when the cue ball is not rotating at the moment of collision. Although I'd love to go into more detail, I'll stop here; this is one of my favorite topics as I've been shooting pool for ~10 years.

The "right angles" mentioned by the son, does refer to 90 degree angles (i.e.: complimentary angles). Simply put, the angles of the exiting paths of both balls, with respect to the initial force vector, will ALWAYS combine to form a 90 degree angle. To see for yourself, draw two circles in contact with each other. Next, draw an initial force vector (any angle you wish the shot to originate from) passing through the point of contact between the circles - this is originally through the center of the cue ball, however it can be tranposed on a parallel path, through the point of contact. Now draw a line through the centers of both circles (this will also pass through the point of contact). Finally, draw a line perpendicular to the last one, passing through the point of contact between the circles. The line drawn between centers will be the exit path of the ball being hit, and the cue ball will exit at a 90 degree angle from it. So, as long as you can visualize the cue ball's exiting angle (from the initial force vector), you should be able to foresee its path by using the bumpers like mirrors (angle in = angle out). Of course, all of this is an ideal situation, with no spins, friction, drag, etc... and the only situations where it does not apply is a perfect head on shot, or a perfect shot to the side of the target ball where there is contact but no movement. This is because one of the balls will not move, thus no angle, and the other will continue to travel on the initial force vector.

Substitute the word "correct" for the word "right" and you'll have a perfect game and the "right" answer!!!! Seems to me that they are rightly interchangable... "Right On!" "Bloody Right Mate!" "Correct-a-mundo!" "Alrighhhhht!" (If you can't have fun... then have at it...!)

Some of you understand physics and some of you understand pool - to a degree. I don't think I've read a reply from anyone who understands both. I've been shooting pool for a little over 40 years. I've been studying and using physics for about 38 years. The angle of incidence equals the angle of reflection - there are too many factors that affect the path of the object ball after being struck by a cue ball. In "perfect" collisions, all (I think) answers using physics are essentially correct, except for the 180 degree crap. A line drawn perpendicular to the line bewteen centers (at the center of the cue ball) is the path the cue ball will take - in a perfect world. There is friction between the cue ball and the cloth, same with the object ball. There is also "sliding" friction between the cue ball and the object ball. That results in a small vector in the initial path of the cue ball. In pool, that's called "throw." I think I'd like to challenge any of you to a game or two - if you'll take out the garbage for a month...

Hey guys, could this be a grammer problem? Perhaps the statement should have read " the correct angle, not the right angle" Gary W.

"A line drawn perpendicular to the line bewteen centers (at the center of the cue ball) is the path the cue ball will take - in a perfect world."

No, I don't think in a perfect world that would happen.

For the sake of vocabulary I will refer to the perpendicular line that bisects the center-to-center line at the moment of contact between the cue ball and the ball that it strikes as the tangent line. The tangent line will be tangent to the circumference for both balls at that contact point. The tangent line will also be parallel to the surface of the pool table.

I agree with the angle of incidence is equal to the angle of reflection, but at no time will the cue ball rebound at a right angle (perpendicular) to the tangent line. There is no angle of attack that the cue ball can make contact with another ball and reflect perpendicular to the tangent line – in, as you say, a perfect world.

The angle of incidence is the angle that is made between the initial vector of the cue ball and the tangent line. The angle of reflection is the outgoing path for the cue ball. The angle of reflection is the angle that is made from the exit vector of the cue ball to the tangent line.

I know, this is getting pretty wordy, but at least I have not started using the word shall!

I kind of think this is what you meant, but the way it was worded isn't clear. Then again, I am not sure everything I wrote is clear, but it is the best I can do without adding pictures. Again, this is a description of pure elastic collisions in a frictionless world. And I have one year of pool experience 38 times. ;-)

I think you're mis-interpreting what I said... The path of the center of the cue ball is parallel to the tangent line. The line drawn perpendicular to the line between centers - at the center of the cue ball is parallel to the tangent line and, hence is the path of the cue ball.

Maybe I am misunderstanding what you mean by a tangent line.

I am thinking of a tangent line as a straight line that has one point along the line that is coincident with the circumference of a circle, but no part of that line lies inside the circumference nor can the coincident point be either end point of the line.

In the case of two spheres that are in contact with each other (one point on each sphere's circumference is coincident) with a tangent line that is common for both spheres, any point along that tangent line will be equidistant from the two spheres' centers. The tangent line will also be perpendicular to the line that connects the two spheres' center points and will bisect that center line.

is the same on either side of the Atlantic. I am from Ireland and live in the US and I know the son is talking about the angle of the elbow when setting up a shot. It's obvious the son had good tutelage and the father is a crap player :D

Well his son is refering to if you strike the cue ball with a center ball shot..That is No english or "spin" on the ball. The Cue ball will come off the object ball at a right angle to the point of impact. Knowing this he is able tell what direction the cue will travel in after the hit and know if it will send the cue ball to a pocket or keep it on the table.

Hi guys, I really enjoyed your discussion on pool, but I've only played pool once in my life and I failed miserably. However, I've played quite a bit with electrons. How would they behave, in comparison to a set of billiard balls, given they posses a negative charge. ????? Jens