opening billard break: sink all 15 balls?

Is it possible to sink all 15 balls on an opening billiard break shot?
Are there any laws of physics that would make this impossible to do? To
your knowledge, has this ever been done?

Russell Shaw
Portland Oregon

I think I’ve seen it done on a tricks-and-tips type video, but I may be wrong. My photographic memory only works on useless trivia.


Russ, I assume you mean from the standard racked opening.
I’ve seen fifteen balls go down on a single shot before, but they were strategically positioned.

Rack’em up!

There are just three balls in billiards.

you are thinking of carom billiards, picmr.

the game that is in question is traditional pocket billiards.

Touche, picmr!

I want you to know, despite Mr. Trout’s attempted nonchalant backpedal, your perceptive comment did not go unnoticed.

“Pocket BILLIARDS??? Oh, you mean POOL!”

POCKET POOL??? he he he

hmmm… i don’t get it.

oh, i see now.

you’re being foolish.

Yes, it’s possible. After the break shot (but while the balls are still moving), lift and tilt the table so all the balls fall in a corner pocket.

What? Against the rules you say? Pfft… I don’t remember any such restriction in the OP. Fine then, wait for an earthquake to shift the table in the same manner.

Oh, now you’re saying it’s a scratch? Well, I say it’s an act of God. Po-ta-to, po-tah-to.

Of course.

No, though chaos theory suggests that it is improbable. I don’t think that a human or machine could reliably reproduce such a shot, but that doesn’t rule out the possibility of random chance.

I’ve never seen or heard of it in a standard 15 ball rack, but I’ve seen it done with a 9 ball (diamond) rack (for what that’s worth).

I’ve seen it done, on TV, but I’m understandably skeptical. Even if it is possible for a highly trained human to do this on demand, it’s got to be a lot harder than making a trick table with a slightly uneven surface, or other such tricks.

I am certain it’s been done before, somewhere at some time, but it’s probably only happened a handful of times ever under controlled circumstances (no shaking the table etc). Joey is right that chaos threories and random chance would jsut about negate the possibility of repeating it every time.

That said, I’ve seen it done in 9 ball a million times (exagerating here…). There are fewer balls and they’re set up in a different way that appears to be more conducive to sinking them all at once. In fact, most good 9 ball players can do it pretty often - that’s why all of the competitions on ESPN are played in sets instead of single games. They end quickly sometimes.

All you really need is a frictionless table with perfectly elastic bumpers. I would have thought you physics types would have lots of those things just laying around. :slight_smile:

the stonecutters assisted homer with this feat using mrblue92’s strategy.

I’ve seen a few nine-ball tournaments on TV, even played in a league, and the most I’ve seen anyone sink on the break is 5. I’d be amazed if anyone has sunk all 15 from a standard rack. There isn’t enough kinetic energy from the cue ball to really mix things up a lot. You’d need to have some balls roll straight into the pockets, others roll slowly in after them, and the balls that get the most energy from the break would have to go off one or two cushions (but no more) and in. It would all have to be too perfect.

And back in the days when people played straight pool, they didn’t even hit the cue hard on the break.

I’ll try to remember to test this next time I play. I’m curious to see if I can hit a rack of 15 and just move each ball far enough to hit a rail.

Consider the variables (I’m sure I’m leaving some off).

(1) Each ball has a slightly different elasticity [16 degrees of freedom]

(2) Each position on each bumper has different elasticity [let’s be generous and assume that this is only 20 degrees of freedom]

(3) The felt has gradients of friction [easily 25 degrees of freedom]

(4) No rack is perfect, the balls can have small gaps between them [32 degrees of freedom (maybe more)]

(5) Angle, velocity, and spin of the cue ball [basically an infinite number of degrees of freedom]
Of course, there are the effects of air currents in the room, temperature fluctuations that can change the elasticity of the balls and bumpers on-the-fly, and many other that we should technically consider…

On the other hand there are 6 holes and quite a few possible solutions that help to reduce the number of degrees of freedom…

I think we’re easily talking odds much greater than a billion to one, but just for the sake of argument, let’s say the odds are only a billion to one… If you set up an experiment where you could control all of the above factors and allowed for a new break every five minutes, you might have to run your experiment for more than 9500 years before you saw all 15 balls go in the pockets…

Just a SWAG… someone check my math… [wink]

It would not surprise me in the least if it has never occured…

Oops! Nitpicking my own post…

#4 is too low because I didn’t consider the effects of multiple gaps and #5 is not infinite because we could surely quantize these effects and rule out some of the extremes, for instance, angles that are perfectly perpendicular to the long side of the table or ball velocities that cause the cue ball to shatter…

But you catch my drift…

Damn, I left out drift from the table being out of level!

No one has EVER sunk all 9 balls on the break in a sanctioned event. Ever. I think the most is six or seven.

Chaos theory tells you that the shot would be unrepeatable, it says nothing at all about how possible it is. For that, you have to look at the physics of a pool rack, the way energy is distributed, and probability.

A 9-ball or 15-ball rack are built so that the force on the center ball is symmetrical. Most of the energy of the cue ball goes into the side and end balls. If the rack is perfect, the 8 ball in a 15 ball rack or a 9 ball in a 9 ball rack will not move at all. Of course, it can be moved through collision with another ball.

But to give you an idea of how unlikely it is, let’s assume that on any break any ball had a 50% chance of being pocketed. If that were the case, then the odds of all 15 balls going down on the break would still be 65,534 to 1. In practice, the odds are much, much higher than that. I’d guess that the chance of all 15 balls being pocketed on the break would be in the trillions to one. That’s assuming you can physically get them all to move far enough to hit a rail, and that would take a phenomenally hard break in an 8-ball rack.