Breakdown of Earl Strickland Shot

Chili Palmer

AzB Silver Member
Silver Member
With no backspin or forward spin it will travel right along the tangent line (as we've been discussing here). So if my tangent line is right and the CB has no forward spin (like this shot), it can't hit closer to the side pocket than about two diamonds away (where the tangent line points).

Maybe you meant "down rail" from where the arrow points...?

pj
chgo

The two would miss the side pocket to the right and hit the side rail about 2-4" to the right of the side pocket. Meaning, your tangent line does not look correct to me, which means the CB would not travel the path you drew.
 

Patrick Johnson

Fish of the Day
Silver Member
The two would miss the side pocket to the right and hit the side rail about 2-4" to the right of the side pocket. Meaning, your tangent line does not look correct to me, which means the CB would not travel the path you drew.
OK, thanks for the explanation.

pj
chgo
 

ShootingArts

Smorg is giving St Peter the 7!
Gold Member
Silver Member
more like the factors involved

I agree this thread is useful. Your last sentence above is very encouraging!

There are a couple of inaccuracies in what you have said, and there are also some inaccuracies in the Barry Stark video. While his credentials as a Snooker coach and all of his advice on the topic is excellent, lets not pretend he is a physicist. The problem with both your and Barry Stark's descriptions is with the use of the word "momentum". If you are aware of the idea of conservation of momentum, then you know that all of the energy required to make the object ball move is LOST by the cue ball. And then some. The cueball begins with a whole lot of kinetic energy, and of different types. "translational" kinetic energy, or momentum, is only part of the energy that the cueball has. It also has rotational kinetic energy. At impact, a LOT of things happen. Energy is lost to heat, sound, friction against the cloth, etc. When the conservation of momentum equation comes into play, the rotation of the cueball is not so much a part of that equation. The equivalence in the conservation equation is between the energy LOST by the cueball and gained by the object ball. So to say that the cueball "still has momentum" in its initial direction is not correct. The altering of its initial direction IS the energy imparted to the object ball. If the masses of the two balls are equivalent, then the new trajectory of the cue ball is ALWAYS on a line parallel to the line tangent to both balls at the point of contact.

At this point, it has nothing to do with the cueball's "momentum" where it goes. It has to do with its remaining rotational kinetic energy, This energy gets converted into friction and an altering of the path of the cueball from the tangent line. (For sure, there is a conservation of rotational energy as well. Some of the cueballs rotation is imparted to the opbject ball, some is lost as friction, and some is converted into translational kinetic energy, while still some remains as rotational kinteic energy as the cueball rolls to a stop.)

When you can think about the separation of all of these forces and their vectors, and the interplay between these different types of energy, I think this topic will become more clear to you. I would recommend in the future, however, to get your physics info from physics textbooks rather than from Snooker coaches. I'd likewise avoid the physics books for pool instruction.

KMRUNOUT


Energy never goes away but it is converted into forms of energy that people rarely consider including the heat and sound you mentioned.

One question I always have about pool balls, I know they flex on impact but how much do they flex? I have seen this mentioned but I have never seen it quantified. I have made caroms where I had to shoot hard enough to have a sliding cue ball then the object ball came off the other ball and went about five feet into the pocket. I have never seen these shots vary noticeably from what two perfectly round objects that didn't flex would create so I have to assume ball flex is a nonissue although I have seen it claimed that the main sound we hear when balls collide violently is the phenolic snapping back.

DIIK! It is interesting to puzzle over these things on the Net but then some pool hall bum that thoroughly understands how and is clueless about why will beat me like a rented mule!

Hu
 

Bob Jewett

AZB Osmium Member
Staff member
Gold Member
Silver Member
... One question I always have about pool balls, I know they flex on impact but how much do they flex? I have seen this mentioned but I have never seen it quantified.....
When two balls collide you get flat spots on both balls. If you prepare the contact point properly, you can see the size of the flat spot afterwards. It is about a quarter inch for a fairly firm shot.

A little geometry will give you how much each ball compressed as well as how far forward the cue ball moved while it was in contact with the object ball. The result for each is much less than a millimeter.

You can also calculate the size of the flat spot if you know the contact time of a ball-ball collision. Wayland Marlow did that time measurement (with a fairly simple apparatus consisting of a battery, resistor and capacitor along with wires and a DC voltmeter) and described it in his book "The Physics of Pocket Billiards".
 

ShootingArts

Smorg is giving St Peter the 7!
Gold Member
Silver Member
I miss CAD a lot of times.

When two balls collide you get flat spots on both balls. If you prepare the contact point properly, you can see the size of the flat spot afterwards. It is about a quarter inch for a fairly firm shot.

A little geometry will give you how much each ball compressed as well as how far forward the cue ball moved while it was in contact with the object ball. The result for each is much less than a millimeter.

You can also calculate the size of the flat spot if you know the contact time of a ball-ball collision. Wayland Marlow did that time measurement (with a fairly simple apparatus consisting of a battery, resistor and capacitor along with wires and a DC voltmeter) and described it in his book "The Physics of Pocket Billiards".



Bob,

I made my living as a draftsman/designer for awhile, using ACAD from 2.X to 13 and Intergraph PC for a few years. Particularly with Autocad I could draw these things out in less time than it takes to talk about them. Even though it is times two as I assume both balls flex equal amounts the amount a 2.25" ball has to flex to cause a quarter inch flat spot isn't much. Then there is the question of snap back while in contact which might offset this compression somewhat.

One example I think of is a one pocket shot. Two object balls within a few inches of each other on my opponent's side of the table near the side pocket and positioned so there was no bank. Looking, I saw an easy carom if the cue ball was skidding into the object balls. A very firm but not break speed hit with no allowance for anything but the collisions of perfectly round objects centered my pocket. With a pool shot we can never be sure exactly how it was hit but seemingly the things like slight compression and rebound are nonissues.

I always wonder a little about pool balls. Even a cheap set I picked up years ago with very little playing time on them plays much better than house balls even at a quality pool hall. One issue is that the pool balls have gotten mixed over the years so the factory weight matching is gone. I can't help wondering if old balls are as round as new ones or if all of the hits have affected their roundness. As always, more questions than answers!

Hu
 
Top