Is this possible ??
- Thread starter SHORTY WRIGHT
- Start date
Woodshaft
Do what works for YOU!
Judging from the wear on the pool balls, I imagine they're quite "sticky". If that's the case, you should be able to throw the 7-ball in by using a ton of almost masse left english on the cue ball. You have to shoot down on the cueball to keep it's speed down so that all that english transfers thru the 4-ball and into the 7-ball to throw it left. So maybe lol
Thanks
Friend just showed me a similar shot. If the gap is small enough you can throw a ball. From this view and that cue ball direct direction I'm unsure
but looks like to me if you hit the first object ball on the right side cutting into the left side of the ball to be thrown with the very very small gap its possible.
This one is close but it looks to me like the shot could probably be made if the balls are frozen, you hit it in such a way as to produce the absolute maximum throw possible, and the pocket shelves are not too deep since this one is going to be close (these shelves don't look all that deep so I think it would go here but it's close). It also helps that the balls here appear to be a bit dirty or at least not new or highly polished, and the dirtier or rougher/less polished the balls, the more they will throw.
Frozen balls throw much better than non frozen balls, but balls within about a millimeter of each other or a hair more will still throw, just not near as much as if they had been frozen.
A few other factors to consider are that a sliding cue ball produces the most throw (as in a stun shot, but this is only really important when throwing a single object ball that the cue ball is directly hitting but that isn't the case here since you are throwing the further of two object balls that are frozen together), and the slower the shot speed the more throw that will occur, and half ball hits produce the most throw with thinner or thicker hits providing proportionally less throw (all of these factors are influencing the cut induced throw, aka CIT).
You will also get throw by using side spin, known as spin induced throw (aka SIT), because some of the spin from the cue ball will transfer (in reverse) to the object ball, and in some cases (especially if they are frozen and you are hitting down the line of all their centers) a bit can even transfer from that first object ball to a second object ball (where it of course reverses again). If I recall correctly the amount of spin that produces the most throw is roughly half the maximum amount of side spin possible or so, but Dr. Dave has a page and video somewhere that gives a more precise and correct answer but it is not maximum spin as might seem intuitive.
Hit in their most ideal manners, cut induced throw can produce a higher maximum amount of throw than spin induced throw can. The maximum amount of CIT throw that will be possible will come from a half ball hit and at a speed just barely hard enough for the object ball to reach the pocket, but if you were trying to throw a single object ball instead of two that are frozen together a sliding cue ball also helps to create the maximum throw but here is doesn't matter so much because the 2 is "sliding" into the 7 whether the cue ball was sliding or not.
It might sound intuitive that if you combine the maximum amounts of CIT and SIT at the same time that you would get the most throw of all but that is actually counter productive and will actually reduce the amount of throw you get. Generally you want to use one of the other (CIT will get a little more throw than SIT will so choose CIT if you need the absolute maximum amount possible). If you want or need to use both for positional or other reasons then only use half of the maximum amount of each or some other ratio that doesn't exceed 100% in total but know that any combination of the two will create a little less throw than if you had only used maximum CIT alone.
So for this shot, since it is already a question as to whether it is even possible or not, if you are going to try to throw the 7 ball in you already know you have to attempt it in the way that creates the maximum amount of throw possible. This means using only CIT and hitting the 2 ball in such a way that it hits the 7 at a half ball hit angle, which in this case means that you are hitting the 2 ball at nearly a half ball hit angle as well, and hit is at the slowest possible speed that barely trickles the 7 ball to the pocket.
I've found that as a rough ball park rule of thumb you can figure that the maximum amount of throw you can get with two object balls frozen together like this is about a diamond's amount of throw over the length of the table, but that varies a bit depending on how dirty the balls are, how humid it is, the type of cloth and cloth conditions, etc so you can can adjust your estimate slightly based on the conditions at hand once you have the experience. This can help you estimate how much maximum throw you can get over some shorter distance such as here with this shot where you could extrapolate to figure you might get somewhere approaching an inch of maximum throw at this distance. As with everything else in pool, somewhere on Dr. Dave's site he goes into fine detail regarding all the intricacies of every type of throw.
How to Play Pool - Dr. Dave Pool Info
Instructional videos, articles, and tutorials for learning how to play and teach pool. All secret tips of billiards are revealed.
billiards.colostate.edu
yeah, that.
Agree with some of the others - if the balls are frozen then the 7 definitely goes if you hit the right hand side of the 2. This is known in some places as a "reverse jacky" where you should strike the opposite side of the ball compared to if they were unfrozen with a sizeable gap. Counter-intuitive to a lot of beginnners (including me - realising this added wheels to my game).