break acceleration
- Thread starter gergs
- Start date
Does he bet???
good one, I'm still laughing
depends....
let's say you hit the cueball 19 mph on the break. let's also say, you are not very good at squatting whitey, and it occasionally flies off the table. ok, now go with me here, the table you are playing on is surrounded on 3 sides by a large precipice.....the cueball would initially decelerate upon leaving your cue, but after contacting the rack and flying off the table, it would accelerate to 21.81818181.....lol, nevermind, this post is a stretch even by my standards.
let's say you hit the cueball 19 mph on the break. let's also say, you are not very good at squatting whitey, and it occasionally flies off the table. ok, now go with me here, the table you are playing on is surrounded on 3 sides by a large precipice.....the cueball would initially decelerate upon leaving your cue, but after contacting the rack and flying off the table, it would accelerate to 21.81818181.....lol, nevermind, this post is a stretch even by my standards.
I get that, but it would acccelerate to nearly 22 mph falling into the precipice in my example....lol
I get that, but it would acccelerate to nearly 22 mph falling into the precipice in my example....lol
Accelerating down the precipice at a rate of 32 feet per second squared - one second to 22 MPH.
The cue ball would go from 0 to max speed in the distance it takes the tip to compress as much as it can and the shaft to flex as much as it's going to for the given force. Once those take place, the ball is off like a shot as both rebound. After that, it slows all the way to the rack. The guy is an idiot. Quit arguing with him.
If you overspin the cue ball, then the cue ball will accelerate after leaving the tip. I'm not sure if it's possible to do this on an extremely hard shot such as a power break. But if "break shots" include a soft break, then certainly it is possible.
Here's Dr. Dave's video on overspin: http://www.youtube.com/watch?v=0WIYecrJ5ec
Here's Dr. Dave's video on overspin: http://www.youtube.com/watch?v=0WIYecrJ5ec
He couldn't be more wrong. The cueball immediately starts slowing down the instant it leaves the tip. What force could be accelerating it *after* it leaves the tip?
KMRUNOUT
Excepting for the added minute amount of speed added as top-spin decays into velocity;
The CB starts to decelerate within 2 millisecond of contact with the tip--that is essentially immediately.
The CB starts to decelerate within 2 millisecond of contact with the tip--that is essentially immediately.
You might be able to do it on a shot like the break shot, but if you are interested in getting max speed into the cue ball you don't want to waste any energy in spin. You need to hit very near center.
On a break shot, the cue ball is normally in the air for the first couple of feet or even all the way to the rack. While it's in the air, it's not losing speed by rubbing on the cloth, although there is a little air resistance.
As I recall, the last time I calculated it, the cue ball would lose about 1-2% of its speed if it did rub on the cloth on the way to the rack (as opposed to be airborne). The faster you shoot, the less energy will be lost because it takes time for the friction to act on the ball and the shorter the time the less the friction can slow the ball and start top spin.
Yeah. But won't it lose some of its potential horizontal vector if it leaves the table? And then there's the loss as it "digs in" and skids on impact with the table. Airborne can also cause a vertical component on impact with the OB if the timing isn't exactly right. That's lossy, too.
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The CB is still decelerating while in the air.
