I'm looking for hard numbers, not a "quantity vs. quality" debate. I'm pretty sure I've run across this before on the web, but after searching various ways for almost an hour now, I can't come up with much of anything but tutorials on using english.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Sloppy Pockets
- Start date

I'm looking for hard numbers, not a "quantity vs. quality" debate. I'm pretty sure I've run across this before on the web, but after searching various ways for almost an hour now, I can't come up with much of anything but tutorials on using english.

I can't imagine what knowing the RPMs (or RPMs per foot of travel) would be useful for. The useful info is how much the rebound angle off the rail changes, and you can just shoot test shots to know that.

I'm looking for hard numbers, not a "quantity vs. quality" debate. I'm pretty sure I've run across this before on the web, but after searching various ways for almost an hour now, I can't come up with much of anything but tutorials on using english.

For instance, maximum* spin/speed ratio changes the rebound angle about 3 diamonds cross-table on my equipment (that's about 37 degrees, in case you're curious).

pj

chgo

*Maximum without "enhancement" using braking draw.

Last edited:

I'm looking for hard numbers, not a "quantity vs. quality" debate. I'm pretty sure I've run across this before on the web, but after searching various ways for almost an hour now, I can't come up with much of anything but tutorials on using english.

Post from ENGLISH in 3, 2, 1...

You can easily hit the cue ball with enough follow that it is rolling smoothly on the cloth from the start. Similarly, you can play with enough draw that the top point of the cue ball is stationary as the cue ball moves away from you (until the cloth begins to take away the draw).

I'm looking for hard numbers, not a "quantity vs. quality" debate. I'm pretty sure I've run across this before on the web, but after searching various ways for almost an hour now, I can't come up with much of anything but tutorials on using english.

If you go very close to a miscue, you can get slightly more spin than that.

If you know how fast the ball is moving, then it's easy to calculate the RPMs. Break speeds go to 30MPH or so, but that's hitting the ball in the center, more or less. Figure about 15MPH with full follow.

You can hit as far to the side as to the top or bottom, so the RPMs there will be similar.

As mentioned on Dr. Dave's site and as has been discussed here before, if you want to increase the side spin to speed ratio, you can mix in some draw with the side spin. That slows the cue ball on the way to the target but the side spin remains. Depending on how far off-center you feel you can hit, for maximum effect you want to hit at 4:00 or 8:00.

I can't imagine what knowing the RPMs (or RPMs per foot of travel) would be useful for. The useful info is how much the rebound angle off the rail changes, and you can just shoot test shots to know that.

For instance, maximum* spin/speed ratio changes the rebound angle about 3 diamonds cross-table on my equipment (that's about 37 degrees, in case you're curious).

pj

chgo

*Maximum without "enhancement" using braking draw.

Actually, it's for academic reasons that have to do with tip-ball, ball-ball, and ball-rail frictional contributions, not to help me with position play. The threads on swooping got me thinking about the role friction plays in all aspects of the game. Just some mental noodling around is all I'm doing.

I'm trying to find the surface speed of the revolving CB, and for that I need to know RPMs. I know it will vary depending on how hard I hit the ball, and how far the tip is offset from center. I'm just looking for a ballpark range for various speeds and offsets. I guess I'll just have to go with what Bob mentioned and figure it out on my own on a case-by-case basis. I'm just being lazy and was hoping there were charts somewhere.

For example, in a frictionless environment, how much spin will initially be imparted to the CB with an 8 MPH hit two tips off center with an 18oz. cue and a tip with a COR of 0.75? High school physics stuff, nothing fancy for this boy.

Post from ENGLISH in 3, 2, 1...

Lol! I can hardly wait.

If you want to calculate it yourself, the formula is very easy....I'm trying to find the surface speed of the revolving CB, and for that I need to know RPMs. I know it will vary depending on how hard I hit the ball, and how far the tip is offset from center. I'm just looking for a ballpark range for various speeds and offsets. I guess I'll just have to go with what Bob mentioned and figure it out on my own on a case-by-case basis. I'm just being lazy and was hoping there were charts somewhere.

Let:

R == radius of the ball,

W == spin rate in radians per unit time (1 revolution per second = 2pi rad/sec = 6.28 rad/sec),

V == speed of the ball.

b == tip offset (same units as the radius - the ratio b/R doesn't need units, of course)

The surface speed of the spinning ball (the product of R and W) at the ball's spin equator is then given by:

RW = (5/2)(b/R)V

Example:

If struck at the miscue limit (b/R = 1/2), the formula becomes:

RW = (5/4)V

If the ball is traveling at 16 mph, the rotational surface speed is then 20 mph. To convert to feet/sec, multiply that by 1.467. To convert to inches/sec multiply by 1.467*12 = 17.6.

From the first formula, the spin rate in radians per unit time is:

W = (5/2)(b/R)V/R

So if the ball's speed and the subsequently calculated surface speed are in units of mph, to convert to revolutions per second, first multiply by 17.6 to convert to inches/sec, then divide by the radius (1.125 inch) to convert to rad/sec, then divide by 6.28 to convert to revolutions/sec. Putting it together, if the ball's speed is in mph, its rotational surface speed at the spin equator in inches per second is:

RW = 17.6(5/2)(b/R)V = 44(b/R)V

and its spin rate in revolutions per second is:

W = (17.6/6.28)(5/2)(b/R)(V/1.125) = 6.23(b/R)V

If you want its spin rate in radians per second, leave out the division by 6.28 in the above:

W = 39.1(b/R)V

Hope that's clear enough.

Jim

Last edited:

Dr. Dave and I came up with a measured value many years ago, but I'll be darned if I can remember it. It's almost certainly in one of his technical proofs. I may (or may not) have the time to search for it.I'm also interested is how much sidespin wears off as the ball rolls on the table surface. Again, I realize that the cloth type, condition, tightness, as well as the ambient temperature and humidity, will all factor into this. Just looking for how to calculate a baseline number.

There is a formula, of course, but wouldn't it be easier to just assume some initial speed?For example, in a frictionless environment, how much spin will initially be imparted to the CB with an 8 MPH hit two tips off center with an 18oz. cue and a tip with a COR of 0.75? High school physics stuff, nothing fancy for this boy.

Jim

To see it dance and spin.

Sometimes I get 9 rails,

It helps me get a win.

I spin it hard with outside

On every hanging ball

It glides along the rails

To see what else will fall.

Whitey every now and then

Will loft into the air

And on a nearby table

It will land with reckless care.

My powerful big stroke

Isn’t welcome at the hall.

I left to many cue ball dents

In the sheetrock on the wall.

If you want to calculate it yourself, the formula is very easy.

Let:

R == radius of the ball,

W == spin rate in radians per unit time (1 revolution per second = 2pi rad/sec = 6.28 rad/sec),

V == speed of the ball.

b == tip offset (same units as the radius - the ratio b/R doesn't need units, of course)

The surface speed of the spinning ball (the product of R and W) at the ball's spin equator is then given by:

RW = (5/2)(b/R)V

Example:

If struck at the miscue limit (b/R = 1/2), the formula becomes:

RW = (5/4)V

If the ball is traveling at 16 mph, the rotational surface speed is then 20 mph. To convert to feet/sec, multiply that by 1.467. To convert to inches/sec multiply by 1.467*12 = 17.6.

From the first formula, the spin rate in radians per unit time is:

W = (5/2)(b/R)V/R

So if the ball's speed and the subsequently calculated surface speed are in units of mph, to convert to revolutions per second, first multiply by 17.6 to convert to inches/sec, then divide by the radius (1.125 inch) to convert to rad/sec, then divide by 6.28 to convert to revolutions/sec. Putting it together, if the ball's speed is in mph, its rotational surface speed at the spin equator in inches per second is:

RW = 17.6(5/2)(b/R)V = 44(b/R)V

and its spin rate in revolutions per second is:

W = (17.6/6.28)(5/2)(b/R)(V/1.125) = 6.23(b/R)V

If you want its spin rate in radians per second, leave out the division by 6.28 in the above:

W = 39.1(b/R)V

Hope that's clear enough.

Jim

African or European?

Sent from my SM-G900W8 using Tapatalk

If you want to calculate it yourself, the formula is very easy.

Let:

R == radius of the ball,

W == spin rate in radians per unit time (1 revolution per second = 2pi rad/sec = 6.28 rad/sec),

V == speed of the ball.

b == tip offset (same units as the radius - the ratio b/R doesn't need units, of course)

The surface speed of the spinning ball (the product of R and W) at the ball's spin equator is then given by:

RW = (5/2)(b/R)V

Example:

If struck at the miscue limit (b/R = 1/2), the formula becomes:

RW = (5/4)V

If the ball is traveling at 16 mph, the rotational surface speed is then 20 mph. To convert to feet/sec, multiply that by 1.467. To convert to inches/sec multiply by 1.467*12 = 17.6.

From the first formula, the spin rate in radians per unit time is:

W = (5/2)(b/R)V/R

So if the ball's speed and the subsequently calculated surface speed are in units of mph, to convert to revolutions per second, first multiply by 17.6 to convert to inches/sec, then divide by the radius (1.125 inch) to convert to rad/sec, then divide by 6.28 to convert to revolutions/sec. Putting it together, if the ball's speed is in mph, its rotational surface speed at the spin equator in inches per second is:

RW = 17.6(5/2)(b/R)V = 44(b/R)V

and its spin rate in revolutions per second is:

W = (17.6/6.28)(5/2)(b/R)(V/1.125) = 6.23(b/R)V

If you want its spin rate in radians per second, leave out the division by 6.28 in the above:

W = 39.1(b/R)V

Hope that's clear enough.

Jim

And in the end who ultimately really cares when it comes to actually playing the game.

If you want to calculate it yourself, the formula is very easy.

Let:

R == radius of the ball,

W == spin rate in radians per unit time (1 revolution per second = 2pi rad/sec = 6.28 rad/sec),

V == speed of the ball.

b == tip offset (same units as the radius - the ratio b/R doesn't need units, of course)

The surface speed of the spinning ball (the product of R and W) at the ball's spin equator is then given by:

RW = (5/2)(b/R)V

Example:

If struck at the miscue limit (b/R = 1/2), the formula becomes:

RW = (5/4)V

If the ball is traveling at 16 mph, the rotational surface speed is then 20 mph. To convert to feet/sec, multiply that by 1.467. To convert to inches/sec multiply by 1.467*12 = 17.6.

From the first formula, the spin rate in radians per unit time is:

W = (5/2)(b/R)V/R

So if the ball's speed and the subsequently calculated surface speed are in units of mph, to convert to revolutions per second, first multiply by 17.6 to convert to inches/sec, then divide by the radius (1.125 inch) to convert to rad/sec, then divide by 6.28 to convert to revolutions/sec. Putting it together, if the ball's speed is in mph, its rotational surface speed at the spin equator in inches per second is:

RW = 17.6(5/2)(b/R)V = 44(b/R)V

and its spin rate in revolutions per second is:

W = (17.6/6.28)(5/2)(b/R)(V/1.125) = 6.23(b/R)V

If you want its spin rate in radians per second, leave out the division by 6.28 in the above:

W = 39.1(b/R)V

Hope that's clear enough.

Jim

Thanks, Jim. That's exactly what I was looking for.

On the spin wearing off as the CB travels, I always hear about this, but I don't see a huge difference when I actually shoot. I can put a ton of side spin on the ball as I send it down to the opposite short rail, and the spin still seems to create a lot a angular change as it rebounds off the rail, even at slower speeds.

Who cares that you don't care? And what does you not caring have to do with actually playing the game?And in the end who ultimately really cares when it comes to actually playing the game.

Jim

You're very welcome.Thanks, Jim. That's exactly what I was looking for.

Yes, sidespin does wear off fairly slowly. I think the reason people believe it dissipates much more quickly is that it's hard to discern the sidespin as it gathers topspin (from the cloth).On the spin wearing off as the CB travels, I always hear about this, but I don't see a huge difference when I actually shoot. I can put a ton of side spin on the ball as I send it down to the opposite short rail, and the spin still seems to create a lot a angular change as it rebounds off the rail, even at slower speeds.

Sorry, I still haven't begun the search for a number to put on it.

Jim

Last edited:

Nobody and nothing. Posting in a thread you don't care about just to say you're not interested is about as lame as it gets.Who cares that you don't care? And what does you not caring have to do with actually playing the game?

Jim

pj

chgo

This seems to be true on napless cloths like Simonis 860, but when I try such slow spin shots on a snooker table with napped cloth, the side spin is gone by the time the cue ball goes its 10 feet to the far cushion. You have to really load the cue ball up to have some left.... On the [side] spin wearing off as the CB travels, I always hear about this, but I don't see a huge difference when I actually shoot. I can put a ton of side spin on the ball as I send it down to the opposite short rail, and the spin still seems to create a lot a angular change as it rebounds off the rail, even at slower speeds.

The friction that causes the side spin to wear off is referred to as "boring" friction. It seems reasonable that it would increase with the size of the patch the cue ball rests on. The other two kinds of ball-cloth friction are sliding and rolling.

I'm looking for hard numbers, not a "quantity vs. quality" debate. I'm pretty sure I've run across this before on the web, but after searching various ways for almost an hour now, I can't come up with much of anything but tutorials on using english.

Not sure how many RPM's are possible but here's a bunch in about 30 sec.

https://www.youtube.com/watch?v=NraM1M_4fV4

Guess if you could count those and double it.....

This seems to be true on napless cloths like Simonis 860, but when I try such slow spin shots on a snooker table with napped cloth, the side spin is gone by the time the cue ball goes its 10 feet to the far cushion. You have to really load the cue ball up to have some left.

The friction that causes the side spin to wear off is referred to as "boring" friction. It seems reasonable that it would increase with the size of the patch the cue ball rests on. The other two kinds of ball-cloth friction are sliding and rolling.

Interesting. So, it's kind of like a drill bit, where a larger diameter bit meets a lot more resistance at the interface with the material it's boring into?