How much of a cueball is useable normally?

[PoolBum]

I'd say 110%.

 

[Geosnooker]

Just curious how you have a 10 foot stop shot? Only possible if you are playing on a 10 foot table, both the cue ball and object ball are at opposite diagonal corners, almost at the drop off points!

almost 14’ on a Snooker table.

However, I don’t try a stop shot without some type of side over 6’ or so if the object ball is near the pocket. I ‘might’ get a longer one but not worth the risk of scratching and giving up a ball in hand. Same with a draw from that distance...not worth the risk.

 

[BC21]

For sure, I saw a video of him stacking balls once.

I tried for decades. He did a demo at the pool room, and stacked it first try.

I went crazy. Starting trying again but could never get it.

Look for little imperfections, nicks or scratches. Stack them on those.

 

[gregcantrall]

And another thing.

If the cue ball ever lands on that flat side, even Mike Massey and Dr. Dave can't draw the ball.

Be careful, wouldn’t bet against the engineers. I am already visualize ing a tip like a floor scraper with progressive pitch with the leather at a v on the leading edge. So as the wedge lifts the 1/2 ball the spin is applied. Ahem, it is a very little known advanced shot.:wink:

 

[MitchAlsup]

Just curious how you have a 10 foot stop shot? Only possible if you are playing on a 10 foot table, both the cue ball and object ball are at opposite diagonal corners, almost at the drop off points!

He could be playing a OB sitting within an inch of the shelf of a corner pocket with the CB within an inch of the opposite corner pocket.

Math: 9 foot table length 4.5 feet wide::

D = SQRT( 9**2 + 4.5**2 )
= 10.06 feet.

So, mathematically, it is just barely doable on a 9 foot table.

 

[jimmyco]

He could be playing a OB sitting within an inch of the shelf of a corner pocket with the CB within an inch of the opposite corner pocket.

Math: 9 foot table length 4.5 feet wide::

D = SQRT( 9**2 + 4.5**2 )
= 10.06 feet.

So, mathematically, it is just barely doable on a 9 foot table.

9.316 feet

 

[straightline]

Lol. I've actually used hockey pucks to show how aiming pool shots has nothing to do with spherical shapes, other than the fact that being spheres makes them move easier (roll).

Kids in the Philippines play this skittle pool thing; must be good training.

I'd say 110%.

I gotta say 99% (means everything except the actual mathematical part on the surface) is unused.

 

[BC21]

Kids in the Philippines play this skittle pool thing; must be good training.

I'll have to look that up....sounds interesting.

 

[Bob Jewett]

I'll have to look that up....sounds interesting.

Carrom.
https://www.youtube.com/watch?v=YNvX-C5Kueo

 

[Black-Balled]

Can believe nobody admits to only using the outside of the cue ball.

 

[ChrisinNC]

He could be playing a OB sitting within an inch of the shelf of a corner pocket with the CB within an inch of the opposite corner pocket.

Math: 9 foot table length 4.5 feet wide::

D = SQRT( 9**2 + 4.5**2 )
= 10.06 feet.

So, mathematically, it is just barely doable on a 9 foot table.

A 9 foot table is only 100 inches (8’4”) nose to nose and no more than 109 inches (9’1”) from diagonal corner pocket drop off points.

 

[Brookeland Bill]

I understand it will depend on speed, but for a normal shot (not slow roll or break), how much can be used, say on a stop shot of 10 feet, with bare minimum speed at the lowest possible contact point on the cue ball? 70%?

Here we go again.

The earth is flat.

Pi R Square.

My sister is also my daughter.

 

[Patrick Johnson]

9.316 feet

That’s right for a 50” x 100” playing area - but balls sitting at those corners would drop into the pockets. Making the playing area 48 x 96 gives a maximum shot length of 9.09’ (9’ 1-1/8”).

pj <- but who’s counting?
chgo

 

[BC21]

Carrom.
https://www.youtube.com/watch?v=YNvX-C5Kueo

Well there ya go! Thanks, Bob. Looks like a fun game.
I'd prefer using a little cue stick rather than flipping with my finger. I've never seen this. I know hockey pucks don't slide too easily on a pool table.

 

[straightline]

I'd prefer using a little cue stick rather than flipping with my finger.

Here's the Filipino version. Looks like he's using the big disc. :grin:

https://www.youtube.com/watch?v=pJtvlolCbh4

 

[BC21]

Here's the Filipino version. Looks like he's using the big disc. :grin:

https://www.youtube.com/watch?v=pJtvlolCbh4

This is exactly what I was trying to explain to a few people about aiming when I used hockey pucks to prove my point, that it's all done with the equator/circumference of the balls, meaning there are no special spherical aiming mysteries or secrets. Being a sphere simply allows a ball to roll, instead of slide like a disk. With fractional aiming it's all done using the diameter of the ball (or the diameter of those discs), rather than the actual circumference.

Thanks for posting the link.:thumbup:

 

[fastone371]

I tell people to visualize the Cue Ball like a Clock with 12 Hour Marker from 12 moving clock wise to 11.

So I am say 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and last 11. Also visualize DEAD CENTER where the Clock Hand Attach.

Then I mention 1/2 Cue Tip out of Center has 12 Contact Points, then I mention 1 Full TIP out of Center is another 12 Contact Points, last position is 1-1/2 Cut Tips out, another 12 Contacts Point out of Dead Center.

Long & Short is 37 TOTAL CONTACTS POINTS.

Most beginners get the idea, but I also tell then to start out small, it's tought at first to hit all 37 POINTS.

Dead Cener, 3, 6, 9, and 12 is tought at first. CENTER, HIGH, RIGHT, BOTTON, and last LEFT.

Are those clock positions AM or PM???? :grin::grin:
Sorry, I guess this social isolation stuff is starting to wear on me.

 

[Patrick Johnson]

But seriously, folks. If the OP meant how far from center you can hit, the answer is that the tip can contact the ball about half way from center to edge. The exact distance depends a little on the quality of the chalk and the tip. As Mike Page illustrated in a YouTube video, the more accurate your delivery, the farther out your safe region will extend.

In case the OP's actually interested in the real answer to the question, Bob gave it here - and the pic below illustrates it: you can't hit outside the largest white circle (1/2 way from center to edge of the CB) without miscuing.

Another way of saying the same thing is "only 1/4 of the CB is 'usable'" - because only 1/4 of the total "2D" area as you look at the CB is inside that circle (3/4 of the total area is outside it).

pj
chgo

 

[Cron]

In case the OP's actually interested in the real answer to the question, Bob gave it here - and the pic below illustrates it: you can't hit outside the largest white circle (1/2 way from center to edge of the CB) without miscuing.

Another way of saying the same thing is "only 1/4 of the CB is 'usable'" - because only 1/4 of the total "2D" area as you look at the CB is inside that circle (3/4 of the total area is outside it).

pj
chgo

View attachment 544596

Thanks, I forgot it's a sphere on paper (it was VERY 2d).

I have this equation to help find the lowest contact point regardless of miscue, but if someone has a 10mm tip and states it's shaped as a dime, is the tip actually the complementary shape of a dime? It could be, be that seems wrong for some reason when it's modelled (it seems very flat).

 

[3kushn]

In case the OP's actually interested in the real answer to the question, Bob gave it here - and the pic below illustrates it: you can't hit outside the largest white circle (1/2 way from center to edge of the CB) without miscuing.

Another way of saying the same thing is "only 1/4 of the CB is 'usable'" - because only 1/4 of the total "2D" area as you look at the CB is inside that circle (3/4 of the total area is outside it).

pj
chgo

View attachment 544596

What does that image look like with say a 12mm tip & max LH English.
Seems I'm further out that your outer circle but I can't lay my tip on there to see what it would look like.