Pool Ball Terminal Velocity

Mr. Bond

Orbis Non Sufficit
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Silver Member
Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)

And how many feet would it take to reach it?

And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)
 

Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)

And how many feet would it take to reach it?

And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)

{edit: ~100 mph (NOT 60 mph per my previous} by my rough calculations (cheat sheet).
 
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Linwood

AzB Silver Member
Silver Member
Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)

And how many feet would it take to reach it?

And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)

Your question does not really make sense.

Terminal velocity is usually defined as the velocity in air, how fast is the fastest it will continue to fall.

The velocity of an object falling toward earth from "very far" with no air, assuming they both start with zero relative velocity, and ignoring effects of orbit and other planets, is about 11.2km/s. That is also the velocity (ignoring air) that you need going up to not come back down.

In theory (ignoring air) the distance to reach escape velocity falling is infinite, in practice you get pretty close within a few thousand, or hundreds of thousands of miles, depending on your definition of "close".

Now terminal velocity as normally defined (falling in air, how fast and how long to reach a steady speed) is a more interesting question, and I no longer remember enough math to figure it out. And of course it might vary by whether it was clean, or had some chalk left on it. :rolleyes:

But I think it's fair to say that the zero-air speed is a bit faster than anyone can break. Most regular, fairly dense objects' terminal velocity tends to be in the 80-200 mph range (one estimate I saw for a baseball is 95mph), so I would think a pool ball might be comparable.

I found a very old article saying a break shot is under 30mph, so I think they are slower than air-slowed terminal velocity as well. Article here. I can't vouch for its accuracy, google gave it to me.
 

Don Owen

AzB Silver Member
Silver Member
Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)

And how many feet would it take to reach it?

And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)
If I'm not mistaken, terminal velocity is the velocity when air resistance equals weight. Therefore, if there is no air resistance then "terminal velocity" looses its meaning or perhaps it's equal to escape velocity, around 25000 mph.
 

mchnhed

I Came, I Shot, I Choked
Silver Member
No wonder I'm such a bad player......
I don't know any of this Pool Stuff!


Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)
And how many feet would it take to reach it?
And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)
~ 60 mph by my rough calculations (cheat sheet).

If I'm not mistaken, terminal velocity is the velocity when air resistance equals weight. Therefore, if there is no air resistance then "terminal velocity" looses its meaning or perhaps it's equal to escape velocity, around 25000 mph.

Terminal Velocity Calculator
Gravitational Acceleration: 32ft/sec/sec
 

BC21

Poolology
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Silver Member
Using the terminal velocity equation below, and assuming a normal air density of 1.27kg/m3, and a drag coefficient of 0.5 for a sphere.....

V = √(2mg/pAc)

m = mass of CB = 170.1g
g = gravity = 9.8m/s2
c = drag coefficient = 0.5
A = area of falling object = 25.66cm2 (convert to meters = 0.002566m2)
p = air density = 1.27kg/m3 (covert to grams3 = 1,270g/m3)


Plug it all in and get 45.24m/s, or 101.2mph. Nobody can break the balls that fast!
 

Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
Using the terminal velocity equation below, and assuming a normal air density of 1.27kg/m3, and a drag coefficient of 0.5 for a sphere.....

V = √(2mg/pAc)

m = mass of CB = 170.1g
g = gravity = 9.8m/s2
c = drag coefficient = 0.5
A = area of falling object = 25.66cm2 (convert to meters = 0.002566m2)
p = air density = 1.27kg/m3 (covert to grams3 = 1,270g/m3)


Plug it all in and get 45.24m/s, or 101.2mph. Nobody can break the balls that fast!
This is correct.
 

EtDM

AzB Silver Member
Silver Member
Using the terminal velocity equation below, and assuming a normal air density of 1.27kg/m3, and a drag coefficient of 0.5 for a sphere.....

V = √(2mg/pAc)

m = mass of CB = 170.1g
g = gravity = 9.8m/s2
c = drag coefficient = 0.5
A = area of falling object = 25.66cm2 (convert to meters = 0.002566m2)
p = air density = 1.27kg/m3 (covert to grams3 = 1,270g/m3)


Plug it all in and get 45.24m/s, or 101.2mph. Nobody can break the balls that fast!

I got results that were pretty close, but I used a drag coefficient of .1 for a smooth sphere, as found here:

https://en.wikipedia.org/wiki/Drag_coefficient

I think it's also important to note that due to his proclivity for historical info, Mr. Bond might also be interested in the terminal velocity of a carom ball. It seems the difference in diameter would actually up this figure a bit.
 

8ballr

Banned
Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)

And how many feet would it take to reach it?

And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)

Terminal velocity for falling objects is about 9.8 m/s squared...whether a marble, a piano or banana.
http://indianapublicmedia.org/amomentofscience/ground-golf-bowling-ball/

When you break, you are not limited by terminal velocity...superman could break the balls at 100x terminal velocity...but the cue ball would explode. :)
 
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Agent 99

AzB Silver Member
Silver Member
Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)

And how many feet would it take to reach it?

And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)

For me, the moment my opponent picks up the cue ball and throws it in my direction things become terminal .. :yikes:
 

Linwood

AzB Silver Member
Silver Member
Terminal velocity for all objects is about 9.8 m/s squared...whether a billiard ball or a cannon ball.

9.8 m/s^2 is the earth's gravity acceleration at the surface, it's not a velocity at all, much less a terminal velocity.
 

mchnhed

I Came, I Shot, I Choked
Silver Member

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Cornerman

Cue Author...Sometimes
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Acceleration to a to terminal velocity...which for a cueball or a marble or just about any object in a free-fall position is about 195 km/h.

Still, no. Terminal velocity is mass and shape dependent. And air drag dependent. We're not talking about vacuums.
 
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