Posted by admin | Posted in pool tips | Posted on 04-03-2010
Tags: billiard, billiards, design, dining table billiard table, king's table billiards, pool, table, table billiard, table billiards

What is the rolling resistance coefficient of a bakelite ball on a worsted surface? (billiard ball on table)?
I’m modeling the movement of a billiard ball on a billiards table.
The ball is made of bakelite and is 156-170 grams, and has a diameter 5.715 centimeters, and is moving on a worsted surface.
I’ve got the drag figured out now, and what remains is the rolling resistance. However, to calculate the rolling resistance, I need the rolling resistance coefficient, and I can’t seem to find it anywhere. Can anyone help me with this?
Interesting question.
It doesn’t sound like that is something you can easily look up, but you might be able to determine it yourself if you do the proper experiment – and assuming you have the basic materials, like the ball, and the worsted surface (billiard table), a stop watch, and a way to measure angles. Oh, and you would have to tilt the pool table to a known angle.
I assume rolling resistance is the same thing as the coefficient of rolling friction.
The basic experiment would be to tilt the table to a known/measured angle, and allow the ball to roll a known distance down the table and measure how long it takes.
You can compare that time to the time it would take for a frictionless object to slide the same distance down the table. This is calculatable from physics.
I believe the difference in the times can be used to calculate the actual coefficient of rolling friction.
Let me see if I can go through the equations.
Say you have a pool table tilted at angle theta, and you let the ball roll a distance d:
With no friction, the ball would slide (not roll) with an acceleration equal to g*sin(theta)
The time it would take to cover the distance would be:
d=1/2*at^2, so t= (2*d/(gsin(theta))^0.5
The work done by gravity would be W=F*d = m*gsin(theta)*d
With friction, you would expect that the ball would take longer to travel the same distance since some of its energy is being lost to friction. If you measure the ball’s time (t’) to travel the same distance under friction, you can calculate its acceleration as the following:
a = 2*d/(t’)^2
The energy used to simply move the ball this distance is then:
W = F*d = m*a*d = m*(2*d/(t’)^2)*d = 2*m*d^2/(t’)^2
The difference between this W and the frictionless W should be the energy lost via friction, and the fraction of energy lost would be given by:
coefficient of rolling friction = [mgsin(theta)*d - (2md^2/(t')^2] / [mgsin(theta)*d]
= [gsin(theta) - 2d/(t')^2]/{gsin(theta))
This should be the fraction of energy the ball loses due to rolling friction as it moves across the table. You can then try to incorporate that into your model.
You might try this experiment at different angles to see how constant it is. I would expect you wouldn’t have to tilt the table much, and that lower angles would give you more realistic results than steeper angles.
Oh, and I’m not sure what you meant by already having calculated drag. Is that air friction? or does that somehow incorporate some of the friction losses included in this experiment – make sure you don’t double count the friction forces.
It would be interesting to see how that turns out for you.
Hope this helps.
Oh, and if you can’t realistically tilt the table, you might be able to make a quarter circle ramp to roll the ball down smoothly onto the table and measure how long it takes to travel a specified distance. Again, since the initial energy of the ball would be its potential energy mgh, you can easily calculate its energy, and thus initial speed on the table by using 1/2m(v0)^2 = mgh, and use that to calculate how long it should take with no friction to travel the distance.
Then if you measure the actual time it takes for the ball to roll that distance, you do the same type of calculation to determine how much energy was lost due to friction:
Fraction of Enrgy lost due to friction = [1/2m(v0)^2 - 1/2m(d/(t'))^2)]/[1/2m(v0)^2]
This method is a bit trickier, since you don’t have the constant source of gravitational force moving the ball, so you get an “average” loss over the distance d, and you will lose some energy as the ball rolls along the ramp before it gets onto the table – Not sure how significant off the top of my head, but you might have to account for that.
Anyways, give it a shot.
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