Register to reply 
Bowling physics... 10 pin bowling... 
Share this thread: 
#1
Oct2610, 03:05 PM

P: 2

I'm an avid bowler... and trying to figure out the reduction in force into the pins when using a 14lb ball as opposed to a 15 lb ball as opposed to a 16 lb ball. I am reasoning that a lighter ball has less impact power (as in getting hit with a 3 lb brick as opposed to a 3 lb feather ball)
Obviously the hardness of the balls is the same, as is the ability to compress/absorb energy. I know that I can throw the 15lb ball harder than the 16, and the 14 pounder harder still, but I'm wondering how MUCH harder I need to throw it to have the same force into the pins at impact (hand and wrist action, directional change and revolutions assumed to be equal) I know that F=MA, but there really isn't any acceleration going on. The ball is a roughly constant speed into the pins (though it would slow down as it's going down the lane) What is the formula for calculating the relative force into the pins of different weight balls? Chris 


#2
Oct2610, 05:25 PM

P: 11

I'm not entirely sure about this, but my physics teacher told me if it's constant acceleration and only gravity is acting upon it, then you should fill 9.81 m/s as A.
So the mass needs to be in Kilograms (I think?) and then 9.81 so: F = 6.8 Kg * 9.81. Hopefully I was correct and was able to help out. 


#3
Oct2710, 06:46 AM

P: 2

no, that unfortunately doesn't help. I'm looking for the speed at which a 15 lb ball must hit the pins in order to have the same force as a 16 lb ball. If they're both thrown the same speed obviously the 16 lb has more force because of the extra mass.



#4
Oct2710, 10:31 AM

PF Gold
P: 691

Bowling physics... 10 pin bowling...
The ball is so much heavier than the pins that I can't see the weight of the ball making any significant difference when hitting the pins. The difference, I assume, is in the point where the ball starts gripping the surface of the lane. A heavier ball will grip sooner. If I hazard a guess: probably a heavier ball must be thrown at a larger velocity, if it is to hit the pins at the optimal angle. As I said, this is just a guess. I wonder whether there are statistical data about this. What if some people have set up equipment that finds the velocity of each throw, to see if the differently heavy balls are thrown at different velocities? I think such inquiries, if they exist, will be the best source of information. 


#5
Oct2710, 07:53 PM

P: 42

I personally enjoy bowling from time to time, but i would also think that a heavier ball would need to be thrown at a greater velocity.



#6
Oct2710, 09:36 PM

P: 7

Lets start with energy
kinetic energy is given as Ke=0.5mv^{2} m is the mass and v is the velocity you throw the ball. you can play around with this equation to see how changing the mass velocity and energy affects everything else. if you throw each ball as hard as you can it would have the same energy but as the mass increases the velocity decreases. the ke can be found with the potential energy in the hight of your backswing. Potential energy (Pe) is mgh where m is the ball's mass, g is acceleration of gravity, and h is how high you hold the ball at the back of your throw. most of this Pe is converted to Ke. (energy lost to heating the air with friction, minimal) since the friction between the lane and the ball is very small the final kinetic energy will only be slightly less than initial kinetic energy. as for forces you can the force on the pin by looking at impulse (momentum change). F=ma which is the same as F=m∆v/∆t since a=∆v/∆t now we have F=m∆v/∆t you know the mass and with our Ke equations above you know the velocity at impact using the collision equation v_{p}=(2m_{b}/m_{b}+m_{p})v_{b} the time is simply one over the velocity (1/v_{p}) best if i give an example say you get your 8 kg bowling ball and throw it from a height of 1.5 meters. mgh=0.5mv^{2} (8)(9.8)(1.5)=0.5(8)v^{2} v=5.4 m/s so the 8 kg ball hits the pins at 5.4 m/s. the stationary pin of mass 1.5 kg will now go through an elastic collision v_{p}=(2m_{b}/m_{b}+m_{p})v_{b} v_{p}=(2(8)/(8+1.5))(5.4) v_{p}=9.09 m/s F=m∆v/∆t F=(1.5)(9.09)/(1/9.09) F=123.9 newtons i think... 


#7
Mar2111, 03:07 AM

P: 4

Assumes the transfer of energy from the bowling ball to the 1st pin to be 35% (max allowed by USBC). Assumes the percentage energy transfer to be same regardless of the weight of the ball. Assumes energy loss due to noise/heat is negligible. Assumes the energy lost as the ball traverse the lane to be the same regardless of the weight of the ball.
Conservation of energy: KE_{ball} = 65% * KE_{ball} + KE_{pin}. So KE_{pin} = 35%*KE_{ball} You want the same pin action; eg same KE_{pin}. KE_{pin} = 35% * KE_{16lb ball} = 35% * KE_{14 lb ball} => mass_{16lb}*v_{16}^{2} = mass_{14lb}*v_{14lb}^{2} => v_{14lb}^{2} = (mass_{16lb}/mass_{14lb})*v_{16lb}^{2} => v_{14lb} = sq root(mass_{16lb}/mass_{14lb})*v_{16lb}. Thus you have to throw the 14 lb ball 6.9% faster than the 16 lb ball to get the same pin action. If you throw a 16 lb ball 17 mph (average speed), then you need to throw the 14 lb ball 18.1 mph. For a 15 lb ball, you need 3.2% faster speed; eg 17.6 mph. To get faster ball speed, you can either power it yourself with your arm at the release point or you can increase the height of your backswing. PE = mgh So you need to increase the height by m_{16lb}/m_{14lb}. eg, you need to increase the height of the backswing by 14% for the 14 lb vs 16lb 6.7% backswing height increase for 15lb vs 16lb. 


Register to reply 
Related Discussions  
Physics Related to Bowling  General Physics  2  
Alley Bowling  Pin Physics  Classical Physics  0  
Bowling Physics  General Physics  1  
My new bowling balls  General Physics  8  
Bowling on an incline  Introductory Physics Homework  4 