Exactly what I was thinking. Matching your ball's rg to your game is also important. If the rg is too high going into the break point, then the ball could over react at the break point if the back ends are very snappy. I had several balls over the years that my window for using them was very small because the rg was too high and the ball would over react on the back end, if I dulled the ball it would burn too quick. A friend of mine would bowl with the ball and have all the room in the world, but for me it would go from too snappy to burning out too soon in a few frames. My friend has a higher rev rate and less side rotation than I had which is why it worked for him.

Pre-note: high RG and low Rg refer to pin drilled 6 3/4" from PAP and pin on PAP, respectively


Ok, I've come up with some questions:

Working backwards, you state that the ball speed is reduced at the same rate for both balls. Does this mean that the TKE would also reduce at the same rate? This would case the RKEs to increase at the same rate, which doesn't seem to work - the higher RG ball should resist a change in rotational energy more than the lower RG ball? Does the RG also factor into the translational energy? Or does the higher RG counteract the slower rev increase, keeping the equation balanced, so that the same increase in RKE for the 2 balls translates to 2 different increases in revolution speed?

Moving to the foul line, given the exact same delivery, the total energy delivered to the balls would be the same, and the speed would be the same (right? still the same translational force applied to the same mass). The high RG ball would have a slightly lower rev rate. Would the TKE and RKE be the same for each ball?

And some deductions:
From this:
I am getting that given the same release and ball, a high RG drilling will have a longer hook phase than a low RG drilling. So (in general, not just stable drillings) if your ball is rolling out before it hits the pocket, going to a higher RG drilling of the same ball could make the hook phase last longer, causing the ball not to roll out. And conversely, if the ball was hooking/skidding through the pins, you could change to a low rg drilling of the same ball to get a quicker hook phase, which might get the ball to enter its roll at the headpin.

That would be the case. As you probably know, the equation for TKE is is 1/2mv^2. The equation for RKE is 1/2Iw^2. I is the moment of inertia, which factors into the equation for RG. RG=sqrt(I/m), so a bigger RG means a bigger moment of inertia. w is the angular speed of the ball (essentially rev rate, but measured in different units.) Both drillings lose the same amount of speed and TKE from the equation 1/2mv^2, so both balls must make up the same amount of RKE from the equation 1/2Iw^2. Since the moment of inertia for the high RG drilling is bigger, it needs a smaller increase in rev rate to make up for the amount of TKE that is lost.

Mark

Awesome.

You should really write a book about the physics that relate to bowling, and how they do. Or a small paper, or something. I may try to organize some of this and put it up on CMGBB if you don't mind.

This is getting way off the original topic, but how much energy do you think is lost as heat created by the ball's response to friction? Is it significant, and would the difference between a ball at 500 and a ball at 4000 be significant?

I'm not sure about writing a book, but if you want to put it on your site, I would not mind.

I said that I would crunch some numbers, this all applies to the total kinetic energy given at the point of release. I assumed the RG's that you gave earlier, 2.46 and 2.51 and a 16 lb ball. I also assumed a ball speed of 18 mph, along with both balls being released with the same rev rate, first 200 rpm, then 400 rpm, and finally 600 rpm. Since the both have the same speed, the will both have the same TKE, but since one ball has a higher RG, it will have a greater RKE and a greater total energy TE=TKE+RKE. How much greater? Well, at 200 rpm, the 2.46 RG ball has 6.05 Joules (J) of RKE, but the 2.51 RG ball has 6.44 J of RKE, so the increase in RKE would be 6.06%. Sounds good at first, but the TKE of both balls would be 236 J. So, for the 2.46 RG ball, TE=242.05 J and for the 2.51 RG ball, TE=242.44 J for an increase of 0.161%.

At 400 rpm, the 2.46 RG ball has 24.30 J of RKE, but the 2.51 RG ball has 25.88 J of RKE, so the increase in RKE would be 6.11%. Note that since the rev rate is squared in the equation for RKE, doubling the rev rate will make the RKE 4 times greater, so the RKE will effect the TE more. So, for the 2.46 RG ball, TE=260.30 J and for the 2.51 RG ball, TE=261.88 J for an increase of 0.603%, still not a big increase.

At 600 rpm, the 2.46 RG ball has 54.67 J of RKE, but the 2.51 RG ball has 58.23 J of RKE, so the increase in RKE would be 6.11%. Note that since the rev rate is squared in the equation for RKE, tripling the rev rate will make the RKE 9 times greater, so the RKE will effect the TE even more. So, for the 2.46 RG ball, TE=290.67 J and for the 2.51 RG ball, TE=294.23 J for an increase of 1.21%, a bigger increase again, but not anything to get really excited about.

So, if both balls have the same rev rate, the difference in TE for high and low RG drilling is very small. Plus, we know that with the same amount of lift, high RG ball will have less revs than the low RG ball, so this increase will be negated by the lesser rev rate. So, it's really not about total energy, it's about how you want the ball to react to match up to lanes and your release.

I'll address your other question about friction a little later, gotta give a quiz now.

Mark

Words, words, words

Seriously though, this is the stuff I love about this site. You get people from all walks of life who have different knowledge about bowling. Some people (like me) who don't know all of the dynamics of the game but know how to get the results that I want. Others know the science behind the game (as listed above) to get the desired result and apply that on the lanes. Great stuff.

A couple of relevant articles to this discussion:

http://www.bowl.com/articleView.aspx?i=13372&f=21
http://www.bowl.com/Downloads/pdf/Specs/08ballmotionstudy.pdf

I've been thinking about this one a little bit, it's really a tough question to answer due to the number of variables involved here. What I can say is that a 16 lb ball going at 18 mph has a TKE of 236 J, where at 15 mph, its TKE is 164 J. This is a reduction of 30.5%, so that would be the absolute maximum that could be lost to heat, at least in that scenario. In reality, it will be less because the increased rev rate and increased RKE is some of the reason that the velocity and TKE becomes smaller, the rest of it is lost to heat. Without knowing the actual increase in revs and RKE, there is no way to say for sure how much energy is lost as heat. Now, with a ball tracking system, both the rev rate and velocity could be measured at release and at the pins, making it very easy to find out.

I'd think the difference between the two would be significant. It depends on the coefficient of friction, I know that the coefficient will be greater for 500 than 4000, but again without measurements, an exact number would be tough.

Mark

According to the specs, there is no minimum COF. The max is 0.320, for a 13lb ball, at medium humidity and a surface grit between 300 and 3000. Unit is not specified - does the number make sense in the default unit? Some specs are listed in SI, and others in metric...

Appendix E lists the test method: http://www.bowl.com/Downloads/pdf/Specs/EquipManual/2008Appendixes.pdf



...

*cg does homework*

Wikipedia says there is no unit. Judging by the example list they give, .32 is actually a pretty low coefficient of friction.

Wikipedia is right, the coefficient of friction is the ratio of the friction force opposing the motion to the normal force that is perpendicular to the opposing force, the normal force is the force that the lane applies upwards on the ball. Since you are diving a force by a force, the units cancel out, which you probably read on Wikipedia. On a level surface, the force of gravity (weight) pulling down is equal to the normal force.

Once you know the coefficient, which only depends on the surfaces in contact with each other, finding the friction force is easy. For the minimum ball mass of 13 lb listed in the ball specs, that would be (13 lb)*(0.320)=4.16 lb. So, the ball experiences a frictional force that is almost one third of its weight, which is a decent amount. Also, the friction force will increase as the weight of the ball increases from 13 lb to 16 lb.

Important to note here that they tested out sliding friction because the ball was not allowed to roll. When the ball has rolled out, it's no longer sliding and the coefficient of rolling friction takes over, which is much less than the coefficient for sliding friction. That's why a ball that has rolled out will deflect more, it's not driving anymore because the friction force is greatly reduced.

Mark