Post by: Magic Carpet on April 12, 2008, 02:19:42 AM
If you have two identical balls with one exception, one has the lowest rg allowed and one has the highest rg allowed. Both balls are drilled with the same layout and the pin is 3 3/8 from the PAP. Both balls have enough differential to flare at least 4 inches.
If a bowler throws the high rg ball with 350 revs off his hand and then throws the low rg ball the exact same way, will his rev rate increase to even 351 revs with the low rg ball?
Ron Clifton
Looking for perspective
Post by: JohnP on April 12, 2008, 11:02:45 AM
Both balls will come off the hand at the same RPM. The low RG ball will stop skidding earlier, so the RPM's will increase earlier as it goes into its roll. The higher RG ball will skid further and "rev up" later. I will guess that the lower RG ball will hit the pins at a higher rev rate. -- JohnP
Post by: Juggernaut on April 12, 2008, 12:55:33 PM
The high RG reading and the low RG reading of any given ball are predicated on where the pin ends up after drilling. If both balls were drilled with the pin 3 3/8 inches from the pap, both balls will have the same RG.
The only way that comes to mind to get a higher AND lower reading from the same ball is to move the pin either towards the track or towards the pap.
Now, you could take two balls with identical COVERSTOCKS in identical preps and install different weightblocks in them to achieve this, but not with totally exact duplicate balls with duplicate weightblocks.
Now, if in theory you COULD do this experiment, their would have to be some type of formula to allow for the calculation of force exerted upon the ball divided by the force needed to impart spin on the ball. It would probably have to include a variable for allowance of the fact that you would have to allow for the minute difference between spin properties of a high RG object vs a low RG object, ie.. how much difference it takes to get them started in the first place and how much force is left after you get them started and how much total force is available in the beginning.
I believe JohnP has a point, except that, at the point that the skid stops and the roll starts, doesn't then the RPM's become a function strictly of the balls initial velocity? I mean, if the sliding is done, and the ball is now rolling, doesn't it roll at the speed at which it is traveling down the lane?
If it is rolling, it can't be skidding, but, if it is turning faster than it is moving, it has to be skidding, even if only slightly, doesn't it?
If so, then it doesn't necessarily have to pick up RPM's after it picks up its roll, it could also LOSE them, couldn't it, depending on the initial trajectory speed ( velocity)?
--------------------
What did you expect, something PROFOUND?
Good transactions list in my profile
My Bowl.com member page (http://"http://members.bowl.com/SearchUSBC/ViewMember.aspx?prefix=2243&suffix=4831")
Edited on 4/12/2008 12:59 PM
Post by: laufaye on April 12, 2008, 06:28:43 PM
--------------------
Laufaye
Post by: laufaye on April 12, 2008, 06:38:26 PM
--------------------
Laufaye
Post by: laufaye on April 14, 2008, 09:55:49 PM
Here is my take, lets say the lenght of the pattern not too long, meaning both balls balls are able to go through the ball motion, and complete the axis migration. In this case, both balls should be able to accelerateover 350 RPM and then slow down in the roll out phase. The only difference I see is the lower RG ball will get in to a roll earlier, but acceleration ability should be the same.
Whats your take Ron?
--------------------
Laufaye
Post by: shelley on April 14, 2008, 10:20:18 PM
For the same energy (work), because the RG changes, the force you are generating changes, and so the speed at which you can move your fingers, and hence the ball, changes; you are acting on the effective mass (what the RG measures) of the ball at different distances (2.43" vs 2.8"), which will change the torque you are applying, which changes the amount of force you are generating, which changes the speed at which the same energy will move your fingers.
Is it linear? A 15% increase in the RG gives a 15% reduction in revs? Probably not. Do your muscles respond to the additional workload demanded of them in a linear fashion to exactly counteract that? Doubt it.
The basic point is that there are entirely too many forces and systems interacting within and around your body to say for certain. Considering a simple physical system that solely includes the ball and some "ideal" fingers, it requires more energy to put 350 revs on a high-RG ball than a low-RG ball.
SH
Post by: Magic Carpet on April 14, 2008, 11:48:46 PM
laufaye
you are right I could have said it better. AFTER DRILLING one ball has a higher rg than the other.
I could have also said that both balls are drilled in their highest rg position, but one is 2.43 and the other is 2.8 for the sake of the first question.
JohnP and Juggernaut
The ball has just left the hand not really headed vey far down the lane yet.
shelley
You are getting close but not quite there.
You said "It will require more energy, more torque to apply 350 revs to the high-RG ball compared to the low-RG ball"
We are applying the same torque to each ball, and that torque from the hand produced a rev rate of 350 RPM off the hand with the high rg ball. Since rg is really a measurement of “resistance†to torque the question was; is there enough difference in rg between a 2.43 ball and a 2.8 ball for the bowler to achieve 1 additional rev off the hand.
Ron Clifton
Looking for perspective
Post by: laufaye on April 15, 2008, 01:09:57 AM
quote:
Of course if you are asking if both balls start at 350rpm, will the low rg one increase its rotational speed throughout the ball path given the same rg differential and drilling AFTER it has been released, the answer is no. An object can not CREATE energy on its own.
When the ball is not skidding, in roll phase before roll out, the ball actually accelerate, my answer to that will be yes.
--------------------
Laufaye
Post by: Ishmael on April 15, 2008, 08:33:11 AM
quote:
If a bowler throws the high rg ball with 350 revs off his hand and then throws the low rg ball the exact same way, will his rev rate increase to even 351 revs with the low rg ball?
Yes, the rev rate will increase with the lower rg ball.
Post by: KDawg77 on April 15, 2008, 08:37:49 AM
--------------------
Ken
Videos at http://www.putfile.come/k-dawg77
http://www.myspace.com/lefthandedhammerpride
http://khlthe2nd.bowlspace.com
http://members.bowl.com/FindAMember/memberView.aspx?mp=418&ms=2006&s=2006-2007
Post by: shelley on April 15, 2008, 08:54:50 AM
quote:
I don't see how rev rate changes based on the data presented. Rev rate is all about the bowler's inputs to the ball. You can rev a plastic ball at the same rate of a particle ball. The difference is when the ball will pick up its roll versus skiddung.
The issue is the force, the energy required to create those 350 revs. When the ball has a low RG, less energy is required to increase its rate of revolution from 0rpm to 350rpm. Less required energy means less work at the release, less effort from the bowler. If they put the same amount of energy into the release as with a higher RG ball, they'll create more revolutions and the low-RG ball will have a higher rev rate than the high-RG ball.
Does it matter? In both cases, you've imparted the same amount of energy into both balls. The high-RG ball stores that energy with fewer revs than the low-RG ball.
I hate the ice skater analogy, but it's applicable. The classic version says that the girl spins faster as she pulls her arms in closer (lowering her RG). Conservation of energy means she's storing her energy in more revs but there's less energy in each rev. The skater with her arms outstretched is storing more energy in fewer revs.
The coverstock makes no difference. Particle, pearl, solid, polyester, urethane, who cares? Can you throw any of those with the same amount of revs? Of course. Will the high-RG plastic ball have more energy than the low-RG particle ball, both spinning at the same rev rate? Yes.
SH
Post by: batbowler on April 15, 2008, 09:07:35 AM
P.S. That's why I'll use a lower rg ball on wetter lanes and higher rg on drier lanes.
--------------------
"Train a child up in the way he should go and when he is old he will not turn from it."
Post by: KDawg77 on April 15, 2008, 09:42:30 AM
--------------------
Ken
Videos at http://www.putfile.come/k-dawg77
http://www.myspace.com/lefthandedhammerpride
http://khlthe2nd.bowlspace.com
http://members.bowl.com/FindAMember/memberView.aspx?mp=418&ms=2006&s=2006-2007
Post by: n00dlejester on April 15, 2008, 10:00:30 AM
Post by: Magic Carpet on April 15, 2008, 10:52:06 AM
I didn’t say the bowler threw both balls at 350 RPM, I said he applied enough force to rotate the high rg ball to 350 RPM off his hand. If he applied the same force to the low rg ball will that same amount of force produce at least 351 RPM.
Too many of you are thinking about the ball way down the lane. I am just asking about off the hand, just a few inches not feet.
In the case of the figure skater mentioned above, if it was a bowler that spun the figure skater and she had her arms out stretched (high rg position) and she turned 350 RPMs would she not spin many more revs if on the next spin she pulled her arms in (low rg position) and the bowler applied the same amount of force?
Ron Clifton
Just looking for some perspective
Post by: n00dlejester on April 15, 2008, 11:29:33 AM
Post by: nospareball on April 15, 2008, 02:03:18 PM
quote:
n00dlejester
I didn’t say the bowler threw both balls at 350 RPM, I said he applied enough force to rotate the high rg ball to 350 RPM off his hand. If he applied the same force to the low rg ball will that same amount of force produce at least 351 RPM.
Too many of you are thinking about the ball way down the lane. I am just asking about off the hand, just a few inches not feet.
In the case of the figure skater mentioned above, if it was a bowler that spun the figure skater and she had her arms out stretched (high rg position) and she turned 350 RPMs would she not spin many more revs if on the next spin she pulled her arms in (low rg position) and the bowler applied the same amount of force?
Ron Clifton
Just looking for some perspective
Shelley is correct, with the same amount of force applied the low rg ball will have more revolutions. The figure skater analogy is a good one. The skater applies force to enter a spin with her arms extended, and spins at lets say 50rpm. When she brings her arms in (lowering her RG) she doubles her spin rate to 100rpm. All that bringing her arms in did was lower her body's RG, no extra force was applied to the spin! So the force stays the same in the equation for both RG values of her body.
If you apply this to a bowling ball with the RG values having a difference of .4" the results will be less dramatic, but follow the same pattern. So yes, the low RG ball will have more revolutions off the hand. 1 more? I don't know, you'd have to do the math.
--------------------
-Clint
Post by: shelley on April 15, 2008, 02:42:27 PM
quote:quote:
n00dlejester
I didn’t say the bowler threw both balls at 350 RPM, I said he applied enough force to rotate the high rg ball to 350 RPM off his hand. If he applied the same force to the low rg ball will that same amount of force produce at least 351 RPM.
Shelley is correct, with the same amount of force applied the low rg ball will have more revolutions.
Boo-yah! Ding-ding-ding-ding-ding! Go me! It's my birthday!
SH
Post by: n00dlejester on April 15, 2008, 03:03:48 PM
Post by: 1MechEng on April 15, 2008, 03:22:13 PM
Rg is a measure of the approximate center of mass of an object from a rotational axis. If the mass of the balls are the same, but the Rg is different, then the rotational moment of inertia (I = k * M * Rg^2) will be different. Lower Rg means lower moment of inertia.
Torque is a time-derivative of the angular momentum of an object. Torque is equal to the moment of inertia (I) times the angular velocity (w).
If the same torque is applied, the lower Rg will produce a lower moment of inertia. Thus, the angular velocity will be higher.
QED
--------------------
======================
Dan
======================
Engineering * Bowling = a fun and practical application of rotational kinematics.
Bowling Nerd Herd (TM) Member
Post by: livespive on April 15, 2008, 03:32:30 PM
quote:
Here's the physics:
Rg is a measure of the approximate center of mass of an object from a rotational axis. If the mass of the balls are the same, but the Rg is different, then the rotational moment of inertia (I = k * M * Rg^2) will be different. Lower Rg means lower moment of inertia.
Torque is a time-derivative of the angular momentum of an object. Torque is equal to the moment of inertia (I) times the angular velocity (w).
If the same torque is applied, the lower Rg will produce a lower moment of inertia. Thus, the angular velocity will be higher.
QED
--------------------
======================
Dan
======================
Engineering * Bowling = a fun and practical application of rotational kinematics.
Bowling Nerd Herd (TM) Member
I agree, I just think that the question should have been stated in terms of the of the release being the same, not the RPM. Stating that the RPM's were the same off of the hand has forced the item that you wanted to be a variable, to be a constant.
--------------------
Eric T. Spivey, P.E.
Visionary Test Staff Member
http://www.visionarybowling.com
http://www.maysbowlingandbilliards.com
Post by: dR3w on April 15, 2008, 03:39:37 PM
(I corrected the units with the help of a fellow poster)
Do balls with different RGs have different initial ball speeds?
I read on a several web sites that the RG of a round object is equal to the square root of the Moment of inertia divided by weight.
If I understand this correctly then RG = (IM/BallWeight)^0.5. where the term ^0.5 means the square root for you non-engineering folk.
Thus: IM^0.5=RG*BallWeight^0.5 or IM = RG^2 * Ball Weight
In typical US units, IM or Moment of Inertia has the units lbm-ft^2 (apparently these are in lbm-in^2.)
It is strange to think that Rg has units of feet (again in inches not feet), If that is the case, then I would think that the manufacturers would have to specify the units of RG, because the number would be different for US units and SI units.
An Rg of 2.6 in would be equal to 0.06604 meters. Since ball manufacturers in the US give us ball weights in pounds, then I will assume for this exercise that RG is measured in US units and thus lbm-in^2
So lets say you have a 15 lb ball with an RG of 2.6, thus the IM would = 2.6^2 * 15 or 101.4 lbm-in^2 . The equivalent Moment of Inertia for a ball with an RG of 2.45 would be 90.0375.
For a bowler who puts 300 RPM on a typical throw, let us figure out the Torque that he is applying.
Since Torque = IM * alpha then alpha = Torque/IM. Unfortunately alpha is radial acceleration, and not radial speed or RPMs.
So what is the difference between acceleration and speed? Well lets just assume that in our experiment that the bowler applies a torque, and accelerates the ball from 0 RPM to 300 RPM (as it leaves his fingers) in one tenth of a second. Thus alpha = ( Initial Speed - Final Speed) / time or (0-300)/0.1 = 3000 rpm/second or 50 revs/second^2 ... x 2 pi = 314 rad/s^2.
Thus Torque for the 2.6 Rg ball would be ... (101.4 lbm-in^2) * (314 rad/s^2) = 31,730.08 lbm-in^2- rad/ sec^2
Since US units are really screwy, let me tell you that typically torque is force through a distance or lbf-ft. 1 lbf = 32.2 lbm-ft/sec^2 So torque is 1 lbf-ft = 32.2 lbm-ft^2/sec^2
After all conversions, the torque would be 6.87 lbf-ft
Since the torque that we apply on the ball is the same, for a good bowler with consistent release, regardless of the ball moment of inertia, let us see if we get a different speed for the lower RG ball.
With T = 6.87 and IM = 90.0375 then alpha is 56.308 revs/s^2 = 353.79 rad/s^2.
With the same 0.1 time of force application, then the final velocity leaving the bowlers hand would be ... 337 RPM.
So an RG diff of two balls with 2.45 to 2.6 would equate to revs of 337 to 300. Thus a 6.1% RG diff is equivalent to 12.3% increase in RPMs
Now we both know that a bowling ball is a little more complicated that this. The RG is not fixed, and depending on where you put the pin, and how you drill an extra hole, you do vary the moment of inertia. But in general, I feel fairly confident about these calculations.
One thing I wasn't sure of at first was using 0.1 seconds as the period of time to accelerate the ball, but I feel that this is ok. If I had chose 0.2 or 0.5 or any other number it really wouldn't matter in the calculation, as long as a I used the same number for both balls. The ball is accelerated only while the force is being applied.
I guess I could be talked into thinking that the period of time that it takes to accelerate a lower RG ball would be quicker, and thus would leave the persons hand sooner, but I don't think that this is the case either.
Small reference below:
Q: Please could you define for me the Radius of Gyration in the context of a Flywheel? Thankyou.
A:
The radius of gyration is defined as (I/M)^1/2, or the square root of the moment of inertia divided by mass. The moment of inertia is an important thing to know when solving problems that have to do with how things rotate. The equation for moment of inertia is different depending on the shape of the object, but for a flywheel (basically a solid disk), it's I=1/2MR^2, where M is the mass and R is the radius of the flywheel. Putting this into the equation from before, we find that the radius of gyration for a flywheel is R*(1/2)^1/2, or 0.707*R.
The units of moment of inertia are mass*distance^2, so if you divide by mass and take the square root, you get something with units of just distance (a.k.a. the radius of gyration). Conceptually, the radius of gyration is the distance that, if the entire mass of the object were all packed together at only that radius, would give you the same moment of inertia. That is, if you were to take the entire mass of the disk-shaped flywheel with some radius and pack it into a narrow donut whose radius is the flywheel's radius of gyration, they'd both have the same moment of inertia. With the same moment of inertia, they will behave very similarly when you spin them.
--------------------
dR3w
"I did nothing. I did absolutely nothing, and it was everything I thought it could be. "
Edited on 4/16/2008 10:53 AM
Post by: batbowler on April 15, 2008, 03:43:25 PM
--------------------
"Train a child up in the way he should go and when he is old he will not turn from it."
Post by: Ahhbach on April 15, 2008, 03:47:37 PM
You all have 'Blinded me with science'
Ahh
--------------------
'I'm Partial to the Fugue' - Radar O'Reilly
Post by: livespive on April 15, 2008, 03:59:28 PM
--------------------
Eric T. Spivey, P.E.
Visionary Test Staff Member
http://www.visionarybowling.com
http://www.maysbowlingandbilliards.com
Post by: chitown on April 15, 2008, 04:25:03 PM
If ball makers get to the point where they can generate additional revs just from tweaking the core some how they will make a killing in sales! lol
I can picture 10 years from now the ball makers will be selling bowling balls that produce additional 100 rpm on top of what the bowler can create on his own. lol
Post by: Ishmael on April 16, 2008, 08:33:58 AM
quote:
I would have to say there wouldn't be any difference. I think a low RG ball revs up quicker but not more than a high RG ball.
Say what? Your gonna have to explain that one...
Simple test. Take your spare ball (usually 3 piece = high rg) and throw it and watch the revs. Then take your strike ball (lower rg) and throw it and watch the revs. The strike ball is going to have a higher rev rate than the spare ball if you use the same release. This is really basic physics. Nothing to argue about.
Post by: nospareball on April 16, 2008, 10:55:29 AM
quote:
I would have to say there wouldn't be any difference. I think a low RG ball revs up quicker but not more than a high RG ball.
If ball makers get to the point where they can generate additional revs just from tweaking the core some how they will make a killing in sales! lol
I can picture 10 years from now the ball makers will be selling bowling balls that produce additional 100 rpm on top of what the bowler can create on his own. lol
The revving up of a ball that you see down the lane is caused by the migration of the spin axis from a higher RG value to a lower one. The RG differential also has an impact on how quickly a ball will rev up, as does how you drill it. The USBC did a study on axis migration and I believe it was found that in most cases the axis will migrate to a lower RG as the ball flares. The affect is the same as the figure skater bringing her arms in during a spin.
--------------------
-Clint
Post by: shelley on April 16, 2008, 02:59:15 PM
quote:
The revving up of a ball that you see down the lane is caused by the migration of the spin axis from a higher RG value to a lower one.
Not to mention that has absolutely nothing to do with the original topic.
SH
Post by: JPratt on April 17, 2008, 03:38:27 PM
I have a pretty high rev rate so it was easy to pick up the revolutions using only the first 15 feet on the lane. I alternated balls every shot playing the same angle and feel through the front. With the lower rg ball, 4 of the 5 balls came in at 4.5 revolutions in the first 15 feet with 1 ball at 4.75. The higher rg ball came in at 4 revolutions every time. Based off of my speed the average RPMs for each ball would be : Low RG ball - 471 High RG ball - 428.