Top weight and the rg of the ball?
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- JBelschner
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Top weight and the rg of the ball?
Does the top weight of the ball change rg of the ball? The numbers given for the rg and differential, do they apply to the ball or to the core? If the ball has a defined rg, doesn't that change with a higher top weight? A higher top weight would mean the core is sitting closer to the pin thus making the ball more cover heavy. More cover heavy means that the ball should have a higher top weight. I read the ball motion study and know that weight is meaningless; however, how much change is created by having a higher top weight?
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Re: Top weight and the rg of the ball?
I believe the RG is a function of the density of the core. While I'm sure the positioning of the core DOES affect RG, the fact that the offsets are in the vicinity of mm's (1mm = 1/25" - ish), any change in RG is likely to be miniscule at best. Certainly in the realm of accepted margin for error.
Am I missing something of did you say the same thing, two different ways? I don't get it?JBelschner wrote:A higher top weight would mean the core is sitting closer to the pin thus making the ball more cover heavy. More cover heavy means that the ball should have a higher top weight.
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Re: Top weight and the rg of the ball?
I think you guys are talking two different things. Density and proximity. I doubt if they have to be related or equal.
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Re: Top weight and the rg of the ball?
JBelschner wrote:Does the top weight of the ball change rg of the ball? The numbers given for the rg and differential, do they apply to the ball or to the core? If the ball has a defined rg, doesn't that change with a higher top weight? A higher top weight would mean the core is sitting closer to the pin thus making the ball more cover heavy. More cover heavy means that the ball should have a higher top weight. I read the ball motion study and know that weight is meaningless; however, how much change is created by having a higher top weight?
RGs are measured for the WHOLE ball. Cores don't roll well. They're not round! TW does NOT matter to RG. The RG is measured on a chosen axis. The axis goes ENTIRELY through the ball. If it is moved along an axis, what becomes more on one side becomes less on the other side. The whole does not change, therefore, the RG does NOT change.
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- JBelschner
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Re: Top weight and the rg of the ball?
I think I worded that question wrong, but that is the answer I was looking for. I had a little too much fun after bowling last night. Post from last night are a little fuzzy.
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Re: Top weight and the rg of the ball?
Understandable!JBelschner wrote:I think I worded that question wrong, but that is the answer I was looking for. I had a little too much fun after bowling last night. Post from last night are a little fuzzy.
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Re: Top weight and the rg of the ball?
So if you have a 0 top weight ball and a 3 oz top weight ball and lay them both out with a 5 " pin to pap with the same VAL and same Drill angles, they will have the same RG ???Mo Pinel wrote:
RGs are measured for the WHOLE ball. Cores don't roll well. They're not round! TW does NOT matter to RG. The RG is measured on a chosen axis. The axis goes ENTIRELY through the ball. If it is moved along an axis, what becomes more on one side becomes less on the other side. The whole does not change, therefore, the RG does NOT change.
Let's assume that the balls are legal and the Cg is in the same position on both balls.
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Re: Top weight and the rg of the ball?
Yes, within the standard deviation of .004".J_w73 wrote: So if you have a 0 top weight ball and a 3 oz top weight ball and lay them both out with a 5 " pin to pap with the same VAL and same Drill angles, they will have the same RG ???
Let's assume that the balls are legal and the Cg is in the same position on both balls.
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