Author Topic: Question about static weights (Read 1926 times)

tjj300 · « **on:** September 01, 2005, 01:54:36 AM »

Are static weights measured as though the imbalance is on the surface of the ball? By that, I mean if you took a one ounce weight and placed it on the outside rim of an empty ball scale, it should measure as one ounce static weight. Am I correct on this?

Goof1073 · « **Reply #1 on:** September 01, 2005, 11:37:55 AM »

You have to think of static weights as the comparison between two hemispheres of a round object...split the ball in half utilizing:

- Your grip centerline...you can compare the halves and get side weight.
- Your grip midline...you can compare the halves and get finger / thumb weight.
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-Chris: DJ's Pro Shop : Auburn, MA

tjj300 · « **Reply #2 on:** September 01, 2005, 11:43:40 AM »

I understand that, but isn't there a lever arm effect in there somewhere? I mean 1 oz. one inch away from center will read differently than 1 oz four inches from center, or does that have no effect?

Edited on 9/1/2005 11:35 AM

Mashed · « **Reply #3 on:** September 01, 2005, 01:34:16 PM »

You are correct that there is some type of lever arm at work. But thing to consider is that the 1 oz of weight is not a point mass, more or less it is distributed evenly through out the side of the ball. Further more the 1oz of weight compared to the total weight of the ball is small. Because of this, the amount of difference between being more towards the center of the ball or more towards the cover is negligble.

tjj300 · « **Reply #4 on:** September 01, 2005, 05:27:54 PM »

Thanks, what I'm thinking about is scale related. Seems to me if you are not interested in total weight of the ball and only interested in the statics, then you should be able to make a simple pivot scale that just compares one side of the ball to the other.

Billy Ray · « **Reply #5 on:** September 01, 2005, 06:31:42 PM »

quote:
Thanks, what I'm thinking about is scale related. Seems to me if you are not interested in total weight of the ball and only interested in the statics, then you should be able to make a simple pivot scale that just compares one side of the ball to the other.

That is what the new jayhawk scale (like the ones the USBC uses) does I believe. Pretty expensive.

--------------------
Billy Ray
www.raysproshop.com
"Let Us Help You Become More Competitive"

tjj300 · « **Reply #6 on:** September 02, 2005, 09:52:30 AM »

Didn't they used to use a go/no go type mechanical scale for checking in balls at the Nationals up to Knoxville? Some type of a gimbal-like scale, anybody know what that was?

T-GOD · « **Reply #7 on:** September 03, 2005, 10:29:07 AM »

tj,

quote:
Are static weights measured as though the imbalance is on the surface of the ball?

Yes..!! Consider it to be a point on the surface of the ball.

The CG is marked as a point on the ball. The ending CG, after holes are placed int he ball, deals with the static weights and can be found as a point on the ball.

Here's a link on how to find it and/or how to mark it... http://lane1bowling.com/tech/cg.html This shows you how to find it in an undrilled ball.

For a ball already drilled, draw the first line perpendicular thru your axis point, instead of thru the label, and follow the rest of the steps. =:^D

tjj300 · « **Reply #8 on:** September 03, 2005, 11:38:37 AM »

Thanks,T-GOD. Still interested in that gimbal scale thing that they used to use at the ABC's. Anybody?

Author Topic: Question about static weights (Read 1926 times)

tjj300

Question about static weights

Goof1073

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tjj300

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Mashed

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tjj300

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Billy Ray

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tjj300

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T-GOD

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tjj300

Re: Question about static weights