# Weighing a bowling ball project

1. Jul 8, 2014

### newguy2000

In bowling, the ball will have a heavy spot marked as the center of gravity (cg). This will have an additional 1oz to 5 oz of weight to compensate for when the fingers are drilled into the ball.

In bowling a dodo scale is used to measure the difference in static weight of the bowling ball.

This link provides some diagrams and better wording

http://www.jayhawkbowling.com/Pro_s_Corner/Balancing___Weig/balance.html

My project is using two digital scales to measure the difference in weight between the top half of the ball compared to the bottom half.

So far I have two digital scales. A board with the center marked to hold the ball inplace. I have feet to put under the board that will touch each scale. After placing scales side by side with the ball directly centered over the two scales on the board I placed the feet 5" from the center of the ball on either side. Everything is centered and balanced. The problem is where the feet are located dictates the amount of force applied to the scale and thus affects the weight shown.

Besides standard trial and error of moving the feet to get the correct distance I was curious how to do this correctly.

I am guessing the formula for torque is involved but thats as far as ive gotten. T= L x F

Further the way this works is I have the heavy spot of the ball facing one scale. Zero out the scales and rotate the heavy spot 180 degrees to the other scale to get the difference marked on both scales.

Thanks for the help.

2. Jul 9, 2014

### Baluncore

Welcome to PF.

Consider a short board with a central hole to locate the ball, about half the ball diameter.
Put a pair of pointed feet fixed on one side resting on the first scales A.
Fix a flap hinge that can remain close to 90° on the other side resting on the second scales B.
Set it up, then zero both scales. Then seat the ball in the hole at the centre of the board.
With rotation check the total of the two scales remains constant, giving the weight of the ball.

As the ball is rotated, the eccentric weight will be transferred from one side to the other. Model the ball as being a point mass. The readings on the scales can give you the position of the point mass relative to the baseline. If the base line is length s and scale readings are A and B then the position of the balls centre of mass will be; s * B / (A+B) along the baseline from scale A.

3. Jul 11, 2014

### newguy2000

Need a little help visually on the hinge part of this design. Heres a sketch of what I have put together so far.

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4. Jul 11, 2014

### Baluncore

Redraw attached.

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5. Jul 11, 2014

### newguy2000

Got it. Just curious what is the point of the hinge?

Thank you

6. Jul 11, 2014

### Baluncore

It is a knife edge. It stops the sideways forces on the scales while maintaining the distance between the legs.
It might not be necessary if the top of the scales were free to slide sideways slightly.