- #1
newguy2000
- 3
- 0
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 that's as far as I've 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.
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 that's as far as I've 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.