cubc said:

Can someone that has 2 of those balls weigh both and post the outcome please?

Red Circle - about 8 years old, lots of play: 5.7 oz

Measels 1 - about 2 years old (bought separate): 5.9 oz

Measels 2 - about 1 year old (Aramith Set): 5.9 oz

Aramith Rempe training cue ball: 5.7 oz

Old Aramith set (mid 80s) (gobs of play): 5.6 oz

*all of the old Aramith balls weigh 5.6 oz, go figure*

I have little faith in the "cue balls lose significant mass" theory. They may lose some mass, but I think it would be hard to lose 5%-10% without being noticable. Any such significant loss should be readily apparent - either visually or through play. Losing 10% would take a cue ball from 5.9 oz to 5.3 oz. This would push the interaction way down the inelastic collision scale.

IMO, unless there are chips out of the ball, they should be within a couple percent of new weight through their life cycle. But what do I know...

With a .6 oz difference in weight the cue ball would naturally rebound about 5% of its initial speed -> making the 90 degree rule tricky at best. But with a lighter cue ball, draw would be much improved.

My rudimentary calculations show that a .1 oz difference in ball weight will result in 99.1% "accuracy" of play. That is, if the cue ball weighs 5.8 oz and the OB weighs 5.9 oz, the OB will have 99.1% of its "perfect" forward velocity (head on hit), while the cue ball will have a .9% rebound (again head on).

For .2 oz difference, its 98.2%

For .3 oz difference, its 97.4%

For .4 oz difference, its 96.5%

For .5 oz difference, its 95.6% and

For .6 oz difference, it's 94.6%.

Using these percentages seems to show that 90 degree rule becomes the

89.2 degree rule for a .1 oz lighter cue ball

88.4 degree rule for a .2 oz lighter cue ball

87.7 degree rule for a .3 oz lighter cue ball

86.9 degree rule for a .4 oz lighter cue ball

86.0 degree rule for a .5 oz lighter cue ball

85.1 degree rule for a .6 oz lighter cue ball

Although this is not a true mathematical calculation, it is a rough approximation that should be close enough for government work.

I think I could play at .1 or .2 difference and feel ok. But at about 5.6 oz, I think I could tell enough to need a new cue ball.

-td

[edit: someone with a calculator please fix this math

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