I didn't, but now I have a video. Check it out:
Thank you again for letting me use your excellent illustration.
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Throwing Frozen Object Balls Question
- Thread starter MattPoland
- Start date
I didn't, but now I have a video. Check it out:
Thank you again for letting me use your excellent illustration.
A major point in the video is that the situation where the effect comes into play is very rare because usually you can use cut to help the throw and the throw you can get from the double spin transfer is slight -- much smaller than CIT. You have to work really hard to isolate the effect.
A situation that is similar is when you have three balls in line. Can the third ball be thrown? If so, which way? This situation will also be rare but may happen if the rack is not well broken at full-rack games. At least the demo is easier than for the two-ball situation which requires a precise hit to avoid normal throw.
soyale
Well-known member
i was waiting for an example of using cut and spin to manipulate the direction. Otherwise, great video
Agreed.
I filmed this and almost included it, but it didn’t make the final cut since it is even more rare and since the 3rd ball goes straight using sidespin only (although, I guess this is good for people to know). The transferred spin after two straight, frozen balls is pretty much down to zero.
I filmed a bunch of these but decided to not include them since the results were really not interesting. Straight cut is much more effective than spin to throw the 2nd ball. I guess I could have included examples where you can’t cut it quite enough and use spin for assist; but again, these situations are rare.
Thanks.
You did a great job! Broke it down to its components, structured it as a quiz, revealed the answer, and shown it all in practical on-the-table circumstances.
A couple weeks ago I was telling Kim one of the players on our team was going to hit one of these on the wrong side while he was on the ball before the combo. She didn’t get it. He did and all I could do was laugh some. Neither understood why it went the wrong way. He’s one of them semi-cocky 3’s. Could’ve called a timeout, but it was a great lesson for him.
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Pubo
Active member
I believe OB2 always heads in the same direction as CB spin. Here is the reasoning. With a head-on collision of the CB onto OB1 with sidespin, OB1 receives some friction impulse, say its horizontal (Using MattPoland’s diagram ) component is J. Then using the linear momentum equation, OB1 has horizontal velocity J/m, while (angular momentum equation) it has angular velocity 5/2J/(mr). OB2’s direction is entirely determined by the relative sliding between OB1 and OB2. Now the point on OB1, touching OB2, has translational velocity J/m, and some velocity provided by the rotation, which is 5/2 J/(mr) * r. Since OB1’s induced spin and throw has opposite direction (when viewed from the point of view of the contact point on the surface), the combined velocity is (5/2-1)J/m = 3/2 J/m, always in the sidespin (induced from CB spin) direction of CB. I feel like this is the correct analysis. So OB2 always throws in the same direction as CB spin, it’s just a matter of how much CB2 throws in the direction of CB spin.
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