Skip to content

Is there a transmission rotation between the balls? (HS video)

by DBK on Май 28th, 2011

Yes, it is transmitted. It’s a proven fact. In the original sources, Ron Shepard, Robert Byrne, etc. such transfer is evaluated as “slight, 5 or maximum 10% of the amount rotation”. However this video raises doubts about this assessment.

If we calculate at what angle turn the stopped cue ball when the Object ball turn on one turnover, it turns out no less than 25, even more than 30%. A Riddle? 🙂

  1. mac permalink

    A ball cannot transmit more than 50% of its spin. Dr Dave haz a website that explains the math.
    Alltho, a ball with zero spin, hitting a ball say halfball, can make the objekt ball spin (collizion induced spin), in which case the transmitted spin % iz i suppoze 1000000000000000plus%, ie infinite%.

  2. Thanks. Yes, there are such data from Dr. Dave. Not even 50%. But 5 / 14 x 100 = 35.7%, from Dr. Dave site.

  3. mac rynkiewicz permalink

    The upper limit of 50% iz i am sure correct.
    A ball with a small amount of spin and zero topspin (zero rolling, ie 100% skidding) hitting a 2nd ball fullball, will allways giv half of its spin to the 2nd ball (if balltoball frikition iz sufficient).
    And in no case can it giv more than 50% of its sidespin. Hencely, my upper limit of 50% iz correct. In fakt i would say that a skidding ball hitting fullball will transmit 50% of its sidespin 100% of the time, if friktion iz sufficient 100% of the time.

  4. mac rynkiewicz permalink

    I said hitting fullball can transmit 50% of its sidespin. I shood hav said up to approx 5deg off fullball.
    A cueball with lefthand spin (and with pure stun) hitting a redball just right of fullball will transmit 50% of its spin to the red — ie it loozes 50% of its own spin. This kan/will happen if…….
    (1) if there iznt too much initial spin (ball’to’ball slippage must stop during impakt),
    (2) if ball’to’ball slippage stops at the same instant az impakt stops (ie if the initial spin iz not too great, not too little),
    (3) if at the end of impakt both balls are travelling in the same direktion (ie if the initial impakt angle iz not too great, not too little, probably 5deg off fullball i think).
    Sorry for my error — i hope that this explanation iz satisfactory.

  5. mac rynkiewicz permalink

    Yes, az u say, Dr Dave haz calculated an upper limit of 35.71% transmission of side for a fullball impakt.
    But i feel sure that Dr Dave wouldnt say that 35.71% iz the upper limit for transmission of side — koz az i say, the limit iz 50%.
    And this 50% iz achieved when the impakt iz say 5deg off fullball, and when the qball’s spin iz in effekt checkside not running side, ie usa pool players would call this uzing inside english, not outside english.

  6. mac rynkiewicz permalink

    Put it this way. For a straight-in pot-shot, what iz the max possible transmission of side.
    The answer iz 50%.
    Koz for a straight-in shot, uzing side, u have to contact the objectball 5deg off fullball, koz the objectball will be thrown 5deg off-line (the 5deg depends on the friction, if more friction then more than 5deg).
    U could say that the 50% iz made up of 35.71% due to transmitted side, and 14.29% due to collizion induced side. Fair enough.
    Anyhow let us say that for a straight-in shot u appear to get 50% transmitted side.

Leave a Reply

Note: XHTML is allowed. Your email address will never be published.

Subscribe to this comment feed via RSS