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I Roulette Prediction Algorithm Problem

  1. Oct 12, 2018 at 5:17 AM #1
    The roulette prediction algorithm is described here: http://pingless.co.il/roulette.pdf

    And I have a problem with the formula attached
    eq2.jpg
    Where teta θ is position, velocity and acceleration of the ball. g is garvity force. r is the ball radius from center of the roulette. a is stator incline.
    According to my calculations:
    TrimCoefficient = -1/-34.13 Inch/Sec^2
    ballVelocity = 120 Inch/Sec
    Gravity = 386.09 Inch/Sec^2
    Rrim = 14 Inches
    statorIncline = (28/180)*3.14;
    Trim = TrimCoefficient*(ballVelocity+math.sqrt( (Gravity/Rrim)*math.tan(statorIncline) ));
    I get Trim=3.6 and it should be as I think (if you count from the start if the ball spin) Trim=18 at least.
     

    Attached Files:

    • eq2.jpg
      eq2.jpg
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  2. jcsd
  3. Oct 12, 2018 at 5:34 AM #2
    And also if someone can help me to solve why the units don't fit
    If you substitute all the units you get -S^2/I*(I/S + 1/S) = S+S/I
    And it should be seconds only why you get +S/I
     
  4. Oct 12, 2018 at 9:09 AM #3
    Angles should all be in radians.
    ##\dot\theta## should be in radians per second, but you're using a value in inches per second. And similarly for ##\ddot\theta##, it should be in radians/s^2 but you're using inches per second squared.

    Since I suspect you are using linear velocity and linear acceleration at the rim, I think you need to divide by the radius to get the corresponding angular velocity and acceleration.
     
  5. Oct 13, 2018 at 3:26 PM #4
    Hey thank you for your note,
    this is my calculations after the radians / sec change:
    Rrim = 14
    TrimCoefficient = -1/(-30/Rrim)
    print("TrimCoeff: "+str(TrimCoefficient))
    ballVelocity = 88/Rrim
    print("ballVelocity: "+str(ballVelocity))
    # (65 - 88) / (1.36 - 1)
    Gravity = 386.09
    statorIncline = (28/180)*3.14;
    Trim = TrimCoefficient*(ballVelocity+math.sqrt( (Gravity/Rrim)*math.tan(statorIncline) ));
    print("Trim Trials: "+str(Trim))

    and the result are:
    TrimCoeff: 0.4666666666666667
    ballVelocity: 6.285714285714286
    Trim Trials: 4.719797943795565
     
  6. Oct 13, 2018 at 3:42 PM #5
    Your original velocity and acceleration were 120 and -34.13, respectively. In your new calculation you are using 88 and -30. I assume you really did intend to work the calculation with different parameters the second time?

    That formula seems to be the correct implementation of the one you quoted. Are you happy with the results?
     
  7. Oct 14, 2018 at 12:38 AM #6
    Hey thank you, but i'm not happy from the results, Trim should be 18 as you can count from this movie
    Because Trim is the time the ball is falling from the rim it completes about 11 circles and it takes 18 seconds and the formula says it takes only 4.7 seconds
    What else could be wrong here
     
  8. Oct 14, 2018 at 1:25 AM #7
    Hey I think the acceleration is wrong because -30 inch/sec should stop the ball at speed 90 inch/ sec in 3 seconds do you agree? So the actual acceleration should be -5-10 inch / sec
     
  9. Oct 17, 2018 at 7:19 AM #8
    [​IMG]
    This equation for the ball position of the roulette game taken from here: http://pingless.co.il/images/APCCAS08.pdf#page=3

    This is my result:
    Rrim = 14
    ballPosition = (30/180)*3.14
    ballVelocity = 88/Rrim (Convert inch/sec to rad/sec)
    ballAcceleration = -8/Rrim (Convert inch/sec to rad/sec)
    Tdefl = 18
    ballAngle = float(ballPosition) + (ballVelocity*Tdefl) + 0.5*(ballAcceleration*Tdefl*Tdefl)
    ballAngle = 21 Radians
    What should 21 Radians mean?
     
    Last edited: Oct 17, 2018 at 8:17 AM
  10. Oct 17, 2018 at 7:53 AM #9

    jedishrfu

    Staff: Mentor

    My guess is that the 21 radians means that the ball has orbited the wheel 3.34225 times. If its starting point was 0 degrees N then it would be ~123 degrees to the left of 0 degrees when it stopped.
     
  11. Oct 17, 2018 at 9:40 AM #10
    Hey but actually it's not make sense the ball orbited the wheel only 3.34 times it should orbit 11 times and more as you can see from this movie:
     
  12. Oct 17, 2018 at 10:31 AM #11

    berkeman

    User Avatar

    Staff: Mentor

    Duplicate thread closed for Moderation...
     
  13. Oct 17, 2018 at 1:02 PM #12

    berkeman

    User Avatar

    Staff: Mentor

    Threads merged and re-opened. @shlomi123 -- Please do not start multiple threads on the same subject. Thank you.
     
  14. Oct 18, 2018 at 2:00 AM #13
    Hey I still have a problem with this formula FORM1.jpg
    Which is from here: http://pingless.co.il/roulette.pdf#page=3

    (1) I don't know from where in the wheel I should put the x-axis (zero angle) and (2) I don't know to define the θ(0) where the ball starts the orbit, please help me.

    << Mentor Note -- post edited to remove inappropriate content >>
     
    Last edited by a moderator: Oct 18, 2018 at 8:41 AM
  15. Oct 18, 2018 at 11:39 AM #14

    jbriggs444

    User Avatar
    Science Advisor

    As I read the initial graphic on that link, ##\theta## is the angle of the ball from the center of the wheel as reckoned in non-rotating coordinates. It does not matter what angle you take as ##\theta(0)## as long as you measure both starting and ending position using the same reference.
     
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