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4/27/05 1:37 AM
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Jacob Lamb
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Edited: 27-Apr-05
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Say you're playing roulette at a casino... When the guy spins the ball, it stays up in that groove until it slow down enough to start falling... If you clocked the ball speed when it starts falling, would it be pretty consistent every time?? Does the ball start falling at X amount of speed every time or does it vary? Seems like it would fall once it slowed to a certain speed... right?
4/27/05 2:07 AM
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RGoodfellow
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Edited: 27-Apr-05
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I think the actual math involved is rather complex, but it is possible to predict where the ball is likely to land. Check out the "How to Win at Roulette" article: http://gaming.unlv.edu/2004_03_21_blogarchive.html
4/27/05 2:21 AM
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Jacob Lamb
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I'm not talking about where the ball will land... I want to know if the ball will always fall at that certain speed or if it varies...
4/27/05 2:28 AM
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Jacob Lamb
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I read the article in that link, but it's wrong... They measure ball speed to predict where the ball will land, but the wheel is spinning too. You'd have to measure ball speed AND wheel speed and then figure it out, which would be really complicated. Maybe they where measuring both, but the article only mentioned them measuring ball speed.
4/27/05 10:12 PM
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Jacob Lamb
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Edited: 27-Apr-05
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ttt
4/28/05 1:58 AM
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Power Paw
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Edited: 28-Apr-05
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http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html "Any motion in a curved path represents accelerated motion, and requires a force directed toward the center of curvature of the path. This force is called the centripetal force which means "center seeking" force. The force has the magnitude" "Note that the centripetal force is proportional to the square of the velocity, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle." "Centripetal force = mass x velocity^2 / radius" so it seems that the ball requires a certain force to keep it "pinned" against the wheel at a certain distance, if the force drops (because of the velocity drop) then the ball will start to fall as well
4/28/05 4:37 PM
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qeySuS
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Edited: 28-Apr-05
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I'm not 100% sure how a roulette game is built (never played it) but judging from movies i've seen :) I'm guessing there's a small "hill" in the edge that then ends in the red/black grooves. if this is so then you will need to calculate the effective force that is pointing downward in that slope, and once that force is greater then the centrifugal acceleration then it should start dropping. Just when that is would depend on the roulette wheels radius since centripetal acceleration is given by (if memory serves me), w^2*r (where w is the angular velocity and r is the radius of the roulette wheel). That's then your centrifugal acceleration which gets less as the angular velocity slowes down. Then you need to calculate your effective gravitational force, which depends on how steep the slope of the wheel is, should be given by some angle called theta, so to find the effective g you'd need to multiply it by sin(theta), made a small scetch of it here: I'm not 100% sure but i'd imagine that the centrifugal force would be pointing directly out of that image as well (orthogaonal to g) so you might have to find the x component of it as well by using theta (it should then be pointing in the -x direction while your g_x is pointing in +x). At least i think that's how it's done. This is of course done ignoring friction which i think is safe to do, those surfaces look pretty smooth.
4/29/05 12:57 AM
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Power Paw
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Edited: 29-Apr-05
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qeysus, http://www.roulette-forum.de/index.php?act=Attach&type=post&id=318
4/29/05 4:02 AM
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qeySuS
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Edited: 29-Apr-05
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Damn those guys really went above and beyond :D That's pretty cool.
6/8/05 4:26 AM
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Paul S
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Edited: 08-Jun-05
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"so it seems that the ball requires a certain force to keep it "pinned" against the wheel at a certain distance, if the force drops (because of the velocity drop) then the ball will start to fall as well" That doesn't sound right to me, but whatever.

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