Re: Please advice on this math - motion of roulette ball
From: Narasimham (mathma18_at_hotmail.com)
Date: 12/16/04
- Next message: Narasimham: "Re: Please advice on this math - motion of roulette ball"
- Previous message: luiroto_at_yahoo.com: "Re: Poll: Are PCs Turing Machines?"
- In reply to: stan: "Please advice on this math - motion of roulette ball"
- Next in thread: Narasimham: "Re: Please advice on this math - motion of roulette ball"
- Messages sorted by: [ date ] [ thread ]
Date: 16 Dec 2004 07:08:14 -0800
stan wrote:
> Hi all,
> I hope i'm posting to the right forum, sorry if not.
> I have dug out some equations that relate to the prediction of a
> roulette balls drop off point on the wheel.
> I've need help verifing if these equations are correct or nonsense.
> Thanks in advance to anyone that can help
> I have copied a section below for anyone to have a look at:
> """""""""""""""""""""""""""""""""""""""""""""""""""
> The following mathematics deals with the X,Y and Z axis within the
> confines of a Roulette Wheel enviroment.
>
> Y axis¦N1¦*COS(a)-(mg)*COS(a)=0
> X axis¦N2¦+¦N1¦*SIN(a)+¦mg¦*sin(@)*COS(Y)=m*¦@
> centre)=m*V^2/R=m*[Y')^2]*R
> V=Linear Velocity
> R=Ball Track Radius
> @=Centripedal acceleration
>
> Z axis¦Ffr¦+¦Air Drag¦=m*¦@tan¦=m*Y''*R
>
> Friction Force a This is negative as it is opposing the Z axis
> Air Drag is the force that is equal too:
>
> ¦Air Drag¦= - 0.5*CD*P*TT*r^2*V^2 (TT is pie) this is also a minus
> value!
>
> CD is Drag Coefficiaent
> P is AIr density
> r is the balls Radius
> Z axis is always tangentially directed.
> ------------------------------------------
> After some very simple Algerbraic Transformation and incorporating
the above formulas we get the next differential equation:
>
> Y''=(a+air*R)*(Y')^2=b*SIN(Y)+c*COS(Y)+d (*)
> Where
> a is the determining friction factor(Ie 0.004)
>
> Air =-[0.5*CD*P*TT*r^2*V^2]/m
>
> b=a*g*SIN(@)/R
>
> c=b/a
>
> d=a.g.COS(@)SIN(a)+1)/(R.COS(@)
>
> The ball movement sters to this equation only till the moment when it
loses the contact with the vertical side of the ball track or:
>
> [N2]=0
>
> So the Drop off condition is:
>
> [(Y')^2]*R+g*COS(@)*tg(a)-g*SIN(@)*COS(Y)=0 (**)
>
> Now lets introduce some real values into the equations and see the
> predicted results:
>
> TT=3.14
> g=9.807
> R=0.4
> a=16.7.TT/180 inner slope of stator
> CD=0.47
> r=0.5.21.10^-3 Radius of ball
> P=1.22
> m=9.10^-3 Mass of Ball
> (a)= 0.004 Friction factor for rolling between the ball and the track
> @=0.8 grad Tilt Angle
>
> t0= 0 sec
> t1=30 seconds
>
> These values determin the time interval of 30 sec since the start of
> spinning!
>
> I also calcualted the time the ball loses contact with the vertical
> wall of the ball track, this is when(**) becomes true!
>
> Time till drop off is 17.04 seconds
>
> By this time the ball passes 4935 Grad or 13.7 revolutions from
> start point!At this moment the ball has a velocity of 2.7 Rads/Sec >
or 0.43 Revs per/Sec!
It is a good attempt at modeling Roulette Ball motion simulation.
Basically it modifies banking curve formula tan(th)= V^2/(R g) for
stationary ball on conical track. You can run ODEs on
Maple/Mathematica/Matlab or other simulation software, plug in the
constants and produce a lot of graphs describing the motion. But result
is as good as the constants plugged in, cannot be used to bet as there
is no means to know velocity, or a guaranteed friction value, may be
used to calibrate a given Roulette :)... scientific attitude is to
believe that chances of bet under controlled or experimental conditions
are better than pure blind shots. All that if you can carry laptops
into a Casino..
- Next message: Narasimham: "Re: Please advice on this math - motion of roulette ball"
- Previous message: luiroto_at_yahoo.com: "Re: Poll: Are PCs Turing Machines?"
- In reply to: stan: "Please advice on this math - motion of roulette ball"
- Next in thread: Narasimham: "Re: Please advice on this math - motion of roulette ball"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
