Many people have attempted to beat the game of roulette using computers. . The task is a little more insurmountable than one may first assume, in that not only do you have to take human errors of timings taken into consideration, but also the mathematics involved for varying degrees of tilt of the wheel. Early attempts tried the use of primitive Polynomials, that although gave reasonable accuracy using exact data as the seed for the calculations, taking timings of the ball and rotor and using human intervention would introduce errors that needed to be overcome. Air pressure and temperature was another factor.

What was needed was a method where the equations could be set up as you play, therefore being flexible and coherent to the actual game in play that could be constantly changing every 20 spins or so.

Firstly a study was made by myself and Formulas invented and tried to replicate the exact data that was observed empirically for level and tilted wheels.

Here is some early experimental mathematics, some of the formulas are invented, but lend themselves to the problem at hand.

Some equations describing the ball behavior on a tilted wheel have been invented.

I am very content that the virtual experimental results resemble what we can see in the real world!

This research has been carried out with a purpose to understand what causes the tilt-like behavior and what we can expect from this.

Enjoy it!

 
The following mathematics deals with the X,Y and Z axis within the confines of a Roulette Wheel environment.

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.

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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!


SEE FURTHER>

 

 

Now lets build a graph Y=Y(YO).

If we change the point from where we spin the ball, in ideal level wheel conditions (When the tilt is absent) the ball travels the same velocity, no matter where the ball is spun from along the circumfrence of the wheel, if it is spun with the same initial velocity, hence:

Y=YO+some constant value

Thats if the tilt angle in degrees is the same and we change the YO from Ograd to 360grad at a given initial velocity 2*pie/0.7, then we get the next picture of Y expressed in Gradients.

 

 

Y=Y+4870.53 Grad

Now lets tilt the wheel with a severe tilt angle of 0.8 grad, then we would have the graph shown below. It is obviously non-linear and also has a big leap up between points B and C.

Look at Figures #1- #4.

In general they are all showing just a transformation of any point from which we spin the ball into a point where it drops off. We can see that the majority of spins will result in falling into a narrow sector of the wheel. This is clearly illustrated in Figure#4.

 

 

The probability of the ball dropping off in the drop off sector is:

P=100%*Grey Area/(White Area+grey Area)

in this case;

P~100%*[360-(Cx-Bx)]/360=100%*(360-[320-304]/360=97%

The steep gradient of the graph(on the interval of the initial angles lying between B and C, where there is a sharpe prolonging of the distance which the ball travels) is due to some interesting behavior of the ball around the point where it is ready to drop off.

Lets spin the ball from the point with an angle coordinate equal to;

YO=312grad(Around the middle between B and C)

Then we have the picture of the angle velocity shown in the graph. When the ball is nearly losing its velocity up to the drop off velocity and its contact with the track is almost lost, it manages to pass the "Peak of the hill" and begins to accelerate due to traveling downhill from the Apex.

Its velocity increases which in turn contributes to increase centrifugal forces against the ball track side until it approaches the Peak of the hill yet again, where it loses its orital velocity finally and proceeds to head towards the rotor after leaving the tracks banking.

This can produce increases of distances of ball travel in the region of 0.7 revolutions and greater.

 

 

CONCLUSION!

The modeling has shown that the greater the tilt, the narrower the resulting sector.

The narrowing drop off sector results in hitting some particular diamonds which are located AFTER the sector (as the ball still has to pass some distance from the ball track to the diamond), spiralling inwards towards the rotor.

The modeling also allows investigating the influence of different parameters on various tilts of wheel. One important note to make is that we can reach the same degree of tilt behavior bu either increasing the tilt angle, or by decreasing the friction arounf the ball track. This also means that roulette wheels with less frictional surfaces need less tilt in order to produce very good narrow sectors which can be used in the prediction of the outcome.

I have also found that A 'Critical Tilt Angle' exists on most wheels, which greatly increases the narrowing of the final predicted sector, that is the intervals between B and C sharply decreases. I found in my research that the the 'Critical tilt angle' was around 0.35 grad.