HI Folks

After spending 19 years developing roulette computers successfully, I was contemplating incorporating a blackjack program on the same computer, this will disguise computer roulette play to some degree!

Basically collecting off both!

I wondered if anyone has got the exact gain pertaining to number of decks used, what is the mathematical maximum gain, and what data would have to be known?

Heres some of my mathematics relating to computer ballistic approach, although the graphs are missing, just to show I am not a time waster and know a litte bit of maths:-)

Although without any Graphs, the mathematics may be even more difficult to comprehend, you can email for the associated Graphs!

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.

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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!

The system returns around a 1 in 18 success rate, placing just one chip per spin on one number, this is on most roulette wheels!

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Hope someone can help, may do a trade witrh someone, supply them a copy of the program in return for a decent Blackjack program:-)

Hope to here from someone soon!

mark AT predictroulette.com

Kind regards

Mark Howe