- #1
gtatix
- 19
- 0
I have done a fair bit of research and reading on the subject I am asking about. There is not much - in fact virtually no - detailed information anywhere on the internet about this.
If you have heard of Claude Shannon, Edward O Thorp, Norman Packard, Doyne Farmer and Thomas Bass, then you will know exactly what I am talking about.
Using classical physics and perhaps some chaos formulae, I would like to know how one would go about creating formulas to calculate the probable location of a spinning ball around and then into a bowl. Roulette comes to mind - in fact it is ON my mind.
I know that we would use classic physics formulas for the first part of the problem which is basically a 2-dimensional model. The ball is going around a circle of a specific circumference. The ball continues on this path - decelerating due to friction - and ultimately STARTS to fall inwards.
At this point, we are now dealing with a third dimension as the ball rolls downwards at an acceleration dependant on the slope of the wall and the initial velocity of the ball where it left the rim. It is also still moving slightly forward in the direction it was on before it started to fall. Vectors come to mind here.
Now we could likely be pretty happy with predicting where the ball would hit the wheel at the bottom - but it aint that easy - of course!
The ball is entering a moving wheel at the bottom that has its own circular 2-dimensional motion. For arguments sake - and to make things a little easier - HA - we can assume that this wheel is moving at a constant velocity - no acceleration.
The ball is also LIKELY - but not necessarily - going to hit an obstacle on the way down. There are small bumps around the wheel that are placed relatively uniformly - I forget the name of them as I type - but they have a name. Clearly the name is not important but perhaps the shape is?!
There is obviously a chaos factor and perhaps we don't need to incorporate any extreme formulae to come up with a solution. We don't need to know EXACTLY where the ball should land, we only need to know a relatively general area that would include anywhere from .125 the circumference of the moving wheel at the bottom to perhaps .25 the circumference. And we would only need to be correct a percentage of the time - not 100% but as close to this as possible. I don't know specifically how accurate but let's say 30% of the time.
What I am asking is for as much formulatic information that may be necessary. Definitely some classical physics and likely some chaos and perhaps some probability.
We could also want to look at placing data into a neural network application to come up with some good predictions based on small to medium data.
All input would be greatly appreciated. I will include below some formulae I have dug up. It may have nothing to do with this or it may be wrong.
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 let's 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!
Any and all help appreciated
Cheers
If you have heard of Claude Shannon, Edward O Thorp, Norman Packard, Doyne Farmer and Thomas Bass, then you will know exactly what I am talking about.
Using classical physics and perhaps some chaos formulae, I would like to know how one would go about creating formulas to calculate the probable location of a spinning ball around and then into a bowl. Roulette comes to mind - in fact it is ON my mind.
I know that we would use classic physics formulas for the first part of the problem which is basically a 2-dimensional model. The ball is going around a circle of a specific circumference. The ball continues on this path - decelerating due to friction - and ultimately STARTS to fall inwards.
At this point, we are now dealing with a third dimension as the ball rolls downwards at an acceleration dependant on the slope of the wall and the initial velocity of the ball where it left the rim. It is also still moving slightly forward in the direction it was on before it started to fall. Vectors come to mind here.
Now we could likely be pretty happy with predicting where the ball would hit the wheel at the bottom - but it aint that easy - of course!
The ball is entering a moving wheel at the bottom that has its own circular 2-dimensional motion. For arguments sake - and to make things a little easier - HA - we can assume that this wheel is moving at a constant velocity - no acceleration.
The ball is also LIKELY - but not necessarily - going to hit an obstacle on the way down. There are small bumps around the wheel that are placed relatively uniformly - I forget the name of them as I type - but they have a name. Clearly the name is not important but perhaps the shape is?!
There is obviously a chaos factor and perhaps we don't need to incorporate any extreme formulae to come up with a solution. We don't need to know EXACTLY where the ball should land, we only need to know a relatively general area that would include anywhere from .125 the circumference of the moving wheel at the bottom to perhaps .25 the circumference. And we would only need to be correct a percentage of the time - not 100% but as close to this as possible. I don't know specifically how accurate but let's say 30% of the time.
What I am asking is for as much formulatic information that may be necessary. Definitely some classical physics and likely some chaos and perhaps some probability.
We could also want to look at placing data into a neural network application to come up with some good predictions based on small to medium data.
All input would be greatly appreciated. I will include below some formulae I have dug up. It may have nothing to do with this or it may be wrong.
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 let's 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!
Any and all help appreciated
Cheers
Last edited: