OK, the system as it stands has a potential energy of one quarter turn clockwise. This potential energy is created when you first build it. if you let the whole setup turn it will do so, but so what - you're now at E2.jpg

what your diagram doesn't show is the linkage that provides the upward force to the weights on the left - it's the energy you need to power that linkage that EXACTLY BALANCES the wheel.

you've agreed that all of the energy from the energy from the balls dropping on the right is needed to raisie the balls rising on the right. Once they are doing that, they are no longer applying any force to the wheel itself, so the wheel no longer wants to turn

Thanks, I think I now understand what it is your saying.  My next step is to now try and show what you are explaining using vector diagrams as proof...  If anyone knows any work already done that would help a lot, preferably with formula to show work  

The reason for this is I had a thought some time ago that is similar to this... as of yet I can't think of any reason my idea wont work so now I guess I should bite the buliet and figure out the math...    Who knows... it might be the holy grail or it might be one more 'balanced' wheel 

... as discussed all of the weights on the right are matched by all of the weigts on the left, so the resultant weight is 0, so our torque is

(0-(0.0x0)) Nm

or

0 Nm

Thanks for trying to help but your math is a little incomplete as you made an assumption and then assigned values based on the assumption.  I was hoping to get some schooling in how to calculate all the forces in a simple system like this one that would then give a mathematical result showing balance or out of balance... this way I can then apply it to other systems.

for the right hand path, it's quite easy to calculate the torque at any particular point.

let f be the downward force of the weight due to gravity,
let r be the radius of the wheel
let a be the angle from TDC

the distance (d) of the weight from the axle will be

r*sin(a)

so the torque will be

f*r*sin(a)

Which would vary between 0 and f*r, the average being around 0.6366*f*r, which is actually

f*r*(pi/2)

for your 16 weight diagram, there would be 8 weights on the right hand side, 22.5 degrees apart.

The torque on the system due to the 8 weights would be

f*r*(sin(a)+sin(a+22.5)+sin(a+45)+sin(a+67.5)+sin(a+90)+sin(a+112.5)+sin(a+135)+sin(a+157.5))

which would vary between 5.0273*f*r and 5.126*f*r (approx), the average torque would be 5.093*f*r approx, or 

(8*f)*r*(2/pi)

now, the energy is torque x angle (in radians), so a half turn (which is PI radians) of the 8 weight system gives us

(8*f)*r*(2/pi) * pi

or 16*f*r energy. (if r is in meters and f in newtons, that gives you joules)

The mathematically optimal route of the left hand weights to take is to move to the hub at the bottom (a distance r), and stay there while their spoke rotates to the top, then slide up the spoke to the rim. in order to do that, we will have to push them each time. To move a weights of mass f/g through the distance r against gravity, to get it to the hub, we need to do f/g*r*g work, and to get it to the rim, we also need to do f/g*r*g work. In half a turn, we need to do do this for 8 weights, so we need

8 * ((f/g*r*g) + (f/g*r*g))

or 16*f*r energy. The question is, where do we get it from?

Here are the forces in the vector diagrams that were missing to explain why it doeas not work.

I hope this helps Flyboy.

The enrgetic explanation is very logical and easy. The vector explanation is some more complex.