Maximum Theoretical Power from Unbalanced Wheel

[stgpcm]

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

[Flyboy]

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 

[stgpcm]

yes, the weights on the right are on longer arms than the weights on the left, so there will be more torque - lets say there are 10 weights of 1 Kg and the wheel is of radius 1 metre, and by a miracle of engineering the weights on the left manage to make the upward journey 1 centimeter from the hub.

at the horizontal position, Rr is the resultant weight of the right weight, and Rl is the resultant weight of the left weight the clockwise torque is:

((Rr x 1kg) x g x 1m) - ((Rl x 1Kg) x g x 0.01m)

or

(Rr - 0.01Rl) Nm

Nice torque!

now, what are the resultant weights of the left and right sides.... 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

[Flyboy]

... 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.

[AB Hammer]

Flyboy

So you want to figure out the power? Simply take the advantage of overbalance and do the math. For instance if a wheel has a 100lb advantage of falling weight (this is over the top for balances) and you want to figure out the distance/leverage from the axle and then do your math at, weight times distance to the axle and you will have your basic torque figured out, and go from there.