Maximum Theoretical Power from Unbalanced Wheel

[stgpcm]

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?

[Flyboy]

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

Thanks will have a play with these numbers and see how I go.

[piergino70]

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.