CALCULATING WEIGHT DISTRIBUTION ON A WHEEL.

[Alexioco]

I have made the following calculations for weight distribution....3oclock 100% ,that is maximum distance from centre of rotation .This means at the 12 oclock position the distance is effectively 0 as a line drawn down is right on the centre of rotation.At 1oclock it is 50% of the distance and at 2oclock @90%.

Interesting, also that means there has got to be a lot of work done on the descending side of a wheel, to lift the ascending side of a wheel.

[mapsrg]

The work on the descending side of the wheel in raising the ascending side is provided by the differential in distance from the centre of rotation between the two sides.Similar to a see-saw with people of the same weight but differing positions from the pivot point.The trick is to attain this change in position continously to cause rotation of a wheel.The rotation is caused by gravity but the postioning of the weights must be accomplished by some sort of guides be it mechanical, magnetic or electromagnetic.Overunity  is the goal and it is attainable...

[helmut]

The distance from 7o?clock to 1 o?clock is the same as the diameter.
The way that a weiht is traveling from 1 to 7 is ~1,57 times the diameter.

Perhaps it helps to calculate the reqired force to repoint the weight.

helmut

[mapsrg]

a weight at 12 oclock has no turning effect as is a weight at 6 oclock.A weight at 3 oclock has maximum effect.......also a weight at 2 and 4 oclock approaches this maximum @90%.
The energy required to travel around the wheel from 12 to 6 (output) and 6 to 12 (input) can be calculated using the circumferencial travel distances as part of the equation.
PE=m*g*d in joules per second
For a wheel of 1metre  r the circuference is 6.24 metres .
if we use a mass of 100 grams per weight
if we move the weight out 100mm on one side and in 100mm on the other

for the falling side we would have a new circumference of 1.1*3.14=3.45M
for the rising side we would have a circumference of .9*3.14=2.82M

Therefore
PE=m*g*d
PE=0.1*9.8*3.45
Pe=3.381 joules per sec output

and
PE=0.1*9.8*2.82
PE=2.2636 joules per second input

the difference is 1.1174 joules per second for one weight in a cycle around an unbalanced wheel

If this offset effect was caused by a mounted smaller wheel that is rotated at 6and 12 oclock then these wheels would be 200 diameter and with a 50 mm clearance between them there would be @24 of them.This is 26.81 joules per second.
the energy needed to rotate a 200 diameter small wheel is
PE=0.1*9.8*0.628
Pe=0.6154 joules per second

Therefore we have excess energy of @26 joules per second from this unbalanced wheel.....

[broli]

I don't know how many know what a crossp roduct is but torque is basicly RxF where R is the vector from the midpoint of the circle to the position and F is that gravity vector. This will give you a new vector. The direction of it will be determinded by the righthandrule/corckscrew and the size is the area of the parrallegram formed by these two vectors. This is also why at 3 o'clock you have the biggest torque contribution since that area is then biggest. ILLUSTRATION TIME!



I hope that's a bit helpfull.