Chas Campbell free power motor

[rMuD]

Okay, Hum,
now redraw your drawing, that you have rotated the wheel by
15 degrees, so you have only 5 balls in the left side and 2 balls
at about 4 x times the distance from the axis.
Then count the torques and add them up.
Do you come up with a different value than 3.9 to 3.6 advantage for running ?

if I go down to 15 degrees per place to hold a ball..  then there will be 10 balls on the left side and 2 on the right

[hartiberlin]

@rMuD,
why do you show yellow balls at the right side being just only 2 or 3 times away from the axis ?
There they will never be !?
There will be 2 balls almost 4 times away from the axis at the right side !

I did a 15 degree off center...  I can modify version #2 to show balls there is no balls at 4x at 15 degree offset,

Please rotate your wheel by 15 degrees so it shows the start position as in the Chas video.
There will be 5 balls at the left side and 2 balls at almost 4x distance.
This is the start position.
There we have 2.16 to 1 advantage.
Maybe you can show then further multiple rotation positions so it gets clear where and
when the balls are where.
Please upload a ZIP archive with 24 pictures showing the balls in each 15 degree
rotation.
I will then make an animation out of it, so everyone can see, how the balls go where and when.
Thanks.

[Humbugger]

Okay, Hum,
now redraw your drawing, that you have rotated the wheel by
15 degrees, so you have only 5 balls in the left side and 2 balls
at about 4 x times the distance from the axis.
Then count the torques and add them up.
Do you come up with a different value than 3.9 to 3.6 advantage for running ?

The whole point is, Stefan, that there cannot be any time that there are two balls present on the right side if the outer radius is approaching 4.  Even at 3, there is only a very short time when there are two balls on the right!   My drawing as it stands shows that very short moment in time; a few degrees of rotation at best when two balls are there.

Cannot draw it like you suggest.  Only M.C. Escher can draw that!

If the radius is greater than 3.8763, there will be times when no ball is on the right; there will never be two balls.
At anything above 3.8763, the lower ball is off the wheel BEFORE the upper ball comes on the wheel!  

[hansvonlieven]

G'day all,

Here is a simple diagram that shows that the arrangement as stated does not work.



As you can see the longer the distance from the axis on the right hand side, the smaller the lift on the left hand side. In this drawing I did not even take into consideration the slope necessary to deliver the balls to the right, therefore the actual usable arc is less and gets smaller with distance.

The amount of force available on the right hand side of the lever is directly proportional to the amount of lift available on the other side.

In other words, if it does not work on X=2 it will not work on X=3 or X=4.

Incidentally, the drawing is not to scale, it is only to demonstrate the principle.

Hans von Lieven

[Humbugger]

I find Hans drawing and explanation very confusing.  Maybe that's because I love my own solution and cannot clearly see anyone elses.    I think my explanation is clear.

Humbugger