Sjack Abeling Gravity Wheel and the Worlds first Weight Power Plant

[Omnibus]

Masses m1 and m2 of the walls attached to the arm are negligible. If you want, instead of gluing you may secure the ball to the arm by a screw of the same mass as (m1 + m2). The main point is that the mode of attaching of the ball to the arm makes no difference, the torque will be calculated the same. Only external constraints which will also require change of arm's length (as in our case) will make a difference.

[mondrasek]

Masses m1 and m2 of the walls attached to the arm are negligible. If you want, instead of gluing you may secure the ball to the arm by a screw of the same mass as (m1 + m2). The main point is that the mode of attaching of the ball to the arm makes no difference, the torque will be calculated the same. Only external constraints which will also require change of arm's length (as in our case) will make a difference.

The majority of the masses are not fixed to the wheel at all.  They are leaning against it at a specific angle.  If this angle is not the same as the angle of the mass' CoG to the center of axle of the wheel, the forces you are calculating are also not normal to the arm.  Only the normal component of that force can provide torque.

[Omnibus]

@mondrasek,

I figure if we do all 9 we should get a sine wave the crosses the zero around 19 degrees, the point of equilibrium.

See, but that’s gonna be the result of incorrect calculation considering that how a ball is attached to the arm makes a difference which it doesn’t. So far, calculations have been done by applying correct methodology and the result is in concordance with the crucial criteria for perpetuum mobile -- the mass-axle discrepancy at every position of the whel. We can't change this firmly established fact by trying to twist mechanics and push it into calculations which have no physical meaning.

[mondrasek]

Masses m1 and m2 of the walls attached to the arm are negligible. If you want, instead of gluing you may secure the ball to the arm by a screw of the same mass as (m1 + m2). The main point is that the mode of attaching of the ball to the arm makes no difference, the torque will be calculated the same. Only external constraints which will also require change of arm's length (as in our case) will make a difference.

Omni, I realized I did not understand your diagram.  I thought m1 and m2 were the force vectors you wanted to consider.

So, assuming they are walls of zero mass.  If both are attched to the wheel, both cases are calculated the same, as you say.  But if only m2 is attached to the wheel, and m1 in not, they are NOT calculated the same.  This is our actual case where m2 represents the slots, and m1 represents the guides.

[mondrasek]

@mondrasek,

See, but that’s gonna be the result of incorrect calculation considering that how a ball is attached to the arm makes a difference which it doesn’t. So far, calculations have been done by applying correct methodology and the result is in concordance with the crucial criteria for perpetuum mobile -- the mass-axle discrepancy at every position of the whel. We can't change this firmly established fact by trying to twist mechanics and push it into calculations which have no physical meaning.

To use your own logic, which is more likely to be correct?  Your scatter shot graph of jumbled torque values that follows no pattern?  Or a smooth sine wave that is the signature of a pendulum?