Sjack Abeling Gravity Wheel and the Worlds first Weight Power Plant

[Omnibus]

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.

Who is arguing with the fact that only normal components participate in the calculation of the torque? Of course only the perpendicular components participate in the calculation of the torque and in both cases a) and b) those components are exactly equal as well as the length of the arms, therefore, the torque in case a) is exactly equal to the torque in case b). That's beyond obvious.

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.

How come? I thought you already understood that force F is unaffected by whether or not the ball is loose but leaning on walls (as in a) of my illustration) or glued to the arm (as in b)). F is F, no matter what and it's not applied differently in case a) compared to case b). Center of mass in both cases is at exactly the same place. I thought you already understood that.

[mondrasek]

Please review.

[Omnibus]

@mondrasek,

This is all messed up. Neither m1 nor m2 are moving with respect to the mass-arm system. They are part of it. Especially m1 isn't a guide. And why are you only using one of the components of F to resolve it along the normal of the arm? Doesn't the other component exist and play a role in the torque? Let alone that in the way you're shown it you dramatically underestimate the component normal to the arm by which the effect I calculate becomes even larger.

The way F should be decomposed is straight to the arm's normal. Again the point of you confusion -- the ball doesn't lose mass, hence doesn't lose any part of its force F just because it's leaning against the wall. You seem to think that the wall causes somehow mass to leak and F to not be the same F that should participate in Fn = F sin(alpha), which multiplied by arm's length L would equal torque T = FnL. Not so, though. Like I already said, F is F and it is that whole F whose normal component Fn participates in T = FnL. It is the same in a_ as in b) of my illustration.

[mondrasek]

@mondrasek,

This is all messed up. Neither m1 nor m2 are moving with respect to the mass-arm system. They are part of it. Especially m1 isn't a guide. And why are you only using one of the components of F to resolve it along the normal of the arm? Doesn't the other component exist and play a role in the torque? Let alone that in the way you're shown it you dramatically underestimate the component normal to the arm by which the effect I calculate becomes even larger.

I repeat:  If m1 and m2 are both moving with the arm then yes, both your case a) and b) calculate the same.

But in OUR case m1 DOES NOT move with the arm.  And the correct vector analysis is what I am showing.  Why do you keep bringing up a case where m1, which is analogous to our fixed guides, moves with the wheel?  IT DOES NOT APPLY.  What I am showing you is the correct method for OUR case where m1 does not move with the arm.

BTW, I'm on the road tomorrow and then we are into the weekend.  Might not be posting much.  Have a good one.

M.

[Omnibus]

I repeat:  If m1 and m2 are both moving with the arm then yes, both your case a) and b) calculate the same.

But in OUR case m1 DOES NOT move with the arm.  And the correct vector analysis is what I am showing.  Why do you keep bringing up a case where m1, which is analogous to our fixed guides, moves with the wheel?  IT DOES NOT APPLY.  What I am showing you is the correct method for OUR case where m1 does not move with the arm.

BTW, I'm on the road tomorrow and then we are into the weekend.  Might not be posting much.  Have a good one.

M.

First of all agree that the torque calculated for a) and b) is the same (m1 and m2 are part of the ball-arm system). Before you understand that these two torques are equal it's impossible to continue the discussion.