SMOT! - (previously about the OC MPMM)

[Omnibus]

Eout (CoE not violated) = (mgh1 - (Ma - Mb))

Eout(CoE violated) = (mgh1 + Mb) = (mgh1 + mgh2 + Kc)

[modervador]

@modervador,

?To show that the ball has excess energy at A would require a corroborating measurement of that energy, which has not been done.?

On the contrary, corroborative measurement has been done showing that the ball at C has not just energy (mgh1 + mgh2 + Kc - Ma), as you incorrectly understand, but the whole amount of energy (mgh1 + mgh2 + Kc) ready to be transformed (and actually transformed) into other kinds of energies such as, say, Ma, when the ball returns at A.

The energy at C was never an issue. Your experiment did not measure the available energy at A. Therefore any claim that harvestable energy at A exceeds mgh1 + Mb - Ma has yet to be corroborated by measurement.

The total energy at C is indeed greater than the initial work done to move the ball from A to B, i.e. (mgh1 + mgh2 + Kc) = (mgh1 + Mb) > (mgh1 + Mb - Ma). However, if you extract all that energy, you cannot get back to point B without putting it all back, i.e. the hand will have to expend the full mgh1 + Mb. For the hand to only exert (mgh1 + Mb - Ma) to place the ball at B, the ball has to start at A (or at some point of equipotential). But you can't get to A from C without the ball gaining magnetic potential Ma at the expense of some other energy pool.

We do agree that "the ball at C has. . .the whole amount of energy (mgh1 + mgh2 + Kc) ready to be transformed (and actually transformed) into other kinds of energies such as, say, Ma, when the ball returns at A." The ball has Ma when it ends up at rest at A, which is just as much as it had when it started there.

[tinu]

Sorry but Kc can not be part of Eout for repeated/cyclical SMOT operation.
A big part of Kc goes back in Ma.
I wrote on the above when you said I did not understood and I write here again shortly: if you take (even part of) Kc out of the system, the ball will get stuck in C and it will not reach Ma. I agree Ec=mgh1+mgh2+Kc but this is not Eout.
Comments?
Tinu

[modervador]

@modervador,

Of course, you should agree. It is the entire amount of energy (mgh1 + mgh2 + Kc) the ball has at C that is transformed into other kinds of energy such as, say, Ma, when the ball returns at A closing the loop. Not just (mgh1 + mgh2 + Kc - Ma), as you used to misunderstand, but the entire amount (mgh1 + mgh2 + Kc).

If I ever seriously disputed that the energy at C is (mgh1 + mgh2 + Kc), then I did so in error. However, if you are suggesting that the energy available at A is the same as available at C when the ball makes the B-C-A transition, then you do so either in error or without explanation and without proof.

According to CoE the ball can never have greater amount of energy available to be transformed into other kinds of energy than the amount of energy imparted to it. In the discussed case the ball at C has greater amount of energy to be transformed into other kinds of energy that the initial amount of energy imparted to it. Therefore, the case under discussion here violates CoE.

The problem with that logic is that the energy Ma was imparted to the ball when it was first positioned at point A. It is built into the system. So the energy of the ball resting at A, Ea = Ma. Therefore the total energy at C is no different from the energy of the ball at A plus the energy to go from A to B, namely Ec = (mgh1 + mgh2 + Kc) = (Ea + mgh1 + Mb - Ma) = (mgh1 + Mb). So CoE seems quite safe.

[Low-Q]

I just put this picture here - just to ease the understanding a bit for new visitors..