SMOT! - (previously about the OC MPMM)

[Omnibus]

@modervador,

To cut out the wordiness:

The energy Mb lost along B-C appears out of thin air. There?s no source supplying that energy.

Mb comes as a result of the ball having Ma at initial point A. Part of the magnetic potential is lost when the ball is moved from A to B, i.e. Mb < Ma. This is why Ein does not require full mgh1 to achieve height h1, but it is reduced by the amount (Ma ? Mb), i.e. Ein = (mgh1 ? (Ma ? Mb)).

Thus the Mb part at point B has been accounted for from a known source and its appearance is consistent with CoE.

No, the source of Mb is unaccounted for. Carry out a second experiment, then a third, fourth and so on. Who's supplying Mb lost along B-C? Every time you do an experiment you only supply (mgh1 - (Mb - Ma)). Who's supplying the Mb spontaneously lost along B-C?

[modervador]

No, the source of Mb is unaccounted for. Carry out a second experiment, then a third, fourth and so on. Who's supplying Mb lost along B-C? Every time you do an experiment you only supply (mgh1 - (Mb - Ma)). Who's supplying the Mb spontaneously lost along B-C?

The Mb in the second experiment comes from the Ma part of Ea(final) from the first experiment.
The Mb in the third experiment comes from the Ma part of Ea(final) from the second experiment.
Etc.

Mb comes as a result of the ball having magnetic potential Ma at point A regardless of how it ended up there. Part of this potential is lost when the ball is moved from A to B, i.e. Mb < Ma. This is why Ein does not require full mgh1 to achieve height h1, but it is reduced by the amount (Ma ââ,¬â€œ Mb), i.e. Ein = (mgh1 ââ,¬â€œ (Ma ââ,¬â€œ Mb)), assuming no kinetic energy is carried over from the previous run.

If we are successful at eliminating friction, Ea(final) from the first experiment is enough to get the ball all the way to point B again without any additional Ein for the second experiment. If however we extract all the Ka in Ea(final) = (Ma + Ka) and send it outside the system during the first experiment, then of course there is only Ma left at point A and to move the ball to B for the second experiment requires again Ein = (mgh1 ââ,¬â€œ (Ma ââ,¬â€œ Mb)).

[Omnibus]

No, the source of Mb is unaccounted for. Carry out a second experiment, then a third, fourth and so on. Who's supplying Mb lost along B-C? Every time you do an experiment you only supply (mgh1 - (Mb - Ma)). Who's supplying the Mb spontaneously lost along B-C?

The Mb in the second experiment comes from the Ma part of Ea(final) from the first experiment.
The Mb in the third experiment comes from the Ma part of Ea(final) from the second experiment.
Etc.

Mb comes as a result of the ball having magnetic potential Ma at point A regardless of how it ended up there. Part of this potential is lost when the ball is moved from A to B, i.e. Mb < Ma. This is why Ein does not require full mgh1 to achieve height h1, but it is reduced by the amount (Ma ? Mb), i.e. Ein = (mgh1 ? (Ma ? Mb)), assuming no kinetic energy is carried over from the previous run.

If we are successful at eliminating friction, Ea(final) from the first experiment is enough to get the ball all the way to point B again without any additional Ein for the second experiment. If however we extract all the Ka in Ea(final) = (Ma + Ka) and send it outside the system during the first experiment, then of course there is only Ma left at point A and to move the ball to B for the second experiment requires again Ein = (mgh1 ? (Ma ? Mb)).

Not so. It's not regardless of how it got there (back at A, that is) because the ball got back at A exactly because of violation of CoE.

Again, the ball having energy Ma at A which no source has any contribution for, who supplied the energy Mb the ball loses along B-C. We only supplied (mgh1 - (Ma - Mb)). Who supplied the rest of the energy, that is energy Mb the ball loses along B-C? Say explicitly what you're supposed to say or I'll continue asking the same question you till the cows come home.

Oh, and don't forget. If CoE is to be obeyed the ball upon its return to A must only lose the energy (mgh1 - (Ma - Mb)) supplied to it. In SMOT, however, the ball loses energy Mb in addition to energy (mgh1 - (Ma - Mb)). In SMOT the ball loses more energy than the energy imparted to it. This is a clear violation of CoE.

[Omnibus]

The situation with SMOT is analogous to the following. You look just at any stone as we know them in Nature. A stone lying on the ground suddenly lifts itself up to a height h and then drops back to the ground. Who supplied the energy mgh which the stone lost when it fell on the ground?

This would be a violation of CoE, correct?

[modervador]

Not so. It's not regardless of how it got there (back at A, that is) because the ball got back at A exactly because of violation of CoE.

Again, the ball having energy Ma at A which no source has any contribution for, who supplied the energy Mb the ball loses along B-C. We only supplied (mgh1 - (Ma - Mb)). Who supplied the rest of the energy, that is energy Mb the ball loses along B-C? Say explicitly what you're supposed to say or I'll continue asking the same question you till the cows come home.

Oh, and don't forget. If CoE is to be obeyed the ball upon its return to A must only lose the energy (mgh1 - (Ma - Mb)) supplied to it. In SMOT, however, the ball loses energy Mb in addition to energy (mgh1 - (Ma - Mb)). In SMOT the ball loses more energy than the energy imparted to it. This is a clear violation of CoE.

The ball arrives back at A because a working SMOT must be arranged such that (Mb + mgh1) > Ma; violation of CoE is not required.

On any given run, the source of Ma in Ea(final) is Ma in Ea(initial), which transforms along with Ein = (mgh1 - (Ma - Mb)) through (Mb + mgh1) and (mgh1 + mgh2 + Kc). CoE violation has not been demonstrated, since all these energies have traceable sources. This can be demonstrated for each subsequent run.

As described in previous posts, Mb (which itself ultimately came from Ea(initial) = Ma) is transformed into (mgh2 + Kc) on segment B to C, but (mgh2 + Kc) transforms into part of Ea(final) = Ma + Ka hence contributes to "recharging" the magnetic potential so as to have Ma when back at A.

As described in previous posts, Ea(final) = Ma + Ka = Ma + Ein. Ma cannot be "lost" by the ball without deviating from point A, thus the maximum that can be lost is Ka = Ein = (mgh1 - (Ma - Mb)), but never more. Thus in SMOT the ball does not lose more energy than the energy imparted to it. CoE preserved.