SMOT! - (previously about the OC MPMM)

[Omnibus]

The initial energy (Ma) is certainly taken into account when talking about the energy input to move the ball from A to B, (mgh1 ? (Ma ? Mb)). That's the (Ma) term in (mgh1 ? (Ma ? Mb)).

The initial energy Ma is taken into account when talking about the energy input because it?s there, the way the energy to secure the screws of the machine has been spent. What is discussed is what happens after that. That?s what the balance is all about especially when this initial Ma is restored after every cycle. As seen from your citation as well, the energy at A is only taken into account when changing the initial state of the ball. The energy Ma is never taken into account separately as an initially input energy, as you erroneously suggested to be done in your previous post.

It most certainly must be taken into account when it is claimed that because (mgh1 + mgh2 + Kc) > (mgh1 ? (Ma ? Mb)), CoE is violated, as you say in this passage:

"On the contrary, the energy at C has always been an issue. Like I said several times already, the energy (mgh1 + mgh2 Kc) the ball has available at C to be transformed in other energies is greater than the energy (mgh1 ? (Ma ? Mb)) imparted to the ball. This is in violation of CoE."

No, Ma  in the sense you suggest, that is, (mgh1 + mgh1 + Kc ?Ma) most certainly must not be taken into account. As I already said many, many times, the only energy which must be taken into account when considering the amount of energy the ball has at C available to be transferred in other energies is (mgh1 + mgh2 + Kc).

The matter of the initial state and how to return to it is very much relevant. In the initial state (ball resting at A), the ball has Ma of magnetic potential, indisputably. The system was made that way. That it takes less than (mgh1 + mgh2 + Kc) to move the ball from A to B is a red herring. The system never has more energy than what it started with (Ma) plus what was put in by the hand (mgh1 ? (Ma ? Mb)), consistent with CoE. If you have an actual measurement that shows otherwise, then please present it.

Not correct. In returning at A where the ball has undisputedly energy Ma the ball must lose only energy (mgh1 ? (Ma ? Mb)) if CoE is to be obeyed. When CoE isn?t obeyed, as in out case, the ball obviously has available more than that mentioned (mgh1 ? (Ma ? Mb)) (it has (mgh1 + mgh2 + Kc)) to convert into other energies, including Ma which the ball has at A where it inevitably returns. CoE forbids the ball to have greater amount available for conversion in other energies such as, say, Ma, than the energy (mgh1 ? (Ma ? Mb)) which has been imparted to it. As you see, I will persistently repeat the truth about this situation no matter how inconvenient it appears to you and no matter how you would try to convolute the argument. Experiment proving the above is available.

[Omnibus]

@tinu,

What you're saying is incorrect. Please read carefully my explanation which I need not repeat and understand clearly what is lost, what is imparted and what is total energy. Mb is also imparted energy but out of no source. In order to have Mb at B someone must have imparted that energy by pulling the ball from C to B. There's no such entity (no such "someone") and the energy Mb is available at B out of nothing. That's the gist of the whole point which you have to understand well before commenting.

[tinu]

@omnibus,

I may have several questions during the next posts if you?ll be gentle to provide your answer:
What is total energy in C? Please write it down.

Thanks,
Tinu

[modervador]

No, Ma  in the sense you suggest, that is, (mgh1 + mgh1 + Kc ?Ma) most certainly must not be taken into account. As I already said many, many times, the only energy which must be taken into account when considering the amount of energy the ball has at C available to be transferred in other energies is (mgh1 + mgh2 + Kc).

As I've said before, I don't recall ever suggesting that the total energy at point C was anything other than (mgh1 + mgh2 + Kc). I said that the kinetic energy at C was Mb - mgh2. So once again we quite agree that the total energy at point C = (mgh1 + mgh2 + Kc) = mgh1 + mgh2 + (Mb - mgh2) = (mgh1 + Mb).

The latter, (mgh1 + Mb), is the total energy at point B, which happens to be greater than the energy put in by the hand, (mgh1 ? (Ma ? Mb)) to move the ball from A to B. This is not a problem, for point A already had an initial energy (Ma). Thus total energy at B equals initial energy at A plus energy from the hand:

Eb = Ea(initial) + Ehand = Ma + mgh1 ? (Ma ? Mb)) = (mgh1 + Mb).

In returning at A where the ball has undisputedly energy Ma the ball must lose only energy (mgh1 ? (Ma ? Mb)) if CoE is to be obeyed. When CoE isn?t obeyed, as in out case, the ball obviously has available more than that mentioned (mgh1 ? (Ma ? Mb)) (it has (mgh1 + mgh2 + Kc)) to convert into other energies, including Ma which the ball has at A where it inevitably returns. CoE forbids the ball to have greater amount available for conversion in other energies such as, say, Ma, than the energy (mgh1 ? (Ma ? Mb)) which has been imparted to it. As you see, I will persistently repeat the truth about this situation no matter how inconvenient it appears to you and no matter how you would try to convolute the argument. Experiment proving the above is available.

We agree that when the ball returns to A, it once again has Ma of magnetic potential, as well as other energy which optimally is in the form of kinetic energy. The total of this is therefore Ea(final) = Ma + Ka. This amount Ea(final) is available to convert into other energies, but I note that while the ball remains at A, only the Ka component is available (it must move from A to access the Ma part).

We agree that this Ea(final) comes from the full (mgh1 + mgh2 + Kc) available at point C which converts into other energies, including Ma which the ball has at A where it inevitably returns.

Ea(final) = Ma + Ka = (mgh1 + mgh2 + Kc)

Ka = (mgh1 + mgh2 + Kc) - Ma

Recall from above and earlier posts , (mgh1 + mgh2 + Kc) = (mgh1 + Mb) thus

Ka = (mgh1 + Mb) ? Ma = (mgh1 ? (Ma ? Mb)) = Ehand

Ea(final) = Ma + Ka = Ea(initial) + Ehand.

Is this what you?ve been leading us to all along?

[Omnibus]

Correct. Despite the fact that in both cases that follow the hand lifts the ball from A to B we have:


1) CoE violated:

Ea(final) = (mgh1 + mgh1 + Kc) = (Ma + other energies)

This is observed in the discussed case.



2) CoE obeyed:

Ea(final) = (mgh1 + Mb) ? (mgh1 ? (Ma ? Mb)) = Ma

This is not observed in the discussed case.