SMOT! - (previously about the OC MPMM)

[Omnibus]

For CoE to be obeyed the energy lost by the ball must exactly equal the energy imparted to the ball.

I agree. However, Ehand I believe is not only (mgh1 - (Ma - Mb)). I think it is (mgh1 - B_{(x  + y)} - (Ma - Mb)), where B_x is  exactly the point where the ball is starting to be (slightly) repelled by the magnets, and B_y is the point where the ball is not repelled or attracted to the magnets (The very top of the hill, seen as a virtual example). Both B_x and B_y is between point A and B, and B_y is closest to B. And where the violation comes in I believe is confused with the extra energy B_{x + y} you must apply with the hand in order to place the ball at point B. So there I think you got that other energy from, but not taken into concideration in the original equation.

Any thaughts about this? - [just4fun]Except for telling me that it is not correct[/just4fun]

Cheers

No, Ehand is exactly (mgh1 - (Ma - Mb)).

[modervador]

[excerpted quote]
As seen above, in SMOT the energy Ein = (mgh1 - (Ma - Mb) = Ehand imparted to the ball to acquire energy (mgh1 + Mb) at C is only part of the energy (mgh1 + mgh1 + other energies) the ball loses. In SMOT the energy (mgh1 + Mb) lost is greater than the energy (mgh1 - (Ma - Mb)) = Ehand imparted.  In SMOT the energy which the ball will have when back at A will be the equivalent to the energy (mgh1 + Mb) = (mgh1 + mgh2 + other energies) = (mgh1 - (Ma - Mb) + Ma =  Ehand + Ma. This is because of the "transformation" part of CoE which isn't violated in SMOT. I repeat, back at A the ball will have the equivalent of (mgh1 + mgh2 + other energies) which is the same as (Ma = Ehand) if we want to be facetious and split up energies in such a way to deceive people but the ball will not have energy (Ma) back at A. Energy (Ma + Ehand) at isn't the same as energy (Ma) at A. The ball would've had energy (Ma) back at A should CoE be obeyed. Now that CoE isn't obeyed the ball back at A has amaount of energy greater than (Ma), dishonestly split by @modervador for purposes of deceiving people into (Ma) +Ehand.

Hope this helps.

Now, I'm expecting from @modervador, already very clear as to why SMOT disobeys CoE to open a thread in the Steorn forum and explain to several curious souls there that violation. That's the honorable thing to do.

From what I can tell from several posts, CoE has been said to be violated on two ground:
1. (mgh1 + mgh2 + Kc) > Ein
2. Ea(final) > Ea(initial).

I agree that these equalities are true, but to defend them as violations of CoE, against questioning likely to arise on the Steorn forum, requires nailing down more details.

Let's discuss the idea of "Elost", the energy lost by the ball. Presumably, the ball's loss is our gain, however we can't use the energy unless the ball gives it up. So examining the total energies at points A, B, C and A again, we have:

Ea(initial) = Ma
Ein = (mgh1 ? (Ma ? Mb))
Eb = Ea(initial) + Ein = Mb + mgh1
Ec = (mgh1 + mgh2 + Kc)
Ea(final) = Ma + Ka

While the ball remains at initial point A, it has only potential Ma and no energy can be lost; the ball must move away from A to lose all or part of Ma. The same holds for point B; the ball must move away from B to lose all or part of Mb and/or mgh1. However, once tha ball has accelerated to C, if remaining at point C it has accessible kinetic energy, Kc, which  may be harvested and sent outside the system. Therefore, the ball may lose all of this Kc, but to lose all or part of (mgh1 + mgh2), the ball must move away from C, which is exactly what happens when the ball travels onward to return to A without any of the Kc having been removed in this example.

On return to A, the ball loses (mgh1 + mgh2) and the total (mgh1 + mgh2 + Kc) is transformed into Ma + "other energies", the latter being in the form of kinetic energy Ka, hence Ea(final) = Ma + Ka. At this final point A, the ball can lose only kinetic energy Ka, for to lose all or part of Ma, the ball must move away from A. Therefore, while at final point A, the maximum energy the ball can lose (and we can gain) is Eout = Ka.

With this out of the way, let?s address: Is (mgh1 + mgh2 + Kc) > Ein a violation of CoE? I can?t really say that it is without receiving further explanation. It is clear that (mgh1 + mgh2 + Kc) = (Mb + mgh1) is a total energy which results simply from raising the initial state energy, Ea(initial) = Ma, by the amount Ein = (mgh1 ? (Ma ? Mb)), i.e. (Mb + mgh1) = (Ea(initial) + Ein). Energies are simply transformed as the ball moves (or is moved) from A to B to C, but all energies are accounted for and no sources appear out of thin air. CoE violation would require that energy appear to spring from or vanish into thin air.

Is Ea(final) > Ea(initial).a violation of CoE? Again, I can?t really say that it is without receiving further explanation. It is clear that because of the "transformation" part of CoE which isn't violated in SMOT, Eb is transformed completely into Ea(final), hence
(Mb + mgh1) = (mgh1 + mgh2 + Kc) = Ma + Ka = Ea(final).
Since Ea(final) = (Mb + mgh1) and as already explained (Mb + mgh1) = (Ea(initial) + Ein), there appears no basis to claim that Ea(final) contains unaccounted energies that appear to spring from thin air. Ea(final) is simply Ea(initial) + Ein, which for any positive Ein would mean that Ea(final) > Ea(initial), consistent with CoE?a trivial result.

Furthermore, it is also easily demonstrable (and has been shown) that all of Eout = Ka is a result of Ein supplied to move the ball from A(initial) to B, transforming into other energies on the path B-C-A. Thus barring friction or other losses, Eout = Ein.

If I am to make a convincing case to the Steorn forum that SMOT violates CoE, I need more explanation to counter these arguments that I have put forth. Far better to have the counterarguments already in hand, than to present a case that can be easily shot down by arguments the presenter should have seen coming but cannot answer.

Thank you for your time.

[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.

Therefore, Ea(final) > Ea(initial) contains energies coming out of no source. This is a violation of CoE.

If CoE were not violated Ea(final) = Ea(initial), as shown.

You already understand this although you?re playing as if you don?t. Now you?ve heard it again and are fully prepared to open a thread in the Steorn forum and explain to the curious fellows there where your mistake was and why you?re convinced now that SMOT violates CoE.

It's preferable not to clutter further this thread with seeming misunderstanding and do what's honorable to do right away--explain to some people in the Steorn forum what your confusion was and how you now clearly understand that SMOT violates CoE.

[modervador]

However, Ehand I believe is not only (mgh1 - (Ma - Mb)). I think it is (mgh1 - B_{(x  + y)} - (Ma - Mb)), where B_x is  exactly the point where the ball is starting to be (slightly) repelled by the magnets, and B_y is the point where the ball is not repelled or attracted to the magnets (The very top of the hill, seen as a virtual example). Both B_x and B_y is between point A and B, and B_y is closest to B. And where the violation comes in I believe is confused with the extra energy B_{x + y} you must apply with the hand in order to place the ball at point B. So there I think you got that other energy from, but not taken into concideration in the original equation.

There might be a local minimum that the ball passes through on the way from A to B, however for the purposes of everything that occurs after the ball is let go at point B, it is sufficient to consider only the magnetic potential at B, which has been defined as Mb. Any magnetic potential energy you must put in to get from point Bx to By would have already been given to you on the way from A to point Bx. The net change in magnetic potential from A to B through points Bx and By is (M_Bx - Ma) + (M_By - M_Bx) + (Mb - M_By) = (Mb - Ma), and the net change in gravitational potential is mgh1, hence Ein = mgh1 + Mb - Ma.

I am somewhat uneasy about a phrase such as "the ball is starting to be (slightly) repelled by the magnets", because I like to think that a simple unmagnetised steel ball is only attracted to magnets, not repelled. I think what's really happening is that the ball is attracted to a point that's somewhat "to the left" of the line drawn between the magnets near point B, due to the shape of the field. In terms of the force vectors, it's the same thing, but the different phrasings speak to the underlying cause in different ways.

[modervador]

@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.