SMOT! - (previously about the OC MPMM)

[modervador]

In the observed case, however, the ball at B loses Mb (which transforms equivalently into other forms of energy such as mgh1 and Kc) in addition to mgh1 which it will inevitably lose when back at A. Thus, the ball at B in the discussed case loses, that is, converts into other kinds of energies, the energy (mgh1 + Mb) = (mgh1 + mgh2 + Kc) and not (mgh1 + mgh2 + Kc - Ma) as you incorrectly state. Observed inequality Ein =/= Eout, that is (mgh1 ? (Ma ? Mb)) =/= (mgh1 + mgh2 + Kc) is a clear violation of CoE.

The ball does indeed lose Mb along path B-C, but I assume you meant to say that this "transforms equivalently into other forms of energy such as Kc and mgh2 (not mgh1)".

But I never said that (mgh1 + Mb) = (mgh1 + mgh2 + Kc - Ma). I said that Kc = Mb - mgh2. Thus when you write "(mgh1 + Mb) = (mgh1 + mgh2 + Kc)" I have to agree because

(mgh1 + Mb) = (mgh1 + mgh2 + Kc) = mgh1 + mgh2 + (Mb - mgh2) = mgh1 + Mb.

There are other interesting points in your post that deserve response, but I have work matters to which I must attend.

[tinu]

?I repeat,  if CoE is to be obeyed and Ein = Eout, the ball at B must lose exactly the amount (mgh1 ? (Ma ? Mb)), that is, exactly the amount of energy imparted to it from a known source along A-B.

In the observed case, however, the ball at B loses Mb (which transforms equivalently into other forms of energy such as mgh1 and Kc) in addition to mgh1 which it will inevitably lose when back at A. Thus, the ball at B in the discussed case loses, that is, converts into other kinds of energies, the energy (mgh1 + Mb) = (mgh1 + mgh2 + Kc) and not (mgh1 + mgh2 + Kc - Ma) as you incorrectly state. Observed inequality Ein =/= Eout, that is (mgh1 ? (Ma ? Mb)) =/= (mgh1 + mgh2 + Kc) is a clear violation of CoE.?

Oh, I think I understand it now...

Kc=Mb-mgh2 (where Mb>0 and Mc=0). Thus, it results that Ec (total energy in C, kinetic and potential) is Ec=Mb-mgh2, which is not equal to the imparted energy mgh1+Mb-Ma, hence CoE violation?
That?s correct in the first part but who says the two energies have to be equal, in the first place?!!!
It?s easy to see that Ec>mgh1+Mb-Ma. Does it mean CoE violation?
Yes and no.
Yes in the same manner as a rock falling onto your feet. Indeed, the rock falling once may seem energy out of nothing. Your leg gets broken.
No, because is you want to have a potentially ever leg-breaking stone, you still need to bring the stone at the initial height. So, instead of breaking your own leg once (which is not wise, by the way ), better put a spring down-there and the stone will jump back to the same height as it was at the beginning. But then you have no free energy anymore and no CoE violation.

In short, C is a potential fountain which was not there in the absence of SMOT. Kc>mgh1+Mb-Ma but if you extract that Kc energy, the ball will never leave SMOT anymore.

?The energy (mgh1 + mgh2 + Kc) equivalent to (mgh1 + Mb) the ball has at C exceeds the energy (mgh1 (Ma ? Mb)) which has been imparted to the ball. CoE does not allow such excess of energy.?

I have to disagree with this also. If I have to raise a weight from the second to the third floor, at any time I may choose to secure the weight with an elastic wire and let it drop. Any external observer will see the weight falling to the ground floor (or bellow; you name how deep it falls) and being brought back by the elastic force. Like in SMOT, energy computation will show that the weight-elastic wire system is having at times more energy that I put in it in the first place. But there is no CoE violation. (And that surplus energy can be extracted once at the cost of the weight remaining bellow the first floor, where I started?)

I guess this story is now daylight clear to me and very close to its end.
But I?m ready to listen if there is more.

Many thanks,
Tinu

[Omnibus]

@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).

The fact that the ball has energy (mgh1 + mgh2 + Kc) to transfer (and it is actually transferred) into other kinds of energies such as, say, Ma, when it returns where it started, that is at A, which is greater than the energy (mgh1 - (Ma - Mb)) imparted to it is in violation of CoE. 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.

[Omnibus]

@tinu,

Read carefully the explanations. You haven't understood it correctly.

[tinu]

@tinu,

Read carefully the explanations. You haven't understood it correctly.

Ein=mgh1+Mb-Ma.
But please also write the explicit formula of Eout.
Thanks,
Tinu