SMOT! - (previously about the OC MPMM)

[tinu]

@ modervador,

Very good post! It pretty much summarizes most of what was told on the subject.
And glad to see you here!

I?d pretty much like to hear your thoughts on the following:
1. Ka=mgh1+Mb-Ma is also the energy hand supplies. Hence, in the absence of friction and any other loses and also by using a bend& appropriate track, SMOT will, of course, self sustain. But no gain of energy.
2. Why do you think omnibus sees a gain of energy in SMOT? (The answer is not clear to me to these days?)

Welcome again,
Tinu

[Omnibus]

@Modervador,

You have stated this argument before but I?ve already told you that it is wrong because you don?t realize that the energy Mb the ball has at B (in addition to the energy mgh1 it also has at B) and which transforms equivalently into (mgh2 + Kc) at C, according to the not-violated here ?transformation? part of CoE, has appeared out of nothing.

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.

It is interesting to note that the ball is spontaneously recharged with the energy Mb, available at B for the ball to spontaneously transfer in other kinds of energy (such as the mentioned mgh2 and Kc), when the ball completes the loop at A. Thus, upon a subsequent loop the researcher again encounters the same surprising fact, namely, that the ball at B has spontaneously available to transform into other energies (and actually transforms it) greater energy (mgh1 + Mb) than just the energy (mgh1 ? (Ma ? Mb)) he or she imparted to it. That isn?t the case if CoE were obeyed?if CoE were obeyed, the ball at B would have transformed into other energies only exactly the amount (mgh1 ? (Ma ? Mb)) the researcher has imparted to it.

Hope from the above you can also figure out why the fact you seem to emphasize that when back at A the ball has again energy Ma available to it, provided a subsequent loop is to be carried out, isn?t a proof that CoE is obeyed.

[modervador]

You're maybe correct, but the gravitional force at point A is too much for the ball to move upwards to point B, where the ball in addition is met by an extra downforce due to the small area of repulsion before it arrives at B. However, if point A was much closer to the magnets, it might be lifted a little bit, but then the the ball also would be more sensitive to the magnets in point C as well, and maybe gently moved/rolled in direction of, but below, point C (?). As the magnets are tilted a bit upwards in point C, the flux density the ball "feels" might be most neutral in x-direction right between but below point B and C (?)

I however think that Omnibus is right IF the ball would be lifted by itself from point A to point B, but that will never happen, or what? That is the point where I get a little bit confused.

Hi Vidar, I hope that I might relieve what part of the confusion I may have caused.

The case I was discussing at the end was mainly mental exercise that is of no practical use except for academic entertainment. It is the particular case where Ma = Mb + mgh1. This means the ball can theoretically get from A to B with no net change in potential. You correctly point out that whether the ball can do this in practice depends on whether the pull from the magnet towards point B can overcome the pull of gravity while at point A. This is why I stated that one condition is that "the track at point A is horizontal or gently sloping, such that the ball experiences little or no net downward force by gravity unmatched by upward force of the track." The track keeps the ball from falling straight down and the trajectory from A to B is largely horizontal. One way to do this is have point A to the left of B instead of under the magnets, and of course the track from C to A will have to be shaped accordingly; this would differ somewhat from the picture that we're used to seeing.

The ball gains kinetic energy on the way from A to B as it loses magnetic potential. If there is any "speedbump of potential" along the way, the ball can surmount it only if the bump is lower than the ball's current kinetic energy when it encounters it. It's possible, I think.

Omnibus is correct that "Ma is the work necessary to be done to move the ball in question from C to A? against the magnetic field regardless of whether this case I described can be practically achieved or whether the ball goes from A to B without help from the hand. The phenomena are independent of each other.

[modervador]

@ modervador,

Very good post! It pretty much summarizes most of what was told on the subject.
And glad to see you here!

Iââ,¬â,,¢d pretty much like to hear your thoughts on the following:
1. Ka=mgh1+Mb-Ma is also the energy hand supplies. Hence, in the absence of friction and any other loses and also by using a bend& appropriate track, SMOT will, of course, self sustain. But no gain of energy.
2. Why do you think omnibus sees a gain of energy in SMOT? (The answer is not clear to me to these daysââ,¬Â¦)

Welcome again,
Tinu

Thanks Tinu. This is an interesting forum.

I agree with point 1, and am glad you finished the train of thought for me.

As for point 2, I believe the current reason is that Omnibus does not recall his own statement, "Ma is the work necessary to be done to move the ball in question from C to Aââ,¬Â. which leads him to say things like "However, if the same amount |(mgh1 ââ,¬â€œ (Ma ââ,¬â€œ Mb))| of energy is imparted to the ball and the ball doesn't settle with B as an apex but, as experiment shows, instead reaches another apex C then, obviously, when the ball returns back at A the ball loses amount of energy different from the amount |(mgh1 ââ,¬â€œ (Ma ââ,¬â€œ Mb))| imparted to it."

For him to leave out the gain of magnetic potential (which is also loss of kinetic energy) along path C-A is certainly not conventional physics and, in my recent memory, it has not been explained beyond the bald assertion that it is so. Nor has it been proven by his experiment, which can be simply explained by conventional physics which predicts the ball will go from B to C to A by itself if Mb > mgh2 and Mb + mgh1 >= Ma. To show that the ball has excess energy at A would require a corroborating measurement of that energy, which has not been done. The claim is extraordinary, the proof is sub-ordinary.

[Omnibus]

@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 (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. CoE requires that only the imparted amount of energy (mgh1 ? (Ma ? Mb)) should be the available energy to be transformed into other energies such as, say, Ma, when the ball returns at A.