SMOT! - (previously about the OC MPMM)

[Omnibus]

Cool. So where's your closed looped SMOT then?

The A-B-C-A loop we're discussing is a closed loop.

[Omnibus]

If CoE were obeyed, the ball starts at A with energy (Ma) it gains Ehand = (mgh1 - (Ma - Mb)), not other energies. Not other energies. Not other energies, you idiot. And ends up at A losing Ehand, ending up with energy (Ma). CoE isn't violated. This, however, isn't what's happening in SMOT.

We all agree with that. So far, so good.

In SMOT the ball starts at A with energy (Ma), gains energy Ehand = (mgh1 - (Ma - Mb)). And ends up at A losing (mgh1 + mgh2 + other energies). (mgh1 - (Ma - Mb)) =/= (mgh1 + mgh2 + other energies). SMOT violates CoE.

The "other energies" part, is quite confusing. By my point of view, "other energies" is confused with "detour". The ball is travelling a longer distance when not going directly from B to A - and the hand is virtually placing the ball slightly beyond the top of a hill so it will not fall back directly to point A. OFCOURSE!!

And as we all know, it does not require energy to move mass angular to gravity or magnetic fields. So the ball is just taking a detour via C before coming back at point A. And the reason why the ball is actually coming back to point A isn't because of "other energies" but because of the energy initially imparted to the ball by hand - namely slightly beyond the top of the hill.

There is no "other energies". CoE is obeyed.

PS! I already know that I'm, by your opinion, cluttering this thread with nonsense and heresy, so you don't need to comment that. WE ALL KNOW IT ALREADY.

Cheers

That's incorrect. Read carefully my explanations.

[Low-Q]

If CoE were obeyed, the ball starts at A with energy (Ma) it gains Ehand = (mgh1 - (Ma - Mb)), not other energies. Not other energies. Not other energies, you idiot. And ends up at A losing Ehand, ending up with energy (Ma). CoE isn't violated. This, however, isn't what's happening in SMOT.

We all agree with that. So far, so good.

In SMOT the ball starts at A with energy (Ma), gains energy Ehand = (mgh1 - (Ma - Mb)). And ends up at A losing (mgh1 + mgh2 + other energies). (mgh1 - (Ma - Mb)) =/= (mgh1 + mgh2 + other energies). SMOT violates CoE.

The "other energies" part, is quite confusing. By my point of view, "other energies" is confused with "detour". The ball is travelling a longer distance when not going directly from B to A - and the hand is virtually placing the ball slightly beyond the top of a hill so it will not fall back directly to point A. OFCOURSE!!

And as we all know, it does not require energy to move mass angular to gravity or magnetic fields. So the ball is just taking a detour via C before coming back at point A. And the reason why the ball is actually coming back to point A isn't because of "other energies" but because of the energy initially imparted to the ball by hand - namely slightly beyond the top of the hill.

There is no "other energies". CoE is obeyed.

PS! I already know that I'm, by your opinion, cluttering this thread with nonsense and heresy, so you don't need to comment that. WE ALL KNOW IT ALREADY.

Cheers

That's incorrect. Read carefully my explanations.

First you must explain:
do you mean:

...at C is only part of the energy (mgh1 + mgh1 + other energies) the ball loses. In SMOT the energy (mgh1 + Mb) lost is greater...

or:

And ends up at A losing (mgh1 + mgh2 + other energies). (mgh1 - (Ma - Mb)) =/= (mgh1 + mgh2 + other energies).

Btw:

In SMOT the energy (mgh1 + Mb) lost is greater than the energy (mgh1 - (Ma - Mb)) = Ehand imparted

If the SMOT loose more energy than the energy imparted by the hand, how can it violate CoE?

The ball is still going detour, OR "other energies" = Ehand.

Cheers

[Omnibus]

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

[Low-Q]

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