SMOT! - (previously about the OC MPMM)

[hoptoad]

I like to explain things, about how the real world work. I like to provide examples on how a device will work based on common sense, basic physics, and experiments. It would be nice if you could do the same.
Vidar

@Vidar
Don't hold your breathe. Omnibus is only capable of semantics and pushing his own opinions using the "Goebels" effect.

[Low-Q]

I like to explain things, about how the real world work. I like to provide examples on how a device will work based on common sense, basic physics, and experiments. It would be nice if you could do the same.
Vidar

@Vidar
Don't hold your breathe. Omnibus is only capable of semantics and pushing his own opinions using the "Goebels" effect.

I think so. The Joseph Goebbels effect is used by many people when "The truth is too terrible for most to conceive, and the lie is too comfortable". Omnibus are by my opinion still wrong about his claims regarding the discussed topic.

[Omnibus]

I like to explain things, about how the real world work. I like to provide examples on how a device will work based on common sense, basic physics, and experiments. It would be nice if you could do the same.
Vidar

@Vidar
Don't hold your breathe. Omnibus is only capable of semantics and pushing his own opinions using the "Goebels" effect.

I think so. The Joseph Goebbels effect is used by many people when "The truth is too terrible for most to conceive, and the lie is too comfortable". Omnibus are by my opinion still wrong about his claims regarding the discussed topic.

Learn physics first before labeling people.

[modervador]

Focus just on the magnetic field and recall that the magnetic field force changes with distance unlike the force of gravity which is practically constant at all distances in our case. Then, recall what potential energy is. This is energy of the position isn't it and amounts to work necessary to move the ball from one position to another in the conservative field? Ma is the work necessary to be done to move the ball in question from C to A while Mb is equal to the work to move it from C to B, correct? Now recall what's work--force time distance, isn't it? Because of the changing magnitude of the magnetic field force Ma and Mb are integrals of that force time distance, correct? Thus, although at A (which is more distant than B from C) the force of the magnetic field is smaller than at B the integral in question at A is greater than that at B. Thus, in going from A to B the ball loses magnetic potential energy because Ma > Mb. Does this make it clearer?

The above quoted analysis is correct. Its intent was to define the magnetic potential and explain why it can be non-zero even when the ball feels little force attracting it to the magnets, but not necessarily to explain why Ma > Mb, which is handled elsewhere.

As Mr. Entropy has explained in this post,
http://www.overunity.com/index.php/topic,3833.msg70708.html#msg70708 ,
point B, defined by Omnibus as the input of the SMOT ramp, is outside a field null, because the must ball feel an attractive force which draws it from B all the way to C. Because point A is farther than point B from the nulls near the ends of the magnets, positive work is required to move the ball from B to A, thus Ma > Mb.

We must note that in the above linked post, Mr. Entropy defines zero magnetic potential as where the ball is separated from the magnets by infinite distance, and a negative value everywhere else. Omnibus defines zero potential as occurring at point C, and positive everywhere else. These definitions differ merely by a constant; Omnibus sets Mc = 0 while Mr. Entropy would set it as some negative number. By either definition, moving the ball from C to A, from C to B or from B to A requires a particular amount positive work against the magnetic field, respectively Ma - Mc, Mb - Mc and Ma - Mb, or simply Ma, Mb and Ma - Mb when Mc=0.

In addition to Ma > Mb, as a result of Omnibus? successful demonstration of the ball climbing the ramp from B to C and falling from C to A, we also know that Mb > mgh2 and Mb + mgh1 >= Ma. If these inequalities were not true, the ball would fail to rise all the way to C, get stuck at C, or fail to fall all the way to A.

Having started at point B and arrived tt point C, the ball has lost magnetic potential Mb while gaining gravitational potential mgh2, and this net loss in potential is reflected in a gain in kinetic energy,

Kc = Mb - mgh2.

Now, as Omnibus has clearly explained above, ?[focusing just on the magnetic field] Ma is the work necessary to be done to move the ball in question from C to A?. Thus after the ball has rounded the bend at C and arrived at A, it has gained magnetic potential Ma and lost gravitational potential mgh2 + mgh1. The kinetic energy of the ball at A will then be the sum of the kinetic energy at C and the net loss in potential along C to A,

Ka = Kc + mgh2 + mgh1 - Ma
= Mb - mgh2 + mgh2 + mgh1 - Ma
= mgh1 + Mb - Ma.

This is of course exactly the potential energy at point B, i.e. the energy that was expended by the hand to lift the ball from A to B, as has been explained in numerous previous posts.

An intriguing circumstance is when 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. It is then theoretically or perhaps even practically possible to locate point A where Ma = Mb + mgh1. In this case, when the ball traverses the B-C-A segment, it arrives with Ka = mgh1 + Mb - Ma = mgh1 + Mb - (Mb + mgh1) = 0, however because Ma > Mb, the ball will nevertheless experience an attraction towards B. The hand must supply mgh1 + Mb - Ma to place the ball at B, but mgh1 + Mb - Ma = 0 in this case. So the ball can theoretically traverse A to B with no help from the hand, and the ball again at B could again traverse B-C-A and the SMOT could self-sustain, as long as nothing were to cause the ball to arrive at A with less than Ka = mgh1 + Mb - Ma worth of kinetic energy.

[Low-Q]

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.

Br.

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