Magnet motor in Argentina

[Omnibus]

Thanks a lot for the translation, _GonZo_. This is what I understood so far:


Denote by Pn the starting point of the arm br and by Z1 the magnetic force which the rotor arm generates by repulsion.

In the external ring (the stator), the magnets that can be raised are denoted by M1, M2, M3, etc. with their respective repulsive magnetic forces F1, F2, F3, etc.

It is essential to consider that forces Z and F are directly related to the size of magnets and their composition. Thus we end up deducing that Z = F + F +  F = 3F.

The magnets that are lifted (the stator magnets) are cut at an angle alpha which can be varied in order to control acceleration, magnetic friction, and speed. The experiment shows that when increasing alpha the magnetic friction becomes greater, the greater initial acceleration and the lower the constant speed and vice versa.

On the other hand, the angle beta at which the rotor magnets are cut and the location of this cut allows to control direction of rotation.

Varying alpha and beta angles allows for the deviation of the lines of force. Thus, when increasing these angles the deviation of said lines is considerably greater which leads to the increase of the magnetic force. Conversely, when decreasing these angles the lines of force are turned aside.

Denote by Q the forces opposing the motion of the rotor. These opposing forces are friction, rozamiento (?), gravity, weight, etc.

At the starting point P1 the rotor arm is at rest generating force Z1. On the side of the rotor magnet M4 is raised by a mechanical system so that its repulsive magnetic force does not influence the rotor arm; the rest of the magnets Ms are all down.

As Z1 is opposed by F1, F2, F3 as well as by part of F15, F16 a net repulsive magnetic force X is generated equal to Z1 + F + F + F = X which when Z1 is replaced leads to X = F + F + F + F + F + F = 6F.

In conclusion, we observe that X >>> F4 (F4 is generated by M4; note also that F4 < F due to its displacement) and also that X >>> Q thus the rotor arm is forced to move to position P2 where M4 goes down mechanically while now M5 goes up thus repeating the whole process and allowing in this way the rotor arm to move infinitely due to the always present net force X = 6F ? F4 ? Q > 4.

[hartiberlin]

If you have a look at this picture to the above explanation,
the rotor is turning CLOCKWISE in this picture below here.
Now the question is:
When does M4 go up again to the rest of the stator magnets being in a row? ?

If the left front of the rotor is still at M5 position or if the right side of the rotor
has completely passed M4 already ?

So how many stator magnets are down altogether in each position ? Only one or 3 alltogether ?
(M4, M5 and M6 alltogether or just one of them ?)



Is the force calculation then correct as is stated above ??

[Omnibus]

The picture, as I understand it, seems to show the rotor being against M1, M2 and M3. They push on the rotor and cause it to move to the left were magnet M4 should be. Magnet M4 is raised, however, and doesn?t oppose the motion. Just when the rotor gets snugly against the M4 position, magnet M4 is dropped but magnet M5 is raised and so on.

[Omnibus]

Something like dangling the carrot in front of the donkey ...

[hartiberlin]

Okay, so only one magnet is down at a time,
so when M5 is going down M4 will be pulled up again ?
But then M4 has to be pulled up against a big counter force as the
rotor is still full frontal against M4 !Is this right ?
This will need a pretty big force to pull M4 again up !

Is the calculation still right then ?