Magnet Motor from Argentina, part2

[2tiger]

Hi
Here some information from Torbay?s patent script.

Page 6
...logramos obtener una fuerza de rechazo magnetico que llamamos X y que sera igual a Z1+F+F+F = X  despejando la ecuacion nos queda X=F+F+F+F+F+F=6F.
Por otro lado Fq=F concluimos en que X>>>F4 (F4 es la F generada por M4) (aclaramos que F4<F debido a su desplazamiento y que X>>>Q por lo que obligamos al brazo rotor a desplazarce hacia la posision 2 (P2), por sistema mecanico M4 baja y se eleva M5 repitiendo todo el proseso anterior, de esta manera el brazo rotor se desplaza infinitamente (Pn+1)
                         X=6F-F4-Q=4F+
....  

translation:
... we obtain a repeling force we call X and the amount would be Z1+F+F+F = X = 6F.
(Z1 = magnetic force of the rotormagnet (covering 3 of the statormagnets), F  = mag. force of ONE statormagnet)
On the other side we have Fq=F and X>>>F4(where F4 is the force generated by M4) (also we explain that F4 < F because of its desplaced position (lifted up) and X>>>Q (where Q include al mecanical losses, so the rotor has only one way to go, to Pos2(P2).
By a mechanical system(push down system) M4 will be pushed down and M5 will be lifted up (regard the order - first he pushed down and then he lifted up!!!) so that the process starts again.
                         X=6F-F4-Q=4F+

Well, as I read my physics book well, a magnetic force can only exist by an interaction between a magnet and another magnet or ferromaterials in other words only when you create a mag. flux.
So I see a mistake an his first formula   Z1+F+F+F = X = 6F.
The repeling force X MUST be 3F, there is no other way.
F4 is smaller than F. But why???. When he says F4 is the force created by M4, he surely means the lower repeling force of the lifted up stator No. 4 (perhaps there is an atraccion force, I don?t know!) so it is a positive value non a neagativ.
Also he supose that all mecanical/friccion losses Q are equal F. So all of you who have a model of this motor report that there is enough torque for lift up a stator, but no enough to feed  a pushdown mechanism, i think that Q must be bigger than 1F, perhaps 2F.
And where in the formula is the forcevalue to push down or lift up a stator?

From my point of view his last formula on page No. 6 must be the following:

              F4=+1F       Q=-2F    ->     X=3F+F4-Q=2F+

The only way to get a continiusly movement is, that you push down a stator ONLY with 1F in order to get 3F repeling force for start the cycle again.
But then you only have a selfrunner   -  NO OU, sorry!

Any kind of critical replies are welcome. Thanks.

Cu
2Tiger

[eavogels]

Perhaps, since 3 stator magnets are in front of the stator, that is 3 times as big, he calculates the rotor force (Z1) equal to the 3 forces of the stator magnets. Z1 = F+F+F
/Eric.

[2tiger]

Perhaps, since 3 stator magnets are in front of the stator, that is 3 times as big, he calculates the rotor force (Z1) equal to the 3 forces of the stator magnets. Z1 = F+F+F
/Eric.

Z1=F+F+F but ONLY when you have these three stator magnets in front of the rotormagnet, otherwise you cannot talk about a magnetic force.

[Mica Busch]

This goes in addition to Tiger's comment, of lifting the stator ahead of the three already lifted. I recall seeing a model where the rotor was a complete circle, North/South along its diameter. Right now we are working on repulsion ONLY and half a rotor, but if the other half was present, providing an attractive force, no rigging would be needed to pull the stators back down, the attractive side would do that for us. All we would need to be concerned with is lifting the magnet ahead of the repulsion zone, to let the magnetic "bubble" we are squeezing around move. Thats the ONLY force to contend with, LIFTING a stator out of attraction so the repulsion side can rotate to alignment...

Just a thought 

[gn0stik]

Hi
Here some information from Torbay?s patent script.

Page 6
...logramos obtener una fuerza de rechazo magnetico que llamamos X y que sera igual a Z1+F+F+F = X  despejando la ecuacion nos queda X=F+F+F+F+F+F=6F.
Por otro lado Fq=F concluimos en que X>>>F4 (F4 es la F generada por M4) (aclaramos que F4<F debido a su desplazamiento y que X>>>Q por lo que obligamos al brazo rotor a desplazarce hacia la posision 2 (P2), por sistema mecanico M4 baja y se eleva M5 repitiendo todo el proseso anterior, de esta manera el brazo rotor se desplaza infinitamente (Pn+1)
                         X=6F-F4-Q=4F+
....  

translation:
... we obtain a repeling force we call X and the amount would be Z1+F+F+F = X = 6F.
(Z1 = magnetic force of the rotormagnet (covering 3 of the statormagnets), F  = mag. force of ONE statormagnet)
On the other side we have Fq=F and X>>>F4(where F4 is the force generated by M4) (also we explain that F4 < F because of its desplaced position (lifted up) and X>>>Q (where Q include al mecanical losses, so the rotor has only one way to go, to Pos2(P2).
By a mechanical system(push down system) M4 will be pushed down and M5 will be lifted up (regard the order - first he pushed down and then he lifted up!!!) so that the process starts again.
                         X=6F-F4-Q=4F+

Well, as I read my physics book well, a magnetic force can only exist by an interaction between a magnet and another magnet or ferromaterials in other words only when you create a mag. flux.
So I see a mistake an his first formula   Z1+F+F+F = X = 6F.
The repeling force X MUST be 3F, there is no other way.
F4 is smaller than F. But why???. When he says F4 is the force created by M4, he surely means the lower repeling force of the lifted up stator No. 4 (perhaps there is an atraccion force, I don?t know!) so it is a positive value non a neagativ.
Also he supose that all mecanical/friccion losses Q are equal F. So all of you who have a model of this motor report that there is enough torque for lift up a stator, but no enough to feed  a pushdown mechanism, i think that Q must be bigger than 1F, perhaps 2F.
And where in the formula is the forcevalue to push down or lift up a stator?

From my point of view his last formula on page No. 6 must be the following:

              F4=+1F       Q=-2F    ->     X=3F+F4-Q=2F+

The only way to get a continiusly movement is, that you push down a stator ONLY with 1F in order to get 3F repeling force for start the cycle again.
But then you only have a selfrunner   -  NO OU, sorry!

Any kind of critical replies are welcome. Thanks.

Cu
2Tiger

Until someone builds the motor to his patent specs we cannot take our test results and apply them to his equations. Every single replication attempt has had something not quite right about it according to his patent, ie wrong magnet pole alignment, no shaped stator magnet, not enough stator magnets, etc.

To me, the fact that we have gotten any results whatsoever is amazing. We did have one person here who claimed to have a slow turner, and he said he built it shabbily, and it needed refinement. Once we build one to specs, THEN we should modify and experiment, until then, we cannot speculate about output power based on our own experimentation.

Just an observation.

I was however curious about the results of eric's design, since it appeared to have so much torque. (it shook the table it was on).