Yes, seems
kukulcangod
has also the wrong stator magnets.
You must turn them by 90 degrees towards the rotor axis with their
big surface...otherwise the forces are too weak !

Hi all
Reading your posts I notice that you have all the same problem.
That the created torque is enough to lift up a statormagnet but is too weak to bring them down after passing them.

So just an idea based on this.
I think that the kinetic energy of the rotor created by the repulsion ist not high enough to do both.
What would happen if you build a unit with a bigger radius so that the rotor magnet has a longer "way" to run "free" and store more kinetic energy in the rotor?

And - if I remember well you all are trying to liftup the first stator just in front of the stator. Torbay uses a rotormagnet that covers three statormagnets at a time.
Have anyone try to lift the third statormagnet in front of the rotormagnet? I think that this would be the better timing. The donkey will run faster if he see the carot more far away as he has it hanging down right in front of his nose.

Please tell me if I?m complete wrong.

CU
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




 

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.

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).

 

Hello Everybody,

Sorry I have not gotten the promised simulation files up online yet. I just spoke to Maran about this today and I will post the dxf's of the pieces I have currently setup. I was trying to wait until I could find the time to put the rest of the screws and bolts into the model but it appears that I am not finding that precious window I need to do it. So I'll at least post the pieces I have made so far and just update the files as I finish it up. The only thing I don't have modeled are the brackets to mount the rollers on top of the stator pieces but I'm assuming that most people would just make some makeshift ones out of sheet metal anyway so I won't stress about that too much. I am not at my normal computer right now so I will post the existing part files tomorrow afternoon.

God Bless,
Jason O

Here are some more screenshots: