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New electric motor principle

  1. Nov 5, 2003 #1

    Please, take a read to


    It's an study based on the Butch Lafonte's equilibrium motor.
    Please, forget the claims, and take a look under classical physical principles.

    Basically, on normal motors we use the electric energy to impulse the motor, and get similar mechanical work. On this device, the iron cores do the mechanical work, and the electric energy on coils "hide" the iron cores at proper timing, using electric energy, but almost without doing work.

    The simulations give a possitive result.

    Could you tell me what do you think?

    Forget the regrets about me, or if it's free energy. Just see the physical fact that a magnet between two north poles (or two south poles) could feel no oposition nor atraction to rotate properly disposed. It could just feel no effect, and also it could cause no backEMF effect on the coils.

    I wish you could see this fact with clear eyes. Sure you have more knowledge about physics than me, and i hope you could see what i'm talking behind this knowledge to explain if it works like spected, and not using other physics concepts to opose the results with the thermodinamics laws, the enthropy, the losses, the "build-it", etc...

    If you answer that the method explained can't work, just tell me how do you get that conclusion. If you talk about all the laws that could restrict the working, please, talk me about how they affect this system specifically, and if you see a solution to avoid that problem, please, let me know.

    If i expose this method to "physics people", they will use physics to go against it, not to analyze it.

    If i expose this method to "free energy people" they will say it sure works, using physics or not.

    So am I alone about this concept?.

    Please, i need REAL physics people that uses physics knowledge like a tool, not like a weapon.

  2. jcsd
  3. Nov 5, 2003 #2


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    If the electrical "control" signal is essentially counteracting the pole magnets it must do work to accomplish this. It requires work to build a field, wether it pulls on the rotor or not, does not matter.

    Perhaps this motor will spin, what ever it does it will consume electrical energy therefore, No free lunch.
  4. Nov 5, 2003 #3
    Hello Integral. A lot of time without talking with you.

    The work to build the fields also occurs on normal motors, isn't it?

    So normal motors have the same problem to create the fields, and also, receive the back EMF, and then need extra work:

    input work = Work to build fields + backEMF + mechanical work + losses

    On the new configuration, you must do work to build the fields, but you have not backEMF, and the mechanical work is caused by the iron cores, not by the coils:

    input work = Work to build the fields + losses

    so the efficiency looks much greater. Dont you think so?

    Could not the work of the atracting cores be greater than the electric work to build the fields?
  5. Nov 5, 2003 #4


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    It has been a while hasn't it!

    I really do not have sufficient information to judge the effiency.

    I must fall back on the old standard, build one. Then tell us what effiency you get.
  6. Nov 6, 2003 #5


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    No. To me it looks like you are doing the same amount of work in a different place.
  7. Nov 7, 2003 #6
    On normal motors, there is no limit (in principle) to the work you can do. If you input more electric energy, the rotor will do more mechanical work.

    On the new system indeed, there is an output work limit given by geometry and materials, not by the electric energy input. Given a geometry of the motor, there is a value of electric energy that is optimum to get the backEMF cancellation. At this point, the creation of the fields has almost no opposition to be created.

    From the point of view of the electric energy on active coils, they have not to do work to move the rotor (this is done by the off coils iron cores), and also they don't receive any work or opposition to create the fields in form of back EMF.

    So my question is:

    Why the work the iron cores can do to the rotor magnet MUST be equal to the work to build a field inside this iron cores without opposition?

    I think the work of the iron - magnet atraction phase is not related with the work of the active coil - iron magnetization phase.

    From the point of view of the active coils phase, it's like if the rotor magnet where not there. Only the iron cores phase "see" the magnet, so the work on one phase may be different than on the other.

    On one phase you've got a magnet been atracted by two iron cores, and two coils that seems not to be there (because the EMF should cancel).

    On the other phase, you've got two coils magnetizating (or magnetizing) two iron cores, and a magnet rotating that seems not to be there (it's rotating by inertia and the atraction of the other phase, and the coils don't give torque).

    So the electric energy is used like iron cores interruptor, when the energy is off, the cores are visible to the magnet. When current is on, the cores are not there to the magnet, that turns free.

    I hope you see what i mean, but tell me more specifically what do you think. Do you think the work must be equal?. Why?.
    Last edited: Nov 7, 2003
  8. Nov 7, 2003 #7


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    Yes, at this point, its efficiency would likely equal that of a typical motor.
    Because blocking a field with another field uses the same amount of energy as harnessing that field. You keep saying "without opposition." There IS opposition.
    Therein lies your error. The iron core isn't going anywhere and its field isn't going anywhere. Its always there and must be dealt with.

    from your site:
    Call it whatever you want, but you still have to factor it into your efficiency calculations.
    Last edited: Nov 7, 2003
  9. Nov 9, 2003 #8
    To get the work on the rotor magnet, you've got to create an specific H or B value with current.

    To get that value of H, a current I must run through the coils, but what V you need? H is I dependent, but not V, so finally, the output work depends only on the electric current specifically needed by coils construction, but the V is not relevant.

    So, to me, it seems that output work will be the same (by iron cores atraction) but the input power can be choosen (you only need a specific current value).

    I'm sure you don't think so, so i wait your answers.
  10. Nov 9, 2003 #9


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    If you decrease voltage, you'll need a higher amperage to get the same amount of work. Of course, if you don't give it the voltage needed to produce the correct work, the power output of the motor will decrease accordingly.

    Cala, one thing I've never understood about guys who come up with such ideas is the tenacious refusal to test the ideas. Some people spend years - decades even - developing an idea, while not spending the days or weeks required to test it. Its almost as if they subconsciously know the idea is flawed (though some are actual frauds, there are a lot who really seem to believe in their ideas).
    Last edited: Nov 9, 2003
  11. Nov 10, 2003 #10
    Go to the patent office website and find #512,340 by Tesla. (I'd put a link, but the site is not working now and I don't know what to type, I usually just go to the site, find the page, and copy the URL.) It describes a coil that at resonance has a big self-capacitance and little or no self-inductance.
  12. Nov 10, 2003 #11
    Russ, you said:

    "If you decrease voltage, you'll need a higher amperage to get the same amount of work. Of course, if you don't give it the voltage needed to produce the correct work, the power output of the motor will decrease accordingly."

    When you said that you were thinking on a NORMAL MOTOR, where the input work is related with the output work. On the new device, the output work is done by the iron cores, and the input current (not work) modulates the timing.

    The output work is not estracted from the electric energy input work (V and I) as on normal motors. It's stracted from the iron cores, and modulated by electric current (I) only. The voltage is not relevant, because the magnet will turn without opposition (no work) at the specific I needed.(At that specific current, you could use what V you want, because the magnet will turn almost without opposition, and then it needs no work to keep turning).

    On normal motors, the input energy (V and I) is what causes the output work. Normal motors are only TRANSDUCERS.

    The work on the new device is estracted from iron free magnetization, and the magnet is removed with no opposition, so no work, but a current is needed (output work depends only on the electric current). The elements that provides the output work are in some way "different" than the elements that "excite" the device.
    The new device is a kind of "ELECTRIC TO MECHANICAL TRANSISTOR".

    I think i've explained myself.

    Why do you think the input work is related with the output work on the new device?
    Last edited: Nov 10, 2003
  13. Nov 10, 2003 #12
    I think he was reffering to the fact that that the work is voltage*current*time so if you decrease the voltage you'll need a higher current to do the same amount of work.

    What do you mean work is extracted from the iron cores? Does and iron core produce work by itself and you need to extract the work from it?
  14. Nov 10, 2003 #13


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    Sorry, no. Like I said before, you can call it whatever you want, but your "control signal" does work and must be included in calculations of work.

    And coincidentally [sarcasm] if you calculate how much work is being done by that "control circuit" you will find it is slightly more than is being ouptut in mechanical work by the shaft.
    It is by definition. If you didn't input any work, your motor would stop spinning.

    Let me put it another way: you can call the input work whatever you want and even pull it out of your efficiency calculations if you want as long as you are clear about what you are doing (otherwise it would be fraud). But you won't ever sell a single motor because for anyone who would ever use an electric motor, the input work matters. Your motor offers those people no improvement over existing motors.

    Another example: I work with air conditioners. The coils have a certain efficiency and work output. But the fan motor gives off heat that takes away from the capacity of the air conditioner (the air conditioner has to cool itself). Even though the cooling coils are outputing a certain amount of work, that work lost to cooling the motor has to be taken into account in the capacity and efficiency ratings of the entire SYSTEM. Your motor efficiency calcuations work the same way.

    I recommend learning some thermodynamics. Thats where the rules for this sort of thing come from. And you must play by the rules if you want to be in the game.
    Last edited: Nov 10, 2003
  15. Nov 10, 2003 #14


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    Yes, thats pretty much it. People have been trying for centuries how to extract work from magnets without any input (aka free energy/perpetual motion). And for centuries they have failed. This failure was named "the first law of thermodynamics" by one such failed inventor.
  16. Nov 10, 2003 #15
    Hello Guybrush.

    When I say work is done by the iron cores i mean that a magnet near iron tends to magnetize it, and the magnet then feels atraction, this atraction goes higer meanwhile the iron goes magnetizing, and finally the magnet put itself in front of the iron core. There is work (W) due to the force of attraction (F) and the movement of the magnet (D), so W=F*D (in reality, we got to do this calculus with torque equation, i know).

    And from the other hand, yes, you're right, russ was refering to this fact: W = V*I. This formula is also applicable to the new device, but he didn't see that once you've got the specific current needed, you get the output work, independently of the input work (or V*I ratio). It doesn't matter if you increase the voltage more, the output work will be the same, because the coils are not the source of the output work. The source of the work is the magnetization proccess of iron.

    The input you need here to get the mechanical work is not electric WORK, but electric CURRENT only. If you input the specific electric current needed, it doesn't matter if you apply 5 V or 5.000 V, the torque will cancel in the same way, and then, the magnet will turn exactly with the same velocity, with 5 V or with 5.000 V.

    The output work is limited by the material and geometry of the cores.

    The input electric current value depends on this geometry, the number of windings and coil length...

    The electric potential value has no influence in this working principle.

    So you can choose the work output - electric current ratio you want (choosing the correct materials and dimensions), and also you can do this proccess at the V value you want.

    That's why i think the output work could be greater than the input work, because the output work is given by the iron-magnet atraction (depending on the core's geometry), and the input work will be V*I (and you can take the V you want, only I is limited by the geometry that i talked before and the geometry of the coils).

    I'm not saying that this law of nature (W=V*I) is not working on the new device. I'm telling that this electric input work could not be equal to the output work on this device configuration.
    Last edited: Nov 10, 2003
  17. Nov 10, 2003 #16

    When you say that the input work is related to the output work by definition, it's true to normal motors, because this normal motors uses the electric energy to impulse the rotor, and then the input work is the same (or more, counting the looses) as the output work, BY THE DEFINITION OF THIS PROCCESS,BUT IT COULD BE NOT LIKE THAT IF YOU CHANGE THE PROCCESS.

    The new device doesn't use electric work to impulse the rotor, so the input is not directly related to the output by definition. Of course, the output work MUST become from another source, and i think that the gaining on power becomes from the magnet magnetic field.

    Take the example of a transistor:

    A transistor is an element of 3 conections: The input signal, the output signal, and another terminal, usually connected to a specific potential.

    You can get an output signal with more power than the power the input signal carries. The input signal only has to activate the device (the transistor) to "launch" the proccess. The extra energy of the output signal comes from the source of potential.

    I think the new motor is like a transistor, but uses an electric "input signal", a magnetic "potential source" to do the work, and a mechanical work is the result or "output signal".

    On the new device, the magnet is acting like the potential source on a transistor, the electric current is the input signal that launch the proccess, and the output signal is the mechanical work.

    On a transistor, if you take into acount the input and the output, there is a gain of power. If you take into acount the input, the output, and the source, no gain is taking place.

    That's exactly what i think is happening here. If you take the input electric work, and the output mechanical work, there is a gain. If you take the electric input, the mechanical output and the magnetic field of the magnet as the source, then maybe there is no gain.

    I think the gain on the output power comes from the modulation of the iron-magnet atraction phases.

    I think you can remove the magnet from the iron cores without work, because you're not removing them in fact. The next cores and the inertia do the removing, you are only "hiding" some cores with the input power. You're not making repulsion, nor atraction, you are doing both (and that is to say you're doing none), so finally, that electric current is doing NO WORK.

    If you have one man pulling from a rope, and another man pulling from the same rope exactly with the same force on opposite direction, Are they making work?. The answer is no.
  18. Nov 10, 2003 #17
    A transistor will follow the rules of thermodinamics: It will offer more output than input signal meanwhile it is connected to a power source on the 3º terminal. When the power source is empty, the transistor could not work.

    I wonder how many times a magnet can be atracted by an iron rod if you put the magnet and the rod at the same distance again...
  19. Nov 10, 2003 #18


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    Thats the definition of efficiency of ANYTHING, not just a motor, cala. Like I said, do it however you want, but you won't convince people that their power bill is irrelevant.
    Again, call things whatever you want, but I doubt you will convince anyone anywhere that a voltage times an amperage doesn't equal a work/energy. Thats another definition and its an important one since thats what your electric bill is based on.

    It would also appear that you don't understand what an inductor is and how it relates to your energy input.

    It also appears that you don't understand how the motion of a metal core on each side of a magnet will be symmetrical - and from that you can deduce the mechanical work done on one side is exactly equal to the mechanical work done (or not done but cancelled out by the electrical work) on the other side. Thats a simple kinetic/potential energy transition. Magnetic potential energy is converted to kinetic energy (rotation of the motor) then the return of that kinetic to potential energy is blocked by your "control signal." By the symetry alone you can calculate the "control signal" must do as much electrical work as the magnets did mechanical work.
    Essentially unlimited of course. Try as you might though, you won't ever get them back to the starting point without an input work exactly the same as the energy gained when they came together.
  20. Nov 11, 2003 #19
    Cala, you can't have CURRENT without VOLTAGE. Use superconductors if you want but there's still gonna be a small VOLTAGE drop on the conductor. That means that you're doing WORK. Keeping those iron cores magnetized requires WORK, and if you ignore it you battery will run out eventually and you motor will stop.
  21. Nov 11, 2003 #20
    although I'm not sure what a 3º terminal is (probably my english is not so good), I suppose you're refering to the fact that the transistor can amplify a input signal. This is done because the the transistor has a source and it takes energy to amplify the signal from there.
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