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Re-Inventing The Wheel-Part1-Clemente_Figuera

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  • seaad
    replied
    Simulation = bad

    -->MM

    ""Seead, Your simulation are not good to see OU effects. You can not expect to get OU results using computer programs ""

    And the sim. gave us a clear picture of that Inductances are better than Resistors for the G-part even if we did know that before somehow.

    Thanks MM!

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  • seaad
    replied
    Originally posted by hanon1492 View Post
    Seead, Your simulation are not good to see OU effects. You can not expect to get OU results using computer programs which have implemented (compiled) the equations for conservation of energy
    I know that but other ideas can be emerged as to achieve sinus from coils on a toroid e.g.
    And you don/t burn your fingers on the soldering iron.
    Last edited by seaad; 10-14-2016, 01:39 PM.

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  • hanon1492
    replied
    Originally posted by marathonman
    If the currant is taken below half way induction will fail from the declining electromagnet because the pressure was not kept between them. the magnetic field will take to long to build up so induction will fall to the rising electromagnet to 50 % output.

    if currant is not taken below half way the out put will be 100 % Electric field.

    this was passed to me and i verified it through own tests to be true. this is the major reason people are getting bad results.

    NEVER below half way.
    This is a piece of good info. I must confess that my test were mainly going below that half way. Maybe the reason for my little success

    MM, I see in your post #1333 that current is always going toward the toroid. If that is the case that same current must get out of the toroid always through the brush toward the battery. This is to obey the continuity equation in circuits ( law of conservation of charges ). That was my point that I could not see the current going back to the electromagnet (recycling effect)

    Seead, Your simulation are not good to see OU effects. You can not expect to get OU results using computer programs which have implemented (compiled) the equations for conservation of energy, and whose results must follow that equations. Tests are our only way to discover new frontiers. Your test is fine. I think that you were reaching 0 volt or negative to feed the electromagnets with that setup. Maybe you could have got even better results going above 0 volts, as Figuera did.
    Last edited by hanon1492; 10-14-2016, 12:55 PM.

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  • seaad
    replied
    Figuera test from February 100(+)% ?

    Figuera test from February 23 2016 100(+) % ? (Free Energy | searching for free energy and discussing free energy)

    My test set up. See picture below. (A)

    My best belived results are 99-104 % (output W/input W x100). RL adjusted to best efficiency about 1/5 ?? of max. out power. This result must be very good since the input coils have both ends open. The primary coil bobins are close or amost close to each other with the secondary inside in the middle. The core does not overhang (much) beyond the coil body. If so it gives a poorer result. Duble core rods, RL max Watts, and transformer laminations gives a poorer result. The contraption response is as a normal transformer when loaded to max output, with rising Ampere and decreasing phase angle, but with much higher input phase angle.

    I have done my best in terms of instrument readings and used only Volt readings from the same probe or instruments to both input and output to minimize measurement errors. I have also tried to use the same instrument scales. Regarding measuring the phase angles I have no possibility to doublecheck this. I have to trust Velleman's reputation (PCLAB2000SE) .

    Pic. (A): Signal generator. First a step up transformer, T1; 200Hz- 200kHz and split output phase 0deg. and 180 deg. . (Not absolutely necessary. Later I used parallel-coupled 'electromagnets' P1, P2 and just flipped one over.) T2;
    P1, P2 primary coils. ext. 44 x 39 x 30 mm int. 21 x 15 mm. Random wound -->, <--, -->, <-- both the same CW, CCW. 0.35mm (AWG 28-27) Cu. Inductance with core; about 80-100mH, 19 Ohm.
    S1 secondary 100 turns, 1,3 mH (as a small compact lump) 0.35 Cu direct wound over a single radio antenna ferrite rod 10 x100 mm. Load; RL+ RLi about 115 Ohm about 1/5 of maximum power out!!

    Input freq; about 5kHz. Input voltage V1, V2; about 6,7 Volt Input current; about 2 mA over Rpi1,Rpi2. Phase angle; 83,2- 83,5 deg. (cos ~= 0,1) Output; about 0,65 Volt .

    Arne
    Attached Files
    Last edited by seaad; 01-06-2017, 01:12 PM.

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  • seaad
    replied
    &quot;Figuera&quot; sim LT-spice XVII FLIP FLOP

    Simulation with about 67% efficiency. Switch pulses with 50% duty cycle.
    The same N,S,y transformer parameters. Load 130 Ohm.
    Arne
    Attached Files

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  • seaad
    replied
    Figuera simulation LT-spice XVII

    Figuera simulation with about 51% efficiency.
    More pics.
    Arne
    Attached Files

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  • bistander
    replied
    Originally posted by marathonman
    The currant is only traveling through both halves when at Set N or Set S high.
    That is where I did my analysis: brush at point 8, and then same applies at point 1.

    However, if you have different currents through N and S, does that not set points 8 and 1 at different potentials on the toroid and then current would flow in the half opposite the brush?

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  • bistander
    replied
    Toroid part G

    Originally posted by Ufopolitics View Post
    Bistander,
    ...

    Now about your comment above about Part G...

    Can I ask You if you are considering -at all- the fact that Part G is not connected directly to negative supply terminal?
    Sure Ufo, I considered the whole primary circuit including the supply. I'll post two of MM's images for easy reference.





    Originally posted by Ufopolitics View Post
    But instead, its negative connection derives from both sets of primaries coils in series?
    Yes, the negative connections go from the N and S primary coils to points 180 apart on the toroid winding (represented by 8 and 1 respectively).

    Originally posted by Ufopolitics View Post
    Wouldn't the fact that Part G component is connected right between the two sets of coils, which also have a resistance as an impedance?
    PartG is used as a voltage (or current) divider to proportion the current from the source to the primary coils N and S. It is desired to have a certain min/max relationship of current through N and S, where N is max when S is min and vice versa.

    At 12:00 (point 8) on the toroid is where primary N connects, so when the brush (being at supply positive) is at point 8, maximum current goes to N. It is desired that at this point, minimum current will flow to primary S which is connected to point 1 on the toroid.

    Current must flow through the toroid winding between point 8 and point 1 to have continuity to supply positive via the brush. There are two paths in the toroid winding between 8 and 1, the right side and the left side, which are in parallel with each other and of equal lengths and turns. So to figure the impedance of the toroid winding (8 to 1) we have 2 impedances in parallel, each with equal resistance and inductive reactance. Because the inductance of the 2 coils is linked, the mutual inductance must be considered. Because the current direction is opposite in the 2 halves of the toroid winding, that mutual inductance essentially cancels the inductive reactance for the equivalent impedance.

    Therefore, the potential difference between point 8 and point 1 will for all practical purposes be zero. This means the current will not be a minimum in primary S as desired, but essentially maximum, the same as in primary N.

    I only looked at this one position (brush at point 8) because this case has the 2 toroid winding halves equal and the parallel combination of 2 equal inductance coils with mutual inductance is easily calculated. Obviously the same is true for brush at point 1. All the other points in between I am unsure and not about to attempt those calculations.

    I see the same results looking at it another way using the flux to figure inductance. I mentioned this to you before. When you have half of the toroid mmf CW and the other half equal and CCW, resulting flux is zero and therefore the inductance of the coils on the toroid is zero. As the brush rotates around the toroid, the potential changes and mmf (Ampere-turns) are unlikely to be equal and opposite so there may be inductance in the coils and impedance causing unequal currents in primary N and primary S.

    My take anyway,

    bi

    edit: Weird thought. What if you wound the left half of the toroid backwards from the right side half? Mutual inductance would add instead of subtract.
    Attached Files
    Last edited by bistander; 10-14-2016, 03:45 AM. Reason: additional thought

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  • hanon1492
    replied
    Ufo,
    Your sketch is wrong. 9 is the same as 8 (paired in the rotary commutator 8 with 9), 10 is paired with 7, 11 with 6, 12 with 5, 13 with 4, 14 with 3, 15 with 2, and 16 with 1. Note the connection of contacts inside the rotary commutator. This is what made the continuous increase-decrease of each signal when approaching or leaving each side tap with the electromagnets. Follow the brush contact along a whole turn.

    Seead, If the graph is of the signal feeding the electromagnet Ufo is right about that simulation. Your system must avoid going into negative voltage in the signals feeding the electromagnets. Always must stay over 0 volt. We are feeding DC and modutating it with more or less resistance. It can never go below 0 volt. If your graph is representing the output voltage from the induced coil then is right.
    Last edited by hanon1492; 10-13-2016, 11:53 PM.

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  • seaad
    replied
    Originally posted by Ufopolitics View Post
    the signal can not even reach zero point, (always on positive side) much less passing zero like that getting 50% negative values.
    UFO Sorry I don't understand what you call "SIGNAL" among things here? English is not my native language pls explain in another way.
    Last edited by seaad; 10-13-2016, 11:08 PM.

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  • Ufopolitics
    replied
    Part G Sequence...

    Just time to upload this image...


    [IMG][/IMG]

    Later


    Ufopolitics

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  • Ufopolitics
    replied
    Originally posted by seaad View Post
    The same sim. as in my post #1265 but the G-part consists here of pure resistors with different values. Scope shot => OUT (y)
    Less power factor 38% and output voltage 135V p-p.

    Only a 1/3:rd of OU
    Arne

    Seaad,

    That sim is not good, the signal can not even reach zero point, (always on positive side) much less passing zero like that getting 50% negative values.


    Take care


    Ufopolitics
    Last edited by Ufopolitics; 10-13-2016, 09:47 PM.

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  • Ufopolitics
    replied
    Originally posted by bistander View Post
    Hi there MM,

    I noticed back in your post #1298 you wrote about your study of inductance. Very good. Because you rely on inductance and inductive reactance in your partG design, I suggest you take your studies further and learn about mutual inductance. Mutual inductance will have some influence on the primary coils but because they are on an open core, the coupling will be low to moderate. However on the toroid the coupling of coils will be nearly perfect, k = .98-.99.

    When the wiper brush on the toroid winding is positioned at 12:00 noon or #8, and is connected to primary coil N passing maximum current that coil, it is also connected to the primary coil S via 2 halves of the toroid winding in parallel. On the surface, one would assume the reactance of the toroid coil will reduce the voltage at 6:00 or #1, the connection to primary coil S thereby allowing only a low current flow to it. However due to mutual inductance between the 2 halves of the toroid coil, the effective inductance of the toroid coil between 12:00 (8) and 6:00 (1) is essentially zero. Therefore the reactance of the toroid is just the resistance, which is very low. So the voltage difference between 8 and 1 is very small and the currents to primary N and to primary S are not too different.

    I tried to explain this to Ufo in post #1260 but it flew past him. Maybe you'll pick up on it. But I don't think the use of a toroid in either method will yield the results you guys expect. But seldom is my advice taken, so I standby and see what develops.

    Good luck,

    bi
    Bistander,

    I am sorry if I did not paid attention to that post you are referring to, but at this point I am not working on part G, since -like I posted previously-, I do not have the toroid as of now...besides, I have consider to start testing a Model with resistors first...However, I believe the starting point should be our Primaries first based on its resistance value plus the power source we would be using....then measure its magnetic response capabilities plus its field length with our input...etc,etc...so whenever we have a pretty good primary...only then I would reproduce multiple units...THEN, and only then...they (all primaries) all will dictate the whole iron mass of my future Toroid Part G.

    Now about your comment above about Part G...

    Can I ask You if you are considering -at all- the fact that Part G is not connected directly to negative supply terminal?

    But instead, its negative connection derives from both sets of primaries coils in series?

    Wouldn't the fact that Part G component is connected right between the two sets of coils, which also have a resistance as an impedance?


    Take care



    Ufopolitics

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  • seaad
    replied
    LT spice XVII simulation G-part ONLY RESISTORS

    The same sim. as in my post #1265 but the G-part consists here of pure resistors with different values. Scope shot => OUT (y)
    Less power factor 38% and output voltage 135V p-p.

    Only a 1/3:rd of OU
    Arne
    Attached Files

    Leave a comment:


  • hanon1492
    replied
    You said that your mentor gave you some hints and you guessed and reasoned the rest of the part G design. I really trust your mentor although, as bistsnder, I can not get to understand why he chose you as his partner. Maybe you should ask him about the exact design.

    I still can not understang how your part G design gets "filled and emptied" of current from/to the electromagnets through a single wire in each half cycle without reversing polarity in the electromagnets!!! Maybe your wiring is not right. But I do not want to keep arguing about this same point again.

    This an old image from you in OU forum. Please explain the current reversal in each wire every half cycle without reversing current in the electromagnets....
    Last edited by hanon1492; 10-13-2016, 07:31 PM.

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