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Lockridge Device - Peter Lindemann

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  • Bill H
    replied
    Originally posted by Michael Kishline View Post
    Greetings emfimp,

    Maybe you could use a set of Government Motors slip rings. I pick these up at a motor rewind shop for $10 when I built my Kromrey Converter, it's already mounted on a insulator that's molded onto the rings. You could slot these into a commutator with a power hacksaw or the right cutting tool.

    Could save a lot of time?

    Merry Christmas, Mike

    [ATTACH]7293[/ATTACH][ATTACH]7294[/ATTACH][ATTACH]7295[/ATTACH]
    Nice Slip rings Mike. Also the delco brush holder is a nice piece to work with.
    Just watched my video this morning, I recieved it on thursday but this morning early was my first opportunity. Thanks Peter, very inspiring and educational.
    Slip rings can also be had from any Automotive electrical rebuilder. Alternators Parts,Slip rings, Lucas,Bosch,Delco,ford,valeo,paris rhone,hitachi,leece neville,ducellier,nippondenso,prestolite and mitsubishi is a good source and reference as well.
    Not sure of the direction of my build yet but will be on the watch for components.
    Bill H.

    Leave a comment:


  • Michael Kishline
    replied
    Commutator prototyping

    Originally posted by emfimp View Post
    Goreggie - good idea! A distributor has all the needed components, & little modding to do.

    Keep in mind you may want to change your pulse width for optimal running - you may be able to attach copper strips to do that on a distributor - not really sure.

    I'm choosing to build a commutator using a 3/4" piece of copper pipe - fill it with JB weld, let harden, then drill the proper sized shaft hole using a friend's lathe. (through the JB) I'll JB a collet to that, so that it can be slipped on & off the shaft. The brushes will be copper strips (in the beginning), and the "off" portion of the copper pipe will just have a piece of electrical tape over it. Fairly crude, but allows easy adjustability via the tape.

    Luv 2 hear more ideas!
    Greetings emfimp,

    Maybe you could use a set of Government Motors slip rings. I pick these up at a motor rewind shop for $10 when I built my Kromrey Converter, it's already mounted on a insulator that's molded onto the rings. You could slot these into a commutator with a power hacksaw or the right cutting tool.

    Could save a lot of time?

    Merry Christmas, Mike

    GM Slip Rings 4.jpgGM Slip Rings 5.jpgGM Slip Rings 3.jpg

    Leave a comment:


  • emfimp
    replied
    Yes, you're right - JB does have metal in it. I was recommended it by someone who has tried this, & apparently there are not enough metal flakes for it to actually conduct.

    I'll post pictures & results as soon as I have it made.

    Happy holidays to everyone!

    Leave a comment:


  • goreggie
    replied
    Commutator lockridge device

    Originally posted by emfimp View Post
    Goreggie - good idea! A distributor has all the needed components, & little modding to do.

    Keep in mind you may want to change your pulse width for optimal running - you may be able to attach copper strips to do that on a distributor - not really sure.

    I'm choosing to build a commutator using a 3/4" piece of copper pipe - fill it with JB weld, let harden, then drill the proper sized shaft hole using a friend's lathe. (through the JB) I'll JB a collet to that, so that it can be slipped on & off the shaft. The brushes will be copper strips (in the beginning), and the "off" portion of the copper pipe will just have a piece of electrical tape over it. Fairly crude, but allows easy adjustability via the tape.

    Luv 2 hear more ideas!
    Thanks for the reply yes your right, the pulse width is everything I'm going with brushes Mark R. Merry Christmas

    Leave a comment:


  • dragon
    replied
    Doesn't JB weld have metal in it? Is it conductive? I plan to follow a similar path for the commutator. I'm going to try some fiberglass resin for the filler. It holds up quite well with heat, to a certain point, I'm not sure how it will handle the pressure of the brushes dragging on it. It will be good enough to run some tests and if all works well a more sophisticated unit can be made up later.
    ________
    Halfbaked
    Last edited by dragon; 05-11-2011, 11:08 AM.

    Leave a comment:


  • emfimp
    replied
    Goreggie - good idea! A distributor has all the needed components, & little modding to do.

    Keep in mind you may want to change your pulse width for optimal running - you may be able to attach copper strips to do that on a distributor - not really sure.

    I'm choosing to build a commutator using a 3/4" piece of copper pipe - fill it with JB weld, let harden, then drill the proper sized shaft hole using a friend's lathe. (through the JB) I'll JB a collet to that, so that it can be slipped on & off the shaft. The brushes will be copper strips (in the beginning), and the "off" portion of the copper pipe will just have a piece of electrical tape over it. Fairly crude, but allows easy adjustability via the tape.

    Luv 2 hear more ideas!

    Leave a comment:


  • goreggie
    replied
    Commutator

    Hi while working on the commutator would a car distributor work [ATTACH][/ATTACH]or brushes? Mark R. Thanks for any reply
    Last edited by goreggie; 04-03-2011, 11:07 AM.

    Leave a comment:


  • Peter Lindemann
    replied
    Voltage Divider

    Hi Folks,

    Sorry for the confusion with the "resistor in series with the multimeter" test. Perhaps that was a bad example. We all know that an applied voltage will be distributed across a set of resistors proportionately. This is a given. That said, I would also like to say that I do not believe that this is the same phenomenon as either "the dissipation of electricity" or a true "voltage drop" in a circuit component.

    The "charging of a capacitor" is the transfer and storage of electricity, and a resistor in series with this process simply regulates the rate at which current flows. The total quantity of electricity lost in this transfer (dissipation) is very low. On the other hand, a transistor is a circuit component that has a "true voltage drop" across it, which represents a real loss of potential during conduction of current.

    Since I do not wish to debate these issues here, I would just like to thank EMCSQ for helping to promote clarity in the thread, and move on.

    Peter
    Last edited by Peter Lindemann; 12-24-2010, 05:40 PM.

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  • dragon
    replied
    I think we can assume, at least for the device/circuit being experimented with, that the 48 ohm resistance in the bulbs shouldn't cause a decrease in voltage to the cap being charged.

    My tests using various resistance bulbs showed the cap would charge to the voltage of the battery. The only difference was that the higher resistance bulbs slowed the charge rate. There is a potential difference until the cap is charged.

    I guess the next test might be to determine if and/or how much energy the bulb(s) are converting to heat. Is all the energy actually going to charge the cap....?
    ________
    Mary Jane
    Last edited by dragon; 05-11-2011, 11:08 AM.

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  • rave154
    replied
    EMCSQ,

    i can confirm your statement re:- meter internal resistance against external resistor in series as a test recomended by Peter.

    My meter in DC mode is listed as having an internal impedance of 10 MOhms....

    measuring a battery directly the voltage was 12.55V, i then placed a resistor of 1MOhm in series with the meter and the measured voltage was 11.4V.

    this appeasr correct since, total resistance that voltage of 12.55V is across = 10M + 1M =11M

    therefore since we are measuring across only the 10M of the meter the voltage should be.... ( 12.55 / 11 ) * 10 = 11.4V which is exactly what i got on the meter.

    Leave a comment:


  • EMCSQ
    replied
    Its in NRW Kreis Heinsberg

    Leave a comment:


  • erfinder
    replied
    Originally posted by EMCSQ View Post
    Hello
    have not received the 'Electric Motor Secrets, Part 2' yet (Germany), and therefore cannot talk about it.
    Where at in Germany are you from?

    I ask because I live in Germany, and am always looking for like minded individuals.... in meine nähe....

    Regards

    Leave a comment:


  • EMCSQ
    replied
    Originally posted by Peter Lindemann View Post
    Dragon,

    Yes, you must run all of these little side experiments to see the truth of what is going on in a circuit like this. Another simple test is simply to test the voltage of a battery with a multimeter, and then add a 1K resistor in-line with the test probes and test the voltage of the battery again. As long as your contacts are good and clean, the voltage should read exactly the same as without the 1K resistor.

    So, the question is: how much "voltage drop" is there across a resistor?

    If the answer is "NONE" as it appears to be, then it becomes clear that we can light the light bulb (resistor) if there is "some current" and a "potential difference" across the bulb. No energy (or at least very little) needs to be dissipated to light the light!

    Keep up the great work!!!!

    Peter
    Hello

    did I miss something? The internal resistance of a multimeter in Volts measuring mode is a few Megaohms ; means infinite. So the voltage drop across the 1K resistor is practically zero,because no current flows.
    If you put a cap in series, for the first moment a current will flow to load the cap and then it diminishes to zero. These are basics.
    For to replicate a FE device one needs a Recipe how to do it. A have not received the 'Electric Motor Secrets, Part 2' yet (Germany), and therefore cannot talk about it.

    Lockridge Device: if this is what the african girls did, there are coils use for the motor and other coils for the generator principal. and both function generator / motor alternate. Generator and Motor don't work simultaneously ; the flywheel is the energy buffer; very similar to the bedini watson machine.
    Last edited by EMCSQ; 12-23-2010, 05:41 PM.

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  • emfimp
    replied
    Possibly this effect is because, as Tom Bearden said, the dipole of a battery (or any source most likely) is not 'destroyed' until the electron moves through the entire circuit & then re-enters the battery?

    Thus a true voltage drop occurs across the source, not a circuit component?

    Leave a comment:


  • woopy
    replied
    Hi Peter and Dragon

    OK lets going on and thanks for the simple experiment with a 1Kohm resistor

    And now something more to think

    good luck at all

    Laurent
    Last edited by woopy; 05-16-2011, 09:34 PM.

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