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Lockridge Device - Peter Lindemann

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  • Peter Lindemann
    replied
    Excellent

    Originally posted by dragon View Post
    Ok... feeling a little stupid for asking the question now... I just did a simple test with a battery, bulb and cap. The bulb, as Peter stated, is just a balast - a delay. If the bulb was connected to ground there would be a voltage drop but with it in series with the cap there isn't. All the energy has to flow through the cap so it's simply a time variance. Back on track now. Odd how simple things clutter the mind while complicated things seem so simple...
    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

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  • dragon
    replied
    Ok... feeling a little stupid for asking the question now... I just did a simple test with a battery, bulb and cap. The bulb, as Peter stated, is just a balast - a delay. If the bulb was connected to ground there would be a voltage drop but with it in series with the cap there isn't. All the energy has to flow through the cap so it's simply a time variance. Back on track now. Odd how simple things clutter the mind while complicated things seem so simple...
    ________
    Ipad cases
    Last edited by dragon; 05-11-2011, 11:08 AM.

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  • dragon
    replied
    Actually that was a "lazy number" I pulled out of my hat, I didn't want to take the time to calculate the % of weight in the solid flywheel at the time, just wanted a general idea of what it might store.

    I've noticed in an AC circuit a similar outcome. The DC throws me a little in this arrangement. I guess a few test circuits will confirm it one way or the other...
    ________
    Cheap Herbal Vaporizers
    Last edited by dragon; 05-11-2011, 11:07 AM.

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  • Mark
    replied
    Goreggie, I sent you a PM.

    Thanks
    Mark

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  • goreggie
    replied
    Originally posted by Mark View Post
    Hi Goreggie

    I purchased the starter today and was wondering what you would charge me for the end plates with the bearing installed. You can private message me or post here.

    Thanks Mark
    Mark I will have the starter end bell done end of this week,[ATTACH]:yahoo: [/ATTACH] can shipped 1st of next week to you. Right now I'm working on a way to fire the motor and will know more on this in a few days Reggie
    Last edited by goreggie; 04-03-2011, 11:07 AM.

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  • emfimp
    replied
    Thanks for the formula! Is there a particular reason you're specifying 2 inches as your mean radius for the flywheel?

    Great questions on the electrical side of the setup as well - here are my unfinished thoughts on the matter...

    Charge, of course, is not 'lost' in the circuit to the best of my knowlege, but you raise a good point - is the potential (or voltage/dipole) 'lost' or dissipated as it passes through the ballast?

    One thing I've noticed in playing with caps is that if you take just the charge side of the circuit (gen through ballast to cap, no further) & use it to charge the cap, there doesn't appear to be an energy loss through the ballast. The cap will charge to the full voltage of the gen, while there is a production of light & heat also. The key element here (imho) is time - the ballast slows the charge rate, so you must allow the appropriate time to charge the cap.

    So if we add in the motor side of the circuit, then to me the question is one of time - the higher the draw of the motor (determined primarily by pulse rate), then the higher the apparent 'loss' we may see over the ballast?

    Good questions - hopefully someone has a more definitive answer!

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  • dragon
    replied
    I'm using WV^2/64.32 where W = weight in lbs V = velocity of the wheel at it's mean radius in feet per second. The 64.32 is simply 2g or 2 x 32.16.
    ________
    PORNO
    Last edited by dragon; 05-11-2011, 11:07 AM.

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  • emfimp
    replied
    Dragon, what formula do you have for the amount of stored energy in a flywheel? I tried some calcs of my own, but I'm not finding the right numbers.

    Thank you Peter for sharing the depth of info that you have on this subject - truly inspiring!

    Daniel

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  • goreggie
    replied
    Originally posted by Mark View Post
    Hey Guys

    I'm ready to get to work here and wondering how many others have purchased the starter suggested by Peter. I've partially torn my apart to understand how it operates and have run some power through it. In its stock form its not very efficient and I think were going to have to make a lot of modification in oder to have a decent drive motor.

    The first thing we'll need is to replace both end plates with ones that have bearings in them (Goreggie we need your help here).

    Looking at the stator winding I think we need many more turns on them to increase the magnetic field. It also looks like the stator windings are wraped around a big piece of iron that srewed to the side of the case. Wouldn't it be beneficial to remove this piece of iron?

    I also have been thinking of ways to collect the spike off the rotor. So far I can only come up with an extra set of brushes behind the powering ones but this maybe very tricky.

    Just some of my thoughts so far

    Mark
    Mark, I designed addional plates last Sat for the starter we should be done...in the next day or two
    if this works both will be available Reggie my cell is 951-300-3371 if you want to call me any time Thanks Reggie

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  • dragon
    replied
    Playing with the numbers for the capacitor size I think there is a flaw in the cap voltage. If you use 3 100 watt bulbs in parallel you have a resistance of 48 ohms and making approx 2.5 amps, assuming the generator is producing the 240 volts there will be a 120 volt drop across the resistance leaving 120 volts for the cap. In order to get the full 6Joule charge + you would need a cap in the range of 833uf. Is that correct?

    The cap discharge into an inductor( motor windings ) would be in the range of 200+ amps according to the formula V*(C/L)^.5 . I measured 151uh on the motor I'm using so.. 120 volts * ( 833uf / 151uh )^.5 = 281 amps. The normal run with the motor provides aproximately 9.5 ft lbs, if it is enhanced to 3 times that then 28.5 ft lbs might be injected into the flywheel per pulse resulting in a 2.85 ft lb continuous assuming a 10% rotation or 1.42 ft lb at 5%. The required continuous torque to produce 300 watts including losses in the generator would be approximately .85 ft lbs which the 5% mark would clearly provide enough of a boost to maintain itself.

    I'm machining a flywheel that is 6.75 inches in diameter and I've approximated the finished weight to be at or around 22 lbs. At 3000 rpm it should store 942 ft lbs per sec of energy ( about 1.6 hp ). Getting the energy into it may need a "pre starter" to bring it up to speed then flip some switches to start the motor/generator action.

    One thing I seem to have confused myself on is the actual input requirement to the loads ( bulbs and cap ). Is it only 300 watts or is it much higher including both loads ?

    I'm still a week away from getting all the parts made to assemble the test unit to begin the tweaking or balancing process.
    ________
    Zoloft Class Action
    Last edited by dragon; 05-11-2011, 11:07 AM.

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  • goreggie
    replied
    Back EMF

    Originally posted by radiant1 View Post
    Reggie
    Do you know of a working model of mini-romag, or romag?
    Alan
    The Mini-Romag explanation ? Here is the link Reggie

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  • Mark
    replied
    Hey Guys

    I'm ready to get to work here and wondering how many others have purchased the starter suggested by Peter. I've partially torn my apart to understand how it operates and have run some power through it. In its stock form its not very efficient and I think were going to have to make a lot of modification in oder to have a decent drive motor.

    The first thing we'll need is to replace both end plates with ones that have bearings in them (Goreggie we need your help here).

    Looking at the stator winding I think we need many more turns on them to increase the magnetic field. It also looks like the stator windings are wraped around a big piece of iron that srewed to the side of the case. Wouldn't it be beneficial to remove this piece of iron?

    I also have been thinking of ways to collect the spike off the rotor. So far I can only come up with an extra set of brushes behind the powering ones but this maybe very tricky.

    Just some of my thoughts so far

    Mark

    Leave a comment:


  • poii
    replied
    Back EMF

    Hi all

    I Just couldn't keep lurking on this one so here goes:

    Thank you Peter for posing the question.

    IMHO it seems that Back EMF is about the most misused term on the forum. I know I have had some fuzzy thinking about it, So after reading a bunch of Wikipedia pages and watching the last part of the video for the eighth time (even Peter stumbles with it @ 1:14:22 in the video) I think I got it.

    Back EMF is the voltage that opposes the applied voltage in conventional motor topologies. It is generated by the coils moving through the magnetic field, and it is the inverse of the "back" torque in conventional generator topologies.

    Back EMF is often confused with flyback voltage. In fact is actually flyback's opposite because Back EMF is what limits the initial current in an inductor. (Help me out here, have I got this right? ) The expanding magnetic field generates a voltage opposing the applied voltage limiting the current, just like in a motor. Even Wikipedia's flyback diode article misuses it. (scope shot)

    HTML Code:
    http://en.wikipedia.org/wiki/Flyback_diode
    Unlike the M&M equations in the Lockridge video the flyback is a forward voltage not back.

    Maybe we can all agree on something unambiguous to call it like flyback voltage or inductive collapse energy ?


    Deepest Thanks to All Who Make This Forum Rock
    Last edited by poii; 12-21-2010, 09:31 AM.

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  • Michael Kishline
    replied
    Hi Peter,

    My best understanding is that every motor has a counteracting generater effect in voltage and every generator has a counteracting motor effect in voltage.

    So if +12V are applied to a motor and the counter generating effect is -3V then the motor only gets to run on +9V and is only 75% efficient.

    If a generator is spun to +12V output and is only 75% efficient than you have a counter motoring effect of -4V from the generated +16V.

    This is my understanding.

    Fantastic DVD Peter, so concise and easy to understand.
    I also found the Basic Electricity 5 volume series first edition from 1954 that you recommended at the conference.

    Thank you Peter for your time on this forum and all you do this brew's for you!

    P1010855.jpg

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  • everwiser
    replied
    Time, geometry, and the application of excess energy to force the magnetic fields to remain in motion.

    It's time because BEMF is a reaction which implies the action it's reacting to took place earlier in time.

    It's geometry because all "typical" motors are built to inhibit/weaken the BEMF so that the operation requires less applied energy to sustain the running speed.

    Without the constant application of extra energy the BEMF will balance the field and cancel its movement.

    Leave a comment:

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