Electric motor ▲ AC/DC 2

[bmuscotty88]

What the hell?

[cadentavaricegrtrj]

The big Oil corporations are trying their best to stop free energy ideas from spreading to common ppl.

We need to put an end to this corruption ,start generating your own electricity now.

Visit LT-MAGNET-MOTORdotCOM and get the blueprints . Join the Revolution!!

[louis12346]

Johnny are you still in the bath room ? what are you doin n there ? OMG !

[ncbookz]

this guy might be dead soon he will get shocked for sure , mark my words he does not respect electricity  pray for him he is too careless

[dragonxxx53]

Special Olympics electrical theory

[kmarinas86]

Haha. Very witty.

[kmarinas86]

Go home Special Olympics hater.

[kmarinas86]

Thousands of people have shown more interest in this video compared to my others.

Yet, my recent videos demonstrate a 32" fan putting out much more than the little winkling in video above.

When I get to it, I CAN SHOW that this fan I now have is more powerful than the ceiling fan in my room (at the highest setting).... And that is when the voltage is down TWENTY PERCENT (at 241 volts) with only 3X POINT OH TWO AMPS. Prepare for the evolution....

[marsL001]

shove a new battery in the multimeter

an almost empty battery gives false readings

trust me ;-) succes

[fenasi2]

This guys speech have put me off watching "You Tube"

Get out of that room and get a life Dick

[kmarinas86]

Hey I was on vacation out for a week. You are telling me to get out of the room? Hah!

[electrofly23]

Dude! you need 3 things:

1: a tripod

2: an education

3: de-caf!

[gargara]

ahahah

[javierToronto]

jaja!! get a book buddy!!

[jonjetzone]

It's really difficult to understand what you are trying to explain, probably talk less but brief and direct to the point and your camera is just as "jungly" as your narration. Good luck to your experiment.

[bendor101]

Never discount the possible.

[Ticklemytummy]

I read thru all the comments and have to agree, you never placed a load on the output thus not showing it can power something useful - like a real fan. The coil itself cannot be considered a "load" as it is the device producing the power. Make a new vid (with a tripod) and plug a small light into the thing.

[kmarinas86]

"I read thru all the comments and have to agree, you never placed a load on the output thus not showing it can power something useful - like a real fan."

It has now. Unfortunately most of my views go to this (outdated) video. A lot of it has to do with the popularity of the videos that are uploaded around the same time as this one. That's a real shame because now the decoy is being seen, not the much, much better system (motor, not generator) I have now!

[Pattonash]

You could open the nearest window, grab all the equipment and just throw them all out at once.

Then, get up on the sill and jump out the window after them. Problem solved

Make sure you are on, at least, the second floor of a building, otherwise it wont work

[kmarinas86]

Dumbass.

[srpitu]

me too thinks like him....your record it ....very bad for eyes....sorry...but the thurth its the thurth....o como sea.,.. me la pela, grabas de pena y la bobinas son de broma.

[kmarinas86]

He wasn't talking about the camera. But you're right, the camera is really bad.

[Snowman374th]

I've said it once I'll say it a hundred time's: YOU CAN LEAD A HORSE TO WATER BUT YOU CAN'T MAKE IT DRINK THE WATER!

SO! GOOD JOB MAN! You can run many things with this after it's modified.

[hsigamj]

Don't be critical with this man, at least he is honest in his investigations and should be appluaded where as you sit on your duff offering collegent advice void of any sensitivity to the human spirit. Thats why You scientists always end up twisted dissatisfied angry old farts, that miss out on life impressesd only by your own critical spirit and prideful errogance .

[intelite]

hmmmm, seems like if it was perfectly balanced he could produce resonance. :)

[tonytvtech]

uu uuu uuu dude get off drugs and go to school and learn that free energy devices dont need batteries

[youser23]

no - my multimeter is even worse. it was just a funny german "saying"

[youser23]

Wer misst, misst Mist! ;-)

[kmarinas86]

I hate to say this youser23, but I agree with that statement. ;-) My multimeter sucks. It can't measure the power factor! .... But this inability is typical of most multimeters, and there's little I can do about it.

[dodoshlodo]

rectify the output to get a dc output and run it parralel to a high voltage neon bulb, i've got a 90V one.

I am getting similar output from my Bedini motor. high 140V spikes and I can barely measure any current. This is not normal electricity. Tesla called it Radiant energy. from whati can understand its kinda like static. This energy is extracted fromn the vacuum with no current. Infact when u apply current u kill the radiant, so dont b surpised by no current..

Keep it up

[oneinunity]

Cos(phi)=0.00000.... LOL!

[kmarinas86]

"Cos(phi)=0.00000.... LOL!" - You're probably right. If what you said weren't the case, the coil would be pretty hot by now.

Now taking your comment too literally, one might assume you were referring to real power/apparent power = 0. But the motor runs while real power is not zero. Not suprisingly, the batteries' voltages drop over time, eventually requiring recharge.

Currently I am trying to refine my ideas about how to make a motor more efficient than this: thinner wire; fewer & wider turns.

[oneinunity]

Well, actually my comment does not fully make sense. Cos(phi) only applies to sinusoidal voltages and currents, but something comparable happens here. At the moment that the AC voltage is maximal the current is zero, while at maximal current the voltage is very low. Cos(phi) is a measure for the time shift between sinusoidal voltage and currents. In this case the voltage and current are not sinusoidal, but there IS a time shift, comparable to a cos(phi) that's very close to zero.

[expsenthil]

connect a bulb and see whether the bulb glows

[strammekabouter]

Back to school my friend!

[kmarinas86]

Literally! I'm going to summer school tommorow (July 7) at my University, and it's no joke!

[execk2]

I want my 5 minutes back.... come on now!

[Pitcrew78]

Dude.

you were measureing the AC milli amps. That is 1/1000 of an amp. So if it read 1, that is one thousandth of an amp, times the AC voltage. Which is .2 watts

[kmarinas86]

Maybe you are right, if the meter is busted.

But if it is not busted, the video at 5 mins and 0 sec should be telling me the root mean square amperage. If I square that value I should be getting the average squared value of current. If I knew the resistance of the load as a function of amperage, I could determine the real power usage. But really, the charge that is being drawn from the leading batteries varies only proportionally to the current drawn from it, not to its square.

[kmarinas86]

The ohmic resistance of the wire and the spark gap is not constant and is a cause and effect of the variable current usage in the wire. This acts like a changing load which causes a changing drop in battery voltage. The fact that voltage drops when you add a load means that the power in is less than the battery's voltage with no load times the current from the batteries with a load. This, the impedance of the load, helps to reduce input power at a given voltage input.

[kmarinas86]

When the current drops in the inductor, the voltage drop due to inductance is negative. This voltage drop equals inductance * change in current / change in time, and it corresponds to change of magnetic flux opposite of that which was produced by the current going forward. As a result, the magnetic field inside the loops, by the loops, contracts, while the magnetic field by the permanent magnet grows.

[kmarinas86]

When the voltage drop due to capacitance of the circuit is greater than the voltage input from the batteries, then both the voltage drops due to resistance and inductance can be negative, meaning that current can increase in the opposite direction and polarity than was coming from the batteries. That is the basis for spark plugs used in automobiles. Interestingly, a high end spark plug can use a capacitor instead of a inductor in order to produce a more efficient burning spark.

[kmarinas86]

For a series of *different* numbers, such as 1, 2, 3, 4, and 5, the root mean square of the numbers is always greater than their average. Anyone who has taken math knows that the average of the series stated is 3. The squares are 1, 4, 9, 16, and 25, the average of which is 11. Thus the root mean square value of varying current is always greater than the its average value. Therefore, the AC reading of current must be higher than the DC reading of current. The current in the video is 20 mA DC...

[a68rebel]

Ramble on....and on....and on....and on....

[nasanction]

I found my self laughing from phrase to phrase. This was a comedy right? (o)(o)

[billybbob18]

hook up an oscilloscope so we can see the Hz.

Those readings are irrelevant without knowing the frequency of the AC voltage

[kmarinas86]

That might take a while. In the mean time, I already know that what the multimeter is reading is not anywhere close to a sinusoidal voltage so I'm pretty sure that there is no where close to one frequency only. It's not like this can be hooked up to the wall to produce 60 or 50 Hz, so its not a way of producing suitable AC power. I have heard recently that the Lutec machine was supposed to do something like that, but... =P nyah nyah

[MAGNA550]

For all of you working on overunity projects. Check out the specs on your meters. Most digital multimeters will only read AC voltage and Current accuratley from 50hz to 400hz. 50hz for europe, 60hz for U.S and 400 hz for aircraft. Outside that frequency range it's anybodies guess as to what the voltage or current is. The Extech 411 meter in this video has a range of 50hz to 400hz. Maybe that's why everyone thinks they are getting more out than they are putting in. Use an analog meter.

[olzzn]

I really don't like your voice, so unfortunly i can't watch this video...

[TheB3e]

Wouldnt the load be the spinning magnet?

[kmarinas86]

The word "load" is kinda of loaded. :)

There are electrical loads.

There are mechanical loads.

Then the rotor itself could be considered a load, if all you were trying to do was to spin the magnets.

Different approaches have their own advantages and limitations. Unfortunately you have to know the assumptions behind them to properly apply a method. A method may give meaningful answers in one context but may be useless in another.

en.wikipedia org/wiki/Network analysis (electrical circuits)

[Aquar1uZ]

You cannot measure amps without putting an actual load on your circuit. Place a bulb in there and measure again. Nothing odd happening here.

[kmarinas86]

The resistance of my coil is over 200 ohms. It gets even higher when I break the circuit. I'm not sure how this would be any less of a load than a light bulb having the same resistance.

[kmarinas86]

And could you even believe the readings? If the AC amps shows 1 amp through a 200 ohm coil (4000 ft of 28 AWG), then the voltage should be above 200 volts. Since I am new to electronics, thought of me producing 200 volts with 16 AA batteries sounded kind of ridiculous. I immediately knew it could not be from the batteries - that would be impossible. If that's not weird then I don't know what weird means. Surely understand that I cannot draw 200 watts from 20 volts unless I am drawing 10 amps :P.

[Aquar1uZ]

Sorry I was not clear, a constant load. The coil is a resistor agreed, but you need a constant load like a fridge light or something. Easy enough to try? If you still get COP > 1 after that the world will bow to you and I will want to be your best friend :-)

[kmarinas86]

What I should do is connect the multimeter in parallel with the battery pack in a way that bypasses the commutator, which I have not done yet. The multimeter should have a built-in resistor that measures the voltage drop without drawing too much current.

[krishields]

Actually you can, but you will blow a fuse doing it....

Short the battery with the meter terminals, Amps climbs and burns out the fuse and heats the hell out of the battery, basically killing it.

In other words, you have to use the meter as the load, which doesnt measure anything, it just shows how much the meter can handle :)

You can hook it direct to the generator to find out max load of the generator before it dies, if your meter can handle more current than the gen puts out.

[kmarinas86]

That's why I'm suppose to use the jack that's used for the voltage and ohms only. That jack has a much higher resistance and so will draw very little current. But with that I will be only measuring voltage not amps since I don't know the internal resistance of the voltage/ohm testing jack. Measuring current requires that I use the 20A port with an external resistor since the 20A port does not include it. I'm not sure of the difference between the approaches in terms of accurate measurement.

[AdminOnDuty]

Here's the problem. Your AC current reading was dead shorting your AC output. When this happens your output voltage drops to practically nothing. So what you have done is give us a no-load voltage reading and a max-load current reading.

Sorry to burst your bubble. :(

[kmarinas86]

How do I take a max-load voltage reading? What I have been doing is taking the AC voltage readings with the probes in parallel and the currents in series with the windings. I did DC voltage readings both ways, in parallel and in series. Tell me what to do.

[AdminOnDuty]

Your output windings need to drive a consistant load. For instance, a small nightlight. You can then measure the voltage across the nightlight and the current through the nightlight. In a lab we would take all four readings simultaneously and have them recorded. The input voltage and current and the output voltage and current. This way you could compare watt for watt for any time interval. This could be accomplished with a 4 trace digital scope and two 0.1 ohm sensing resistors.

[kmarinas86]

Why should I observe over 200 volts AC going in, or going out, or being dissapated, at no-load anyway? Where do you think that comes from? The batteries? Hell no!!!

Why should I observe over 1 amp of AC going into [or from!] 4000 feet (>250 ohms) of 28 gauge wire with 21 volts of batteries?

Why should I observe AC that is greater than the DC that the batteries are supplying?

Why couldn't some of the power be dissipated in the coil itself? Wouldn't that make it a load?

[AdminOnDuty]

Please don't misunderstand me, I am trying to help. We are both on the same side here. I applaud your novel approach to using ribbon cable for your coil. The resistance of your coil becomes part of your 'Source' impedance and does have an effect on your limits as listed in your next post. There is a point in the rotation of your magnet when the load drops to zero, the induction field collapses and the back EMF raises to voltages over 200V. However this little current associated with it. The ...

[AdminOnDuty]

...actual A/C current you are measuring is the coils attempt at recharging the batteries during the back EMF cycle and while the current is flowing the voltage is clamped near battery voltage (20V). It is when the batteries can no longer take anymore current that you witness the back EMF climbing to 255V. But this is similar to static electricity. If you are using rechargeable batteries and good bearings on your rotor your motor may run indefinintely - at least I think we should test it out.

[renderrob]

arrrrg... please, get a tripod, editing software (free apps are available), script your presentation so you don't have to talk so much, clean up the device so you don't have probes randomly disconnecting, and don't read the multimeter while you are filming it.

[kmarinas86]

I also need a new camera.

[renderrob]

I think your camera does really well, especially compared to the average quality of youtube videos. I think having camera stability (on tripod) and having your presentation planned out would be the most beneficial changes. Tripods can be small and cheap, all you really need from it is stability.

[kmarinas86]

I think all I need is a tissue box, but thanks anyway! :) +1 +1

[KillallTVs]

keep it up man. DO IT!

[kmarinas86]

Thanks! Recently I found that AC Power actually is a product of 3 things: The RMS voltage times the RMS current and the power factor of the circuit. The power factor is resistance divided by impedance. The impedance is RMS voltage divided by RMS current. So the AC power of the circuit is really the resistance of the circuit times the RMS current squared. The wire is 270 ohms (40x100' of RC40 ribbon cable from Altex(.com)). Mathematically, RMS of V and I >= Average of V and I. AC power>=DC power.

[brahmamurti]

very good work!! :)