More Watts Out Than In? P2

Pulse 1 section of a round coil and collect any extra energy. part 2

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Tags:

• Freeenergy
• Overunity
• Science
• Technology
• motors
• Tesla

[camster6]

So what we need to do is get rid of the bridge and calculate rms on the coil output to determine true current. There is also a formula to calculate the current of the periodic wave form.

Perhaps a revisit to this experiment is needed. Great comments thanks.

[rob790]

You CAN NOT use P = IV in this situation! That equation is strictly for DC cases. You obviously have a very non standard AC type of current and voltage going on with your resistor. Power is better measured with the proper equipment where P = V^2/R where V^2 is averaged over the time of measurement. In other words, you need specialized circuits to properly measure the power. Sorry, but your calculations are worthless.

[camster6]

The input is pulsed DC, the output is that pulsed DC and BEMF bridged to DC. The first resistor is on the Battery the second on the bridge. The resistors only see DC as the electron only flows one way through both resistors. Perhaps I did not explain that part well enough, sorry for any confusion.

[stivep1]

what you need to do is to connect light bulb at the end and measure luminescence .

Than connect the same light bulb at the input and measure luminescence as well and compare.

Than

take DC circuit and check luminescence for two instances based on previously calculated power.

while checking luminescence put light bulb in the tube to insure the same distance to the light strength meter.

I do not like approach of rob790 . IGNORANT

Get a lab. I did it.And I never stated I'm the best.

[SuperDisbelief]

Are there complex formulaes? I guess it is easy to calculate things wrong. Did you put a load on the putput (something that really uses the power)? That might be the missing piece.

[kapierenundkopieren]

Get a scope with RMS reading... that circuit is similar to a joule thief and i bet there are more spikes which your scope doesen't see.

[Mushketer888]

big thanks for vid!

[fgleich]

why not increase the output number of turns to get a greater output voltage, then add a filter capacitor to the output to make a DC supply, and then use that as the supply for the entire circuit ? ditch the battery and use the output generated, if that worked, you would know it's the real thing.

[blaargenheimer]

Have you tried turning up the juice a bit to see if you can get enough back through to run the 555 timer?

[klexn]

I agree with KokomoJ0.. Measure DC in then convert it to a clean DC out and measure your power differences. Don't use a resistor on your input or you will need to add it to your power consumption calculations. it's much easier to measure DC power in and out any day of the week.

[Drevtoobe]

This clip is way way off the mark. I hope that you have been going up the learning curve in the two years since it was made.

[openmindmbj]

Also, consider compensating for the values due to the waveform. (ex. If it were a sinusoidal waveform, rms expressed by amplitude x sqrt of 2, would be sufficiently accurate.) Or better try computing power per pulse.

You could use the oscilloscope to read values of V & I, & the displacement. Thank you for sharing. I hope this helps.

[openmindmbj]

Camster,

You have complete resources for your experiment. I would suggest you remove the resistor in sec A.Take input voltage, V1 and current, I1 readings including the displacement angle between V1 & I1. Take output voltage,V2across the resistor in sec B & current, I2 in series, include displacement angle. Then express your computations in vectors. Very basic yet very effective.