Longitudinal Wave Experiment to demonstrate Overunity

[magpwr]

Please put an E element with a gain of one as a differential voltage sense across the light bulb load.  Insert  a current sense resistor in series with the bulb.  Follow that resistor with an E element with gain set to 1/R_{CURRENT SENSE}.  DC couple those E element outputs to your scope  Once the voltage and current look sane, then connect a X1 multiplier to get your instantaneous power.

hi MarkE,

I got even better way to show OU.I am using  "1" Ohms resistor at output instead of using 250watt x3  120volt bulb to demonstrate the capability of this circuit.
The spike is nearly 100volts with 1 ohms resistor.
This time i would have to change fuse to 5Amp so my guess it is somewhere below 5Amps.

6.6volts x 5Amps=33watt input with around 98watt output for 1 ohms load."This is just an estimate"

The interesting part the output frequency did not deteriorate much with 1 Ohms load.

--------------------------
Latest update-I have attached single spike waveform using 100v/div in the 200ns range for reference using 1 ohms resistor.

[MarkE]

hi MarkE,

I got even better way to show OU.I am using  "1" Ohms resistor at output instead of using 250watt x3  120volt bulb to demonstrate the capability of this circuit.
The spike is nearly 100volts with 1 ohms resistor.
This time i would have to change fuse to 5Amp so my guess it is somewhere below 5Amps.

6.6volts x 5Amps=33watt input with around 98watt output for 1 ohms load."This is just an estimate"

That simplifies things a bit.  Now what you need to do is Tie an E element across the resistor, and then feed that to both inputs of a X1 multiplier.  Finally, you need to put that through an integrator and watch the level change from cycle to cycle after operation has stabilized.  Multiply the difference in the integrator output over the course of one cycle by the operating frequency to get the true power out.

[gyulasun]

Hi magpwr,

Sorry to chime in,  would like to ask whether the Multisim circuit simulator uses ideal coil and capacitor models?
I mean when you pick a coil symbol and place it and its Properties window opens, can you choose a loss factor or series resistor value to approach the properties of a real life coil?

If there is no such feature in the simulator, then you have to connect a low Ohm value resistor in series with each of your coil symbols in the schematic to avoid using ideal coils.  (You do not have supraconducting wire to make your coils, do you?)

Considering a 12 kHz "middle" frequency [(4+21)/2] for your circuit, and considering a practical Q=100 unloaded quality factor for all your coils, the series loss resistances would come as follows:

for the 220 uH coil, X_L=16.6 Ohm so r=16.6/100= 0.166 Ohm please connect such resistor in series with each 220 uH coil,

for the 500 uH coil, X_L=37.7 Ohm so r=0.377 OHm  please connect such resistor in series with each 500 uH coil,

for the 1 mH coil, X_L=75.4 Ohm so r=0.754 Ohm please connect such resistor in series with each 1 mH coil

In fact, if the simulator uses ideal capacitors too, then a loss resistance ought to be introduced for them too, estimating the resistor values similarly for them like for the coils.

I do hope that your actual build will give similar extra performance like the simulator forecasts...

rgds,  Gyula

[magpwr]

Hi magpwr,

Sorry to chime in,  would like to ask whether the Multisim circuit simulator uses ideal coil and capacitor models?
I mean when you pick a coil symbol and place it and its Properties window opens, can you choose a loss factor or series resistor value to approach the properties of a real life coil?

If there is no such feature in the simulator, then you have to connect a low Ohm value resistor in series with each of your coil symbols in the schematic to avoid using ideal coils.  (You do not have supraconducting wire to make your coils, do you?)

Considering a 12 kHz "middle" frequency [(4+21)/2] for your circuit, and considering a practical Q=100 unloaded quality factor for all your coils, the series loss resistances would come as follows:

for the 220 uH coil, X_L=16.6 Ohm so r=16.6/100= 0.166 Ohm please connect such resistor in series with each 220 uH coil,

for the 500 uH coil, X_L=37.7 Ohm so r=0.377 OHm  please connect such resistor in series with each 500 uH coil,

for the 1 mH coil, X_L=75.4 Ohm so r=0.754 Ohm please connect such resistor in series with each 1 mH coil

In fact, if the simulator uses ideal capacitors too, then a loss resistance ought to be introduced for them too, estimating the resistor values similarly for them like for the coils.

I do hope that your actual build will give similar extra performance like the simulator forecasts...

rgds,  Gyula

hi gyulasun,

Thanks for your invaluable input.I have applied the losses as you suggested for the inductors by connecting resistors in series.

I'm not sure how to apply loss for capacitors but i simply put it as 5% tolerance.There is a remark mentioned as mica/teflon capacitor in properties.

This time the maximum observed spike is reduced from 180volts which was near the bulb filament breaking point to around 140volts spike for the 120volts bulb.

The output frequency is reduced as well to around 3x of input.

[poynt99]

hi MarkE,

I got even better way to show OU.I am using  "1" Ohms resistor at output instead of using 250watt x3  120volt bulb to demonstrate the capability of this circuit.
The spike is nearly 100volts with 1 ohms resistor.
This time i would have to change fuse to 5Amp so my guess it is somewhere below 5Amps.

6.6volts x 5Amps=33watt input with around 98watt output for 1 ohms load."This is just an estimate"

The interesting part the output frequency did not deteriorate much with 1 Ohms load.

--------------------------
Latest update-I have attached single spike waveform using 100v/div in the 200ns range for reference using 1 ohms resistor.

How are you coming up with your input and output numbers? It's not a good idea to estimate power from spikey wave forms.

Doesn't MS have a power probe?

I would urge you to watch this video and perform the same power measurement both on your input and output.

http://www.youtube.com/watch?v=kYliPwbWInc