Confirming the Delayed Lenz Effect

[gotoluc]

Here is the identical setup but connected to the Grid

15uf series capacitor, 1 Ohm Shunt, 10 Ohm Load @ 6.70vac

If this was Saturation then where are those Peaks?

Luc

[gotoluc]

yeah, this is what i remembered seeing - a YT video by user 'selfonlypath' showing results obtained with his 'Transverter' setup (2x MWO trafos modified & connected together, then driven with an 'H' bridge)

  http://www.youtube.com/user/selfonlypath    eg., see 'Transverter 1'


at a combination of resonance and with the i/p volts just passing thro' a threshold, the o/p sine wave 'snaps' into a more triangular waveform with elongated peaks

the snapshot below captures the moment when the o/p changes from a sine to a 'triangular' wave with peaks

cheers
np

Thanks nul-points for posting this video once again. It was posted on page 14 by plengo: http://www.overunity.com/index.php?topic=11350.msg301689#msg301689

This great researcher explains and demonstrates that one needs an accurate timing circuit to recover energy from a Reactive circuit.

I have said and I believe we are subject to the same.

Anyone interested should study this technique and post their ideas of a recovery/recirculating circuit.

Thanks for sharing

Luc

[CRANKYpants]

THIS POST WAS EDITED

under load it's at 1,130ma
Luc

IS THIS 1.13 mA OR IS IT 1.13 A?
IT WOULD BE NICE IF WAS 1.13 mA BECAUSE YOU EFFICIENCY WOULD BE 32297.3%  .

OTHERWISE INPUT CURRENT = 720 mA AND PRIMARY DC RESISTANCE = 0.3 OHMS
IF PRIMARY POWER FACTOR IS INDEED 0.0 THEN Pp = 0.216 WATTS WITH AN EFFICIENCY OF ONLY 2078.2%. 

IF THE PHASE SHIFT IS 81 DEGREES (AS I SEE IT) PF = 0.15 AND IF Pcurrent = 727 mA AND Pvoltage = 112 V THEN

Pin = 112 x 0.727 x 0.15 

= 12.7 WATTS AND Pout = 4.49 WATTS AND EFFICIENCY = 35.4% 
A PHASE SHIFT OF 87 DEGREES OR PF = 0.052 GIVES AN EFFICIENCY OF 105%.

IF THE DC INPUT CURRENT IS 1.13 A AND THE DC VOLTAGE IS 12.3 VOLTS THEN THE EFFICIENCY IS 32.3%

PLEASE EXPAND YOUR TIME AXIS SO:

1) WE ONLY SEE 1/2 OF THE SINE WAVE (180 DEGREES) AND THEN
2) EXPAND THE VOLT/DIV SO THE LINES ARE STRAIGHT UP AND DOWN AND THEN
3) MAKE THE YELLOW LINES EQUAL ON BOTH SIDES.

NOW WE CAN CALCULATE THE PHASE ANGLE CORRECTLY.

ALSO NOTE POSITION OF RED "POWER" SINE WAVE IS SLIGHTLY HIGHER ON TOP AND NOT QUITE EQUAL FOR A PF OF 0.0 - CAN YOU TWEAK IT WITH ANOTHER CAP VALUE?

CHEERS
T

[gotoluc]

Hi Thane

it is 1.13 Amps and your math seems to be correct.

I always learn new stuff as I go so there is nothing wasted (other than power )... I'm only going to get better at this.

Thanks for everyone's input

Luc

PS hey! nice scope shot. That is the ultimate one. We'll get there

[gotoluc]

Hey Thane,

have a look at these Shots

I decided to re-test my GTL-90 compared to the Toroid and the GTL-90 is better

First shot is GTL 90 with 22 ohm shunt, 1uf series cap, 10 ohm load @ 0.42vac

Second shot is Toroid with 22 ohm shunt, 1uf series cap, 10 ohm load @ 0.42vac

Let me know what you think

Luc