Thane Heins Perepiteia Replications

[samedsoft]

Dear THANE,

   YOU ARE REAL MAN! THANKS FOR ALL !!!!

    THIS IS HISTORICAL ACCOMPLISHMENT FOR BTT (Bifilar Thane Transformer)



   Thane, what form of waveform do you have on secondaries when you pulse primary with square pulse?



   Also have you tried to use DC source and pulse it at 60 Hz and drive a lamp?

   So bipolar pulsed DC (As in H-bridge) may drive transformer ha?

   Then the voltage on DC battery will go down very very slowly... You can also start with a cap and apply controlled feedback to source...



    Is primary also bifilar wound?



   AGAIN YOU ARE MAN OF THIS WORLD! REAL REAL HERO!

   GOD BLESS YOU AND FAMILY!

   Say hi to your wife.

    NT,TR!

[minde4000]

Thane,

Back to perpetia folks. To get acceleration you must use a MINIMUM of two I core coils sharing mutual flux path (ring/laminates) OR a single E coil?

OR you can achieve acceleration with a SINGLE single wire wound I core HV coil?

Please advice

Regards Minde

[CRANKYpants]

Thane,

OR you can achieve acceleration with a SINGLE single wire wound I core HV coil?

Please advice

Regards Minde

SINGLE WIRE WOUND HV COIL WORKS - BUT A CLOSED MAGNETIC CIRCUIT (I.E. TWO COILS OR A C CORE) GIVES BETTER ALL ROUND RESULTS.

IF YOU DON'T START POSTING YOUR BEAUTIFUL WORK (PICS) AND INSPIRING OTHERS HERE I AM CHANGING YOUR NAME TO "MindeMiser"! 

T

[CRANKYpants]

TODAY I GLUED DOWN A NEW PRIMARY AND AS EXPECTED THE CURRENT DROPS - POWER FACTOR IS MORE LIKE 0.05 BUT THEN THINGS JUST START GETTING CRAZY... EACH ONE OF THOSE LITTLE MARKERS ON THE X AXIS = 90/13.5 = 6.66 DEGREES SO THE POWER FACTOR IS COS(90 - 6.66) PARALLAX ERROR ADJUSTED
T

July 11th, 2009 Data

Primary Input Voltage = 58.8 volts
Primary Input Current = 0.017 amps
Power Factor = 83.3 degrees / 0.116
Primary Input Power = 115 mW

Secondary Output Voltage = 3.1 volts
Secondary Load = 27 ohms
Secondary Output Power = 356 mW   

Efficiency = 310%

[LarryC]

TODAY I GLUED DOWN A NEW PRIMARY AND AS EXPECTED THE CURRENT DROPS - POWER FACTOR IS MORE LIKE 0.05 BUT THEN THINGS JUST START GETTING CRAZY... EACH ONE OF THOSE LITTLE MARKERS ON THE X AXIS = 90/13.5 = 6.66 DEGREES SO THE POWER FACTOR IS COS(90 - 6.66) PARALLAX ERROR ADJUSTED
T

July 11th, 2009 Data

Primary Input Voltage = 58.8 volts
Primary Input Current = 0.017 amps
Power Factor = 83.3 degrees / 0.116
Primary Input Power = 115 mW

Secondary Output Voltage = 3.1 volts
Secondary Load = 27 ohms
Secondary Output Power = 356 mW   

Efficiency = 310%

Outstanding T,

Great work, on both the transformer and your presentation, you seem to be surpassing the big T.

But since you just created one of the top OU devices in current history, can you disclose more detail before it possibly disappears or just to make sure it doesn't disappear.   

Is it an optical delusion or is the right secondary larger than the left?

Also, can you give out the permeability of the secondary and primary core and what is the wire gauge and ohms of each of the secondaries?

Thanks, Larry

PS: Thanks for the latest low angle shot of the transformer. It really shows why the reluctance would be much less even if it had the same permeability.