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Overunity Machines Forum



PhysicsProf Steven E. Jones circuit shows 8x overunity ?

Started by JouleSeeker, May 19, 2011, 11:21:55 PM

Previous topic - Next topic

0 Members and 16 Guests are viewing this topic.

poynt99

Quote from: JouleSeeker on February 10, 2012, 01:31:37 PM
Agreed, Mags.  And I like to see more than one method used to measure power out, as a check.

My question has always been this:

Why must the load be a diode or LED? Can the load not be a pure resistance?

.99
question everything, double check the facts, THEN decide your path...

Simple Cheap Low Power Oscillators V2.0
http://www.overunity.com/index.php?action=downloads;sa=view;down=248
Towards Realizing the TPU V1.4: http://www.overunity.com/index.php?action=downloads;sa=view;down=217
Capacitor Energy Transfer Experiments V1.0: http://www.overunity.com/index.php?action=downloads;sa=view;down=209

Magluvin

Quote from: poynt99 on February 14, 2012, 07:24:16 PM
My question has always been this:

Why must the load be a diode or LED? Can the load not be a pure resistance?

.99

Actually, I wouldnt use a diode as a load in any case. What for?   :o

I am also not keen on using leds as indicators of power. I think we talked this the other day. ;]

And, just above, on this page, I posted this, "Either way, if using the current resistors, measuring the resistor for power out would be the best bet."  8)

http://www.overunity.com/10773/physicsprof-steven-e-jones-circuit-shows-8x-overunity/msg312165/#msg312165

;)

Mags

poynt99

Quote from: Magluvin on February 14, 2012, 08:40:06 PM
Actually, I wouldnt use a diode as a load in any case. What for?   :o

I am also not keen on using leds as indicators of power. I think we talked this the other day. ;]

And, just above, on this page, I posted this, "Either way, if using the current resistors, measuring the resistor for power out would be the best bet."  8)

http://www.overunity.com/10773/physicsprof-steven-e-jones-circuit-shows-8x-overunity/msg312165/#msg312165

;)

Mags

Yes, I saw the recent posts.

I assumed that the "current resistor" is still in series with either a diode or LED.

Did I assume incorrectly?

.99
question everything, double check the facts, THEN decide your path...

Simple Cheap Low Power Oscillators V2.0
http://www.overunity.com/index.php?action=downloads;sa=view;down=248
Towards Realizing the TPU V1.4: http://www.overunity.com/index.php?action=downloads;sa=view;down=217
Capacitor Energy Transfer Experiments V1.0: http://www.overunity.com/index.php?action=downloads;sa=view;down=209

Magluvin

Quote from: poynt99 on February 14, 2012, 09:06:25 PM
Yes, I saw the recent posts.

I assumed that the "current resistor" is still in series with either a diode or LED.

Did I assume incorrectly?

.99

Yea, but I still objected to using led brightness as an indicator of power when comparing 2 circuits. Depending on the circuits, sure one may drive the led at 20ma and another at 30ma, then we could say  something about the power out by saying one is brighter than the other thus more power. But I would count on the measurement before relying on the led brightness to evaluate how much power. So I  (pre)agree,  no led needed. ;]

Mags

Mags

poynt99

So if you agree that no LED or diode is required, please post a diagram of the circuit. Please also indicate which resistor is the load.

Has anyone tested the circuit without a diode/LED? Does this then still conform to the Professor's circuit and claim?

.99
question everything, double check the facts, THEN decide your path...

Simple Cheap Low Power Oscillators V2.0
http://www.overunity.com/index.php?action=downloads;sa=view;down=248
Towards Realizing the TPU V1.4: http://www.overunity.com/index.php?action=downloads;sa=view;down=217
Capacitor Energy Transfer Experiments V1.0: http://www.overunity.com/index.php?action=downloads;sa=view;down=209