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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

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JouleSeeker

Back to the relativity puzzle:
Xee2:
QuoteAccepted answer is Feynman's virtual photons traveling backwards in time,

I don't that will work in this case, xee, because we can move the loops arbitrarily far apart and get the effect.  Virtual photons are limited by the Heisenberg Uncertainty Principle  -- which also applies to Jmmac's question"

Quote from: jmmac on June 13, 2011, 06:27:47 PM
Hi Professor,

This is a bit too much for me but interesting... 2 questions:

- The conservation of moment must happen in 'real time' or is it possible to have delays because of the propagation times?


Momentum non-conservation is allowed for a short time for virtual particles per Heisenberg, delta-momentum*time < h-bar, where h is Planck's constant.  This is an exceedingly short time, for this set-up, compared with 0.1 n-seconds.  Nope, Heisenberg won't help solve the puzzle.


For your other question and gyula's, let's simplify the experiment to this:  one short current pulse in A, then A off.  As the field reaches B (we know when this will be, from t = separation D/speed-of-light c), B receives a current (from the outside current source, not eddy because the source is otherwise high-impedance) so it receives a jolt, a push.  THAT field from B propagates to A, but A is OFF by then (again essentially zero eddy currents due to high impedance).

Your diagram with the water flowing up and down is interesting, NerzhDishual , but not a clear violation of Conservation of Momentum -- because the ball going up impedes the water coming out of the hose upward, so more water will flow downward to compensate.  There is (clearly) water moving both up and down -- whereas in this thought experiment, A launches a field (which moves both left and right) which pushes B to the right due to the momentary current in B, but there is nothing pushing A to the left (i.e., no compensating force). 

I think - and may be wrong, but I can't see anything pushing A to the left.

JouleSeeker

Quote from: NickZ on June 13, 2011, 10:01:04 PM
  I think think I'm showing more wear and tear.

Nick -- are you referring to the "cement cell"?  could you be more specific - what is showing wear and tear?
Thanks -- honesty is what we need to make solid progress.

About antennas -- " I have seen some pretty amazing experiments from respected individuals that show size and shape of the antennas are very important when designing the unit. " 
YES!  but for the relativity-puzzle-thought-experiment, two simple loops will suffice for the discussion, because that's all we need for the simple electromagnets, one pushing on the other and not being pushed back...

If the action were INSTANTANEOUS, instead of limited by the speed of light, we would not have this conundrum...  But it is an important effect, the limit imposed by the speed of light.  (Thanks Albert!)

MeggerMan

Quote from: JouleSeeker on June 14, 2011, 12:17:53 AM
Best result tonight:
n ~ (1.366**2 - 1.258**2) / (2.0**2 - 1.89**2)  = 0.7 = 70% (conservative)

Now, this is with matched caps for input energy and output energy, with the energy in the LED "thrown away", as it was lit visibly.  Hope you're following what I'm doing here -- achieve a low Pin while LED is still lit visibly (quite bright), then get a FIRST CONSERVATIVE ESTIMATE for Pout (or Eout in this case).
Hi JouleSeeker,
I have wound a toroid that comes out as about 119uH and 118uH per coil.
I have 2N2222 transistors but not the MPS2222 - I think they are equivalent.
The rest of the parts I have in stock.
It seems your 8x OU could be a measurement issue, but discharging/charging a cap seems a very fair way of comparing in to out.
What you might want to consider is the adding some 0.1uF caps in parallel with your input and output caps to ESR (effective series resistance) of the electrolytics.
Look forward to your next test results.
Thanks.
Rob

JouleSeeker

Quote from: MeggerMan on June 14, 2011, 08:36:23 AM
Hi JouleSeeker,
I have wound a toroid that comes out as about 119uH and 118uH per coil.
I have 2N2222 transistors but not the MPS2222 - I think they are equivalent.
The rest of the parts I have in stock.
It seems your 8x OU could be a measurement issue, but discharging/charging a cap seems a very fair way of comparing in to out.
What you might want to consider is the adding some 0.1uF caps in parallel with your input and output caps to ESR (effective series resistance) of the electrolytics.
Look forward to your next test results.
Thanks.
Rob

Yes, thanks MeggerMan, I look forward to your results also.  I agree that "discharging/charging a cap seems a very fair way of comparing in to out" with the caveat that we are capturing only a portion of the output energy.

Over at OUResearch, laneal makes this observation:
QuoteThanks professor for sharing those measurements.

So with a 10mF capacitor in place of R0, the voltage rises to 1.58V from 1.30V.
For a red LED, the forward voltage drop is about 1.67V (I measured mine with a DMM, please replace it with your own measurement).
Therefore, the power spent on the LED is: C * deltaV * Vdiode = 10mF * (1.58 - 1.30) * 1.67 =  4.676mJ.
The total energy stored in the cap is  (1.58**2 - 1.30**2) * 10mF /2 = 4.03
So, total output is: 4.676+4.03 = 8.706mJ.

Total input: (2.54**2 - 2.25**2) * 10mF/2 = 6.9455

Therefore n = 8.706 / 6.9455 = 1.253

Hey, that's more like an OU  :) But clearly it needs an accurate measurement of your Vdiode.

P.S.: my calculation shows that as long as the Vdiode > 1.04125V, we will have n>1.

Quote[more from laneal: ]
Quotes Prof:  Best result tonight:
n ~ (1.366**2 - 1.258**2) / (2.0**2 - 1.89**2)  = 0.7 = 70% (conservative)

For this best case, if Vdiode > (1.366V+1.258V)/2 = 1.312V, then n > 140%.
Vdiode is the voltage drop over the diode at the moment when the capacitor is being charged.

For this best case, if Vdiode > 0.667, n>1.
Of course, this is still a conservative computation of n, as we have not calculated power wastes in the transistor and toroid.

Here is my reply --

QuoteHere I attach a photo of the set-up, using one cap for Ein and the second cap to capture some of the Eout.  Two DMM's read the voltage, one on each cap.
 
   Also a screen shot of the voltage across the LED, while powered by the Ein cap alone, with the second cap charging (from approx 0 volts, starting voltage).

The measurements read:
Vmean 40 mV, which is approx what I read with the DMM across the LED...  but --
Vpp  4.84V
Vrms 680mV
Vtop 3.12V -- this is the voltage in the forward direction, the direction of the current flow allowed by the LED

So you tell me, you asked for the voltage across the LED -- but which voltage does one use??

MeggerMan

Quote from: JouleSeeker on June 14, 2011, 10:02:19 AM
Yes, thanks MeggerMan, I look forward to your results also.  I agree that "discharging/charging a cap seems a very fair way of comparing in to out" with the caveat that we are capturing only a portion of the output energy.
Hi JouleSeeker,
The other thing I thought of is that the spike voltage will be suppressed to a degree by the output capacitor and this may be pulling down your output gain.
So the effect could rely on the sudden surge and capturing it would be a real challenge.
One way may be a synchronized switch using a mosfet that cuts in near the peak of the spike, perhaps using the "avalanche" process. Then dump this into a capacitor.