PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[ltseung888]

Dear Prof. Jones,

I believe that I might have performed the conclusive experiment on sound resonance.  Please see
http://www.overunityresearch.com/index.php?topic=588.msg17563#msg17563
for details.

The experiment absolutely confirms that placing 1,2,3,4 resonance boxes at appropriate locations and distances from a constant sound source will create louder sounds.  The extra sound energy must come from the environment - led-out or brought-in the already available kinetic energy of the air molecules.

This fact will have great impact on all our research - including multiple LCR resonance.

Lawrence

[albertouno]

Good morning again Steven and Nul-Points:  (I am on East Coast time)

"Have a great vacation, alberto.
Can you explain how your plotted "the discharge curve of a battery to estimate the energy collected"? == I'm interested.

(I can imagine draining a battery into a large cap and using E = 1/2 C V**2, except it may be difficult to calculate the loss of energy due to "leakage" from the cap...)"


Interesting info Nup-Points.    T0 explain, I was charging a very depleted battery with a Bedini circuit and trying to estimate what energy I was getting out.  Just connected the charged battery to a small resistor (R), and measured voltage decay vs time in discrete increments.   Don't have a plot handy, but compared the battery discharge voltage to a simple RC exponential decay curve on an Excel plot. Noticed very similar shapes.  If I assume that the battery has a capacitive equivalent circuit, I can calculate that capacitor value as follows:

C=T/(R*(ln(V1)-ln(V2)).  Where V1 is starting voltage, V2 is ending voltage at time T.  I can then use the usual energy equation for the capacitor and V1 to estimate what the battery starting energy must have been.  Not sure if this is valid, but I did get some interesting results, and repeatability for different charging times. (The cap value is usually in farads).  Hope this helps.  Note, I read that most alkaline batteries are rechargable (and certainly cheaper), and I have also noticed this.  I may try this test on the subject oscillator circuit.

Best Wishes,
Alberto

[nul-points]

hi Alberto

it's interesting that you say you were starting with a very depleted battery in your (Bedini) charge tests:

my results with discharging a reasonably well-charged battery seemed to suggest a mostly linear voltage decline with a fixed resistive load over the working range

however, as you can see from the example graph above, which  i just picked at random, once the level of charge has decreased beyond the working range then you do see a more exponential decrease in terminal voltage

could it be that the batteries in your test were not getting charged into the working range level?

there is another possibility - ie. that your batteries were fully charged

i noticed when discharging fully charged cells/batteries that just before entering the working range the terminal voltage discharged more quickly, so that would be another possible position for a more capacitor-like discharge curve

you can see a bit of that characteristic on the example graph below - the results of discharging an 8.4V NiMH: the 'linear' discharge range is sandwiched between two more 'exponential' slope regions (although the extremes were not recorded in this graph)


of course, i guess you may have been using Lead-Acid batteries - whereas i was using NiCds and NiMHs, so this could account for the difference in characteristics we've observed

i look forward to learning of the results which you and Steven get from your charging tests with the blocking oscillator

thanks
np


http://docsfreelunch.blogspot.com

[albertouno]

Nul-Points:

Thanks for the info.  I believe that my depleted (alkaline) test batteries were starting well below the normal working voltage range, as you pointed out in your first curve example.  In fact, I tried to make sure they were quite depleted before charging them.  By accident, this made the comparison to a capacitive discharge obvious and easy to calculate.  Is it better to do the test with depleted batteries for this reason?  Does just the voltage difference accurately measure the energy input to a more fully charged battery?

Alberto

[JouleSeeker]

  Thank you NP and Alberto for the comments on batteries -- very helpful plot, NP.

NP: "- then discharge them through a known resistive load " -- I have found when I do this with rechargeable AA's, that when the battery voltage gets to roughly 0.9V, that the battery drops in voltage rapidly -- with a constant resistive load (about 100 ohms).  Then, when the load is removed, the voltage on the batt climbs back up to close to what it was...  I.e., it is difficult to discharge one of these batts to say 0.6V using a constant R... 

Anyway, I discharged 4 of these in parallel through about 100 ohms for approx 20 hours, then let them sit for about 10 days.  The voltage on each was close to 1.0 V as it turned out and quite stable after this "rest".  I took the two with lower voltage and put these in series as Vinput in my circuit (simplified circuit shown above on this page), and the other two in parallel located in series with the LED so as to be CHARGED (see circuit above). 

I will report results after about 20 hours of running.  Here are the starting voltages:

Vinput (2 AA's in series) = 1.9317 V

Voutput (2 AA's in parallel) = 1.0334 V

both measured using the same Keithley DMM.

One important change -- I found that the LED still lit up quite brightly when I cranked the potentiometer (Rb) up to about 0.9 Mohms, megaohms.  (When I stop the run, I will measure the resistance more precisely.)    I was surprised that from roughly 300kohms to 900kohms, I could see no noticeable drop in light-intensity from the LED (by eye).  So I left Rb at the high resistance, to cut down on drain from the input batteries while keeping the output (LED) fairly steady.