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PhysicsProf Steven E. Jones circuit shows 8x overunity ?

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

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xee2

Quote from: JouleSeeker on May 30, 2011, 11:26:27 AM
As noted earlier, I am trying to find caps that do not leak so fast, or at all.  Any ideas on this would be helpful.

All electrolytic capacitors have internal resistance that drains energy and causes voltage drop. My 10,000 uF caps drop from 5.5 volts to about 5.0 volts in about a minute. The best capacitors for holding charge are silver mica caps.

However, this should not be a problem, since a circuit with 8x power gain should be adding power much faster than it is being lost in the capacitor.

Congratulations on getting the circuit to self run. That is a big step towards showing that it is over unity.

I like to perform reality checks. If your circuit is producing 8x power gain, then putting 1/2 watts in should give 4 watts out. To test if this is happening you could use an 1/8 watt resistor as a load and see if it gets very hot with 1/2 watts input. It should if it is really getting 4 watts into it.




nul-points

hello Steven


welcome to the wonderful world of 'alternative energy'  - a minefield of measurement issues and previously-uncharted system behaviour!


i am not yet convinced that a true 'overunity' electrical system should always be capable of self-sustained operation with only a capacitor as it's main energy 'buffer'

so, for example, i was unprepared for the moment in Romero's video when he disconnected the battery and the system continued to run without any apparent impact of this (using only a capacitor and rotor momentum as its short-term energy buffers) for a further 15 minutes or so, until he switched off the device

obviously, if, as appears to be the case with Romero's device, you have a circuit which is capable of self-sustained operation without requiring a battery, then all well and good - but this is only a confidence booster for us in being able to accept what is claimed - ie. 'i'll believe it if i can see it'

the objection of 'measurement error' is immediately redundant if, for example, a group of people witness a 'powered, heavier-than-air, contraption' run along the grass, take to the air, and perform a circuit of Kitty Hawk airspace!


i don't feel 'uneasy' about claiming 'overunity' which still depends on the presence of a battery (otherwise i wouldn't be performing the cell experiments recorded at the blog linked below!) - a battery is after all, in some sense, just a rather longer term energy 'buffer'

obviously the main difference between a battery and a capacitor is that (we believe) a battery is largely a 'chemically' produced charge separation, whilst (we believe) that a capacitor is largely an 'electrically' produced charge separation

and it's because of the possibly more complex micro-scale processes at work in a battery that i can imagine that it's possible for a battery to play a significant role in achieving 'overunity' within a particular system (ie. the battery may have to be considered as just one of many components within a particular 'overunity' process)


therefore, if we can accept that it is still 'ok' for an overunity system to require a battery, we just need to account for it in our burden of proof

surely the battery-related equivalent of the 'self-runs only from capacitor' type demonstration is this:

the system is measured to perform a significantly greater total amount of work than the previously measured average Watt-hour capacity of that battery (measured using a conventional dissipative load, eg. a resistor, or heater, etc)

in other words, a more formal test along the lines of your informal test:  'i left the system running overnight, with no measurable drop in battery voltage'

the confirmation of 'overunity' in a battery-dependent system can either take the form of more energy converted in the same time (ie. higher continuous power out than drawn from the battery) or it could be just that the system is capable of sustaining a certain power level for significantly longer than the Watt-hour capacity of the battery (where 'significantly' longer may also be 'indefinitely' longer, effectively)


since you have made your initial findings on a system which includes a battery, why not make your next step to be a 'batteries included' style test? (using a suitably small capacity battery for convenience!)

this will either confirm or deny your instantaneous measurement results

if the new test results prove positive, then it would be interesting to move on to a 'capacitor only' style test and see if this also provides confirmation - or if instead it produces another anomaly (eg. 'overunity with battery' does not necessarily imply 'overunity with capacitor')


of course, if the battery style test does NOT provide confirmation of the 'instrumentation' results - then see the 1st paragraph of this post!  ;)


looking forward with interest to your next steps!
np


http://docsfreelunch.blogspot.com


"To do is to be" ---  Descartes;
"To be is to do"  ---  Jean Paul Sarte;
"Do be do be do" ---  F. Sinatra

JouleSeeker

Hey, I appreciate your comments Xee2 and Nul-points. 

Meanwhile, a quick test today:

Quote from: poynt99 on May 30, 2011, 01:13:02 PM
Professor,

Under ideal conditions, the scope method is accurate.

What I am suggesting is this; if the scope and DMM methods do not agree, one of them must be wrong. DC power sources have a power factor of 1.0, therefore heavily averaging the current and voltage measurements is not only the best way to measure the INPUT power, but it is the easiest and most accessible.

.99

I got 10 mW in on one early sj1 system (see reply #1 above), using the Tek 3032.  Itsu got 40 mW input power using the DMM method on his system.

Limited time today (holiday w/ family) -- but I did a quick test, another way to measure Pin.
Four 10K uF caps, to run the sj1 circuit.  By measuring the volts before and after 30 seconds on the caps, I can calculate input power easily.

delta-E = 1/2 C(Vi**2 - Vf**2) ,  Pin = deltaE / delta-T  , 30 seconds.  C = 40mF.

Start, Vinitial = 1.385V  , Vfinal =  1.255V

So delta-E = 6.8 mJoules.
and Pin = 6.8/o.5min = 13.6 mW,   pls check my math.

in reasonable agreement with the Tek-scope measurement under similar conditions, 10 mW
(see reply #1 for the data, Pin on the left).

Again, the Tek3032 is distant from here, so I can't do the two measurements within minutes, but I think this tends to verify the scope method.

I would ask Itsu to do the same thing on his sj1 circuit, and compare with the dual-DMM method.  I like to check things out, especially when measurement methods appear to disagree.

TinselKoala

I am afraid I'm touching a tar baby here, but... here goes.

First, I am glad that you are using caps, because that does give you an accurate way of estimating the ENERGY that you are inputting to the circuit in a given amount of time. Please for the moment forget about POWER and especially "mean power".
Energy is not power and power is not energy. Very high power multiplication factors may easily be achieved in oscillating circuits with no gain in energy. And I think we are all clear that it is ENERGY that is the important parameter when claims of Overunity or COP>1 are being made.

Now.... you can measure the energy output of your JT by integrating the INSTANTANEOUS power curve over a time period. There is no need to get any kind of average power reading, in fact this is a major (and common) error.
If your scope can only do the one math function at a time, then you must do the integration manually. There are several ways to do this. First, get away from the habit of displaying so many cycles on the screen that they are uninterpretable. Display only 3 or 4 complete waveforms, or even a single one.
OK, so now you display, say, two complete cycles of the instantaneous power curve. Overlay a piece of tracing graph paper on the screen and trace out the curves carefully. The integral of this curve is the VOLUME occupied  by the surface defined by the vertical dimension (the inst. power value) and the horizontal dimension (time). Using the scope's graticle and the horiz and vert settings, calibrate your little graph paper squares. (they will be in Joules). Then count up the area of your waveform.... and don't forget to multiply that by enough to fill up your known 30-second input energy from the caps.

Compare and contrast.  You are comparing Energy IN, using the correct calculation you have shown above, over a 30 second period, with the Energy OUT, which is integral(VxI)dt, from 0 to 30 seconds. Only if Energy OUT exceeds Energy IN is there any reason to get excited at all.

No "average power" or especially "RMS voltage and current" goes into the calculation at all.

Of course, if your scope will do integration, your problem is solved.

(I get 6.8 microJoules; I suppose you are using "mF" and "mJ" to mean microFarads and microJoules. I am more used to using "m" as "milli" and "u" (like greek mu) for micro.)

EDIT.. Whoops, sorry, my bad... you DO mean "milliJoules". I misread the size of your cap bank, I didn't realize you were using 10,000 uF x 4. Apologies. I accept your 6.8 milliJoules figure.

nul-points

Quote from: JouleSeeker on May 30, 2011, 03:00:12 PM
Hey, I appreciate your comments Xee2 and Nul-points. 
[...]
Meanwhile, a quick test today:
Four 10K uF caps, to run the sj1 circuit.  By measuring the volts before and after 30 seconds on the caps, I can calculate input power easily.

delta-E = 1/2 C(Vi**2 - Vf**2) ,  Pin = deltaE / delta-T  , 30 seconds.  C = 40mF.

Start, Vinitial = 1.385V  , Vfinal =  1.255V

So delta-E = 6.8 mJoules.
and Pin = 6.8/o.5min = 13.6 mW,   pls check my math.

in reasonable agreement with the Tek-scope measurement under similar conditions, 10 mW
(see reply #1 for the data, Pin on the left).
[...]

hi Steven

i think you have an incorrect method for calculating Pin

[(multiple) EDITs: (to clear up my mess!  LOL)
a Joule is a Watt-second - i see you've divided Ein by units of minutes;

Also - thanks to TK for spotting my transcription error!
Steven, apologies - your Ein calc method is good, but Pin needs units of seconds, not minutes]

Pin = 6.87/30 = 0.23mW


another potential issue to note - the cap value can be +/- 10-20%

when doing these calcs, it's wise to measure the cap value

of course, for a 'ball-park' calculation it's not necessary!  ;)

hope this helps
np

PS  i admire your other 'extra-mural' work , investigating & providing low-cost solar cooking solutions for developing countries!


http://docsfreelunch.blogspot.com


"To do is to be" ---  Descartes;
"To be is to do"  ---  Jean Paul Sarte;
"Do be do be do" ---  F. Sinatra