STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[exnihiloest]

No, you're saying that, not I. Don't put words in my mouth.

I didn't put them in your mouth but in "omnibuzz"'s mouth, some one like you.

[Omnibus]

@All,

Are you ready? If so, hold on to your hats. I'll show you a bunch of figures which go beyond the bare bone studies I last posted. Here we're not only having one element but that element isn't even a coil. Take a look at the schematic diagram:

[Omnibus]

@broli,

Hope you'll agree that such a setup settles the proble as to where exactly the current flows. There's no question that the entire current we're measuring with the current flow passes through the active resistance R. Therefore I^2R is indeed the entire obtained power and the integral of that I^2R over time is the entire energy dissipated in the active resistance. Further, you'll also agree that that parasitic capacitance or parasitic inductance for that matter is out of the question here. Clear as a sunny day in June.


P.S. I should have addressed this posting also to @Omega_0, @teslaalset and some other folks I'd love to discuss this but I know they are doing their own research and am not sure they're reading this thread.

[Omnibus]

Now, let's get to the essence of this finding:

By Ohm's law every momentary current should correspond to a given momentary voltage which is determined by the active resistance. This is illustrated in screen shot a) below.

Here's a description of those screen shots presented below:

Oscilloscope screen shots---trace 1 (yellow trace) is the current while trace 2 (blue trace) is the voltage. Each trace is the average of 16 traces taken at the interval of 1s. Each trace typically consists of 3,125 points and occasionally 100,000 point traces were recorded for a check-up. The results from the data processing of the more and less detailed traces show practically the same effect.  a) pure 99.7147+-0.0004Ohm resistor, b) 99.7147+-0.0004Ohm resistor plus 115pF capacitor shown in the circuit schematics I posted above, c) 99.7147+-0.0004Ohm resistor plus 64pF capacitor shown in the schematics above.

However, as is well known, at a given frequency, placing a capacitor in series with the resistor not only changes the voltage but also changes the current in two ways. First, it changes the current but not to the value that should be obtained from Ohm's law using that changed voltage. Second, it changes the phase of the current with respect to the voltage---the current begins to lead the voltage--see screen shots b) and c).

It has not been noticed, however, that, while the output, including momentary output, can only be positive because of the squaring of current, the momentary input can also be negative as well as positive. Moreover, these input values of varying sign are not fixed by some fixed current and voltage values determined by R through Ohm's law, but can be made different for a given R through introducing capacitances C in series with the resistance R (cf. circuit schematics posted above). Thus, conditions exist whereby the integrated values of the input power (the energy) in the presence of a capacitor would be lower than the integrated power in absence of a capacitor, which is given by I^2R, that is, using Ohm's law. In other words, the observed overunity effect is due to saving on the side of the input power compared to what it might have been provided Ohm's law were in effect. The saving of input power can be to such an extent that it may not even be spent but will be returned to the source, due solely to the construction of the device, ensuring conditions for the integrated input power to be of negative slope. The above data confirm experimentally that such possibility to save input power exists. It also confirms the view expressed previously in many Omnibus' writings and posts that constructions, not only mechanical, exist which allow for the production of energy without exhausting any previously existing energy source.

In connection with screen shot a) and the energy-time plot derived from it, which I'll post shortly, notice also that the output curve has been calculated for R measured at room temperature. The current passing through the resistor in absence of the capacitor, corresponding to the applied approx. 30V voltage, heats up the resistor substantially, thus raising its resistance. If that rise in the resistance of R is taken into account, then the integrated powers Pout and Pin will show overunity in the case of a pure resistor as well. This will be the subject of further studies.

As discussed above, by properly choosing the conditions; i.e., controlling the current through choosing the right capacitance, one may not only achieve the production of more energy than the energy spent (seen in the diagram I'll post shortly), but also can reach a state whereby, in addition to the excess heat, the functioning of the device amounts to having energy continuously sent back to the power source--you'll see that in the graph I'm going to post in the submission following this one. That effect can be used to construct a self-sustaining device which will power itself without the need for spending external energy. Therefore, the next step is to manufacture such device.

@broli, please notice the levels here where OU is observed. These require calorimetry of unusually high accuracy and precision hardly available anywhere. Therefore, this question can and should be resolved by purely electrical means. The best resolution for those who are practically minded will be by producing a self-sustaining device. For scientists the data presented here should be enough to convince them in the reality of the OU effect.

[Omnibus]

Here are the promised energy-time plots for the different conditions. This first one is for the above-stated R in series with  a 115pF capacitor: