LightinDarkness wrote:The usage of volt meters to show inputs/outputs is intentionally deceiving, I can't remember why but I was reading a free energy debunking elsewhere and they were explaining why it doesn't actually show what you are looking for to determine if something is producing excess energy. Something about how it doesn't take into account resistance or something, which if it did, would show its using more energy than it generates.
They are using volt meters because its part of the technology cargo cult aspect of it, it fools most people.
Yes, plus volt- and ammeters are cheap, and widely available. For the record:
Many people learned in high school physics, or elsewhere, the relation P = IV, that electrical power is equal to the voltage times the current. Knowing that, it's very easy to take a voltmeter and an ammeter on the input, and a voltmeter and an ammeter on the output, and show some big change in the two, and suddenly, OMG! Free Energy!
But P = IV only strictly works for direct current. When you get into AC, two complications arise.
First, AC values transmit power with different efficiencies. (That's not accurate, but it's a quick approximation of the point.) For a given AC quantity, like a voltage or a current, there is something called its "rms" value, which is roughly its equivalent in DC terms. A simple sine wave, for example, has an rms value that is about 70.7% of its peak value. This is important because a lot of meters commonly available on the market assume that the rms value is 70.7% of peak, and that's not true for a lot of trashy waveforms like the ones you get off of home-built free energy machines. You can get true rms meters, and they're better at this, but still have their limitations. This ties into the comments above about spiky waveforms and so on.
Second, even once you're looking at rms and not peak values, when you are talking about AC, there are three kinds of power involved. Multiplying rms voltage by rms current gives you apparent power, measured in volt-amperes. All that's really good for, on its own, is heating up the wires. When you want to do useful work, you need to know the real power, which is:
P = V(rms) * I(rms) * cos ø
where ø stands for the phase angle between the voltage and the current: zero when they're in phase (that is, both flowing in the same direction at the same time), 180˚ when they're dead out of phase (flowing against one another at the same time), and so on. (The third kind of power is called "reactive power," and takes a sine rather than a cosine in the formula above. It's not useful for much except for magnetizing equipment cores and controlling power grid voltages.) V and I are easy to measure, but ø is not, and depends on a phase angle meter or (hey!) an oscilloscope.
So, supposing that we hook a circuit up, and it shows 120 volts and 1 amp on the input, and 130 volts and 1 amp on the output. So by simple math, 120 volts times 1 amp on the input is 120 W in; 130 volts times 1 amp on the output is 130 W out. Hey! free energy! Except it turns out that we've snuck in a capacitor on the circuit, and capacitors (for technical reasons that I won't explain here) draw current almost 90˚ out of phase with the voltage. So if you actually measure the phase angles, and do the math:
P input = 120 V * 1 A * cos (-85˚) = 10.5 watts, and
P output = 130 V * 1 A * cos (-89.9˚) = 0.22 watts.
Oops. No free energy. In fact, we're losing almost all of the power going into the circuit. But, just looking at the voltage and current, you'd never know it. And that's really why you can't trust people throwing multimeters around to prove these things are producing free energy. I mean, aside from that violating the fundamental laws of the universe as we understand them.
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A correction, while I'm at it. I said above that the QEG waveforms showed an output of 950 watts. This was incorrect: their oscilloscope showed a value of 1900 V peak-to-peak, and I failed to divide this by two (to get peak voltage) in converting to rms. Their measured output is much more like 475 VA (vs. 655 W input), and perhaps even less depending on how they measured their output current. Evidently I finished EE 105 a little too long ago to remember the relation correctly without checking references, and I apologize for the error.
), they'd have TWO watt-hour meters -- one on the input and one on the output -- to measure the POWER goiong in and coming out. However, they won't ever do this because (1) the results will show that it actually CONSUMES power due to electrical and physical resistance and (2) the two meters would tell the machine it wasn't being trusted, thereby casting a pall of negative vibrations over it.