Yikes, there are a lot of problems with this whole attempt to measure here the more I think about it.
Firstly, there is some uncertainty if the time period is 60 seconds or 15 minutes.
Let's assume it's 15 minutes for the sake of argument. Bill would have to set up a control test where he just has a cap and connects approximately the same amount of wire to the capacitor terminals. It's very possible that the big electrolytic cap is selfcharging from the ions in the air. So if the control setup creeps up in voltage also that just about kills the proposition.
Going back to the testing as it is being done now, when you first start the test with the big cap at zero volts, then essentially 100% of the alleged free energy being generated by the device is being burned off inside the device itself. It's a crippling problem for the cap test and the larger the capacitor the worse off things are.
WaveWatcher also mentioned you probably need a 100 Megaohm potentiometer to do the impedance matching test and he might be right. If he is right then the typical 1 Megaohm input resistance of the multimeter would be disturbing your attempt to make the impedance matching measurement. If that's the case you need to do a differential voltage measurement between say a 1 kiloohm potentiometer and Billmehess' device and turn the potentiometer to zeroout the voltage differential. Then measure the voltage on the potentiometer output separately. All that just to measure the opencircuit voltage on what we assume is a superhighimpedance output device.
Then you have to repeat the process, but this time say use a 10 Megaohm resistor across the device, and use the potentiometer and voltage differential method to measure the voltage drop.
You do all that and you have a relatively undisturbed opencircuit voltage measurement, and a relatively undisturbed voltage measurement under a 10 Megaohm load. That gives you the ability to calculate the output impedance of the device with some confidence.
Once you do that you are done. You know the opencircuit voltage and you know the output impedance of the device. That's it, now you know it's power output capabilities under load. All of this is based on the presumption that the input resistance of your multimeter is too "low" and will disturb Billmehess' device.
Before I did any of this, I would do the control experiment with the capacitor to see if the whole thing is just due to hot (ionized) air and the nature of electrolytic capacitors.
But certainly this primitive big electrolytic cap test is junk, the device is seeing a shortcircuit to ground or nearground for a long time and most of the alleged free power is being burned off inside the device itself. Therefore the output power is not really being measured properly and the number crunching is showing power values that are much lower than they potentially are with an impedancematched setup.
The critical thing is to figure out the output impedance of Billmehess' device. If the output impedance is very very high then you have to go to the extreme measures like I outline above.
Anyway, this posting will likely be ignored so what the hey....
MileHigh
« Last Edit: 20110819, 01:21:14 by MileHigh »
