Quote from: captainpecan on December 12, 2008, 05:14:20 PM
I do find it interesting though, that if you lose half your energy simply transferring from cap to cap, it would seem you would lose a noticeable amount more when you use that energy to do work along the way between them. 

CP, in both cases we are doing work, one is a combination of mechanical and electrical (with the motor in series), and the other is purely electrical (only wiring and diode in series).

So in the case with the motor, energy is being expended in heat, in mechanical losses, and defying gravity with the weight. In the simple case with only a cap to cap transfer through a diode and the associated wiring, it is strictly a heat loss.

It comes down simply to energy loss, and energy can be utilized to do mechanical work or generate heat in calories.

.99

Quote from: captainpecan on December 12, 2008, 09:56:34 PM
@99

Good point.  The resistance of the motor is 22 ohms though, which is about 4 times the resistance in the tests. Couldn't tell you the inductance of the motor coils though.  My assumption was the added resistance could skew the results just slightly enough to not follow your work exactly. As I said, it is an assumption. But we all know what happens when we assume.. lol.

Hi CP,

Right . I should clarify that as we increase the resistance and a certain value of series resistance is reached, in relation to both the inductance and the capacitance, the losses will always be about 50%. So even with a series resistance of 1000 Ohms, there would be a 50% energy loss, even if there was a small inductor in series with this resistance.

The resistance "swamps out" the inductor after a certain point.

The inductance in that little motor is just too small to make any appreciable difference to the energy transfer efficiency.

.99

CaptainPecan,

Look at Bedini's work, and what he has to say about capacitors.

One thing I wanted to mention that seems to be overlooked (probably isn't though, I haven't read all 70 something posts).  Anyway, when you place the capacitors in parallel, you are adding the capacitance of the system.  Together they become one capacitor of 9400uF charged to 9Volts.

One thing I was thinking about today was what happens when you add capacitors that are not of the same capacitance.  As Allcanadian said, if you add a 2F capacitor in parallel instead of another 4700uF, what is the voltage in both after the discharge?  I know the ratio will be different, but how do you determine what the voltage ratios are?  Its probably simple, but because the energy stored in the first capacitor is limited, its not initially apparent to me.  Any ideas?

Charlie

Also, I can tell you right now a coil of 22ohms is WAY too lossy.  Try getting a half inch piece of copper pipe and bend it into a coil, and make sure all your leads are large wires and very short.  That should improve efficiency a lot.