innovation_station is the most underestimated inventor on here. He's the tesla of this forum  . Any electrical concept that is brought up, he has experimented some variation of it already . IST, you only need to work on your presentation skills  . I bet the following days will be interesting.

     Here's another little effect you can do.  Take a short length of garden hose and place it on a table and let one end hang over the edge of the table.  Fill the hose with water to get all the air out then connect it to the spigot.  Open the valve.  The end of the hose should move defying gravity.  Did the water going into the hose at the faucet end have to go all the way to the other end of the hose to move it?  Nope just the pressure wave.  Now if the water company drops all it's pressure doing work somewhere else and we open the valve again.  We get a second motion on the hose end and the water company gets some of it's pressure back.  Now if this is done at a high rate of change or frequency and each time the hose lifts or falls it rings a bell.  Things are going to get real loud from the reflection of the pressure wave.  Just a matter of timing.

QuoteWe are talking about conserving all energy within the circuit--a very simple process everyone wants to complicate for some reason.

Then just leave the cap charged and don't do anything with it.!!! If I can's extract power to a heater or a real resistive load, I have not done any useful work. If I can't extract any useful horsepower from the motor what good is it?.

This experiment is so flawed and lacking credulity, I won't even bother to get caught up in the circular and pretzel logic arguments.

But I'm sure IST will be able to explain it all for ya..yada... yada.......Aerobine anyone?

Quote from: TinselKoala on November 17, 2008, 09:35:38 AM
When you transfer between capacitors you are transferring charge, that is, energy, not "just" voltage. The energy on a capacitor goes as the square of the voltage. E=(1/2)(C*V*V) where E is in Joules, C is in Farads, and V is in Volts. SO a capacitor charged to 12 volts has much less than half the energy of the same capacitor charged to 24 volts.
Capacitors do NOT store power, they store energy. Power is the rate at which energy is released over time. Energy is expressed in Joules, which are equivalent to Watt-Seconds. That is, one Joule of energy is one watt of power expended over one second of time. If you scrunch that same one Joule of ENERGY into a tenth of a second, then you have 10 watts of power, for a tenth of a second. And so forth.
Also, capacitance of a capacitor is almost always going to be different than what the label says, sometimes by 10 percent or more.

E=0.5CV^2 only tells how much energy it'd cost to transfer charge to the plates, since action = - reaction, it also equals the energy stored in the cap.
Why does it cost energy? Because voltage-force on the charge works against the stored charge, like loading a spring: the further you press, the more energy is stored, but you have to exert more force to store more and not to be pushed back.

or something like that.

so, since we are dealing with something else than conventional, does E=0.5CVV still apply? Maybe we should consider second field over time derivatives to come up with an extended energy formula?

Quote from: broli on November 18, 2008, 11:07:21 AM
innovation_station is the most underestimated inventor on here. He's the tesla of this forum  . Any electrical concept that is brought up, he has experimented some variation of it already . IST, you only need to work on your presentation skills  . I bet the following days will be interesting.

Yeah my thoughts exactly, he need to post more video's which show his findings.