everything i am showing i have made my own through my RESERCH ONLY

i did not copy a single person and i learned what i have through practial tests

truth is i had this figured b4 i picked up those pattends.... b4 i met erfinder....  thank you...  that was almost 1 year ago already...


it is a vibrator peroid..... it intrupts the flow and catches what comes back and uses it with the next fire to cut way back on inputted power....

ist

JUST DO THE TEST!

sorry thebuzz... 

this is diffrent than the vids but it is desinged to coinside with this tech....

also on that test you will note.....

you can place the reed switch permantally in the proper location ...

you will then use a permant magnet to start it and stop it .....

intresting DONT YOU THINK.....   hummm

now if anyone gets this then you could then take this 1 step further with a resonant coupled coil bifiller .... get the drift.....  maybe tune the coil to 7.5 hz.....

well 


lets see some pics of work   any one ...  or am i wasteing my time ...

ist

@IS
Damn, thats the last time I build one of your circuits, LOL
Im just kidding--it worked fine until my only reed switch,which was weak to begin with went poof. I see what your getting at with this circuit, I needed a little more time to match the capacitance to really get this circuit in the groove but in worked good.
This is why I have always been a commutator kind of guy
You guys will like this circuit but you have to scope the capacitor to see what is really happening in the circuit.

Edit-reduced the capacitance and tuned with a relay and she took off. Nice my inductance was still a little low so the wave form was a bit dampened but it works very good.
Regards
AC

Captainpecan, allcanadian, blind believers and all others that still do not understand:

This is fairly basic stuff.

Dissipation? Yes of course there is dissipation in the coils. As long as there is series coil resistance.

Scenario 1: Connect a 18V capacitor across a coil, and leave it connected for 30 seconds No rotor or motor involved, just a simple real inductor.

Scenario 2: Connect the same 18V capacitor across the same coil that is in series with another identical value capacitor. Leave this connected for 30 seconds. Again, no rotor/motor involved.

If you were to plot the coil current vs. time, in which of the above scenarios will the current fall (after the initial rise) to a given value the quickest? Why?

After 30 seconds, the capacitor in scenario 1 will be almost completely drained of voltage and will contain very little remaining potential energy.

After 30 seconds, the 2 capacitors in scenario 2 will both contain roughly 9V, and contain some potential energy as a result.

Work is a measure of force through a distance so we can not measure work in this setup. We can measure the power through a series resistor though, and power can be converted to work if one so desires. We can place a very small resistance in series with the coil, OR we can simply measure the temperature rise of the coil itself, since it will have its own inherent DC resistance.

In which of the two above scenarios will the coil/resistor temperature rise the most? If you believe that the temperature rise will be equal, that would be a bad start.

In scenario two the temp rise will be less, even if you add the remaining two 9V discharges to the previous temp rise!

Why? The process of transferring charge/voltage between capacitors is a lossy process in itself. The energy (in Watts) in a 9V cap is 1/4 that of one at 18V. The energy (in Watts) of a 18V capacitor discharging into another equal value capacitor is somewhat more complicated, but it will fall somewhere between 1/4 and 1/2 compared to a capacitor sitting at 18V being discharged into a short. Let's use 1/2. So we have 1/2 + 1/2 = 1. Minus some losses, and it is below the power that can be delivered directly from a 18V cap itself.

So you see, even the best possible outcome means you can only get about the same power/energy/work out from the two above scenarios.

BTW, if you want to be completely fair with your present motor setup, you must use a pulse circuit (one-shot) with an adjustable pulse width for the scenario 1 test, and steadily reduce the pulse width until the number of rotations equals what you are getting in scenario 2. Once you have done this, you will find that you have a surprising amount of voltage left on the capacitor in scenario 1 after the pulse. This would I hope, prove to yourself that the test procedure you are presently using is faulty and that you are wasting a lot of energy that you were previously unaware of.

Regards,
Poynt99

There is still one thing that puzzles me about the Young Effect.
Something I haven't seen anyone comment on yet.
Even Captainpecan didn't make mention of it.
What about the energy that is conserved when you charge the capacitor up through the motor? 
I placed the motor between one lead of the batteries and the capacitor and the motor ran for quit some time
before the capacitor was charged up. 
Now we have many turns of the motor and a charged capacitor. Which can then be used to do some more work.
Instead of destroying the charge put out by the battery while running the motor,
we store that energy in a capacitor.  The Larger the capacitor the longer the motor will run before the capacitor is charged.
The dead short of a dead capacitor makes for a good jolt to get the motor spinning,
Then as the motor speeds up the resistance of the capacitor starts to go up as well.
Once the motor is moving at a good pace what would connecting another dead capacitor up to it do?

Perhaps this would have some value?