The Young Effect, my gift to the free energy movement!

[allcanadian]

@captainpecan
Excellent video, It's funny isn't it----- all of them tell us the motor has performed "work" in our motor systems, LOL it has done no work, there is no work involved other than ohmic losses. Whether the motor is fully loaded or free wheeling the motor has done little if any "work" relative to the source. The only reason people have assumed the motor has done electrical work is because of the funny relationship whereby the source discharges at a rate equivalent to the work the motor performs-----change the relationship and you change the rate at which the source discharges  . As you say-- what if there was a way to transfer energy without "misplacing" a great deal of the energy? Much of the confusion revolves around electrical work, power and tangible work, I will tell you electrical work is not equivalent to the tangible work output from a motor and they are two completely different things. When you consider why energy is lost in your two cap/motor experiment you should also consider the capacity of the system or capacitive resistance as well, that is the moment the caps have equalized cap 2 offers a great resistance--- replace cap 2 with a 2 Farad cap and you will see this "resistance" is no longer present . I have proven this for myself and you have made an excellent start towards this goal, keep up the good work you are definately on the right track.
P.S.-- I have found it easier to just ignore the motor as in reality it is nothing more than a time delay ie... it momentarily "stores" energy then releases it and very little electrical energy is actually lost. Solve the "time" issue and you will be well on your way.
Regards
AC

[wattsup]

@AC

Glad to see you are back. Hope that little stint with TheBuzz did not ruffle your feathers and we can all put that in the past and move forward. Yes it is true that the feed will only supply what the motor needs but in this case, I would suspect that the motor needs are way above what the capacitor could deliver in any sustained power supply and I think this is where there is some confusion on what is being seen.

@CP

You know what the capacitors are but do you really know what that dc motor is, how is it made, how is it wound, does it have a commutator, are the commutator brushes wider then one segment, what is the ohms of one segment, how many turns did the motor do when you discharged the cap through it then to the other cap.

Again, from the start of this thread I mentioned when you were using your first rotor with three coils that there is power generation potential there. Same thing applies to your motor if it has stator magnets. If it is the type of motor I am thinking of, some of the coils are not connected to the brushes that energized other coils that started the turning. Those unconnected coils are still turning in a magnetic field and hence they can start getting charged up themselves. Now when you disconnect the cap, if a charge coil happens to now be connected by the brushes, the flyback can charge back your cap. Confused???????

Just put a diode on a bled capacitor and connect it to your motor all in series. Turn the shaft manually and take a voltage reading off the cap. That would be the first variable to look at to find your answers.

I suspect that between the applied energy of the cap that moves the motor and the actual motor stopping, there are some degrees of rotation that happened due to the simple momentum of the rotor and that was enough to put back juice inside both caps.

Caps discharge when you connect them. Inductive coils discharge when you disconnect them so if you kept your voltmeter on the first cap, you should see the voltage drop when you connect the cap and go back up when you disconnect the first cap. The inductive flyback from the motor coil will return faster then you can manually disconnect it. lol

As long as the first cap holds the loop closed, the motor coil will also hold its charge.

wattsup

[innovation_station]

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ist!!

[poynt99]

Simply the addition of the inductor in the circuit has caused less energy to flow from C1 to C2.  Less energy flows, less losses occur. No great discovery there, lol.  Just posting my results so I may understand them better.  Thanks for your insights.

On the contrary...

From the "Capacitor Energy Transfer Experiments" document, page 13:

If we increase L1 to a higher value, C1 will actually turn out to be lower in voltage than C2, meaning more of the initial energy is transferred between the two capacitors.

This is of course provided that the resistance of the inductor is not too high, in which case the losses would be high.

.99

[captainpecan]

@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.