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Old 02-25-2012, 05:31 AM
mbrownn mbrownn is offline
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Join Date: Jan 2011
Posts: 1,637
In this setup, I can see how it will desulphate in the period that the motor wont run by the fact it is working as a tank circuit. This makes sense.

During the running period it makes sense that the current is flowing through the motor causing the motor to have torque with the only loss being the ohmic loss in the motor as the current that has passed through is collected in the second battery. This is in its self proof that running a motor does not consume power other than that consumed in ohmic resistance. I have said this before on many threads and as yet no one has challenged it.

Adding loads to the third battery just reduces the impedance seen by the primaries and motor allowing the motor to run faster. These loads will consume power based on their ohmic resistance. All this makes sense and it is an efficient way to run a motor. The spikes generated by the running motor will also desulphate the third battery but as I see it we are still consuming the same power out of the two supply batteries and they should be getting depleted.

Yes, lead acid batteries have the ability to recover somewhat from short heavy load drains but only as much power can be drawn as is seen when the battery is drained at C20. While they might appear fully charged they should in fact not be. If the batteries are in fact keeping their charge, I am baffled

Does anyone agree, disagree or have another explanation?

I like what you are doing as it proves that a motor does not consume power to make it run. If there is no loss or even a reduced loss from the primaries this needs to be documented as something else must be going on.

Of course the power produced by the motor can be used to power a generator that can feed power back into the circuit and when you do this you will most probably find that there is more energy in the combined output of the generator and that going into the charged battery than what is being drained from the source. It should be possible to charge three batteries in parallel this way while depleting two.

If the motor used was of my design it may be possible to charge many more as we will be having an additional pulse effect charging the batteries plus overall higher efficiency.

I fully believe you have an overunity circuit when a generator is fitted but the source batteries must go down.

To see the overunity you need a high efficiency motor and a high efficiency generator, that is both need to be in excess of 70% efficiency. As the efficiency of the motor and generators go up the overunity will go up.

To make it self running, I would feed the output of the generator to the output battery and use an inverter to feed this back to the source. Once the system is running all the batteries could be replaced with capacitors.

Note the similarity with one of my Lockridge circuits but this would require a special motor. Circuit Simulator Applet
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