Muller Dynamo for experimentalists

[plengo]

Fausto,

Power wise, you are showing less than 40 mW out and 6W in for one set of generator coils, which is less than 5% of the input power consumption! Multiplied by 7 for a full set of generator coils, this would give 280mW / 0.28W. I would not expect your input ammeter to deflect noticeably with lamp load or short circuit for one or two coil sets, as the supply is being very lightly loaded. Try adding the remaining coils and repeat the same test. The load will still only be consuming very low power but you may detect a deflection to the RHS on the meter as more coil sets are added. Its important to think in terms of input and output power / energy consumption, not just amp draw.

Hoppy

Hoppy, correct BUT did you notice that the RPM did not change. Any amount of load will change the speed very easily. I have done the experiment. It is very difficult to have the combination of input power, RPM and having a load and not change any.

I think even though the amount of power is small it is still of great value that indeed the phase "trick" works.

Fausto.

[lumen]

Hoppy, correct BUT did you notice that the RPM did not change. Any amount of load will change the speed very easily. I have done the experiment. It is very difficult to have the combination of input power, RPM and having a load and not change any.

I think even though the amount of power is small it is still of great value that indeed the phase "trick" works.

Fausto.

The problem is shifting the phase reduces the power factor. I think the trick is to phase shift only enough to get the magnet under the core. At this point more current could be produced along with Lenz force. However, if enough shift/delay can be achieved, the Lenz force will have no effect pushing directly on the face of the rotor magnet. Additional shift could be achieved by raising the rail voltage where the dump capacitor engages without additional losses in the PF.

There is likely to be many other parameters involved in this feat since even the ratio of core size to rotor magnet size and rotor magnet spacing would be important and even the spacing of the coils to the rotor could all play into it.

Using this principal you might think of the rotor as having only two long magnets with each magnet covering 1/4 of the rotor and the spaces also 1/4 rotor.

Now as the magnet approaches the coil, it remains invisible due to the stator magnets on the coils. Then very rapidly, it switches polarity. With a resistive load, Lenz forces would instantly repel the incoming rotor magnet and slow the rotor. But due to increased rail voltage and some phase shift, the rotor magnet would be under the coil BEFORE the Lenz forces start repelling.

At this point, any current draw would have no rotor drag as increased Lenz forces could only push on the magnet face.

On exit the rapid polarity reversal would again produce voltage, but again delayed so as to not pull back the exiting magnet, and once out of range, the stationary stator magnet would be left to deal with the Lenz forces and still no rotor loading.

Just a theory!

[nueview]

bolt
i would have thought that you would have expanded upon this to explain other phase chenges and corrections and multi signal coupling but instead you reduced what i was saying to watts are watts sorry i even bothered.
Martin

[Hoppy]

Hoppy, correct BUT did you notice that the RPM did not change. Any amount of load will change the speed very easily. I have done the experiment. It is very difficult to have the combination of input power, RPM and having a load and not change any.

I think even though the amount of power is small it is still of great value that indeed the phase "trick" works.

Fausto.

Fausto, like the input current, the RPM will not change noticeably with the tiny load you had connected. Also, there is a lot of kinetic energy stored in the rotor. As Bolt has suggested get all coil sets connected and tuned and make sure you work your COP out using Power & time (in secs) to calculate Joules. The input to output ratio of energy in Joules will give the COP, not amps or power.

Hoppy

[plengo]

Fausto, like the input current, the RPM will not change noticeably with the tiny load you had connected. Also, there is a lot of kinetic energy stored in the rotor. As Bolt has suggested get all coil sets connected and tuned and make sure you work your COP out using Power & time (in secs) to calculate Joules. The input to output ratio of energy in Joules will give the COP, not amps or power.

Hoppy

It may be true for you not in my setup. I have tested ANY load without those capacitors and the change in RPM is instant and very, very visible. This is the difference in theory and practice. I can make a video showing that to prove what I am saying. Are you only repeating the theory you know or experiments you have done?

Fausto.