Re-Inventing The Wheel-Part1-Clemente_Figuera-THE INFINITE ENERGY MACHINE

[NRamaswami]

@Cadmon:

I will check your suggestion for reversing the middle coil. Earlier Dough indicated that the coil should be NS-SN-NS but I found that it is against nature as a magnet must have a north pole and south pole and it did not work.

Now changing the pole faces in the middle induced circuit is one that I have not thought about. Let me do it tomorrow. However will this not result in the coil getting mageic flux. If you supply DC the primary should keep changing in magnetic field strength to create time varying magnetic field. Without that there is no induction. If you supply DC again you need a bank of 8 batteries to reach 100 volts. Buforn input was 100 volts and 1 amp as Hanon said. If it is DC 100 volts = 1 Amp x 100 ohms. The primary should have had 100 ohms resistance. A 1 sq mm wire has 18.1 ohm DC and that means he used 550 metres of such wires for primary alone.

Again his reported output is 20000 watts. Assuming he used large wires capable of handling 30 Amps it shoudl have required about 650 volts, if it was 40 amps it would have required 500 volts and if it was 50 amps it would have required about 400 volts.

Therefore do you mean to say that the primary was thin wire and the secondary was thick wire. if the secondary was thick wire, then the picture presents a wrong image of the secondary being smaller than the primary. If you have a smaller dia secondary core, at these amps the secondary would sound enormously and the core sizes would have been massive. Primary would have been about 9 inch dia and the secondary 6 inch dia for all these things to work.

the second question that comes to my mind is even if the Primary input is DC, the secondary output is AC..How does that work in combination.

Could you please share your thoughts on this..I'm obliged.

[dieter]

Gyulasun,


Thanks for the hint, I knew about the +- -+ caps, but not the diodes. Actually, you can use an electrolytic cap in proper AC, it does damage and selfheal in every AC cycle, but it needs to have a big enough capacitance. But interesting concept with the diodes, I will study this closer.


After my flop with the "gain trough sparks" mistake, I decided to make some serious measurement, as far as my equipment allows. Even if the spark thing finally was not yet the key, there is no reason to be discouraged. Here's my latest data:


The setup (as seen in the video) consists of:
- a Wall supply, labeled 220V 50Hz , 10.5vdc, 2.9W, max, 7.5VA.


-A Resistor 27 ohm, max. 23 Watt.


-The Generator Coil group (2 prim., 1 sec., on same double-E ferrite core)


- a 4 diodes bridge rectifier and a 1uF 400V unipilar capacitor at the output of the secondary coil, aka Y coil.


When in operation, the ac voltage of the whole circuit is 13 V. The Resistor and the generator are connected in series. The resistor alone has 12 Vac at his pins. The Generator alone has 1 Vac.


The Wattmeter says total dissipation is 6.84 Watt. Surprisingly, the Wattmeter says the same when I connect the Resistor only, without the Generator. I thougt maybe the wattmeter cannot see the generator because coils bring the current out of phase. But when I connect the generator only, the Wattmeter reads 27 Watt (which is why I use a Resistor). Clearly, the coils are not made for such high currents. Two calculations can be made: 27 Watt/13, that would mean 2.1 Watt dissipation when run with 1 Vac, Or: 6.84W/ 13= 0.526 Watt.


Since the supply gets hot and is really not made for 29 Watt, I think the first calculation is wrong . Also, the Generator primaries don't have enough turns to handle such "high power", lots of losses everywhere.


So let's say the Generator dissipates a 1/13th of 6.84Watt, 0.526 Watt.


On the output I have rectified DC, smoothed with the 1 uF Cap. The analog Voltmeter says 25vdc, and a current of clearly more than the max DCmA meter range of 250. The digital meter with the defective AC parts, says 21.3 VDC, 65 mA.  Let us assume the worse, that's 21.3* 0.065= 1.3845 Watt.


So that may be:
In: 0.526 Watt
Out: 1.3845 Watt


Efficiency : 263.21%


To be fair, it is hard do exact measurements in the 1 Vac range with this analogue device. But let's say it was not 1 Vac, but 1.5 Vac, then it would be: 0.789 Watt, that's still 175.4%.


Shortening the rectified output resulted in a voltage drop in the primaries of about 0.5 V, where the voltage of the resistor rised the same amount. The total Voltage remained the same, just like the Watts calculated by the wattmeter, regardless of short circuit of the secondary or no load on secondary.


If I'd add 12 more elements instead of the wasteful resistor, on optimistic calculation would be:
in: 6.84 Watt
out: 17.99 Watt.


As I am an amateur, I am not sure if I measure the Amperes right. Of course the voltage drops almost to zero when the Amps are measured, because this will practicly short cirquit them. Am I doing this wrong?


Regards

[gyulasun]

....
I am not sure if I measure the Amperes right. Of course the voltage drops almost to zero when the Amps are measured, because this will practicly short cirquit them. Am I doing this wrong?
....

Hi Dieter,

Yes I think you are doing the output current measurement wrong. See this link where I show input and output current measurements correctly in the edited Alvaro's schematic http://www.overunity.com/12794/re-inventing-the-wheel-part1-clemente_figuera-the-infinite-energy-machine/msg388286/#msg388286 and this involves using a real load which actually consumes the output current, ok? 
For a real load you may wish to use a 10 or 22 or 47 Ohm etc resistor. If you still have the 1uF capacitor across the DC output, you may wish to increase it to 47uF or 100uF or higher if the setup lets it increase, that is, it would be better to have there a higher value than 1uF. To receive the maximum DC power output (max power match) you may wish to choose a load resistor value which actually reduces the unloaded 25V or whatever DC to about its half value (between 12-13V in the 25V case).

Because you have only the analog multimeter for AC and DC, I suggest using the following measuring procedure:
1) In your switched-on and working setup with a resistor load already chosen for power match and connected, first measure the AC voltage input with your analog AC voltmeter and log it as V_AC.
2) Now insert your analog meter set to AC Amp range in series at one of the AC inputs as I drew in the above link and log the AC current as I_AC (the same resistor load across the DC output is still connected of course).
3) Now put the analog meter to the DC output, namely in series with the load resistance as I drew in the schematic and measure the DC current the load consumes. (Of course when you remove the meter from the AC input side you reconnect the wires there to have a working setup again). Log the DC current value as I_DC.
4) Now remove the meter, reconnect the load to the DC output again and measure the DC voltage across the load with the meter, then log the DC voltage value as V_DC.

From your measured data, the approximate input power P_AC would be V_AC*I_AC, this is a very rough method because we do not know any phase shift between the current and voltage but maybe better than nothing but you do not have a scope to watch and consider the phase shift.
The output DC power, P_DC would be V_DC*I_DC, this would be a more dependable value, close to the truth. So the efficiency would be P_DC/P_AC.

EDIT:   I understand that the series 27 Ohm consumes much more power than your 'transformer' input but in case you wish to deduce its consumption from the total AC power input, you could measure the AC voltage drop, V_R across the 27 Ohm with the analog AC voltmeter when you measure the AC voltage input in step 1 above. Then the power loss in the resistor would be P_R=(V_R*V_R)/27 and please substract this P_R from the P_AC power. Now the efficiency would be P_DC/(P_AC-P_R).  Check the 27 Ohm resistor with your digital Ohm meter too and use that value for the calculation.

Gyula

[Cadman]

@ a.king21
You wrote: "I remember reading your post last summer re your testing of the Figuera/Buforn device.
I wonder, what prevented you from pursuing your research further?
Did you find some obstacle that prevented self running?"

– Life interfered. Major medical issues and then winter hit hard and put a complete stop to everything. It had nothing to do with the device itself. Only recently have I started to get going again. Coincidentally, my 16 pole flat face commutator and brushes arrived today. When weather gets warmer I will be back at it and attempting a more serious sized build of this.

@ NRamaswami
If I remember correctly, I did swap the middle coil direction, and I think it did not matter in my build of this, but I may be mistaken. My build is very different than yours. Please do not put too much faith in my coils or change what you are doing based on my build. When I built it I threw it together with what I had on hand in the garage. Actually I was amazed it worked as well as it did.

You wrote: "However will this not result in the coil getting mageic flux. If you supply DC the primary should keep changing in magnetic field strength to create time varying magnetic field. Without that there is no induction."

– My conclusion (for my build) is that the center coil core nearly always has 100% flux strength. What varies with time is the relative intensity of the north and south poles of the one center core. Each primary coil does keep changing in field strength but as one is reduced the other is increased a like amount.
Faraday's Law: Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be induced in the coil.

You wrote: "Therefore do you mean to say that the primary was thin wire and the secondary was thick wire."
– No, my coils were all the same, identical. Again, please do not infer anything from this. It may or may not be relevant.

You wrote: "the second question that comes to my mind is even if the Primary input is DC, the secondary output is AC..How does that work in combination."
– I am not sure what you are asking here. If you are asking why I have AC output for DC input, I am not sure. In the patents it is said, as I recollect, that for every half revolution there will be a change in sign, which there certainly was, but I can only surmise that it is due to the influence of the predominant primary coil, the one that has the full voltage, and the two primaries are opposite in sign at their end of the center core.

Since you reported such great success using 220 AC I hope you continue to reproduce that particular device. I would hate for my input to derail your efforts.

[dieter]

Gyulasun


Thanks very much for your help! Unfort. my analogue meter can measure only DCmA, limited to 250 mA, so I was stuck at point 2.


But it was a good idea to use a resistor as the load. I measured the voltage across the resistor and used the formula (V*V)/R=W. And that was really disappointing. I tried several resistors, 5,10,15,27 and 1800 Ohm, the results were under 100mW, although inconsistent, so this formula doesn't seem to be precise or reliable. Eighter the formula is crap, or my device is. Yeah, the ups and downs in life... actually downs could be used with a dynamo attached 


I have to test this under better conditions. But  thanks a lot.