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Overunity Machines Forum



E-Orbo replications

Started by Omega_0, May 09, 2010, 09:30:12 AM

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Omega_0

Added second coil.
Rotor is going at 330 RPM on roughly the same peak current. Setup was not optimized for duty cycle or reed position too accurately, as this is just a quick measurement to see if everything is going ok.
Note that the net power is showing bigger hills, which means cemf is kicking in.
(First plot is voltage across two coils, in red, and current, in blue)
I have more respect for the fellow with a single idea who gets there than for the fellow with a thousand ideas who does nothing - Thomas Alva Edison

Omega_0

Effect of adding a diode to short the path of self induction:
The net energy input lowers.

I'm facing problems in measuring the actual current in the diode path. No negative current is flowing in that path after the pulse ends, but the negative voltage is snipped off by the diode, this is strange........Hopefully the circuit is correct.
I have more respect for the fellow with a single idea who gets there than for the fellow with a thousand ideas who does nothing - Thomas Alva Edison

Omega_0

Quote from: Omega_0 on June 19, 2010, 05:04:30 PM
More about material T38.
I'm attaching here the specs of toroids that I'm using in current version and also of the alloy T38 (which I guess is MnZn Ferrite). Its EPCOS part no. B64290L618X38, dimensions : 25.3X14.8X10 coated with blue epoxy.

Its Mu_i is 10000 and H(sat) is around 100 A/m from the datasheet. So from
H = N*I/lc
where N is number of turns, I is current and lc is flux path length, we can get the saturation current for say, 100 turns of coil:

100 = 100*Isat/lc
lc = pi*( 25.3+14.8 )/2 = 63 mm = 0.063 m
Isat = 100*0.063/100 = 63 mA

Or say you want Isat = 1 Amp, which gives N = 6.3 or 7 turns. So you would expect that just 7 turns at 1 Amp or 100 turns at 63 mA shall saturate the core completely and the rotor will not be attracted to the core at all. Sounds like free lunch.....

But, don't be surprised if the core refuses to follow the text book in presence of a strong neo. In my case the coil is taking max 2.8 Amps@100 turns and the core still shows strong attraction for the magnets ! I have no idea why this is.......... anyone knows ???? Isn't the core supposed to "disappear" for magnets when saturated ?

Either the formula I'm using is wrong or something else is happening here. This is an unknown land for me.

Did some more research on this. It turns out that the above calculation will not apply in presence of an external field. The correct equation is:

B = mu(H+M)
or H = B/mu - M
where M is the Magnetization of the core (which behaves as a magnet in presence of a strong neo magnet)

The Hsat taken from BH-curve is meaningless here as it is measured at M=0 (i.e. for non-magnetized core). So my updated view is that the current required to make the core invisible would be sum of currents required to demagnetize the core and re-magnetize it in the direction of current. I'm guessing that this can be huge.
I have more respect for the fellow with a single idea who gets there than for the fellow with a thousand ideas who does nothing - Thomas Alva Edison

Omega_0

Added a pickup coil. It was shorted through a 1 ohm sense resistor. This slows down the rotor. The result is not OU obviously. The setup is not perfect either. But getting close. At this stage a measurement error of 0.004 ohm in input resistance make it OU from an UU. So the next step is to measure Rin reliably.

The efficiency varies but tends to settle down with time. I don't know how to interpret this result. It seems the system is OU for some times.
I have more respect for the fellow with a single idea who gets there than for the fellow with a thousand ideas who does nothing - Thomas Alva Edison

Omega_0

Rather interesting results here.
These readings were taken at low RPMs (around 75 to 150 or so). And by placing a flat pickup coil of ~500 turns very near to the magnets (which lowers the RPM even more).

One would expect the net efficiency to decrease at low RPM, because the power induced in pickup coil decreases. But the interesting thing is, the power wasted in inductive rise and fall in input side decreases and the cemf is nearly 0. So after deducting the heat losses from the Ein, the net energy transferred is very very tiny.

I guess I have totally replicated the Steorn's Eorbo demo now. Learned many things during this built and my sincere thanks goes to steorn team.

There are still many things to resolve, and as I'm using a crude setup and not so accurate instruments, I can't place my 100% confidence on these results at this time.
.
I have more respect for the fellow with a single idea who gets there than for the fellow with a thousand ideas who does nothing - Thomas Alva Edison