STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[Pirate88179]

@CP:

Wow! Paul thinks that you and I are the same person.  His delusions are getting worse.

He even posted on his "website" that we are one and the same.  Of course, if someone calls him on that, he will change it like he always does.  I have the screen shots to prove it.  He wants to sell OU devices.  He is here to find some that he can sell.  Watch out for this guy.

Bill

[teslaalset]

Or it could be due to switching delay in some electronics or relays. Note that the V drops slightly as soon as the current starts. Depends on where the probe is connected.

Anyway, looking at the falling portion (sheet2), L behaves strangely. Perhaps a calculation error somewhere, or I'm doing something wrong and there is another way to get the L from impulse response. Opinions plz ?

@ Omega_O

My opinion is that the delayed current is caused by a very high value of the coil
(I mentioned earlier that a value of about 100 Henry's have to be overcome due to a permeability rise that is very steep).

The voltage drop after the current rises is most likely caused by the internal resistance of the power source (e.g. 0.5V/2A=0.25Ohm)

[Omega_0]

Oh, I see what you mean -- measuring voltage starts slightly before it is actually applied to the coil. Thus, the voltage seen prior to current rising from zero is just measurement of the open-circuit voltage. How will that happen, though?

No idea, because the position of the probe is unknown, may be its an artifact of the probe. Using complicated instruments complicates the things. It can be due to high inductance as teslaalset said above, but 100 H is very high, and we should see some values of 50, 40 30 or 10 H in the curve as it falls, but you can see that it falls from 0.5H or 1 H max.

@teslaalset
What do you think about the calculation of L? Is it correct?
Can we use the equation for static L to compute here ? Because the i=Im(1-e^(-t/tau)) assumes a constant L, IMO.
I have plugged a constant L into the equation with real Im and t values and the curve is not the same. Which means L is definitely varying in the real setup, but its value that I computed may be wrong.

At rising edge we see the L (or mu) drop as there is a magnet nearby and I is rising fast. So as expected it drops to almost 0. On the falling edge, the I is decreasing and there is no magnet near the core, so it must either increase or remain constant. Problem is, with these numbers (in sheet2) I don't see that. Which means either the equations are wrong or not applicable or something else is going on.

Btw, there was a mistake in the calculations, time should be (t-t0) instead of t, which gives L=0.25 to 0.001, but the behavior is same.

[PaulLowrance]

@airstriker
I would disagree, I have built both the steorn and callanan motors and they both  have benefits and issues that need to sorted out. As well both have very little torque and were never designed to carry a load persay, they were designed to recharge the source thus the flywheel effect as you call it is an integral and equal part of the operation of both devices.

The pulse duration can make a huge difference in the orbo torque. Increasing pulse duration might not make much difference, but then suddenly it will come alive. During the last stage of my "tiny orbo replication 1" it accelerated and sounded like a dremel drill. From nearly dead stop to over 7000 rpm's in a few seconds. Also I could place a large thick solid piece of Aluminum ~ a centimeter away from the spinning magnets, and it still ran. I have not done the calculations, but IMO that's a lot of power due to eddy currents relative to the input.

[Omnibus]

@teslaalset.
I think @Omega_0's explanation of the seeming phase shift is the correct one, that it, there is no phase shift between I and V.

@Omega_0,
L undoubtedly isn't constant. However, why should one be concerned as to what L is for the present analysis of the energy balance? That's a detail which doesn't pertain to the analysis at hand.