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

[PaulLowrance]

Am I nuts or is J-L Naudin now showing OU on youtube? Any comments on the lack of apparent current draw while lighting LED's?

No. Unfortunately Naudin's shown everything except input : output measurements.

[happyfunball]

No. Unfortunately Naudin's shown everything except input : output measurements.

Current draw is irrelevant? Isn't that what the meter is showing?

[haithar]

No. Unfortunately Naudin's shown everything except input : output measurements.

what's there?
oh yeah, "Stay tuned, more to come soon..."

we'll hopefully see in the next days

[teslaalset]

Some thoughts on the reason why Orbo is demonstrated by a 10 Ah D cell.
At first I thought they need that kind of cell to provide enough current to power the coils.

Now, on second thought I think that is not the reason.
They use soft metal cores. If one look at e.g. metglas, these have an H_c of a few A/m, let's say 2 A/m.
With a number of windings of 60 per coil you need around 12 - 15 mA to saturate those cores. More current is just a waste of energy.

So, why the 10 Ah?
I think they need such a cell because those can handle a peak charge current of several 10th of Amps.
Such value is being returned if the drive power of the coils is stopped in an abrupt way. The peak current that is occuring at that moment is hugh.
A normal D cell would not be able to absorb such value.

Of course a (super)capacitor would also fit in here.
It's unclear why the don't use those.
Replicators should keep this mind. If you don't use a power cell that is capable of charging with high current values, you could also use a capacitor in parallel with e.g. a more common rechargable cell.

[edit]
example specification of such D cell can be found at:
http://www.batteryspace.com/nimhrechargeablecelld-size12v10000mahflattop50arate1pc.aspx

@ Paul,
I think you are quite right, although a bit on the optimistic side. The coil energy can be re-gained for a great deal. If you look at the integrated power coil on the input power (below). The power level between the start and the end of the curve shows a very tiny difference. This indicates the net energy loss. I estimate 10%.

To refresh memories, the bottom graph shows the energy graph as a function of time, derived from the differential of the integrated power graph.
A large part of the re-gained power is in the power peak, which is very large because at that moment the coil value has changed from a few mH to almost 1 Henry. Energy of a coil that hold current is 0.5*L*I*I.
So, energy jumps linear with the inductance jump and in a very short period of time at the end of the power period of the coil. If inductance jumps from a few mH to almost 1 H, energy jumps a factor of a few hundred times!

The bottom graph is a bit incorrect w.r.t. to the 0 power value just before the peak occurs. There should be some current left in the coil when the power is switched off. So timing of the switch off moment is also very critical.

[edit2]
It also means that the nr. windings on the coil are critical and not just any number can be used. The coil value is critical because the nr. of windings determine the saturation current and the switch off moment.

[happyfunball]

what's there?
oh yeah, "Stay tuned, more to come soon..."

we'll hopefully see in the next days

Was referring to this one  He doesn't seem to be claiming OU but..?

http://www.youtube.com/watch?v=SjDbIrKIVXs