Solid State Orbo System

[exnihiloest]

Please answer the questions in Lumen's post: You skirted around these questions in your reply.

1) At what point exactly did you expect to NOT feel the pulse?

Wrong question as I expect to feel the pulse

2) How does your experiment prove leakage of field? 

A simple probe coil connected to an oscilloscope, near a pulsed toroidal coil. See the trace when it is near saturation.

[exnihiloest]

...
The led's lighting and the EMF induced in the pickup coil has nothing to do with a small amount of flux leakage due to saturating the toroid because the pickup coil is open during the magnetization phase in Naudin's experiment.
...

During the magnetization, energy is stored for a part in the magnetic field of the coil and as internal energy for saturation. A voltage is induced in the pickup coil by the leakage flux (see the oscilloscope trace). There is a leakage flux because the permeability is not uniform along the toroid.
During demagnetization, the leakage flux, which is not small, induces voltage in the pickup coil and current in the load.

If the principle was not a leakage flux but permeability modulation, then the toroid plane could be put parallel to the pickup coil plane. But this configuration doesn't work, as well as your thesis.

[gravityblock]

Where are the measurements about a cooling effect in Orbo system?

Why does this cooling effect bother you?

Did I mention anything about a cooling effect in the Orbo system?  I said the toroid in my replication attempt of an Orbo had no noticeable heating.  The other references to the cooling effect was in regards to the Adams's motor.  For those who claimed to have a cooling effect in their experiments, you will need to ask them for the measurements.  I'm sure you would say the measurements were wrong for whatever reason (This is what you do).

Don't you think a cooling effect, especially in a solid state device such as what Bruce is working on, is interesting?  Of course you do and this bothers you. 

GB

[gravityblock]

During the magnetization, energy is stored for a part in the magnetic field of the coil and as internal energy for saturation. A voltage is induced in the pickup coil by the leakage flux (see the oscilloscope trace). There is a leakage flux because the permeability is not uniform along the toroid.
During demagnetization, the leakage flux, which is not small, induces voltage in the pickup coil and current in the load.

If the principle was not a leakage flux but permeability modulation, then the toroid plane could be put parallel to the pickup coil plane. But this configuration doesn't work, as well as your thesis.

The oscilloscope trace is a straight line in Episode 7.  How can we see this leakage flux in Episode 7?  There are two circuits in Episode 7 and only 1 circuit in Episode 6, so you can't use the oscilloscope trace from Episode 6 for your analysis of Episode 7.  This is how you're intentionally trying to twist the facts by mixing two different experiments of your analysis of a single experiment, and you continue to knowingly do this.

You continue to make references about measurements errors even after being warned several times.  I don't think we have to worry about you're distractions here any longer.

GB

[gyulasun]

Hi Gyula,

I have tried this, and the result is an output of 1.7 volts ac, so what does that now tell us?  It should have been about 12 or 13 volts.  I am running now about 25 or 26 volts steady.

I knew if no one else could figure this out, you would...LOL

Thanks!

Bruce

Hi Bruce,

If you measured 1.7V AC output voltage across an OUTSIDE 49 Ohm resistor, and assuming your output frequency is 500Hz, then see the followings.

You have a series L-R closed circuit, fed from (a hypotetical) ideal voltage generator of 25V at 500Hz, see attachment, where L is unknown at the moment, it is the resultant inductivity of the 6 Brooks in parallel, including their mutual inductance due to closeness and some iron cores you use in or near them.
(By the way, you can measure the resultant L to check the calculation that comes below) and R is 49 + 49 Ohm, one is the copper resistance and one is your outside load across which you measured the 1.7V AC.

So the 1.7V across the 49 Ohm means the AC current in the circuit is I=1.7/49= 0.0347A i.e. 34.7mA

The inductive reactance is unknown at the moment, for I do not know the resultant L of your 6 paralleled coils but here is the formula:
XL=2*pi*f*L    where f=500Hz (I think)

The total Z impedance that the L and the R constitutes in the circuit is:

Z=sqrt(R²+X²_L)

Because we know the AC current is .0347A in this series R-L circuit what the 25V generator maintains, we can calculate the value of Z as Z=25V/.0347A=720.46 Ohm.   This equals with the above formula for Z, hence I can calculate L from it:

720.46=sqrt(98²+4*pi²*500²*L²) and from this, L can be calculated.

The result for L is L= .2273Henry  this is what you can check with an inductance meter, by switching off any power in your circuit, no any load across your output connections, only the L meter, ok?
If you find a significant difference with the L meter from this  227mH inductance, then probably the voltage measurements are way off because I used the 1.7V loaded and the 25V unloaded values you measured.

If you do not have an L meter, no problem, we could assume your measurements and my calculations are more or less ok, this would mean you have a "generator" output that has about the following inner impedance value at 500 Hz output frequency:

Z_out=sqrt(49² + (2*pi*500*.2273)²)=sqrt(49²+713.72²)=714.72 Ohm at 500 Hz.

This is a huge inner impedance for a generator. You can reduce this to as low as the 49 Ohm copper resistance by tuning out the 713 Ohm inductive reactance by a capacitor that also has a 713 Ohm reactance at 500 Hz.
X_C=1/(2*pi*500*C)    C=1/(2*pi*500*713)= 446 nF, try to use a 470 nF (or two 220 nF in parallel) (non-electrolytic) capacitor IN SERIES with your output to tune out the inductive inner reactance of your coils. You may find the output voltage now across a 49 Ohm resistor as the load now from outside to be closer to 10-12V, hopefully.  Do not connect the tuning cap in parallel, only in series with the coil output, then comes the 49 Ohm resistor load.

If my calculations turn out to be way off, then sorry, I used what I have had data available. 

rgds,  Gyula

PS: I used this link for some background in AC calculations and also took the series RL picture from there, the 37 degree phase angle is of course not valid here.   http://www.allaboutcircuits.com/vol_2/chpt_3/3.html