This is not a thread to be confrontational.  The purpose of this thread is to pick it up where it left off the other day without censorship when discussing technical matters.  Anybody is free to contribute but all of you should refrain from personal attacks and name calling.

Here is Luc's last posting:

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Hi everyone,

as promised here is a video demo with accurate input power measurements done by using a DC input to the Dremel (universal motor) instead of AC which is difficult to measure.
The power is supplied by a variac connected to a full wave bridge rectifier and a 1000uf DC capacitor to smooth the DC.

Link to video demo: https://www.youtube.com/watch?v=5TKByKqLV0M

TEST RESULTS:

Input to Dremel on load is 60vdc @ 0.500ma = 30W
Input to Dremel off load is 60.3vdc @ 0.475ma = 28.64W

Power difference is 1.36W of extra power consumption by prime mover when coil is on load

Output is a 25 Ohm load @ 7.84vrms = 2.46W
then we subtract - 1.36W =  1.1W of gained power output which is not supplied by prime mover.


You may of noticed this is not the same coil as the first test since that one was not performing very well (too small), so I made a new one since the first one only had a half a watt gain and I was sure it would of been argued that it's just measurement error.
So I went trough the time, trouble and expense to make this new one so there's no second guessing.
Like I have said, a coil can be made to give more of this effect but more testing needs to be done to understand what are the ideal perimeters as I've seen coils with more gain then this new one but I think it's a good starting point.
A stronger magnet will also give more output. If someone is ready to pay for a larger 1" Diametrically Magnetized magnet, I'm ready to do the test: http://www.kjmagnetics.com/proddetail.asp?prod=RX04X0DIA  please pm me and I'll send you a US address.

COIL INFORMATION:
Coil DC resistance is 2 Ohms.
Coil has 41.56mH with magnet pole in attraction to core and 49.42mH with magnet positioned between poles.
Coil is wound Bifilar but connected in parallel.
I tested the Capacitance between open wires strands and it is 38.46pf with magnet in attraction to core and 38.51pf with magnet between poles.
Coil wire (with enamel) measures 0.8mm and the core is a square Ferrite rod measuring 12mm x 12mm x 90mm long.
I can wind more turns on the coil at a later date to confirm if it give a boost in gain.

This was the first load test and more will be done to find the coils most efficient output (most watts out with minimum effect to prime mover).

Please share your thoughts as to what is contributing to this effect.

Luc

Here are my comments:

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Luc:

You saw an increased power draw when the generator coil is loaded.  Before you were insisting that there was no increased power draw.  You should try to account for your past errors and try to square them away with your readers.

QuoteInput to Dremel on load is 60vdc @ 0.500ma = 30W
Input to Dremel off load is 60.3vdc @ 0.475ma = 28.64W

Power difference is 1.36W of extra power consumption by prime mover when coil is on load

Output is a 25 Ohm load @ 7.84vrms = 2.46W
then we subtract - 1.36W =  1.1W of gained power output without affecting prime mover.

I hate to say it again but you are leading yourself down a garden path.  Also, you are making real measurements here, you should not be ignoring the resistance of the coil.  The coil capacitance measurement is also meaningless and has no affect and should not be considered.

Here are some issues that have to be factored in:  1) You have no idea what the efficiency of the Dremel is.  2) You have no idea if the efficiency of the Dremel will change under different supply voltages, loads and RPM.  3)  You are drawing a conclusion without having enough data to support the conclusion.  4)  You are not correctly relating the waste heat with the "payload" power that goes into the generator.

Here is your real data:

UNDER LOAD:

Input:  30 watts electrical

Generator output:  2.46 watts
Motor output:  27.54 watts heat
Total output:  30 watts

Note:  The motor heat output power is derived by subtracting the generator output power from the input electrical power.  That is how the power is split in your setup.

NO LOAD:

Input:  28.64 watts electrical

Motor output:  28.64 watts heat

The above is the real analysis of your data.  That's all that you can conclude from your experiment with the caveat that the heat dissipation in the coil windings was not accounted for.

QuoteOutput is a 25 Ohm load @ 7.84vrms = 2.46W
then we subtract - 1.36W =  1.1W of gained power output without affecting prime mover.

The above calculation is an invalid calculation.  If you review what I wrote above this should become clear to you.

MileHigh

I will just repeat:  The purpose of this thread is to have a technical discussion without any technical censorship at all.   There is no intention to be confrontational at all with this thread.   Personal attacks and abuse and demeaning comments and name calling are to be avoided.

As long as that is understood, people are free to say anything they want about the alleged "delayed Lenz effect."

MileHigh

Here is a decent posting I made that I think is worth repeating:

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Tinman:

In the first half of my posting I stated that my comments were geneic and not directed at anyone in particular.

Mark often makes great succinct technical arguments and makes graphics and I thank him for that.  Here is my version of a short summary of the issues:

1.  With a resistive load on the generator coil, there is no Lenz delay, it's simply impossible.

2.  With a capacitive or inductive reactive load then there is a Lenz drag during the charging phase and then a push when the reactive load discharges its stored energy for a net gain of zero (ignoring losses.)

3.  What has only been mentioned a few times is the "fake out" Lenz delay.  This is when the generator coil and load together dissipate less power when you are making the Lenz delay test as compared to the original configuration.  Less power dissipated by the generator coil + load equals a higher RPM for the rotor.  This is a no-brainer and many beginning experimenters simply failed to make these measurements.

4.  Changes to the mechanical and electrical configuration of the setup when doing a Lenz delay test can inadvertently change the overall average electrical impedance of the pulse motor and where the power flows in the system.  If the impedance goes down the current draw from the power supply will increase and most likely the rotor will speed up.  If the impedance goes up the current draw from the power supply will decrease and most likely the rotor will slow down.  That's what's taking place in the JLN clip that was linked to.  That is another "fake out" and you have to be on your toes to not hoodwink yourself.

When it comes to #4, you might make a change and the rotor speed will in increase by just a few percent.  Something like simply lowering the stresses on the main bearing while rotating at high RPM could cause an effect like this.

What's been happening over the past two years is that many amateur experimenters when working with their pulse motors will say "delayed Lenz effect" when they observe a speed up in the RPM for whatever reason.  They don't actually investigate the real reasons, they just use the blanket term "delayed Lenz effect" when they see a speed up and they are convinced that they have "replicated the effect."

As a generic shout out, guys and gals, you have to do better than this.  Working an investigation together might help where people encourage each other to get the right answers.  You have to think "outside of the box" and in this context thinking "outside of the box" actually means that you do a proper investigation using standard electronics principles and measurement techniques.  You all can do better if you work together and encourage each other to improve your craft.

MileHigh

Here is a great technical posting bt MarkE that is also worth repeating:

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Anyone who believes in the idea of free energy from a "Lenz delay" should study the graphic below.

Lenz' Law sets the orientation of induced voltage resulting from Faraday induction.
If a load is resistive, induced current is in phase with induced voltage and the resulting magnetic field at all times acts directly against the inducing current.  This is a unity power factor.
If a load is purely reactive, then energy is stored in the load and later returned to the source, and no net energy conveys to the load.  This is a zero power factor.
If a load is resonant, then the inductive reactance and capacitive reactance magnitudes are equal.  The load appears resistive.  In the case of a series L-C, the resistance appears low.  In the case of a parallel L-C, the resistance appears very large.
If a load is partially resistive and partially reactive, then more energy transfers to the load each cycle than is returned.  This is a power factor greater than zero but less than one.

Can making a load reactive unload a driver?  Of course it can:  Less work is done on the load.  In the extreme case the load effectively disappears, along with any useful work that could have been done by transferring net energy to the load. 

Can making a load reactive actually drive the source?  Not net across one or more complete cycles.  A reactive load can only return less energy in any given cycle than supplied by the source.

Can resonance help?  No, at resonance, the load appears resistive as either an effective short circuit across the source, or an open circuit.