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

[Silvije]

It is my understanding that the energy you put into the coil is smaller than the kinetic energy the wheel gains by the attraction of the ferrit while a magnet approaches.
Dutycycle does matter, but you should keep the switch on time short enough to gain energy.

If this what you are saying would be true, than we would have a self running motor..
But as I understand that is not the case..

So what could I possibly do with free energy which is totaly consumed by friction losses and heating? 

[teslaalset]

Ben, OM,
Calculating powers at constant RMP (2500) gives you the power needed to sustain constant speed. In other words: losses by the wheel meet exactly the input power (COP is exactly 1)

Since the input power seems constant even during ramping up, the COP > 1 situation is only occurring during acceleration.
To calculate the highest COP value, we need to know the ramping up time and the weight of the wheel

[teslaalset]

If this what you are saying would be true, than we would have a self running motor..
But as I understand that is not the case..

So what could I possibly do with free energy which is totaly consumed by friction losses and heating? 

The replication setup by Naudin shows that the wheel in not very aerodynamic. If you slow it down by getting electrical energy out of the second wheel (putting receiving coils near the wheel), there will be less air losses, while input power is similar

A lot of optimization has yet to been done. This is only the beginning.

[Omnibus]

Hi Ben,

That's correct. If you slash down the input power, everything else being equal, or increase the rotor weight, everything else being equal, etc. there will be OU. Problem is you touch one variable, for the motor as is, and everything else gets affected.

As for the steady-state, that's exactly where the KE formula applies to. The wheel turns, right? The way the bullet flies, correct? Bullet has a given KE, corresponding to its mass m and velocity v, hasn't it? Same thing with the wheel -- it has mass m and angular velocity. The input energy per second sustains a given angular velocity of that rotor of given mass m. If you din't feed it continuously the rotor will wind down and will lose its rotational KE.

[eatenbyagrue]

The english version is up now :

http://jnaudin.free.fr/steorn/indexen.htm

I tried to calculate input power using the scope shot.
A = 18 Amp (10A/div)
V = 8 V (5V/div)
Duty cycle = 25% (2ms on, 6 ms off)
P = 18*8*0.25 = 36 W
Energy spent in a sec = 36 J

As there is no load, all energy gets stored as kinetic energy of a flywheel.
KE of rotor = 0.5*m*r^2*w^2
= 0.5 *0.25 *0.075*0.075*(2*3.14*2520/60)^2
= 49 J (assuming a 250g rotor)
or 35 J (assuming a 180g rotor)

So the rotor must weigh more than 200g for OU. I did a rough calculation and the weight of magnets alone comes out to be 720g !! (which means an output of 140 J and efficiency of 300%, same as claimed by steorn)
Plz correct, if I made any mistakes.

Hang on fellas, Omnibus and Omega O.  This doesn't quite make sense.

The rotor maintains a constant RPM, so its kinetic energy is not increasing over time.  Yet the motor continues to consume electical energy.

So to me, the kinetic energy in the wheel does not really enter into the analysis of overunity.  We have to measure the load on the wheel, if any, plus friction losses.  That is the work done here, not spinning the wheel.

To give a gross example of what I mean.  Let's say this device was in a vacuum, so no air resistance.  And let's say the bearings involved were truly zero friction.  So it takes zero energy to maintain the spin.  But if the device continued to consume, let's say, 10J per second to keep running, while RPM was not increasing, the device would be clearly under unity, no matter how fast or how heavy the flywheel was.

I am not trying to clutter the thread, but if someone is trying to measure overunity this way, maybe there is a better way.

Please tell me where my logic is wrong, I am trying to understand.