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Author Topic: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM  (Read 1197921 times)

powercat

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #600 on: December 27, 2009, 04:15:45 PM »
Did anyone see this one.
dmmpower.wmv
http://www.youtube.com/watch?v=JAoTrqnZpfg
cat

Silvije

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #601 on: December 27, 2009, 04:56:09 PM »
hi guys,

I have found this:
"The current required to get the temporary depolarization of the magnetic domains of the ferrite is fully independent of the mechanical torque produced on the motor shaft."

as JLN says: "Braking the rotor rotation has no influence on the amplitude of the pulse current measured"

but I ask: what happens with duty cycle of this pulses...

so you see, everything adds... no free energy ;)

k4zep

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #602 on: December 27, 2009, 05:01:52 PM »
Hi OB,

Don't be sorry. Thanks for the effort. Much more things to explore. Steorn is supposedly going to report on heat in the motor in January so we shall see just how much heat versus mechanical energy equals input for them. For my motor, I don't know. But it will be a factor. Anyway, I did a video of the running motor. Here it is.

http://www.youtube.com/watch?v=G-gXQagKSNc

Regards,

Ossie

Never kill the messenger, use the message.  What you are saying is "if we can reduce the resistive heating, we will go OU."

Ben


teslaalset

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #603 on: December 27, 2009, 05:04:00 PM »
hi guys,

I have found this:
"The current required to get the temporary depolarization of the magnetic domains of the ferrite is fully independent of the mechanical torque produced on the motor shaft."

as JLN says: "Braking the rotor rotation has no influence on the amplitude of the pulse current measured"

but I ask: what happens with duty cycle of this pulses...

so you see, everything adds... no free energy ;)

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.

k4zep

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #604 on: December 27, 2009, 05:24:03 PM »
Sorry Ossie, but yours appears to be undeunity (if the unaccounted for heat losses don't turna that around). So, let's review what you have:

Input energy every second:
E = 2.2V
I = 0.3A
duty cycle = 0.48
Therefore, energy spent every secon is:
W = 0.48 x 2.2V x 0.3A = 0.32W or 0.32J every second

Output energy (only rotational kinetic energy):
Mass of rotor = 0.35kg
Radius of rotor = 0.05m
Rotations per second = 180/60 = 3rps
KE = 0.5 x 0.35kg x 0.05m x 0.05m x (2 x 3.14 x 3)^2 = 0.16J

Therefore, efficiency = 0.16/0.32 = 0.5 (heat losses unaccounted for) which is far from OU.

Hi OM,

Something not quite right here.  It shows if we got our input down to .16 J cop would be 1 all other things being the same. Less than .16 J all things still equal, OU.  Steady state also shows IF you made the rotor heavier at same speed, more weight would help the numbers, less friction, etc. would also help the numbers.  The numbers say that all losses in the rotor as stated, .16 J is required to overcome all losses and then the extra energy is resistive losses, heat. 
,
It seems that we are calculating KE at steady state only but not the KE when you turn it on and ramp up and turn it off, ramp down in speed and NO energy is used as it slowly stops, much longer period than the run up and is VERY high OU...Total over time.....Something is not right...........I"ll be the first to say MATH is not my strong subject, I do not think in equations but in pictures of dynamic systems...That and a quarter will get you a cup of coffee......Again this just for a more thorough discussion of these numbers.

To generalize, it would seem to say that we should ramp up to equilibrium, turn it off, use the energy over time when rotor is OU as heck with  ZERO J input,, turn it on, ramp up till nominal RPM, turn it off, generate, etc.  Only use the rotor energy to generate energy in the coast down time eliminating the resistive heating effect loss during that period.

Respectfully
Ben

Silvije

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #605 on: December 27, 2009, 05:30:44 PM »
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?  ;D

teslaalset

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #606 on: December 27, 2009, 05:33:17 PM »
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

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #607 on: December 27, 2009, 05:48:53 PM »
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?  ;D

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

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #608 on: December 27, 2009, 06:04:45 PM »
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

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #609 on: December 27, 2009, 06:19:10 PM »
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.

lumen

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #610 on: December 27, 2009, 06:33:16 PM »
The whole system may need to vary transaction time. On the approach, the rotor needs to store all the energy it can since this is free energy, then when the coil is energized, the magnets need to pull away quickly, faster than the approach. This would unbalance the transaction time and cause an energy gain.


esaruoho

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #611 on: December 27, 2009, 07:11:48 PM »
It's probably the reed switches failing.

tachoman said that the  reed switches arent designed for these types of spikes.

k4zep

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #612 on: December 27, 2009, 07:57:22 PM »
The whole system may need to vary transaction time. On the approach, the rotor needs to store all the energy it can since this is free energy, then when the coil is energized, the magnets need to pull away quickly, faster than the approach. This would unbalance the transaction time and cause an energy gain.

H Luman,

Close, On the approach it is free energy as the magnet is sucked in.  Then upon energization, it needs to coast on by without any loss of speed, hence saving the free energy that was added to wheel in the wheel.  So, there is one  positive vector force speeding up the wheel as the magnets are sucked into the core, another zero or negative vector force depending on efficiency of the coil/core system, a constant negative vector from air drag at equilibrium, a constant negative vector force of the bearing drag and I have to have missed a couple others.
Then too, the simple math. assumes a constant speed when in fact the wheel is constantly speeding up and slowing down but these are small variables.  It would seem that with a load, with lower aerodynamic forces, the efficiency would be higher but again that is a small number.  A nice live dynamic torque meter would be a neat analytical tool here wouldn't it?

Respectfully,

Ben

Omega_0

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #613 on: December 27, 2009, 08:32:18 PM »
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.

You guys have a point, that's why I asked for a second opinion on the calculations, I was not very sure. It seems that we need to take into account the whole time needed to bring the rotor from stop to the max RPM.

So if it takes t sec to get to a speed of 2520 rpm, the input energy becomes
E = 36*t J

This energy is stored as usual as KE, as there is no load. Any input after this will not contribute to increase in KE (This is a strange case, as the input is independent of output here, unlike a normal motor).

We don't have the rotor weight, but lets take worst conditions here. So to be OU, it should take

49 > 36t
or 49/36 > t
or t < 1.36 sec  to reach 2520 rpm

(Assuming a rotor weight of 720g, t = 4 sec). Of course, if you can recover the input energy back into the battery , t will increase.

So now it seems less probable that JLN setup is OU, but the real test will be to load the motor, without which all calculations are mere guesswork and should not be taken seriously :) :)


mondrasek

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Re: STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM
« Reply #614 on: December 27, 2009, 08:41:30 PM »
The way I see it, in the eOrbo, the electrical input to the toroids does NOT drive the rotation of the rotor in any direct way.  Yet the rotor will ACCELERATE.  If F=ma holds true (giggling to myself now...), then we are witnessing a Force, F, causing this acceleration.  And that force is NOT due to the input electrical energy being used to switch the toroids!

I believe the electrical input energy used to switch the toroids is 100% conserved.  You can try to recover it if you like.  But whatever you do not recover is lost in the system as HEAT.  That conversion of electrical energy to heat should be 100% efficient by CoE (again, lol).

The fact that the rotor spins at all is the evidence of OU.  Unless anyone can prove that some of the electrical input energy to the coils is directly causing the force that creates acceleration of the rotors.  If input energy and rotation are completely separate, then it must be OU.

Does the electrical input directly cause the rotors to accelerate?  If so, that relationship should be measureable.  So far, I see a lot of evidence that there is no such relationship.