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

[powercat]

Did anyone see this one.
dmmpower.wmv
http://www.youtube.com/watch?v=JAoTrqnZpfg
cat

[Silvije]

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]

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]

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]

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