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

[IceStorm]

I never said anything about Power or torque in my example.  My example was meant to figure out how long it would take a battery to discharge in a controlled experiment.  It is the same as putting a resistor on a fully charged cap and seeing how long it takes to discharge across the resistor.  This is a good way to figure out how much energy is in the battery so you can make your comparisons.  Obviously you missed the entire point.

GB

Here you talk about 750 rpm , 2 motor with no LOAD on the shaft , one take 3 time the amount of the other one in power.Take your example and look at what i wrote about the rotoverter so Idle speed mean nothing  to evaluate the TORQUE (MECHANICAL POWER) because THERE NO LOAD,

If a motor runs @ 750rpm for 10 hours with drawing a constant current from a battery, then building a motor that runs @ 750rpm for 30 hours from a same type battery is cop = 3 in mechanical energy gained vs electrical energy expended.

You have gained 3 times more mechanical energy than what the battery is capable of producing.  The additional energy gained was mechanical energy and was expended in the additional 20 hours of mechanical motion instead of being converted to electrical energy. 
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GB

And here you talk about the coils in motor who act as load. if both motor is the same, first one is a rotoverter the second one is a normal one (both identical motor spec) , first one can be 50W idle and the second one 300W idle at SAME speed but the TORQUE it can deliver will not be the same because the PF is not the same, on the rotoverter one the VAR is bigger than the normal one, so less current input at the expense of torque.

If both motors have coils with the same load on the battery, then the RPM is relative in both systems and is not absolutely nothing.  A coil is a load on the battery, is it not?  This means both motors in my example have the same load!

GB

Now tell me how you can evaluate if motor 1 is better than motor 2 ? no one deliver any power , they just idle so current input is absolutly nothing.

Best Regards,
IceStorm

[gravityblock]

IceStorm,

I'm not trying to figure out power or torque.  Power correction devices don't lower the amount of energy being consumed by a device, they help the watt hour meter to more accurately calculate the correct amount of energy used by a device.  The watt hour meter does not calculate energy consumed correctly and the end result is the consumer is billed for energy they didn't use.  The total amount of energy actually consumed is the same with or without the power correction device.  Calculating energy in vs energy out is not as cut and dry as you may think.

GB

[Cloxxki]

If mass of a motor is great, and friction low, a slightly better battery or circuit can give hugely longer running times. Time will be spend more running the thing, than power is used to start up.
The other way around, a lightweight motor with high friction, that is more like a load. 10% more runtime, means ~10% more efficiency, or capacity.

[PaulLowrance]

Lets just give Steorn a chance to prove it. If they're a scam, then they have their day coming. If investors have given money without proof, LOL, geez, then that's going to be big lesson for them. I have not given Steorn a dime, and I would *NEVER* sign any NDA.

Supposedly January is the big month they're going to prove it.

[IceStorm]

IceStorm,

I'm not trying to figure out power or torque.  Power correction devices don't lower the amount of energy being consumed by a device, they help the watt hour meter to more accurately calculate the correct amount of energy used by a device.  The watt hour meter does not calculate energy consumed correctly and the end result is the consumer is billed for energy they didn't use.  The total amount of energy actually consumed is the same with or without the power correction device.  Calculating energy in vs energy out is not as cut and dry as you may think.

GB

Again some more non sense, PF CORRECTION do lower the input power , that don't mean the load will get LESS power , that mean the INPUT POWER will be less , look here : http://www.allaboutcircuits.com/vol_2/chpt_11/4.html and go to 2/3, look for :

"This correction, of course, will not change the amount of true power consumed by the load, but it will result in a substantial reduction of apparent power, and of the total current drawn from the 240 Volt source:"

And dont change your non sense by saying "I'm not trying to figure out power or torque" because you should look at this post your wrote, look at what is in bold:

If a motor runs @ 750rpm for 10 hours with drawing a constant current from a battery, then building a motor that runs @ 750rpm for 30 hours from a same type battery is cop = 3 in mechanical energy gained vs electrical energy expended.

You have gained 3 times more mechanical energy than what the battery is capable of producing.  The additional energy gained was mechanical energy and was expended in the additional 20 hours of mechanical motion instead of being converted to electrical energy. 

RPM at idle MEAN NOTHING , i showed you that with the analogy with the rotoverter for 2 identical motor , one with rotoverter modification and the order one normal, at IDLE the speed can be the same but the POWER consumption WILL NOT BE. so discharge your battery as you wich , that will never tell you if you are COP = 3 because THERE NO LOAD on the shaft.
Best Regards,
IceStorm