Re: F.B.D.I.S.S.M - Flux.Boosted.Dual.Induction.Split.Spiral.Motor.
« Reply #203 on: February 24, 2008, 10:27:22 AM » Quote
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Some calculations and observation notes:
I'd like to add that the rotor magnet have a totaly free pass at 165W input per Solenoid in the latest set of tests.
The activated Solenoid time is approx 0.25% of each full rotor turn and this equals 82.5 watts going into the motor.
But the calculated torque/rpm output from a motor using this size of magnets is most probably higher than the input.
The average torque under load is calculated to 3,13 ft-lbs
At 100 RPM under load I get = (3,13*2*3,14*100)/33000 = 0,059 Hp = 44 watt output from the shaft.
At 200 RPM under load I get = (3,13*2*3,14*200)/33000 = 0,119 Hp = 88 watt output from the shaft.
At 300 RPM under load I get = (3,13*2*3,14*300)/33000 = 0,178 Hp = 132 watt output from the shaft.
At 400 RPM under load I get = (3,13*2*3,14*400)/33000 = 0,238 Hp = 177 watt output from the shaft.
At 500 RPM under load I get = (3,13*2*3,14*500)/33000 = 0,297 Hp = 222 watt output from the shaft.
You can see for yourselves. Already at 200 RPM I've got more out than in.
Don't forget how torque behaves in an electrical motor.
At No Load I get Maximum RPM at Zero Torque and I'll get Maximum Torque when stalled at Zero RPM.
And there is a linear relationship between these two points.
Meaning that I'll get Half the Torque when the Maximum Free Spinning RPM is loaded to Half the RPM.
And I do expect to hit a lot more than 400 RPM when the Sticky Spot is passed without loosing any momentum at all.
If I'll hit 1000 RPM at free spinning I'll get 222 watt output on the shaft when loaded down to 500 RPM.
But this is just some calculations on a small motor.
The big one I'm building is calculated to deliver approx 4500W output at 200W input.
4.5kW délivrés pour 400W en entrée c'est impressionnant