Free energy has been here all along ,But some very powerfull ppl don't want you and me to be free from energy costs,Check this free energy magnet motor at LT-MAGNET-MOTORdotCOM ,The revolution begins!
@existance12d: No, that was not my concern at the the time. I believe I will need a lot more wire for overunity. Besides, my aluminium frame is not doing so well. I am considering on redesigning it with wood, but funds are tight, so it may take a while.
You have an inductive load in your circuit, so V*V/R does not equal the power input of your circuit. V*V/R would be about 100 watts (340*340/1155). You calculated R*I*I which does turn out to be around 7 watts (1155*0.08*0.08). That is actually the heat loss in your motor (if you ignore the back spike). If you take 340 volts and multiply it by 0.080 amps, that brings it to 27 watts. When my 32" fan had three coils, I got the same results you did, but now it is 16x more efficient with six coils.
I am already on it. I ripped out some 330uF 200v caps from a used computer supply, and with my JTC I can chage it to 100+v with only 6v input. I have charged up the cap, pulled the cap out of the circuit, and took it to the motor, and it moved a tiny bit. With enough caps I should be able to make this work off of say...12v. Oh yeah. ;)
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Free energy has been here all along ,But some very powerfull ppl don't want you and me to be free from energy costs,Check this free energy magnet motor at LT-MAGNET-MOTORdotCOM ,The revolution begins!
intermitrj 1 year ago
Hi!
Have you got overunity?
existance12d 1 year ago
@existance12d: No, that was not my concern at the the time. I believe I will need a lot more wire for overunity. Besides, my aluminium frame is not doing so well. I am considering on redesigning it with wood, but funds are tight, so it may take a while.
detrix42 1 year ago
Sure would be nice to see what's going on behind the two coils as to better understand your setup.
619568701 1 year ago
You have an inductive load in your circuit, so V*V/R does not equal the power input of your circuit. V*V/R would be about 100 watts (340*340/1155). You calculated R*I*I which does turn out to be around 7 watts (1155*0.08*0.08). That is actually the heat loss in your motor (if you ignore the back spike). If you take 340 volts and multiply it by 0.080 amps, that brings it to 27 watts. When my 32" fan had three coils, I got the same results you did, but now it is 16x more efficient with six coils.
kmarinas86 1 year ago
@kmarinas86
Ok I can see that. oops. Thanks for the correction.
I am working on a "joule thief" circuit to supply the voltage.
I have a joule thief circuit right now taking 6.5v out of a 9v battery, and stepping it up to 85v. I plan on using two 6v lantern batteries
detrix42 1 year ago
You'll need capacitors with the joule thief, to build some amps
cuprocarbon 1 year ago
@cuprocarbon
I am already on it. I ripped out some 330uF 200v caps from a used computer supply, and with my JTC I can chage it to 100+v with only 6v input. I have charged up the cap, pulled the cap out of the circuit, and took it to the motor, and it moved a tiny bit. With enough caps I should be able to make this work off of say...12v. Oh yeah. ;)
detrix42 1 year ago
Beefy JTs.... Motors, Filament bulbs, heater cores, small appliances...
cuprocarbon 1 year ago