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Overunity Machines Forum



Sharing ideas on how to make a more efficent motor using Flyback (MODERATED)

Started by gotoluc, November 10, 2015, 07:11:57 PM

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gotoluc

Quote from: Over Goat on February 24, 2016, 11:33:43 AM
:) so it is looking to be 50% more efficient at least , at least 50% recovery?
and over 50% recovery in the second instance. 
are there tweaks which you could to get this even better, not that this is not amazing enough,
great work Luc

I'm sure Electrical Engineers would not calculate it quite that way ;)
The final concept is to use the recovery power which now seems to be around 50% and put it back as torque in the Motor shaft by using a 2nd assist motor.
Only when that is done will we know if my idea has any value. We are not there yet and before doing that we must get some consistent and accurate load test results with the motor on its own so we have something to compare to once we hook up the assist motor.

Thanks for your supportive post.

Luc

gotoluc

Quote from: Magluvin on February 24, 2016, 12:23:03 PM
Hey Luc

Seems odd that recovery from the motor. with the other motor added, is less, because the switching on times should be longer if the motor is slowed down by the other loaded motor.  Not sure. Maybe Im not fully understanding.  But neat stuff to say the least. ;)

Mags

Hey Mags,

it's only odd because of the conventional way we are use to seeing motor commutation.
The way this switch (commutating) circuit works is, the pulse width stays consistent no matter the rpm, so if we reduce the motors rpm by applying a load we are reducing the frequency of the coils on times, which also reduces the flyback frequency.
In a standard motor commutation, as you reduce the motors rpm the coil on time increases because the motor coils pulse width (on time) get wider as the rotor is turning slower, so the flyback in this case would increase.

Let me know if you have any other questions or need additional details.

Luc

billbailey

Howdy Luc
that was a very nice piece of work removing those 4 areas on your motor shaft.Is there a reason why channel 1 seems to hold the 40 volts thru out  the latest 2 video's,while channel 2 and 3 vary quite a lot? also if you used a different alternator and still used the same 5 ohm resistor would the numbers change?new to this and easily confused :-\

thanks bill

gotoluc

Quote from: billbailey on February 24, 2016, 04:16:22 PM
Howdy Luc
that was a very nice piece of work removing those 4 areas on your motor shaft.Is there a reason why channel 1 seems to hold the 40 volts thru out  the latest 2 video's,while channel 2 and 3 vary quite a lot? also if you used a different alternator and still used the same 5 ohm resistor would the numbers change?new to this and easily confused :-\

thanks bill

Hi Bill,

The Scopes channel 1 is measuring current (across a 0.01 Ohm resistor) of all 3 phase.
The other 3 scope channels are measuring the voltage (input and flyback) across each motor coil.
A different kind of alternator would change the results. Each have their own characteristics as far as voltage and current output at a certain rpm.
If what you are asking is if I hooked up an alternator that produced more output, would the motor sustain it and the answer is no, there would be a point where the load would stop the motor. My best guess is, we may get another 1 or 2 Watts more before the motor stops.

Hope this answers your questions

Luc

gotoluc

Okay, here is the official load tests comparing the stock motor & circuit to the modified motor & new circuit.
I have done my level best to keep the tests as fair as possible.

Link to test video: https://www.youtube.com/watch?v=H55BVbgDW0E


Test results:


Stock Motor

At 730 rpm it consumes 21.85 Watts (no load)

At 725 rpm it consumes 29.29 Watts (with load)
so 29.29W - 21.85W = 7.44W used from input to
to deliver 1.75 Volts on 1 Ohm resistor = 3 Watts


Mod Motor

At 730 rpm it consumes 6 Watts (no load)
with 1.66 Watts recovered from Flyback
so 6W - 1.66W = 4.34 Watts used from input

At 725 rpm
it consumes 16.64 Watts (with load)
with 4.49 Watts recovered from Flyback
so 16.64W - 4.49W = 12.15 Watts used from input
so 12.15W - 4.34W = 7.81W used from input to deliver 1.75 Volts on 1 Ohm resistor = 3 Watts

It is comforting to see the Mod Motor has not lost its electrical to mechanical efficiency with the modifications.
Now the next step is to add the assist motor to see if the flyback is best used to assist the motor or charge a battery.

Any bets?

Luc