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Author Topic: Sum of torque  (Read 122997 times)

Offline dieter

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Re: Sum of torque
« Reply #150 on: February 06, 2015, 09:06:44 PM »
Hi EOW,


did I understand this correctly:
the motor turns the big pulley, the small pulley recovers the energy, and the support rotates due to inertia?


Did you get the gears yet?


BR
 

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Re: Sum of torque
« Reply #150 on: February 06, 2015, 09:06:44 PM »

Offline EOW

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Re: Sum of torque
« Reply #151 on: February 07, 2015, 07:51:32 AM »
Quote
the motor turns the big pulley, the small pulley recovers the energy, and the support rotates due to inertia?
yes, there is no torque on the support IF pulleys don't accelerate.

Quote
Did you get the gears yet?
yes but the energy recover is lower (there is friction and efficiency of generator is not 1 maybe 0.8 or lower) beause w is small and the motor turn very quickly => the force is small, I need a low speed motor and a support that turn at w high enough. I gave an example, the power recover is smal with a small force.
« Last Edit: February 07, 2015, 10:46:16 AM by EOW »

Offline dieter

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Re: Sum of torque
« Reply #152 on: February 07, 2015, 08:22:09 PM »
There may be some low RPM dc motors using built in gears, like eg. a grill motor. I also got one from the mentioned laserprinter, although more of a motor with attached gear set. However defective laserprinters are a great source for gears, I got mine for 5 bucks, was even still working, but I wasn't interested in printing...


Other sources may be Microwave oven plate motor (dc?) or scanner motor, they all need to be low RPM for what they do.


BR


Free Energy | searching for free energy and discussing free energy

Re: Sum of torque
« Reply #152 on: February 07, 2015, 08:22:09 PM »
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Offline EOW

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Re: Sum of torque
« Reply #153 on: February 08, 2015, 11:07:59 AM »
Quote
There may be some low RPM dc motors using built in gears
with gears inside, there is a counterclockwise torque on the support, the motor needs to rotates slowly or the angular velocity of the support must be high.

With:

w (support) = 157 rd/s (1500RPM)
w (motor) = 157*2rd/s and P=300 W, the torque is 1.91 Nm => the force is 19 N if 3R = 0.1 m, R=0.1/3

The generator turns at 6w but the stator of the generator turns at w, so the difference is 5w. The generator can give 5FRwt.

The power won is 2FRw = 200 W

If the efficiency of the motor and the generator is 0.8, the global efficiency is 5/3*0.8*0.8 = 1.06 so it is necessary to use brushless motor (some can have an effciency of 0.9) and without take in account of the efficiency of the electronic card and the belt... maybe 0.95 for the belt and for the electronic card, the global efficiency become 5/3*0.8*0.8*0.95*0.95=0.96. With a motor/generator at 0.9 this could be: 1.21

Maybe the best is to test a basic dc motor/generator and look the efficiency without rotation of the support and turn more and more the support and look at the efficiency, the efficiency must increase more and more like the angular velocity of the support. The small pulley can be smaller, 3R for the bigger and 0.2R for the smaller, the efficiency will be at 29/15*0.8*0.8*0.95*0.95=1.11
« Last Edit: February 08, 2015, 03:15:04 PM by EOW »

Offline dieter

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Re: Sum of torque
« Reply #154 on: February 08, 2015, 05:30:49 PM »
There are also losses in getting the required current to the pulley's motor. As the support is rotating, you maybe need to use brushes. Or you could try inductive coupling if the support has near zero mechanical losses: an additional disc on the support's shaft, maybe 1-2 foot away to prevent magnetic interferences with the dc motor etc. Although the Lorentz force is the only cost in frictionless inductive coupling (other than friction etc.), the implementation would complicate things.


Furthermore, you also got to extract the power from the generator to measure it ...


If the COP is over 1.0 then it may be possible to have everything on the support...


BR


Free Energy | searching for free energy and discussing free energy

Re: Sum of torque
« Reply #154 on: February 08, 2015, 05:30:49 PM »
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Offline EOW

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Re: Sum of torque
« Reply #155 on: February 08, 2015, 06:04:04 PM »
Quote
Furthermore, you also got to extract the power from the generator to measure it ...
It's easy with brushes and a plate with paths of copper, it's possible to use 2 brushes for input and 2 brushes for output. The main problem is to find a motor with a good efficiency.

Offline dieter

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Re: Sum of torque
« Reply #156 on: February 08, 2015, 09:28:06 PM »
Hard enough to find high efficiency dc motors, let alone low RPM without gears... ebay has only high RPM, down to 3600 RPM, as far as I see.


BR


Free Energy | searching for free energy and discussing free energy

Re: Sum of torque
« Reply #156 on: February 08, 2015, 09:28:06 PM »
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Offline EOW

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Re: Sum of torque
« Reply #157 on: February 09, 2015, 04:20:31 PM »
Hi Dieter,

Maybe I can use a stepping motor with a big difference of radius like that I can use a dc motor for the generator. I will look for the efficiency of the stepping motor.


Offline DreamThinkBuild

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Re: Sum of torque
« Reply #158 on: February 09, 2015, 07:10:50 PM »
Hi EOW,

Here is a patent that you may find interesting, it's in French but he gives examples in the drawings on how to match input motor, to gear reduction, to generator rating. He's using servo motors.

Free Energy | searching for free energy and discussing free energy

Re: Sum of torque
« Reply #158 on: February 09, 2015, 07:10:50 PM »
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Offline EOW

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Re: Sum of torque
« Reply #159 on: February 09, 2015, 11:13:05 PM »
Hi DreamThinkBuild,

I speak french, so it's easy for me. Thanks but there is an error in the patent and there is no rotationnal support.

Offline dieter

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Re: Sum of torque
« Reply #160 on: February 10, 2015, 04:14:10 AM »
EOW,


Steppermotors is a good idea to get a very low RPM. But usually they are driven by synthesized Sine waves / Phases. If you can control these shapes PWM wise, then it may be possible to reduce the power consumption.


Otherwise, my experience with stepper motors tells me, the slower they are rotating, the hotter they get (because of unneccessary  DC currents), and heat means loss here.


Also, the rotation is rather vibrating, not very smooth. Not sure if this is a problem.


The SMC800 control card does in fact allow the shaping of the 2 phases, but only in a 2 bit resolution. Like:  0,1,2,3,2,1,0.. not really a smooth sine wave. And that is used to control -12 to +12 volts. I don't know if this card is still manufactured. There may be other, similar ones.


BR


Free Energy | searching for free energy and discussing free energy

Re: Sum of torque
« Reply #160 on: February 10, 2015, 04:14:10 AM »
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Offline EOW

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Re: Sum of torque
« Reply #161 on: February 10, 2015, 01:51:36 PM »
I watched a lot of specifications of stepper motors and I think it's not a good solution. The easiest it's to build a small device and turn it at 750 RPM or more. It's not necessary to have an efficiency greater than 1, if there is no torque on the support (it can be test with w=0 for the support) and if the efficiency is greater with the support in rotation I think it could be ok. For example, if I have an efficiency of 0.2 for the device when the support don't turn, the efficiency must increase of 5/3*0.2=0.33 (if ratio of bigpulley/smallpulley = 3/1 and if the motor turns at 750RPM). Like that a simple test can be done with a drill and 2 dc motor (any efficiency could be ok). The efficiency must increase like the angular velocity of the support (linear law).

The general formula for the extra power is :

Power = F(R1w - R2w)

with:

R1 the radius of the big pulley
R2 the radius of the small pulley
F the force on the belt, it's Pm/(R1*wm), with wm the angular velocity of the motor and Pm the power used by the motor
w the angular velocity of the support

So the extra power is:

Power = Pm*(R1w - R2w) / (R1 * wm)

The efficiency of the device is :

Eff = (Power * ηg*ηb + Pm*ηg*ηb*ηm) / Pm = (Pm*(R1w - R2w) / (R1 * wm) * ηg*ηb + Pg*ηg*ηb*ηm) / Pm = (R1w - R2w) / (R1 * wm) *ηg*ηb + ηg*ηb*ηm

With:

Pm the power consumed by the motor
Pg the power recover by the generator
ηb the efficiency of the belt
ηm the efficiency of the motor
ηg the efficiency of the generator

I don't take in account friction and lost by Joule effect in electric circuit.

For example, with Pm=20W, w=750RPM, ηm=0.6, ηg=0.6, ηb=0.95, R1=3, R2=0.5, wm=4000 RPM, the efficiency is 0.43, without the support the efficiency is 0.34. With a motor with 750 RPM, the efficiency move up to 0.81
If ηm = ηg = ηb = 1, then the efficiency is 1.15 with 4000 RPM and 1.83 with 750 RPM for the motor
« Last Edit: February 10, 2015, 06:16:22 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #162 on: February 11, 2015, 09:59:23 AM »
With the same support, the small pulley don't turn at 6w but at 4w, so the sum of energy is 0.
« Last Edit: February 11, 2015, 12:25:39 PM by EOW »

Offline dieter

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Re: Sum of torque
« Reply #163 on: February 11, 2015, 06:37:19 PM »
Sum of energy is zero? You mean energy gain is zero, llike in efficiency = 1.0 - losses?


BR


Offline EOW

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Re: Sum of torque
« Reply #164 on: February 12, 2015, 01:16:56 PM »
Yes, efficiency=1 (I recover that I lost).

I study this new idea. Without the support3, the efficiency is 1. I will calculate all energies for find the global efficiency.

The motor needs -9FRwt
Torque F6/F9 gives 2R*2w*Ft=4FRwt
Torque F5/F8 gives 4R*2w*Ft=8FRwt
The friction (brake from the pulley2 to support3) gives 3FRwt

The sum is +6FRwt


« Last Edit: February 12, 2015, 09:12:03 PM by EOW »

 

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