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Author Topic: Pulse motor - is this overunity?  (Read 8968 times)

mr2

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Pulse motor - is this overunity?
« on: March 20, 2008, 06:34:50 PM »
I have made a small pulsemotor that is inspired from EV Gray. It is only made to verify my theories. And has a very small energy consumption.
All persons i have demonstrated this motor for, says that i may only done a motor efficiency better from COP 0,2 to 0,4.

Therefore i have disconnected the motor. Replaced it with a 33 ohm resistor. Putted on 0,1 ohms resistors as shunt before my invention and one after the invention. Then i put a oscilloscope on the shunt and measured the RMS voltage.
I have attached a small schematic to show my measurement method.

Pulse rate only 2 Hz.

My question is: IS this overunity? And is it correctly connected/calculated?

At R1 the scope said 3,67mV RMS. With the formula P = U^2/R this is 0,1346 mW
At R2 the scope said 5,23mV RMS. With the formula P = U^2/R this is 0,2735 mW

This gives a consumpion over the load 103% more than the load from battery.
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hartiberlin

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Re: Pulse motor - is this overunity?
« Reply #1 on: March 26, 2008, 05:56:40 PM »
input 440 mW
output 90 mW

Way underunity.
The input is calculated by the 12 Volts x (Voltage at R1 / 0.1 Ohm), so it is 440.4 milliWatts.

Output is 52mA ^2 x 33 Ohm.
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mr2

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Re: Pulse motor - is this overunity?
« Reply #2 on: March 26, 2008, 08:13:34 PM »
Thanks for answer.

Is not the formula P=(U^2)/R correct as powercalculation?

Well anyway...
The voltage is same on both from battery and outside. So then the outside should be calculated as 12 Volts x (Voltage at R2/0,1 Ohm) = 627,6mW?


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Groundloop

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Re: Pulse motor - is this overunity?
« Reply #3 on: March 27, 2008, 12:01:24 AM »
@mr2,

3,67mV = 0,00367 Volt.
5,23mV = 0,00523 Volt.

Input current  I = U/R = 0,00367 / 0,1 = 0,0367 Amp.
Output current I = U/R = 0,00523 / 0,1 = 0,0523 Amp.

Input Watt     P = U * I = 12 * 0,0367 = 0,4404 Watt.
Output Watt    P = R * (I*I) = 330 * 0,00273529 = 0,9026457 Watt.

COP = OUTPUT / INPUT = 0,9026457 / 0,4404 = 2,0496042234332425068119891008174

It seems that you have over unity.

Are you 100% sure you have measured true RMS values?

Groundloop.
« Last Edit: March 27, 2008, 12:25:24 AM by Groundloop »
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hartiberlin

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Re: Pulse motor - is this overunity?
« Reply #4 on: March 27, 2008, 12:27:34 AM »
Groundloop,
not 330 Ohm,
but 33 Ohm,
so underunity !
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Groundloop

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Re: Pulse motor - is this overunity?
« Reply #5 on: March 27, 2008, 12:47:21 AM »
@hartiberlin,

You are right. I lloked at the circuit drawing. The drawing says 330.

EDIT : Very poor drawing but I think it says 33 Ohm.  ;D

So just divide the output by ten = 0,09026457 Watt.  ;D

Groundloop.
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mr2

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Re: Pulse motor - is this overunity?
« Reply #6 on: March 27, 2008, 02:30:54 AM »
Tanks for answers.

First of all:
If R2 "sees" the resistance R3 at 33 ohms, why don't the R1 see it?
R1 is dependent of the R3 as much as R2.

Take away the "black box" and explain it then out of the same values.


I have the same voltage out. I have higher voltage over the shunt-resistors at the output than the input. That says: More amperage at same voltage at the output.
No overunity?
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mr2

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Re: Pulse motor - is this overunity?
« Reply #7 on: March 27, 2008, 02:37:46 AM »
Quote from: Groundloop on March 27, 2008, 12:47:21 AM
@hartiberlin,

You are right. I lloked at the circuit drawing. The drawing says 330.

EDIT : Very poor drawing but I think it says 33 Ohm.  ;D

So just divide the output by ten = 0,09026457 Watt.  ;D

Groundloop.

The drawing and my initial message says 33 ohm. :)

But the R1 is as much dependent of the R3 as R2. Don't know why you both calculate with the 33 ohm resistor, when the shuntresistors measure power on the "same line".

EDIT: btw: the formula says P=U^2/R. Then I have it correct in my first message. :D (Well.. i did know)
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hartiberlin

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Re: Pulse motor - is this overunity?
« Reply #8 on: March 27, 2008, 04:18:12 PM »
@Mr2,
in your first posting you did only
compare the wattage at the shunt resistors but not the
input power from the battery versus the output power at
the 33 Ohm load resistor.
These are 2 very different cases.
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mr2

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Re: Pulse motor - is this overunity?
« Reply #9 on: March 27, 2008, 05:01:24 PM »
@hartiberlin

I understand that. But taking the voltage at both battery and output in consideration, what then?

If my initial drawing is not correct measurement, i could use ordinary amperemeters instead. Easier then. And put a voltmeter at the battery, and at the output.

But out of the shunt voltages groundloop calculated the amperage. I use them in the P = U * I formula.
The measured voltage on battery is 11,62 V, and at output is it 11,6 V.

Then I have:

At R1: 0,0367 A * 11,62 V = 0,4264 Watt
At R2: 0,0523 A * 11,6 V = 0,6067 Watt

Hmm.. THIS must be correctly measured and calculated.

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tinu

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Re: Pulse motor - is this overunity?
« Reply #10 on: March 28, 2008, 12:47:57 PM »
Quote from: mr2 on March 27, 2008, 05:01:24 PM
The measured voltage ... at output is it 11,6 V.

Then I have:
...
At R2: 0,0523 A * 11,6 V = 0,6067 Watt

Either current but most probably the voltage (11.6V) can not be correct because:

on R2: 0,0523 A * 0.1 Ohm = 5.2mV
and
on R3: 0,0523 A * 33 Ohm = 1.7259V

so, on R2 and R3, if current is properly measured, you should have 1.7311V, which is in contradiction with 11.6V.

One question though: what is ?output? actually? R2 or R3? Normally is R3 of course, but from your post (power calculus on R2) it seems that you consider R2 as output.

Cheers,
Tinu
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Groundloop

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Re: Pulse motor - is this overunity?
« Reply #11 on: March 28, 2008, 02:23:58 PM »
@mr2,

Remove the R2 and measure the voltage over R3. Then you can calculate the output.
Measure the voltage over R1 and calculate the input current. Then measure the voltage
over the input of your circuit. (Inside of R1, direct on the input.)
Then you can calculate the input power.

Groundloop.
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