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Author Topic: GENERATOR- YOU DO THE IN/OUT POWER MATH  (Read 72289 times)

Omnibus

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #15 on: April 20, 2010, 04:58:33 PM »
And a heavier than the air device can't fly (as they tought in the XIX Century)... Wow, Santos Sumont was really a rebel...

If E=MC², what happens to so called "conservation of energy"? Becomes conservation of "mass-energy"... Good trick...

No, they didn't. This is a misnomer. Birds were heavier than air in those days too and they were seen to fly back then. That's a wrong analogy for the suppression of progress.

Low-Q

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #16 on: April 20, 2010, 05:12:52 PM »
You might be right about all of that. I am currently constructing a minature full wave bridge rectifier.
I will dump all the AC output into it. Sure there will be some power loss but it then becomes very easy to compare input to output power as both are now DC.  Hi frequency AC is tricky to measure correctly.
It would be way much faster using a long conductor so you can measure the amps away from the magnet. You'll get the result right away. No rectifier is required (Unless I missed something with your rectifier plans).

Vidar

e2matrix

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #17 on: April 20, 2010, 06:04:39 PM »
Hi.

I put a new fuse in my analogue meter.

Input is 9VDC @ 0.075A.
Output is 33VDC @ 0.080A
 
When i measure DC mA, the magnet slows down a lot, is this because it's such low amperage that even measuring it is taking too much away ?


Gary.

That would possibly depend on how close your meter was to the magnet also.  There might be enough metal or copper in it to effect the magnet speed.  And an analogue meter will put more load on a circuit than digital normally would.  But either way it sounds like you've got a winner.  Based on your numbers:
Input=   0.675 Watts
Output= 2.64 Watts
COP = 3.91    :o
 
That's assuming no errors in measurement.  I'd sure like to hear what Tom is getting also.  I do think this concept has good potential. 

mscoffman

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #18 on: April 20, 2010, 07:48:38 PM »
You can’t just switch your meter to Current and measure it. You need to measure
current through a load resistance. But what value load resistor?

Measuring Power correctly;

What you want do is to measure the voltage across a load resistor using a DVM.
Stick the voltage meter reading number into an equation, then adjust the resistor
value until you get maximum Power. Or else use a variable resistor , then measure
the resistance setting after it has been set to maximize P.

Since P=E*I and E=I*R it makes P=E^2/R we want to adjust our load R so
that P (power) is maximal in the equation P=E^2/R. That means adjusting R
around until the calculated P value becomes maximum (not E maximum). We
then don’t need to measure I the current at all.

What one is doing here is matching the source impedance (AC resistance) of
the coil to the load resistance of the resistor. Maximum power is transferred
between the generator and the load when the internal impedance of the generator
matches the impedance of the load. We want to know what the max calculated
P is in milliwatts.

---

I’ve noticed that even in cheapo DVM’s the resistance of the meter in
measuring current can often affect the load. This wouldn’t happen but
the meter cannot display enough leading zeroes (have enough accuracy)
when displaying small currents. So they boost up the resistance so that
the meter gets more voltage to work with…But that insertion resistance
then affects the circuit.

---

Self Running;

What one wants to do now, is down convert your 33 volts or whatever you get
under load that produces max Power. Down convert it to 12 voltsDC across a big
cap then use one of those automobile lighter inverters to step it to 110/220Volts AC.
Then use a wall wart power supply to supply input to your motor…If you do this
carefully and get this right the system should self run. If your auto inverter has an
conversion efficiency of 30% then you need a system COP gain of 3.0 to offset it.
Your particular inverter’s efficiency may vary. If you start the system to run on
utility power then switch it to internal the capacitor will either charge up - overunity
or discharge for long enough time where you can see what is going on.


:S:MarkSCoffman


DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #19 on: April 21, 2010, 02:52:03 AM »
That's really helpful, i know little enough about electronics let alone proper testing methodology.

Toms new video is up now showing 40 LED's being lit (brightly) :

                       http://www.youtube.com/watch?v=i3Cpv4Lxdyw&feature=player_embedded

He's outputting AC until his rectifier components arrive.

Total watts used to light the LED's is 24 and he's only inputting 4.86 ...

My setup can't compare at the moment because my axel is high-friction until my bearings arrive.

I'm off to test with load resistors ;+}


Gary.

Rapadura

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #20 on: April 21, 2010, 03:21:55 AM »
This last video is nice!

What he is using as DC input source? D cell battery?

I hope Tom use a rectifier soon, and see what DC output he can get. Let's see how many LEDs he can lite continuously with DC output after the rectifier.

 Better... Let's see if the DC output after the rectifier can power an exactly equal generator, and the DC output of this second generator can power the first one. Self running is so cool!

Omnibus

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #21 on: April 21, 2010, 08:33:05 AM »
@mscoffman,

Quote
Accurately measuring the power of an AC signal with instrumentation is almost the definition of an impossibility. I don't care who or what is doing it
- I don't trust it.

That's the crux of the matter. If you measure momentary currents and voltages with a good DSO, multiply them and integrate the product over time you'll get very accurate energy values. The slope of that integrated curve is the power. This is how Steorn are doing their power measurements and so far that's the best I've seen in this respect. The signal should be studied as is, avoiding conversions, rectification and so on which are accompanied by inevitable losses distorting the power measurements.

Therefore, on the contrary, I would not trust power measurements of an AC signal with a DVM's.

Of course, once the procedure for correct power measurements is established one has to provide a load for maximum transfer of power from the source, as you explained.

Also, the input power causing the rotor to spin (or in a transformer whatever there is in the secondary coil) should be calculated by subtracting  the Ohmic losses R*I^2 from the momentary I*V. Therefore, for an accurate measurement of the input power (effective input power) it is very important to measure momentary R accurately on the fly, that is, while the device is working, together with measuring momentary I and V. Would be interesting to hear suggestions as to how this can be done.

DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #22 on: April 21, 2010, 01:56:18 PM »
Reading this document :

http://davidsonsales.com/docs_pdf/CoilPitch.pdf

Makes me wonder about the coil-pitch of this setup ...

The angle of the coil sides to the center of the rotor is easily measured, but what about the pole-to-pole angle ?

Is it 180 degrees ?

This document goes into more detail about why this matters :

http://www.bmcoi.com/CatLit/Power/TECHNICAL%20PAPERS/GEN.%20WINDING%20PITCH%20-%20LEKX3115.PDF


Gary.



DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #23 on: April 21, 2010, 03:23:50 PM »
OK, hope i'm doing this correctly.

Input : 12.56VDC @ 0.130A

Output (rectified) : 33.6VDC @ 0.125A

The output is going through a 10k pot, turned all the way up, in series with the analogue meter.

If i've done this correctly then :

Input : 1.6328W

Output : 4.2W

COP = 2.57

It's RPM is 12,000 !

I've been reading up on generator windings and have some ideas, will post when formulated and tested properly.


Gary.


« Last Edit: April 21, 2010, 03:52:45 PM by DeepCut »

e2matrix

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #24 on: April 21, 2010, 03:30:59 PM »
Tom,  I'm no expert here and I'm sure mscoffman knows more than me but it seems if you are going to try for a self runner you'll lose a lot of efficiency with the way he is suggesting.  So I'll suggest a different way and if others see a flaw with my suggestion that's fine as I said I know mscoffman and others here are way more up to speed on this than I am. 
  I'd suggest simply stepping down the voltage from your output with a small transformer that has about 4:1 ratio and then rectifying the output and feeding that to the input - maybe a filter cap on the rectified output too?  At least to me it sounds like it would be a lot less losses that way. 

Rapadura

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #25 on: April 21, 2010, 03:58:11 PM »
Input : 1.6328W

Output : 4.2W

COP = 2.57

Wow! I love COP greater than 2! Because it's not over-unity, it's over-double!

If a generator can really put out COP > 2 then this generator can power itself (self-running), and, at the same time, can power another generator that is exactly equal the first one. And this second generator will also put out COP > 2, then this second generator can power TWO more generators exactly equal the first two. And this two new generators can power FOUR more generators equal the previous four.

Then you will have:

INPUT = Nothing. Zero. Just a self-running generator.

OUTPUT = Four generators exactly equal the first self-running generator.

That's why I have dreams with "over-double" every night! Over-double rulez!

DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #26 on: April 21, 2010, 04:09:56 PM »
Tom,  I'm no expert here and I'm sure mscoffman knows more than me but it seems if you are going to try for a self runner you'll lose a lot of efficiency with the way he is suggesting.  So I'll suggest a different way and if others see a flaw with my suggestion that's fine as I said I know mscoffman and others here are way more up to speed on this than I am. 
  I'd suggest simply stepping down the voltage from your output with a small transformer that has about 4:1 ratio and then rectifying the output and feeding that to the input - maybe a filter cap on the rectified output too?  At least to me it sounds like it would be a lot less losses that way.

Funny, i just tried that.

I made a small coil to step down the voltage and see what current increase i could get.

The voltage went  w a y  down to around 6V and the current doubled to 250mA but the magnet slowed right down.

Obviously it's not a proper transformer because there's no physically connected ferrite-core ...

mscoffman

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #27 on: April 21, 2010, 05:44:32 PM »
The thing about the adjustable load resistor is...in all probability
the behavior of the motor is going to change as you load it down
because you apply mag braking force to it, which is what happens
when work is extracted...So adjusting to maximize the Power will
affect the operation of the device and hence feedback to the input
power required. Now, that feedback may require more input power
(which it would in a typical motor), or it might require less -
because slowing the rotor will force the drive "on" periods apart.

So once you have the design working you have to shift your goals
to maximum power output of a calculated P rather than your inital
design goals to maxize rotor RPM.

You really want to measure both Power(s) at the exactly the same
time. A DVM won't load the voltage (with many megohms meter input
impedence, but the mechanical voltmeter will) And unfortunately the
current ranges will not have nearly the theoretical zero ohms insertion
loss that it should. But generally the resistance of the load resistor
won't change with use...So that is one thing that remains constant.

By finding the matching impedance you find out something useful about
how to design the load utilization circuit. Listen, a diode bridge *will*
wring out most usable power out of even spiked AC signals but
impedance matching (making your voltage converter the same impedance)
as the maximum power point load) is somewhat tricky. Do it wrong and you
risk getting unnecessary ineffciency in the power loop. The ability to build
a pulse transformer built like the JT torroid would be a good capability to
have because a good transformer automatically keeps the E and I terms
of the P=E*I in the same ratio. So you want to step the pulse AC signal
down and then rectify it. Filter it with a big cap. The 12VDC inverter
110/220V may be inefficient but it will help stablize your feedback loop.
It's input power will very much be a function of it's output demand. A
motor COP 2.0 is usable if the inverter has a efficiency of 50% - a nice
experimental capability to have, and hold on to.

BTW in the real Overunity device the power is going to build up, then you
will need a way of disposing of it properly...Also a nice problem to have.

:S:MarkSCoffman

DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #28 on: April 21, 2010, 06:30:04 PM »
Nice one Mark :)

One thing that i feel the need to understand, why can my analogue meter read the mA but my digi can't ?

My digi even goes down to micro-amps, unlike the analogue, but still registers nothing ...


Gary.

mscoffman

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #29 on: April 21, 2010, 07:01:33 PM »
Nice one Mark :)

One thing that i feel the need to understand, why can my analogue meter read the mA but my digi can't ?

My digi even goes down to micro-amps, unlike the analogue, but still registers nothing ...


Gary.

You say you have replaced the fuse, That fuse may have more resistance
than the original, the pass resister inside the meter could be burnt,
changing it's resistance or something inside the meter's IC could be damaged.

:S:MarkSCoffman