PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2024-03-28, 09:11:28
News: Registration with the OUR forum is by admin approval.

Pages: 1 2 3 [4]
Author Topic: MrClean's claimed "self-looping" device with load, and no +Battery connection  (Read 65640 times)
Group: Guest
Mr Clean see's OU in everything he builds, he just needs to measure it the right way to show it.  ;)

But seriously how can people use a function generator or signal generator and not consider the input to it as input to the circuit.
Without the signal generator there is no output so all the consumed energy by the signal generator is input.
If the input power to the Function generator or signal generator would be considered as input then there is no need to
determine if the signal generator is running the circuit. Bottom line is without all the power being used by the signal generator
to produce a signal there is no output so all of the signal generators input is input to the circuit.
That alone would have stopped Ainslie in her tracks I believe. How much power does a sine wave generator consume ?
Call me mad but it looks like a lot more power is being consumed there in that circuit than is being produced.

I've asked a few questions and requested a schematic showing how the input was scoped.

Cheers

P.S. One thing I am confused about, do people consider that if the input  and output sine waves are out of phase by 180 degrees the power factor is Zero or something ?
Or is it that when the secondary is loaded under such conditions they say the primary cannot give power to the secondary ?
Or is it that if the primary current and voltage wave forms are 180 degrees out of phase the primary can't give power to the secondary ?

..
« Last Edit: 2012-11-21, 06:02:07 by Farmhand »
   
Group: Guest
Mr Clean see's OU in everything he builds, he just needs to measure it the right way to show it.  ;)

But seriously how can people use a function generator or signal generator and not consider the input to it as input to the circuit.
Without the signal generator there is no output so all the consumed energy by the signal generator is input.
...

I agree, Fh. And as it is doubtful that these people would be too stupid for considering this point, the answer to your question is that they are lying. The free energy field is full of counterfeiters acting for miscellaneous reasons (money, egocentrism, green ideology, self-deception...) and full of people relaying their falsified speech for similar reasons, even academics among them.

   
Group: Guest
OK, in order to try to understand these claims better, I've performed a test measurement to see if I have it correct in my head, I just learned some things.

I measured the input/output of a transformer. Sine wave input from function generator, 10 Ohms in series with the primary and 330 Ohm resistor load across the secondary.

I used the vrms measurement on my DS1000E Oscilloscope and measured 45.7 mV across the 10 Ohm resistor and 97.7 mV across the input to the primary,
the phase angle between the voltage and current waveform was 11 degrees. So from that I get a power factor of 0.98 by cos(11) = 0.981.
And I get the current by dividing the 45.7 mV by the 10 Ohms to get 0.00457 A then I multiply that by the voltage across the primary which is 97.7 mV = 0.0004464 Watts
I then multiply that by the power factor of 0.98 to get 0.0004375 Watts input power.

To get the output I measured 294 mV across the 330 Ohm load the current was 0.00089 A, multiplied by the 294 mV = 0.0002616 Watts Output.
Not sure if I'm supposed to measure and phase difference in the output.

So I get 0.0004375 Watts input and 0.0002619 Watts output which output divided by input = 59.8 % efficient.

What do I do wrong, if someone wouldn't mind pointing it out.  :-[ I'm very much a learner, I need feedback.

The low efficiency would be due to driving that transformer incorrectly (not enough voltage, wrong frequency), frequency was about 200 kHz.

Oh and the difference in phase between the secondary voltage/current and primary voltage was 6 degree's (I think) Not sure if that actually means anything.

Cheers

I think something is wrong because if I consider the power dissipated by the resistors I get for the 10 Ohm resistor,  0.00457 A x 0.0457 V = 0.0002 W
and the 330 Ohm resistor I get 0.00089 A x 0.294 mV = 0.0002616 W plus the 0.0002 W = 0.0004616 Watts dissipated. But the input was only 0.0004375 Watts.
Must be measurement error, calculation error or OU, I would guess one of the first two.   :)

I suppose I will need to re-do the measurements and the calculations, the numbers won't be the same though because I changed the FG settings and it will be difficult to
get them the same again. I'll use more power anyway.  O0

..
« Last Edit: 2012-11-21, 13:48:38 by Farmhand »
   
Group: Guest
I like stuff like this Farmhand,

Looks like you have account for everything from primary resistance loss to secondary.  You could play around with frequency and loading to see if you can get better number.  As long as you are not reaching MHz or something, I think measurement is good.  60% efficient is poor so we need to learn more about underunity. lol  .00046/.00043 is also curious. You'll become a measurement pro in no time.  :)
   
Group: Guest
OK, this time I get a different and I think more accurate result,

With 143 mV across the primary and 101 mV across the 10.2 Ohm resistor in series with the primary, and a phase difference of 27 degrees.

I get 0.0099 A primary current and 0.001 Watts dissipated by the 10 Ohm resistor. And I get 0.003366 Watts (apparent power) I think.
The real power is 0.00299 Watts after multiplying the apparent power by the power factor of 0.891. So input is 0.00299 Watts

Output is 0.000955 Watts as 313 mV across the 327.6 Ohm resistor.
So the power dissipated by the load resistor (327.6 Ohms) added to the power dissipated by the 10 Ohm series resistor is 0.001955 Watts.

This leaves a difference of  0.001035 Watts which must be losses involved in the core and copper and/or still measurement/calculation error.

Efficiency would be 65.38 %.

Maybe that is better. More accurate.

Cheers  

P.S. It would be very easy if using a wire wound power resistor and very low power levels to pull the trick I show in this video clip.  :D  http://www.youtube.com/watch?v=_By-pWkEdVQ
With very low power levels and a bit of fiddling that trick could look convincing for 1000's of % OU, at least it might look that way to some, if the voltage across the open resistor was kept low
and the resistor a high wattage one relatively to the voltage applied there would be no way to tell except to inspect the actual resistor and arrangement, unless it was somehow shown.

Maybe I could do a demonstration, any requests ? How many thousand % OU do people want to see in the mW to Watts range ?
Unfortunately I threw my modified resistor away and they are expensive, but I will do it if someone wants to see it.

..
« Last Edit: 2012-11-21, 15:34:57 by Farmhand »
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
I had a good chuckle from your video FH, thanks.

I hope not too many of these folks are actually trying to fool us. I truly believe most often the case is that they are simply ignorant of what they are doing and observing, and jumping to preposterous conclusions.

Nice touch with the rooster in the background. ;)  :o
   
Group: Guest
 ;D

I better watch out for you now Farmhand. lol 

Okay, assuming you didn't fake your measurement.  I got a different apparent power than you do.  I multiply current through the 10 Ohm by 143mV and get .001415.  Power factored it and get .00126 watts of real power.

The output I got is .00299.  I think the .0000955 is the output current. 

So total heat you got is .001299 and input is .00126. 

I don't think you have room for other losses.  :D
   
Group: Guest
Hi Gibbs, It looks like you are correct. I made another mistake.  :-[ As I said I am very much a beginner at this, I haven't done calculations like that since I was in college 20 years ago.

So it still measures OU, hmm. 0.00126 input and 0.001299 output. I guess I'll need to video it and make a dodgy claim, hehe just kidding.

But I will video it so I might be told if I am doing the measurements correctly, I'm using 1/4 Watt metal film resistors so they can't be modified.  ;)

Still it is way to close to say it isn't a measurement error. But I'm learning fast by doing it with some feedback, it's like having many teachers. If my mistakes can be pointed out I very much appreciate it.

I guess weather or not people like to be told their mistakes depends on if the person wants to learn or just wants to make claims.

Easy to see for me though that measuring OU is Easy with low power levels, the first transformer I picked up shows it in measurement.

I think you are missing a Zero from the .00299 output measurement I think it should be .000299 for the 327.6 Ohm resistor then add the .001 for the 10 ohm resistor heat
= .001299 total heat and .00126 power factored input a very small amount so I would call it 98 % efficient. Anything that close to 100 % efficient is likely just under in reality.

Thanks Gibbs.
Cheers
   
Group: Guest

The intention is to get you pro in measurement and I think you did great.  Out of college for that long with this kind of strength would list you among the greatest in my OU journey.   Everyone I encounter has special ability that I could learn. 

You pointed out my mistake too so we can call it even. lol  Don't be too cynicism about OU result.  If it's OU, it's God's will.  If it's not OU, I guess it's also God's will. lol 

Cheers
   
Group: Guest
I'll try a different transformer, it looks like we finally came to a reasonable result.  O0

I used 1/4 Watt metal film resistors, they can't be modified.  ;)

Thanks again for the correction. Much appreciated.

Cheers
   
Group: Guest
So I take it the theory is that with a 90 degree phase shift between the primary current and voltage the power factor would be zero which would mean all input power is being returned, I doubt very much the phase difference could remain at exactly 90 degrees while there was an output from the secondary over a reasonable period of time.

Even a very slight repetitive deviation from 90 degrees would mean energy was consumed.
I would measure the input to the function generator but I ruined my two cheap kilo Watt meters by trying to measure the input power to my Tesla coil.
New one is coming.

Oh and my scope is a RIGOL DS1052E I'm just learning how to use the ultrascope software on the laptop to control the scope, lots to take in. I don't understand how I could use the math function of the scope to measure power. I don't want to go too far off topic, maybe I should search a more appropriate thread to post in about learning power
stuff and scope uses.

Cheers 
   
Group: Guest
@fh

Careful measurements and calculus, you are following the right method. As you didn't make this exercise from the college 20 years ago, I'm impressed by your motivation O0!
Imho, even if the measurement process is perfect, a measurement uncertainty is the likely cause of apparent overunity while the output doesn't exceed about 120% of the input power. I already fell in this trap. Resistances, capacitors, measurement apparatus, have uncertainty that is even often exceeding the specifications, question of component aging and lack of calibration. Another cause is the signal shape which can contain a spectrum outside the frequency band of the measurement apparatus, even if it is RMS. My advice is to take a 25% margin before we can expect for some anomaly.

   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
From FH

Quote
I used 1/4 Watt metal film resistors, they can't be modified.  Wink

 I had summer employment in a precision resistor factory, 1965. Raw resistor stock had deposited metal film to a resistance below what we desired. We used special machines that cut the film up to a precise value, but in a spiral which adds a tiny amount of inductance (nH).

You can do this manually. I modify them up to a precise value by using a small file and abrade away some of the metal film. Best not to change the value too much as you will reduce the power handling capability.

For low ohm shunts I wind non inductively with  14 or 16 gauge Manganin or use carbon rod stock in a divider to cancel thermal drift.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Guest
...
You can do this manually. I modify them up to a precise value by using a small file and abrade away some of the metal film. Best not to change the value too much as you will reduce the power handling capability.

I like this speech  :). Where there's a will there's a way, even for making our own components. It reminds me the time when I filed quartz crystals from army surplus to increase their frequency up to values interesting for me.

Quote
For low ohm shunts I wind non inductively with  14 or 16 gauge Manganin or use carbon rod stock in a divider to cancel thermal drift.

Yesterday I have tried to make a resistance of 0.1Ω with a carbon rod from an old battery, to make a current measurement that Verpies asked in the DSRD thread. I failed. The resistance was unstable, very depending on contact points, their surface and the tightening forces. Your other solution of a resistive wire is surely preferable.

   
Group: Guest
Hi Guy's, of course anything can be modified, I just meant not without being obvious, not opened any way .

Anyway Mr Clean is being evasive again, I think in his last video he might have measured the power consumed by one LED and called it input.
He seems to have responded to the requests to clear the matter up by getting some fancy equipment and measuring the power consumed by one LED across the primary as the input, if there was an LED in parallel with the primary I would call it a load If the LED is in series with the Primary I still think he is just measuring the power consumed by that LED the same way as he measured the secondary loads. Doesn't the current through the primary need to be multiplied by the voltage measured on the supply side of the current resistor ?

It all seems a bit strange, like to most people the most important thing is the hype. Actual OU seems unimportant to them, are they all mind controlled or something. Am I crazy. What also seems strange is the way he comes up with the equipment.

http://www.youtube.com/watch?v=UR7Ul5J_HcM

That thread on "that" forum has almost a million views on it and 90% of the info in it is hogwash.

Does anyone have an opinion on the way Mr Clean measures the input power in the latest video ? I'm disregarding everything in the video except the way he measures the primary input power. The outputs he measures across LED's and the wave form is irregular but the primary is sine wave. He doesn't seem to want to give schematics even when they are so simple. I don't get how he is measuring the primary power.

Ahah maybe he is bypassing the current probe with the LED and is only measuring what is not bypassed. meaning the power through the LED is not measured.

Cheers

P.S. Where is the 32 Ohm resistor the scope is set to for WRMS ? IS the resistor inside the scope or probe or something ?
How much is the resistance of a conducting LED ? Does the resistance of the LED vary with different applied voltages ?

Video added.


...

« Last Edit: 2012-11-23, 09:01:16 by Farmhand »
   
Hero Member
*****

Posts: 2600
@Exn
Quote
Careful measurements and calculus, you are following the right method

If no mistake have you made, yet losing you are ... a different game you should play.---YODA




---------------------------
Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
Group: Guest
@Exn
If no mistake have you made, yet losing you are ... a different game you should play.---YODA

Here is the difference between the two cases:
by following "careful measurements and calculus" you know that you are a loser until the victory that your method will show outside any doubt, otherwise you are a loser who always believes he is a winner. Reality or fiction, choose your camp, Comrade!  ;)

   
Hero Member
*****

Posts: 2600
@Exn
Quote
Here is the difference between the two cases:
by following "careful measurements and calculus" you know that you are a loser until the victory that your method will show outside any doubt, otherwise you are a loser who always believes he is a winner. Reality or fiction, choose your camp, Comrade!

I understand your response however for me the quote has another meaning which is why I posted it, maybe a little clarity is needed.

Quote
If no mistake have you made
--You have been perfectly correct in most all of your posts and they are in complete agreement with science ...  you are correct.

Quote
yet losing you are
-- You are doing everything correct and yet you have made absolutely no progress and are no closer to solving the energy crisis than anyone else even if they are completey wrong.

Quote
a different game you should play
-- An intelligent person might conclude, I am right but obviously I must be doing something fundamentally wrong, maybe I have to try something different because I know as a fact that what I am doing is not working. How can you be more right if you are no better than the people you are accusing of being wrong?, you see any claim requires proof and the proof is called progress. Just because you’re not doing anything wrong doesn’t mean you’re doing anything right.

AC


---------------------------
Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
Group: Guest
@fh

Careful measurements and calculus, you are following the right method. As you didn't make this exercise from the college 20 years ago, I'm impressed by your motivation O0!
Imho, even if the measurement process is perfect, a measurement uncertainty is the likely cause of apparent overunity while the output doesn't exceed about 120% of the input power. I already fell in this trap. Resistances, capacitors, measurement apparatus, have uncertainty that is even often exceeding the specifications, question of component aging and lack of calibration. Another cause is the signal shape which can contain a spectrum outside the frequency band of the measurement apparatus, even if it is RMS. My advice is to take a 25% margin before we can expect for some anomaly.



Taking a 25% margin is a large one.  I advice everyone reporting anything above 100% even if you think it's out of calibration, then we'll follow calibration (if any) and update. 



   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
FH,

Unfortunately, unless you are a member of EF, you can not view the attached pictures there. I was a member at one time, but they deleted my account. I have asked to have it re-registered, only to get a deafening silence in response.
   
Group: Guest
Taking a 25% margin is a large one.
...

I agree. It's not the acknowledged rules of the art in physics, but most of us have not at our disposal, professional means nor a full competence for doing proper measurements. My margin is only an expedient, a stopgap.

Nevertheless if someone want report just a bit more than 100%, I suggest that before troubling every one he makes another measurement with a different method and discards the results when they don't fit each other (this should eliminate near 90% of the errors) and he should measure each component and calibrate his measurement apparatus (this should eliminate 99% of the errors).

   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
From Ex:
Quote
Yesterday I have tried to make a resistance of 0.1Ω with a carbon rod from an old battery, to make a current measurement that Verpies asked in the DSRD thread. I failed. The resistance was unstable, very depending on contact points, their surface and the tightening forces. Your other solution of a resistive wire is surely preferable.

Agreed, using carbon rod from a battery is problematic. I used pure carbon rod with screw copper clamps and in a 2:1 voltage divider to cancel thermal drift. I just wanted to see how well it would work. I have some heavy gauge Manganin for high power shunts.

Yes the Manganin wire is a better approach.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Guest
I agree. It's not the acknowledged rules of the art in physics, but most of us have not at our disposal, professional means nor a full competence for doing proper measurements. My margin is only an expedient, a stopgap.

Nevertheless if someone want report just a bit more than 100%, I suggest that before troubling every one he makes another measurement with a different method and discards the results when they don't fit each other (this should eliminate near 90% of the errors) and he should measure each component and calibrate his measurement apparatus (this should eliminate 99% of the errors).



I'm thinking we can start up some research in this area.  We'll do in depth trend analysis from collected data.  The purpose is to check for margin error between theoretical and real time data,  education, anomaly check etc...  Let's start with driving a single inductor with sine waves.  Data is collected on frequency, resistance, coil spec, core type, phase, efficiency ...etc..  The format is like this:

Experiment type: (Scope/sim , sim model...etc..)
Coil spec:  (inductance, turn, diameter, resistance..etc..)
Core type:  (air, iron...)
Series resistor value:  (1, 10 ohms...)

Frequency                           Vsource (RMS)                   Vresistor(RMS)                Phase            efficiency

_________                         _____________                  ___________                _____            __________
_________                         _____________                  ___________                _____            __________
_________                         _____________                  ___________                _____            __________
_________                         _____________                  ___________                _____            __________
........


I think we can also see how good your scope is. lol  I think all sim model would give out the same data under the same condition but it's also good to check.   Core loss is ignored on this.
   
Pages: 1 2 3 [4]
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-03-28, 09:11:28