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Author Topic: Apparent C.O.P. of 1.413469133935024 ...  (Read 11436 times)

Offline DeepCut

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Apparent C.O.P. of 1.413469133935024 ...
« on: November 30, 2010, 02:04:49 PM »
Hi.

My latest test results :

INPUT VOLTAGE : 18.03 VDC

INPUT CURRENT : 0.15 A

INPUT WATTS  = V*I = 18.03*0.15 = 2.7045 W


LOAD : 220 Ohms


OUTPUT VOLTAGE : 29 VDC

OUTPUT CURRENT : V/R = 29/220 = 0.1318181818181818 A

OUTPUT WATTS = V*I = 29*0.1318181818181818 = 3.822727272727273 W


APPARENT C.O.P. = 3.822727272727273/2.7045 = 1.413469133935024

I have a call-out to attend to, when i get back this evening i will try to self-run.


Thanks,

Gary.



Free Energy | searching for free energy and discussing free energy


Offline akunkeji

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #1 on: November 30, 2010, 03:03:10 PM »
What device? Have photos?

Offline FatBird

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #2 on: November 30, 2010, 03:21:37 PM »
Sounds Good DeepCut.

Can you please post some photos and a diagram?

Thanks.

.

Free Energy | searching for free energy and discussing free energy

Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #2 on: November 30, 2010, 03:21:37 PM »
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Offline DeepCut

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #3 on: November 30, 2010, 08:12:25 PM »
Back from work now.

Got to eat then stick the DPDT on and go for self-running.

There's only a 27 ohm difference between the load that's on now and the load that equals the input so i'm hopeful.

It's freezing here in London and once again we are seeing all the old-age pensioners sitting in their homes wearing loads of clothes and gloves because they can't afford extra heating on the government pension.

Makes me sick.


Later,

Gary.

Offline DeepCut

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #4 on: November 30, 2010, 09:37:37 PM »
OK, self-running blew the transistor :(

I'm using a DPDT switch for the self-run test and i added a cap which increased output voltage to around 330 VDC.

Does anyone know if there is a 'standard' and safe way to connect for a self-run test ?


Thanks,

Gary.

Free Energy | searching for free energy and discussing free energy

Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #4 on: November 30, 2010, 09:37:37 PM »
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Offline DeepCut

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #6 on: November 30, 2010, 10:22:38 PM »
Thanks broli i'll try that tomorrow.


Gary.

Free Energy | searching for free energy and discussing free energy

Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #6 on: November 30, 2010, 10:22:38 PM »
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Offline gyulasun

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #7 on: November 30, 2010, 11:48:35 PM »
Hi Gary,

I fully agree with broli to use a voltage regulator against the run-away situation.
Would like to suggest an off-the-shelf regulator that includes all the neccessary circuits shown in the link, it is produced by most big analog IC manufacturers. It is known as the three terminal 7818 linear voltage regulator, like LM7818, MC7818, uA7818 etc. Gives out exactly 18V at any current up to maximum 1A, seems fully cover your input requirement of DC 18V and .15A current.
Here is a link for a data sheet at random, out of many:
http://www.datasheetarchive.com/MC7818-datasheet.html 

One thing to consider:  linear voltage regulators waste power because of the voltage difference between their input and output. Just think about it: they need a minimum of 2.5V higher voltage than the output 18V to be able to work correctly, they are designed like that. (there are special low drop out regulators too that need only .2 or .3V voltage higher than the output but now you have 29V-19V=10V difference (when you load the output with 220 Ohm resistor) and that is what counts now. So the regulator will have a 10V times .15A=1.5Watts loss in heat form inside it, use a small meatal heat sink to defend it from overheating. You will have to consider this loss of course when estimating total power, later if looping is a success, we can discuss how to reduce this loss.
The 7818 regulator IC (any make) can work with a maximum of 35V so your 29V is perfect input voltage for it.
I suggest using still at least a 360 or 420 Ohm loading resistor across your 29V output and then connect the input of the 7818 regulator IC and only then close the loop  i.e. to connect the 7818 output to the input of your circuit, to replace the power supply.  Use a diode 1N4001 or similar in series with the positive output of your power supply voltage to separate  the 7818 output from the power supply voltage. This way the two outputs cannot get connected in parallel, the diode blocks the supply to load the output of the 7818. Hope this all is understandable.

Gyula

Offline Omnibus

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #8 on: December 01, 2010, 12:04:07 AM »
@DeepCut,

I would also like to know how you achieved this. Obviously it is some part of a research you've discussed elsewhere which I've missed. Could you please elaborate?

Free Energy | searching for free energy and discussing free energy

Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #8 on: December 01, 2010, 12:04:07 AM »
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Offline DeepCut

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #9 on: December 01, 2010, 01:09:29 AM »
Gyula, thankyou.

I will grab this one tomorrow :

http://www.datasheetdir.com/TS7818CZ+download

The specs seem appropriate but please correct me if i'm wrong.

I'll set that all up tomorrow and then post results.


Thanks, Gary.

Offline TinselKoala

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #10 on: December 01, 2010, 01:23:20 AM »
Hi.

My latest test results :

INPUT VOLTAGE : 18.03 VDC

INPUT CURRENT : 0.15 A

INPUT WATTS  = V*I = 18.03*0.15 = 2.7045 W


LOAD : 220 Ohms


OUTPUT VOLTAGE : 29 VDC

OUTPUT CURRENT : V/R = 29/220 = 0.1318181818181818 A

OUTPUT WATTS = V*I = 29*0.1318181818181818 = 3.822727272727273 W


APPARENT C.O.P. = 3.822727272727273/2.7045 = 1.413469133935024

I have a call-out to attend to, when i get back this evening i will try to self-run.


Thanks,

Gary.

Before we go much further, let me please point out to you that when you say:
Quote
APPARENT C.O.P. = 3.822727272727273/2.7045 = 1.413469133935024
You are saying it is NOT 1.413469133935025 and NOT 1.413469133935023...that is, you are giving ridiculous precision that you in no way can support with actual measurements.

Please let me remind everyone that your final answer CANNOT POSSIBLY be more precise than your LEAST PRECISE measurement.

That is, in the above experiment, DeepCut, you have measured your input voltage and current to 2 decimal places. This means your intermediate results and your final answers cannot be more than 2 decimal places in precision --- even assuming that your input data is correct in the first place. (Have you calibrated your meters?)

Please, quoting precision to all the digits that your calculator spits out only assures you of 2 things: 1) that you are wrong (that is, the true value is certainly NOT 1.413469133935024), and 2) people like me, who do this stuff for a living, pay less attention to you because we know your numbers aren't credible.

Now, if you like, we can address the issue of your apparent COP of ABOUT 1.41. First, though, perhaps you would like to check up on the difficulties associated with power measurements in oscillating circuits, particularly using consumer-grade DMMs.

For an example, I have a device here that draws 5 to 8 amps at 120VAC from the wall, and produces over 250 kV with a continuous plasma arc that can carry well over that 9 amps. Using your method of simple calculation of COP, what do I get? Does it mean I've made free energy?
Maybe...
http://www.youtube.com/watch?v=PdOhIoA1Z-k&feature=related

Free Energy | searching for free energy and discussing free energy

Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #10 on: December 01, 2010, 01:23:20 AM »
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Offline DeepCut

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #11 on: December 01, 2010, 01:36:26 AM »
@Omnibus

Hi Omni, you haven't missed it, just forgotten a previous discussion we had :)

Here's  a process-list :

1. 18 VDC PSU powering a Bedini-circuit.
2. Bedini circuit linked to the usual bifilar-wound 'drive' coil.
3. A small iron rod inside the drive coil.
4. A diametrically-magnetised cylinder magnet mounted horizontally above the drive coil on a carbon-rod axle.
5. An output coil consisting of 1.15 kilometres of 0.25mm copper magnet wire, this coil is on a separate former that is larger than the drive coil's former and so can be placed around the drive coil.

I suppose it's a pulse-driven induction generator, here are some pictures, please excuse the mess :

http://qvision.pwp.blueyonder.co.uk/pic1.JPG

http://qvision.pwp.blueyonder.co.uk/pic2.JPG

http://qvision.pwp.blueyonder.co.uk/pic3.JPG

My new analogue meter should arrive tomorrow or Thursday then i can measure the current directly also.


Thanks,

Gary.





Offline DeepCut

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #12 on: December 01, 2010, 01:39:52 AM »
Hi Tinsel.

Yes, please excuse my ignorance and shoddy methods, i am learning all the time.

Where i have used only a couple of decimal places is because there was constant variance (!) and i took an average.

*EDIT* and my PSU only goes up to 2 decimal places. *EDIT**

I hope to get a scope next year and to grow into a more mature experimenter.


Thanks,

Gary.

Offline TinselKoala

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #13 on: December 01, 2010, 01:53:54 AM »
Keep up the work, Gary, and don't worry about it. And especially don't let it get you down.
There's a lot to learn, electronics wise, and I'm just a chapter or two ahead of you. I suffer from a traditional education, though, and I even see the merits of a lot of it, so if I get a bit pedantic about things like significant digits and power measurements...well, just consider the source, take what you need and leave the rest.

I'd like to see the circuit you are working with, and I know others would too.

cheers--
--TK

(Often, believe it or not, a good ANALOG meter like the legendary Simpson can give you a better reading (that is, a more honest and stable average) than a typical DMM on spiky oscillating signals. If you can find one surplus, it's a great piece of kit for the FE experimenter. Of course, a fast oscilloscope is even better, but more costly and has a steeper learning curve...)

EDIT to add: OK, I see the description. Thanks ! Good luck.
« Last Edit: December 01, 2010, 02:15:10 AM by TinselKoala »

Offline DeepCut

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Re: Apparent C.O.P. of 1.413469133935024 ...
« Reply #14 on: December 01, 2010, 02:15:23 AM »
Thanks TK.

I think you're probably a few kilos of volumes, rather than chapters, ahead of me !

I suffer from a lack of education, sometimes that has it's merits ;+}

The circuit is the standard SSG circuit, in the black box in the pictures i posted a couple of posts above this one.

I ordered this today :

http://www.rmcybernetics.com/files/pdf/PWM-OCXI.pdf

So i'll probably be offending your electronic sensibilities with some ridiculous coil-pulsing-related posts in the very near future.


Thanks,

Gary.


 

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