Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

User Menu

Custom Search

Author Topic: GENERATOR- YOU DO THE IN/OUT POWER MATH  (Read 72296 times)

FatChance!!!

  • Full Member
  • ***
  • Posts: 197
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #150 on: June 14, 2010, 11:41:23 AM »
INPUT : 9.57 VDC @ 0.83 A = 0.79431 W

OUTPUT : 19.8 VDC @ 0.22 A = 4.356 W

Your input calculation is wrong.
INPUT: 9.57 VDC @ 0.83 A = 7.9431 W
Efficiency: 4.356 W / 7.9431 W = COP 0.54 = 54% efficiency.

Conclusion:
You can save your money for better things.

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #151 on: June 14, 2010, 02:51:59 PM »
Hi.

Sorry i typed it wrong !

Should be .083 not .83.

Sorry i do that i lot i gota be more careful when typing.

INPUT: 9.57 VDC @ 0.083 A = 0.79431

I think we should ALL be replicating this since it's piss-easy to construct as well as cheap, and most of us probably have Bedini circuits lying around anyway.


Gary.

gyulasun

  • Hero Member
  • *****
  • Posts: 4117
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #152 on: June 14, 2010, 03:55:20 PM »
Gary,  you mentioned your output load is the DC resistance of a coil, correct?

Question: Do you have a full wave bridge (or a single diode) at the output and then you use a puffer capacitor and you load the capacitor with the coil?

Or you do not have a puffer capacitor, maybe you do not have a diode either? I wonder.

EDIT: just noticed you wrote DC output voltage so you surely had the diodes bridge, now the question is the puffer capacitor. 

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #153 on: June 14, 2010, 07:16:16 PM »
Hi Gyulasun.

I don't have a load on it.

I wnted to know, purely, the watts in vs the watts out without measuring current with a resistor load bla bla so i did what you said before, read the voltage with a DVM and calculated the current because the coil resistance is known.

I couldn't find the post where you previously suggested a puffer cap, am i right in thinking you put it across the bridge output and also what properties should it have (uF and V) ?

(I am in the process of making a new housing for the magnet assembly so that the entire coil is wrapped in the plane of the shaft to maximise induction, getting the wire tomorrow).

*EDIT* I have a 47uf/35v and a 100uf/10v to hand *EDIT*


Thanks.
« Last Edit: June 14, 2010, 07:56:21 PM by DeepCut »

gyulasun

  • Hero Member
  • *****
  • Posts: 4117
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #154 on: June 14, 2010, 09:49:38 PM »
Yes, you put the puffer cap in parallel with the diode bridge DC output, with the correct polarity.
It is ok that you calculated the output current but:

If you did not use a puffer capacitor when you measured your output DC voltage, then you cannot trust on the the measured 19.8V DC output because of the half sinewave series of output. The capacitor would smooth them out, use the 47uF/35V capacitor. A 470 or 1000uF would be much better but you do not have that at hand in at least for 35V WV.

There can be another "problem":  I would suggest watching the input current whether it changes to a higher value than the .083A WHEN you add the 90 Ohm load at the output (the puffer cap is already there too). Here I assume the 9.57V supply voltage remains stable.
IF the input current of .083A changes to a higher value, then note the new value and then check the DC output voltage across the puffer capacitor (which is equivalent with the output where you connect the 90 Ohm load and note the output voltage, then recalculate, ok.

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #155 on: June 15, 2010, 12:50:21 AM »
OK i put a cap on and the voltage counts down from 25 to 0.6 and statys at 0.6.

What am i doing this for and what does it mean LOL ?

p.s. it is the 47uf/35v cap.

Gary.

« Last Edit: June 15, 2010, 01:25:32 AM by DeepCut »

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #156 on: June 15, 2010, 02:18:55 AM »
OK my fault (as usual !) one of the coil ends wasn't contacting the rectifier properly.

I have rectified the situation (LOL !).

OK.

I have made some adjustments to the system so here is an appraisal of current setup :

1. A DC PSU is powering a Bedini circuit.

2. The Bedini circuit is driving a magnet and the drive coil is now air-cored.

3. The magnet is rotating inside an output coil.

4. The output coil goes through a full-wave bridge rectifier with a 47uF/35V puffer cap.

The coil resistance is 90 ohms.

Results :

INPUT : 9.53 VDC  @  0.070 A = 0.6678 W

OUTPUT : 36.8 VDC

The current is :

I=V/R = 36.8/90 = 0.408

So the output watts is 36.8 * 0.408 = 15.0144.

Apparent COP = 15.0144/0.6678 = 22.48 !

Some of you guys MUST have the time to explore this in parallel with Tom and i, it's just far too interesting not to, surely !

I am buying a DPDT switch or relay tomorrow, this thing has to self-run.

Also, Gyulasun, there is no change in voltage/current at the PSU when the cap is on.

Thanks,

Gary.

FatChance!!!

  • Full Member
  • ***
  • Posts: 197
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #157 on: June 15, 2010, 08:25:05 AM »
Wow, if you surely got COP 22 then a self runner is a must.
And hopefully a small video for us to see.
Otherwise you need to check where the measurement error lies.
Hopefully there is no error this time.

How are you gonna adapt the high output of 36V down to 9V input?
The best way is using a regular buck switch with an efficiency of 95%.
But maybe switched electronics is out of your league?
Then a standard linear regulator, type LM7810, for 10V output could
work fine due to the low input currents of 70mA involved.
http://www.fairchildsemi.com/pf/LM/LM7810.html

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #158 on: June 15, 2010, 01:11:27 PM »
Thanks Fatchance.

Yes i'm a noob with electronics but i can construct simple circuits from schematics.

I'm off to the shop today so will look for the right sort of switch.

I double-checked the figures and i'm sure it's all correct. It's hardly complicated so there's little room for error even for someone like me who rushes sometimes.

Will report back later.


Thanks,

Gary.

Bruce_TPU

  • TPU-Elite
  • Hero Member
  • *******
  • Posts: 1437
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #159 on: June 15, 2010, 01:56:12 PM »
Tom, who originated the idea also has overunity.  He and I write one another often.  He is soon to be working with much larger magnets.  I can not for the life of me figure out why more people are not interested in this.  Tom has given all of the details and Gary has sought to replicate.

Gary,
What size wire are you using for your pick up coil?  How many feet? 

Good work.  Now post some videos and pictures for the forum, please.

Cheers,

Bruce

FatChance!!!

  • Full Member
  • ***
  • Posts: 197
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #160 on: June 15, 2010, 03:48:56 PM »
I believe the reason to the lack of interest is the blurry information.
I haven't found any links to detailed build info and results.
Could you point me to good and specific info how to replicate this device?
I'd love to replicate if I had access to all the details without any hidden parts.

And I have the dough and and resources for a heavy duty replication.

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #161 on: June 15, 2010, 04:49:27 PM »
OK i've taken some pictures and i'll do a video after posting this. My video quality is crap but the pictures are high-res.

What you need :

1. A DC power supply. I have a PSU but obviously a battery could be used.

2. A Bedini circuit. But since we're not charging batteries here you just re-route the wires that would normally go to the battery that would be being charged. As i was able to do this i'm sure i don't need to explain how to anyone here but here's a picture of my old Bedini circuit with the rerouted wires, bear in mind this is not the setup under discussion but just a picture that had the Bedini circuit in it :

http://qvision.pwp.blueyonder.co.uk/bedini.no.charging.jpg

3. A drive coil. This is the small coil you see in the above picture. This coil is bifilar wound, the red wire is the trigger wire and the gold wire is the power wire. When you start the magnet rotating at first by hand, this causes a tiny charge in the red wire via induction, this then activates the transistor in the Bedini circuit to let in the main supply power which goes down the gold power wire and gives the coil a magnetic field strong enough to repel the magnet. My trigger wire is 0.25mm and my power wire is 0.28mm but there is a large tolerance when it comes to the wire gauge. I did 800 turns of both wires. I found an air ore works better than a ferrite core.

4. A rod-shaped neodymium magnet that is diametrically magnetised. This means that rather than North and South being at either end of the magnet, each curved side of the magnet is North and South, it is magnetised through it's diameter rather than it's length. This magnet also needs a hole through it's length so it can be mounted on a shaft. The hole in my magnet is 6mm, here is a picture :

http://qvision.pwp.blueyonder.co.uk/magnet.on.shaft.JPG

I copied Tom and use 6mm carbon rod for the shaft because it's light.

5. A pickup coil. I used 0.25mm wire, same as the trigger wire. My resistance is 90 ohms. I wrapped it around a cut-off pringles can and made holes in the can for the shaft to go through. Here is a picture :

http://qvision.pwp.blueyonder.co.uk/pickup.coil.JPG

I put insulating tape on the bottom of the holes to stop the shaft rubbing against the wire.

6. A full-wave bridge rectifier to turn the AC from the pickup coil into DC.

7. A capacitor across the rectifier output. Gyulasun called this a puffer capacitor. This gave a great improvement to the output.

I put my Bedini circuit into a project box just so it was tidy, here is a picture :

http://qvision.pwp.blueyonder.co.uk/project.box.JPG

The three wires sticking out of the top are, from left to right :

a. From collector of transistor.
b. From emitter of transistor.
c. From outer leg of potentiometer

I'll do a video now and stick it on youtube.


Thanks,

Gary.




mscoffman

  • Hero Member
  • *****
  • Posts: 1377
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #162 on: June 15, 2010, 05:00:37 PM »

...
I wnted to know, purely, the watts in vs the watts out without measuring current with a resistor load bla bla so i did what you said before, read the voltage with a DVM and calculated the current because the coil resistance is known.
...


@Deepcut

You have learned a lot I think, and this is good.

But I have problem with the above measurement scenario.

The output circuit need to be divided into two parts;
the coil that operates on the AC signal and the rectifier
bridge and cap filter that operates on DC signal.
What you are trying to do is use the DC resistance
on the AC side of the circuit and this is *incorrect*.
The resistance the AC signal sees is called (capacitive
or inductive) reactance. The coil is an inductor so
the resistance that the AC signal sees has to be the
inductance of coil combined with the frequency of the
AC signal, which is also the speed in REVS per Second
that the rotor is turning.

The only way to do what you are doing is to know the
resistance that is the same as effective resistance of coil
at the AC frequency. So you can't skip the full DC analysis
of the power. Once you have measured the true DC power
correctly -> then you can calculate the effective AC
resistance of the output coil at frequency and use
that in the future and I guarantee it is going to be significantly
higher resistance then the DC resistance of the coil.

Because you are using a significantly lower effective resistance
in your calculations the actual numbers you calculate are erroneously
showing more output power and therefore a higher COP than you
actually have. This doesn't mean you have no overunity COP as
there may or may not be some left.

Which is why self running device is the gold standard for overunity
energy production.

:S:MarkSCoffman

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #163 on: June 15, 2010, 05:31:55 PM »
Mark.

I don't like you anymore !

Just kidding :)

Damn i knew it was too good to be true ...

So, how do i do the proper calculations ?

Oh, and here is a link to the video, i've just put it up so it may take a minute or two to be there i don't know :

http://www.youtube.com/watch?v=a7LbB7FWfO4


Thanks Mark, for teaching :)

btw i found this, i think it's what you told me so i will learn it :

http://www.tpub.com/neets/book2/4a.htm

those NEETS books are great, nice one US Navy :)

Gary.

« Last Edit: June 15, 2010, 06:01:03 PM by DeepCut »

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #164 on: June 15, 2010, 06:07:19 PM »
Well thankfully the inductive reactance formula is simple.

I don't yet have an LCR meter though :(

I need to get one that measures frequency as well.

This free-energy isn't very free is it ? LOL ...

Can inductance be derived from other known properties maybe ?

Sorry, i'm posting when i should be googling !

I have a job to go to this evening so see you later tonight or tomorrow.


Thanks,

Gary.