Overunity.com Archives is Temporarily on Read Mode Only!



Free Energy will change the World - Free Energy will stop Climate Change - Free Energy will give us hope
and we will not surrender until free energy will be enabled all over the world, to power planes, cars, ships and trains.
Free energy will help the poor to become independent of needing expensive fuels.
So all in all Free energy will bring far more peace to the world than any other invention has already brought to the world.
Those beautiful words were written by Stefan Hartmann/Owner/Admin at overunity.com
Unfortunately now, Stefan Hartmann is very ill and He needs our help
Stefan wanted that I have all these massive data to get it back online
even being as ill as Stefan is, he transferred all databases and folders
that without his help, this Forum Archives would have never been published here
so, please, as the Webmaster and Creator of these Archives, I am asking that you help him
by making a donation on the Paypal Button above.
You can visit us or register at my main site at:
Overunity Machines Forum



Stepping Down a Wimshurst

Started by Foggy-Notion, December 30, 2009, 01:04:27 AM

Previous topic - Next topic

0 Members and 2 Guests are viewing this topic.

Steven Dufresne

Quote from: wojwrobel on April 22, 2010, 03:05:07 PM
so 8 sectors each have 460 A centimeter so total or my capacitor is 3680 square centimeters which is 0,368 square meter
...
so at 1 round per seconds it souled create 8 x 0.0027695 = 0,022156 C/s = A ??? that's a lot, is it right? so its 0.022156 x 50000 = 1107 W ??? well 8 sectors and full round so it will create 8 charges right?

You multiplied by 8 twice, once to get the area and then again to calculate current. So the last part quoted above should be 1 x 0.0027695 = 0.0027695 C/s = A, for a power of 0.0027695 x 50000 = 138 W. When you include the discharge cycle too that's 276 W.

I can't see any flaw in the logic. You get a lot by having such big area with such a small distance between plates i.e. a high capacitance. If you can build it then more power to ya (pun intended)  ;D

And of course you'll need at least 276 W for the motor and HV power supply to power it. I think it'll be the capacitive reactance that'll be your biggest loss: Xr = 1 / (2 x pi x frequency x C). So Xr = 1 / (2 x 3.14 x 1 x 0.000000055) = 2,893,726 ohms. That'll be the resistance your power supply will feel to charge the disk each cycle. Depending on your power supply it may have trouble maintaining 50000V during charging.

Hmmm. But if you make the inductive reactance of your coil and load also have the same value then you'll have done impedance matching and have resonance. Though they're in series. I know better for parallel coil and capacitor.
-Steve
http://rimstar.org   http://wsminfo.org
He who smiles at lofty schemes, stems the tied of broken dreams. - Roger Hodgson

wojwrobel

hello

I have to disagree,

1 round per second - means full round of the disk , so our capacitor will form 8 times per round=second

and by meaning "that's a lot" i meant the power generation at full speed!

let say 1200 rpm = 20rps = 20 x 138 W = 2760W since there is no opposing magnetic field like in regular generators it dont need so much power for motor so let say we use 300W for driving and 160W for supplying HV and 300W for loses so we still have 2000W

and now imagine if you have many generators like this in series it takes 5mm each so in 1m you will have 200 units, a little power plant 1m by 1m that produces  400KW of power in just a "charge mode"

or 800KW in both modes that's a lot

and if I'm right that we A 8 capacitor at 1rps its even more!! eight times more!!

i don't think that anybody can do better!

wojsciech

gauschor

I don't want to be a partypooper, but your calculations in range of Kilowatts are simply unrealistic. What you will receive is in range of milliWatts at best, nothing more, because the impulse your pickup coil receives is too weak. I am not joking, and you will see it by yourself, once you've build your own working device...

Steven Dufresne

wojwrobel,
You say your capacitor will form 8 times per second but 8 times which capacitor? You're mutiplying 8 times an amount of charge that's based on a 0.368 square meter capacitor. But 0.368 includes all 8 capacitors.

gauschor,
I wouldn't doubt the calculations are missing all the negative aspects that'll make it output only milliwatts - though I tried to find some with capacitive reactance. Assuming he could charge up the sectors fully, the motor then has to rotate the sectors away from each other. Ever try to turn a charged Wimshurst machine with the collectors not collecting? It's as if the disks are emersed in thick syrup. The attractive forces between opposing sectors are strong, espectially these large sectors with 2mm spacing. In this case the positive sectors have no "collector" and the negative sectors have a coil with inductive reactance to work against. So I think you're right. Either the output won't be much or the output will be significant but the input will be even more.

But I say, go for it. Sounds like a fun thing to try and you'll learn a lot which can be used toward your next idea.
-Steve
http://rimstar.org   http://wsminfo.org
He who smiles at lofty schemes, stems the tied of broken dreams. - Roger Hodgson

wojwrobel

hello

ok i made a picture to explain, but with 4 sectors, so every 1 round you form 4 capacitors that are total of 4 sectors

imagine that under white paper is positive potential thats charge cycle and between white papers is discharge.