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Author Topic: Stepping Down a Wimshurst  (Read 48136 times)

Steven Dufresne

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Re: Stepping Down a Wimshurst
« Reply #45 on: April 22, 2010, 10:02:30 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

wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #46 on: April 23, 2010, 07:29:16 AM »
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

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Re: Stepping Down a Wimshurst
« Reply #47 on: April 23, 2010, 02:23:25 PM »
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

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Re: Stepping Down a Wimshurst
« Reply #48 on: April 23, 2010, 02:55:07 PM »
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

wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #49 on: April 23, 2010, 05:36:44 PM »
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.

wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #50 on: April 23, 2010, 07:33:28 PM »
one more thing if we want to increase capacity we can change dielectric material to let say mica its 5,4 from plexi 3,4 gain of 60% of capacity!!!

wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #51 on: April 24, 2010, 09:47:46 AM »
hello

ok i recalculated everting and i made a mistake

2mm is not 0.0002 but 0.002 so then our C=0.01221852 uf 0.000000 01221852 F

so its 0.000610926 C
at 1rps =0.000610926 x 8= 0.004887408 A = 244W
at 1200rpm=20rps= 4880W cycle 1
                           
but this mistake makes me thinking if we can use some special film as isolator that have good dielectric constant and strenght it will make big difference in output!!!
wojsciech

gauschor

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Re: Stepping Down a Wimshurst
« Reply #52 on: April 24, 2010, 02:29:13 PM »
@Steven: True, I forgot that with the "thick syrup" maybe causing a stronger influencing and therefore probably causing a stronger induction effect in the pickup coil. This definitely needs a second look I guess.
@wojwrobel: hmm... I am eager to know how it will work out, since I can see you are actually building the device. It is still quite possible I have overseen something in my own experiments, also my construction was not the same as yours. It motivates to start experimenting again if one sees that other lay their hands upon it as well  :)

On a sidenote: beware... the most difficult thing I experienced during constructing devices like that is getting the discs not to wobble  :-\

Steven Dufresne

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Re: Stepping Down a Wimshurst
« Reply #53 on: April 24, 2010, 02:34:54 PM »
Hi wojwrobel,
What, you're not satisfied with 4880W or even 244W?  ;)
I'd suggest you try a small version first with what you have readily available, unless you're already experienced at this making these sorts of things.
-Steve
http://rimstar.org   http://wsminfo.org

Steven Dufresne

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Re: Stepping Down a Wimshurst
« Reply #54 on: April 24, 2010, 02:44:33 PM »
@Steven: True, I forgot that with the "thick syrup" maybe causing a stronger influencing and therefore probably causing a stronger induction effect in the pickup coil. This definitely needs a second look I guess.

Actually, what I was trying to get at was that his motor is probably going to do a heck of a lot of work just to turn the disks. But I'm not certain of that. I'm extrapolating from what happens with my Wimshurst machine, which is a bit of a different beast.

On a sidenote: beware... the most difficult thing I experienced during constructing devices like that is getting the discs not to wobble  :-\

Agreed - groan. Another reason for starting small if you're not experienced at making these things. 1 meter disks can be expensive if you end up having to make a few versions before you get it right.
-Steve
http://rimstar.org   http://wsminfo.org

mscoffman

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Re: Stepping Down a Wimshurst
« Reply #55 on: April 24, 2010, 08:46:33 PM »
@All

Somebody mentioned the number of microamps involved in a Wimshurst
spark...but this current would be at least at 10KVdc+ so that is a lot of
zero decimal points coming from the voltage side of the P=I*E equation.

At the time I calculated Watts to tens of Watts available.   

Don't forget, you can always add an "antenna" wire to your influence
machine device to cause it to affect more of its environment. This
is why I don't think an electrostatic machine like this obeys the
simple conservation of energy laws and is where the plexiglas Bedini
ten-coiler machines gets their extra battery charging ability.

Also back when they were using Wimshurst machines as a power
supply for x-ray tubes they would spin a number of disks on the
same shaft. Capacitive coupling will cut down on the available power
over DC brushes but has some benefits as well.

Don't forget, the target goal for a free energy generator is
~1.5KWatts continuous~ for household use - not megawatts.

---

Book: Modern High-Speed Influence Machines
circa 1922

http://ebook.lib.hku.hk/CADAL/B31428137/

:S:MarkSCoffman

wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #56 on: April 24, 2010, 10:07:50 PM »
hello

all this idea gets me to thinking

if we can charge by induction witout loss of our sorce so why dont we make something stationary that will work at the same princip...

maybe like this?

gauschor

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Re: Stepping Down a Wimshurst
« Reply #57 on: April 25, 2010, 12:43:58 AM »
Very interesting thought; The only problem I can see here is, how both negative & positive potentials are transferred alternatingly and in short time onto the first big capacitorplate.

In detail: at first you fill up the big capacitorplate on the left with a high negative potential e.g. with the help of a wimshurst device. If the capacitor area is big, it takes 1-3 seconds(!) to get it fully loaded. Then after you have loaded it, you need to reverse the potential quickly into a positive one. How could one do that?

wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #58 on: April 25, 2010, 09:49:51 AM »
hello

well i dont think that its gonna take 1-3sek, 1m x1m 2mm apart mica insulator k=7.5 is just 0.0332024 uf and with potential 50kv its gonna charge in ms it is not rc circut that charging time would depend from R.

and how to make sinus 50 kv?
i have think about that but i think mechanical setup plus wimshurt?
wojsciech

gauschor

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Re: Stepping Down a Wimshurst
« Reply #59 on: April 25, 2010, 12:35:41 PM »
It definitely takes more than a few milliseconds to charge it up. I would say at least a half or 1 second for a spark of 50kV+ which is definitely too slow for a power generation device; consider that common wimshurst devices have 2 big capacitors on them (leyden jars), which are nothing else than 2 metal plates for each capacitor. However these leyden jars need some turns of the wimshurst discs to get fully loaded. Only then they have enough different potential to create a strong lightning of 50kV+. If you remove the capacitors completely you can get a lightning arc, which however is only an estimated 2-6kV (from my experience). So in any case the charging procedure is too slow.

And then the problem of the alternation between positive/negative potential is still not solved yet. You could achieve that by shortcutting both negative & positive potentials and then build up only one potential afterwards. But this sounds very inconvenient and would take too much time either.

That said, I am still sure there must be a way to achieve that. Maybe somehow like in the image beloiw although I am afraid that I am missing some points here, or make some logical errors. At least I wouldn't know how to connect the wires to the disc segments if the discs are rotating... :/