Stepping Down a Wimshurst

[gauschor]

Hmmm, looking at this picture it should be possible to create an influence device with the help of magnets, with no touching/frictional contact necessary like in the wimshurst. I'll have to try that out again.

[gauschor]

Hmm, just informational post... I mentioned some posts before Everts electrostatic power generator. I build his device but eventually found out that it will not work, except you have a hi-tech laboratory on your side. Evert claims that the Ether pressure can push electrostatic charge in his rotor device (see his website for more detailled instructions) to create enough current for common usage (current is the thing we miss on all the Wimshurst devices).

In a documentary I have seen that an "Ether pressure" is measurable and happens if you manage to put 2 very very polished plates opposite to each other by using a distance of only micro or nanomillimeters between them. Unfortunately no hobbyist can achieve this small distance, not to mention how precise the rotor and its ball bearings must be that it will not touch the other plates (regarding Everts device). I would say it is impossible to build for mainstream and even a headache for research.

Back to the Stepping down of a Wimshurst: I tried some more things but none gave reasonable or usable results. Yesterday I stepped upon the "Magnet Resonance Amplifier" http://peswiki.com/index.php/Directory:Magnetic_Resonance_Amplifier
I wonder if the Thestatika uses this kind of feature in the middle of their cylinders to create power: a piezo crystal which maybe is pulsed by the electrostatics instead of the high audio signal source? Just a thought... maybe the Methernita also used magnets the way it is shown on the sketch...

[wojwrobel]

Hello

im still sticking to my "charging by induction" video and after giving it some time to settle in my heat i came to this idea (see sketh)

now the question is how big potential sould we use as exiter (rotor) and how big the sectors sould be ? and the szie of machine to make some usable power?

[Steven Dufresne]

Hi @wojwrobel,
The amount of current will depend on at least three things (ignoring the A, losses): the voltage you supply (V), the capacitance between opposing sectors during Cycle 1 (C), and the speed of rotation (T, sort of).

For the capacitance, when two sectors are facing each other they form a plate capacitor, so you can use the standard formula for a plate capacitor to figure that out:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

Once you have the capacitance you can then work on how much current you want. This helps to choose the voltage you'll need to get that current. For a capacitor, C=Q/V, where Q is the amount of charge on each plate in units of coulombs. Rearranging the formula for the voltage we get, V=Q/C.

Current (I) is charge per second in the wire, so I=Q/T. T is time and that's where the speed of rotation comes in. Each time two sectors face each other (Cycle 1) you'll have made Q amount of charge pass through the wire and each time they're no longer facing each other (Cycle 2) Q amount of charge will have gone the other way through the wire. Let's say that only the positive disk rotates and that there are 20 sectors on it. If it rotates at 150 RPM, that's 150 / 60 seconds/minute = 2.5 rotations per second. Multiply by the number of sectors, 2.5 * 20 = 50 sectors/second. Multiply by the number of cycldes, 50 * 2 = 100 cycles/second. We invert that since T is really seconds/cycle (time). So T is 1/100 = 0.01 seconds/cycle.

So first use the formula for capacitance to select your dimensions (area of each sector, distance between opposing sectors, dielectric constant of material between opposing sectors), and then you'll have C.

Choose desired current in amps, I. Choose a rotation speed and work out T like above. Rearranging I=Q/T for the charge, we get Q=IT. So now you know Q. We already saw that V=Q/C. So now you know V. Next, realize that V is stupidly high so go back and choose new values for things until you have a reasonable V.

Something like that anyway.

Oh, make sure you insulate everything very well, otherwise you'll have huge losses due to ionization, decreasing V, and you won't get anywhere.
-Steve
http://rimstar.org   http://wsminfo.org

[wojwrobel]

Hello Steve

OK i did a model on paper for 1m in diameter disks just positive potential

so i made just 8 sectors on it because i think it don't mater how many sectors its still the same capacitance right?

so 8 sectors each have 460 square centimeter so total or my capacitor is 3680 square centimeters which is 0,368 square meter

its made of two disks 1mm each with air bearings , it means that in between sectors are small holes on angle so a bit of air goes in between plates but just a little to reduce friction

anyways its 2 times 1mm so spaced apart 2mm which is 0,0002 m

OK since we have 2 x 1 mm plexi i will use 50 kv as potential and plexi dielectric strength is 30kv/mm so it souled be ok, i know its a lot but there is no other way except going even bigget with disks but 1m is enough

so our capacitance of all sectors is accordind to website that you have provided
is 0.055390623 uF with k for plexi 3.4 if you want to use glass i think its better because k for glass is 6-10 so when 10 it would be 0.16291359999999996 uF so it would be tripled !!!!!!  here is list of dielectric i used http://hyperphysics.phy-astr.gsu.edu/hbase/tables/diel.html#c1
0.055390623uF = 0.000000 055390623F

OK so my capacitor will create charge 1C=1F x 1V  so
0.000000055390623 x 50000 = 0.002769531115 C

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?

well 1C = 1A x 1s so 1A=1C/1s

i don't know lets say its right so this only shows the charging cycle and what about discharging? cycle 2 so it souled be 16 x 0.0027695= 0,044312 C/s = A = 2215W ??
1200 rpm = 20 rps = 20 x 1107 =  22140W ??
what you think ?

wojsciech