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Author Topic: Stepping Down a Wimshurst  (Read 40445 times)

Offline wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #30 on: April 08, 2010, 08:18:18 PM »
hello

I'm new to this subject but i would like to add something

why don't we try to use this machine like in patent US4897592 ELECTROSTATIC ENERGY FIELD POWER GENERATING SYSTEM
http://www.freepatentsonline.com/4897592.pdf

this machine does exactly what we need changes static potential to DC voltage
and the most important thing is that energy required to generate power is just to create a static charge and rotate the disks ,

its something like in this video http://www.youtube.com/watch?v=o0hXYfvnND0
5:10 - 8:00  (this video is so old but every time i watch it it makes me thinking , this is basically free energy concept shown in very understandable way)

and the most important information is that there is no charge exchange so you charge this machine only once so all power it needs is just to rotate

and since there is no collapsing magnetic field like it is in regular power generators the rotation don't need so much power so you can connect many machines like that in series to get more power !!! 

cheers from poland
wojsciech

Free Energy | searching for free energy and discussing free energy

Re: Stepping Down a Wimshurst
« Reply #30 on: April 08, 2010, 08:18:18 PM »

Offline Steven Dufresne

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Re: Stepping Down a Wimshurst
« Reply #31 on: April 08, 2010, 09:33:35 PM »
why don't we try to use this machine like in patent US4897592 ELECTROSTATIC ENERGY FIELD POWER GENERATING SYSTEM
http://www.freepatentsonline.com/4897592.pdf

this machine does exactly what we need changes static potential to DC voltage
and the most important thing is that energy required to generate power is just to create a static charge and rotate the disks ,

Hi wojsciech,
I've tried it, but it's very hard to build properly due to the high rotation speed needed. Hyde said the effect didn't start until around 6000RPM. So far I've been unsuccessful. I'll try again in the future if I don't succeed with what I'm doing now. Here's my webpage about it:
 http://rimstar.org/sdenergy/hyde_generator
and here's the thread here at overunity.com where we've discussed it:
 http://www.overunity.com/index.php?topic=6790.0

its something like in this video http://www.youtube.com/watch?v=o0hXYfvnND0
5:10 - 8:00  (this video is so old but every time i watch it it makes me thinking , this is basically free energy concept shown in very understandable way)

and the most important information is that there is no charge exchange so you charge this machine only once so all power it needs is just to rotate

and since there is no collapsing A field like it is in regular power generators the rotation don't need so much power so you can connect many machines like that in series to get more power !!! 

I don't think this is the principle of the Hyde generator. I think he started out doing that but then at around 6000RPM he started getting 300kV spikes as output, which he then rectified to DC. These spikes weren't what he expected.
-Steve
http://rimstar.org   http://wsminfo.org

Offline wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #32 on: April 08, 2010, 10:47:06 PM »
hello

i understand this as machine that converts high electrostatic potential to lower potential dc ,

but this machine needs external source of electrostatic potential, positive and negative !!! often understand as positive and ground !! which is wrong negative is not a ground !!

like o said its all about CHARGIN BY INDUCTION 

and if you look closely to rectification (fig 6 in patent) it has dual variant rectifier 38A and 38B which meen that plates 82 and 84 will be charged one + second - and vice versa when there is position 72 and 76

so our external potential have to reach for + to disk 26 and 22 and - to disk 24 and 20
and its very far from disks 18 + and disk 16- which supply electric field thats why i A that voltage potential at disk 16 and 18 must be very big i would say 100-150 kv to make it work

please try to understand what im trying to tell you ok

wojsciech

Free Energy | searching for free energy and discussing free energy

Re: Stepping Down a Wimshurst
« Reply #32 on: April 08, 2010, 10:47:06 PM »
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Offline Steven Dufresne

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Re: Stepping Down a Wimshurst
« Reply #33 on: April 08, 2010, 11:51:43 PM »
I understand what you're saying. My ground was negative with respect to the positive - so at first I'd say it doesn't matter. However, you're right because the rest of the device and structure is at ground potential. So the ground/negative plate was 0 volts with respect to it's surroundings. That is a problem.

So a few months ago I modified my Cockroft-Walton voltage multiplier so that it has three outputs: HV+, ground, HV-:
 http://rimstar.org/equip/pos_neg_voltage_multiplier.htm
I tried it again but using the HV+ and HV- instead. Unfortunately my multiplier has ripple. I've found over the years that if you have ripple, even 1 or 2 volts, the ripple gets multiplied if there is ionization going on somewhere. I first learned that years ago with this experiment:
  http://rimstar.org/sdenergy/testa/potsmk1.htm
So this meant that even though my power supply was HV DC with small ripple, the device ended up with HV DC with very large ripple.

So when I try again I will fix my rotors and stators so there is no ionization -no sharp edges. Then the ripple won't matter as much. I might also try and use a different HV DC source that doesn't have ripple.

EDIT: Oh, and when Hyde got his 300kV spikes while the rotors were turning at 6000RPM he was using only 3kV across the exitor plates. This information is from Moray B. King's book "Quest for Zero Point Energy" from his conversations with Hyde.
-Steve
http://rimstar.org   http://wsminfo.org

Offline gauschor

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Re: Stepping Down a Wimshurst
« Reply #34 on: April 09, 2010, 01:38:28 AM »
phew... 6000 RPM is a lot...

Say... does anyone of you know an Influence generator which works somehow like the wimshurst, but has no friction and only 1 disk rotating? The wimshurst and toepler slow down a lot because of the wires of neutralizer etc. touching the segments. These frictional contacts are the biggest problem.
Else one could feedback the electrostatic power (saved into a leyden jar) to the source disk or onto a second disc on the same axis working after the principle of an electrostatic motor and create a perpetuum mobile. There are some inspiring clips on youtube like that http://www.youtube.com/watch?v=YpemKuf6X_c&feature=related

Any builds?


Oh, and another thing: this guy "Evert" has an interesting theory on a possible electrostatic electricity generator http://www.evert.de/eft768e.htm For those of you who speak german, this whole text is also available in german language. I am currently trying to build the small version of his suggested device, although I doubt somehow that much power (if any at all...) will be generated. We will see... Unfortunately he has never build any device himself, so one must build by the rule of the thumb.

Free Energy | searching for free energy and discussing free energy

Re: Stepping Down a Wimshurst
« Reply #34 on: April 09, 2010, 01:38:28 AM »
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Offline wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #35 on: April 09, 2010, 09:18:33 AM »
hello

ok i slept with this and have idea how it work
so the stationary stator 20 and 22 receives alternately positive and negative charges (when 20 is pos, 22 is neg and vice verse, cycle 1 and 2 ) that is possible only when connection 100 is not there!!! i dont now if its mistake or inventor wanted to show that this influence does mater , well if disk 20, 22 dont rotate and 24,26 rotate there cant be fixed connection so i think inventor how charges interact - + - + - +...

neg charge induces pos charge in sector at disk 20 -cycle 1 (neg side)
pos charge induces neg charge in sector at disk 22 -cycle 1 (pos side)
and
neg charge induces pos charge in sector at disk 24 which induces neg charge in sector at disk 20 -cycle 2 (neg side)
pos charge induces neg charge in sector at disk 26 which induces neg charge in sector at disk 22 -cycle 2 (pos side)

and since the pair of sectors at disk 20, 22 receive pos and neg charges at the same time there is in balance of the charges so they wand to be exchanged which is happening thru rectifier creating current

wojsciech
« Last Edit: April 09, 2010, 09:58:44 AM by wojwrobel »

Offline wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #36 on: April 10, 2010, 09:38:06 PM »
hello

i was thinking all day today about this machine and it gets me to beggining !!!

the idea is to conver static charges to electricity right??

and the law of physic says:


An electromagnetic field is an area in which electric and magnetic forces are interacting. It arises from electric charges in motion. Electromagnetic fields are directly related to the strength and direction of the force that a charged particle, called the "test" charge, would be subject to under the electromagnetic force caused by another charged particle or group of particles, called the source.

An electromagnetic field is best understood as a mathematical function or property of spacetime, but may be represented as a group of vectors, arrows with specific length and direction. For a static electric field, meaning there is no motion of source charges, the force F→ on a test charge is F→ = qE→, where q is the value of the test charge and E→ is the vector electric field. For a static magnetic field (caused by moving charge inside an overall neutral group of charges, or a bar magnet, for example) the force is given by F→ = qv→ × B→, where v→ is the charge velocity, B→ is the vector magnetic field, and the × indicates a cross-product of vectors.

A stationary charge produces an electric field, while a moving charge additionally produces a magnetic field. Since velocity is a relative concept dependent on one's choice of reference frame, magnetism and electricity are not independent, but linked together, hence the term electromagnetism.

so why dont we build a machine different that this one !!!!

what you guys think


Free Energy | searching for free energy and discussing free energy

Re: Stepping Down a Wimshurst
« Reply #36 on: April 10, 2010, 09:38:06 PM »
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Offline gauschor

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Re: Stepping Down a Wimshurst
« Reply #37 on: April 10, 2010, 10:27:04 PM »
I think that the charged particles would not move in a circular way (if this is what it's supposed to do, judging from the sketch). Also I wonder how you will charge the particles if there is an insulator between the static charge fields from the Wimshurst and the particle pipe.

I believe what you say is true, that an electric field in motion creates a magnetic field. But for the most part a small electric field itself is simply too weak for getting an usable induction effect. You can try to collect induction current by trying to hold a coil next to the moving segments of a wimshurst device (which are charged too), but you will most likely see it is too weak.

But maybe I also understand something wrong / misinterprete your sketch.

Offline wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #38 on: April 11, 2010, 12:38:59 PM »
hello

i forgot to  mark pump, is there to move particles at the side....

anyway from wikipedia

"The B-field can be defined in many equivalent ways based on the effects it has on its environment. For instance, a particle having an electric charge, q, and moving in a B-field with a velocity, v, experiences a force, F, called the Lorentz force (see below). In SI units, the Lorentz force equation is

F=q(vxB)

where × is the vector cross product.

so our magnetic field will depent directly from q  and v so the greater the charge and speed of moving (particles) the higher our magnetic field will be !!!!! and since we move particles in closed loop there is no limit to speed and we can charge our particle to very high charge because from our wimshurts machines can produce large potential it all depends from size of the disk and 1/3 of size is our potential in kv

wojsciech



Free Energy | searching for free energy and discussing free energy

Re: Stepping Down a Wimshurst
« Reply #38 on: April 11, 2010, 12:38:59 PM »
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Offline wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #39 on: April 11, 2010, 02:45:11 PM »
hello

you know what its a good idea to try moving a coil of wife in electric A !!!!

magnetic field lines and electric field lines are identical but have different properties

so when you move magnetic field thru coil of wire you give when N pole push of free electrons and when S pole pull of electrons right

i have to try this setup

Offline gauschor

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Re: Stepping Down a Wimshurst
« Reply #40 on: April 12, 2010, 11:05:40 AM »
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.

Free Energy | searching for free energy and discussing free energy

Re: Stepping Down a Wimshurst
« Reply #40 on: April 12, 2010, 11:05:40 AM »
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Offline gauschor

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Re: Stepping Down a Wimshurst
« Reply #41 on: April 22, 2010, 12:13:36 AM »
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...

Offline wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #42 on: April 22, 2010, 12:08:12 PM »
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?

Offline Steven Dufresne

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Re: Stepping Down a Wimshurst
« Reply #43 on: April 22, 2010, 03:26:12 PM »
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

Offline wojwrobel

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Re: Stepping Down a Wimshurst
« Reply #44 on: April 22, 2010, 09:05:07 PM »
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

 

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