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4hv.org :: Forums :: High Voltage
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how does a Whimshurst machines work

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Andy
Wed Nov 21 2012, 08:07AM Print
Andy Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Hi
I'am mucking around with a MHD generator and was wondering if I could use the principles of a Whimshurst machines. Just wondering if it transfer a charge or creates the difference it by motion. Can you use one of these machines without the neutral combs.

Cheers
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Ash Small
Wed Nov 21 2012, 08:53AM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
The wikipedia article on Wimshurst machines is here: Link2

Google has dozens of descriptions of how they work: Link2,or.r_gc.r_pw.r_qf.&fp=584b092946b3f909&bpcl=38897 761&biw=1366&bih=634
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Andy
Wed Nov 21 2012, 07:05PM
Andy Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Thanks , still trying to get my head around it.
Would this formula be able to predict the voltage gained from the machine or a mhd

0.5*u*v2 = e*V

u in kilograms
e in coulombs
V in volts
v in meter a second

if the mass was the dialectic can I transfer e across to the left.

Cheers

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Proud Mary
Wed Nov 21 2012, 10:42PM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
There is a Wimshurst calculator here: Link2

Antonio has been a member of 4HV for years.
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Andy
Thu Nov 22 2012, 03:10AM
Andy Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Had a look at the program cheers. Making the sections small gives 22cm in a sector less machine but less amps.
Just throwing a thought out, if you have a charge on the right side comb that negative, when it spins around to the neutral combs , it makes the opposite side negative. When that goes back to the right side comb it adds to the voltage?
Wouldn't the voltage stay the same as the negative charge is the same amount on the sector and comb?

Cheers
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Antonio
Thu Nov 22 2012, 11:06PM
Antonio Registered Member #834 Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644
When a sector touches a neutralizer brush it is under the influence of several sectors in the other disk, and one adjacent sector in the same disk. The charge that it receives from the neutralizer is greater than the charges in these adjacent sectors due to the concentrated electric field. The machine (any electrostatic machine) can be also understood as an array of variable capacitors and switches. See here: Link2
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Andy
Fri Nov 23 2012, 04:36AM
Andy Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Thanks Antonio. So it serials two capacitors(both side of the neutralizer) and in parallel then repeats?
Would a Marx that feeds a second one be the same thing.

Cheers
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Antonio
Sun Nov 25 2012, 12:17AM
Antonio Registered Member #834 Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644
Andy wrote ...

Thanks Antonio. So it serials two capacitors(both side of the neutralizer) and in parallel then repeats?
Would a Marx that feeds a second one be the same thing.

Not exactly. The operation requires the variable capacitances between the sectors in opposite disks, that are the source of the energy. It also involves all the capacitances between sectors and ground and between adjacent pairs of them, and the switching to ground made by the neutralizers, in a quite complicated way. I made a model than can run in a conventional circuit simulator (allowing variable capacitors). There is a picture with the results of a simulation in the page mentioned above. Link2
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