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Author Topic: Wimshurst - let's clear this up  (Read 436 times)

Offline nix85

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Wimshurst - let's clear this up
« on: February 18, 2020, 02:14:11 PM »
How exactly does Wimshurst produce HV? I understand how charge induction happens on opposite discs. Charge is picked up from two pairs of oppositelly charged sectors (two on each side), but these two pairs alone surely don't produce HV.

Are all the sectors somehow adding up in voltage? That sure does NOT seem to be the case, at least i don't see how that would work.

Wikipedia says: https://en.wikipedia.org/wiki/Wimshurst_machine

"The available voltage gain can be understood by noting that the charge density on oppositely charged sectors, between the neutralizer bars, is nearly uniform across the sectors, and thus at low voltage, while the charge density on same charged sectors, approaching the collector combs, peaks near the sector edges, at a consequently high voltage relative to the opposite collector combs."

Wikipedia explanation that sharpness of the combs makes charge density on the adjecent sector higher and thus provides higher voltage makes sense, but i still don't see how many dozens of kilovolts would be generated that way, especially by a single sector pair.

I watched MIT video and this one too https://youtu.be/nA4aCd5qFWs?t=239

Anyone?


Free Energy | searching for free energy and discussing free energy

Wimshurst - let's clear this up
« on: February 18, 2020, 02:14:11 PM »

 

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