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Wimshurst charge leakage

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  • mebikerider
    I have an inexpensive commercially made Wimshurst machine. The charge accumulator parts are built from fairly small diameter chromed steel rod, with some
    Message 1 of 3 , Nov 9, 2012
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      I have an inexpensive commercially made Wimshurst machine.

      The charge accumulator parts are built from fairly small diameter chromed steel rod, with some sections about 0.15 inches in diameter.

      When people estimate the voltage that Van de Graaf spheres can support before they lose charge into the surrounding air, they use an equation like 70KV per inch radius.

      According to that formula, those Wimshurst metal rods, with a radius of 0.075", will start to lose charge when the voltage reaches about 5 kv.

      But I can get 1.5" sparks fairly easily between its discharge gap, so using the voltage estimate of 30 kv per cm, it could be generating over 100 kv. How can that be?

      One thought is that those small diameter parts are being shielded in some way by nearby larger diameter metal parts so their field strength is less than a bare exposed sphere.

      Another thought is that there is leakage from the parts, not large visible sparks but a steady loss, a leakage of charge. Could that be? It is true that to get those large sparks, I have to crank the Wimshurst to a pretty high speed, and I'm wondering if that cranking is needed to feed the charge loss through all the small diameter parts.

      Is there such a thing as silent, invisible charge leakage?
    • Antonio
      ... This would be valid for spheres, or spherical ends of conductors. Straight rods support much greater voltages, as most of the charge moves to the ends. ...
      Message 2 of 3 , Nov 9, 2012
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        --- In VanDeGraaffGenerator@yahoogroups.com, "mebikerider" <david_beals@...> wrote:
        >
        > I have an inexpensive commercially made Wimshurst machine.
        >
        > The charge accumulator parts are built from fairly small diameter chromed steel rod, with some sections about 0.15 inches in diameter.
        >
        > When people estimate the voltage that Van de Graaf spheres can support before they lose charge into the surrounding air, they use an equation like 70KV per inch radius.
        >
        > According to that formula, those Wimshurst metal rods, with a radius of 0.075", will start to lose charge when the voltage reaches about 5 kv.

        This would be valid for spheres, or spherical ends of conductors. Straight rods support much greater voltages, as most of the charge moves to the ends.

        > But I can get 1.5" sparks fairly easily between its discharge gap, so using the voltage estimate of 30 kv per cm, it could be generating over 100 kv. How can that be?

        30 kV/cm is approximately valid between flat surfaces (or large spheres at small distance). Between spheres the breakdown voltage is always smaller. Se here:
        http://www.coe.ufrj.br/~acmq/hvmeasurements/

        Two 2 cm spheres separated by 4 cm spark at ~62 kV.

        > One thought is that those small diameter parts are being shielded in some way by nearby larger diameter metal parts so their field strength is less than a bare exposed sphere.

        This may happen.

        > Another thought is that there is leakage from the parts, not large visible sparks but a steady loss, a leakage of charge. Could that be? It is true that to get those large sparks, I have to crank the Wimshurst to a pretty high speed, and I'm wondering if that cranking is needed to feed the charge loss through all the small diameter parts.
        >
        > Is there such a thing as silent, invisible charge leakage?

        Leakage through the air is always visible, at least above a certain current. The maximum voltage that the machine can reach is limited by leakages, specially between the sectors in the disks. If you look at the machine operating in the dark you can see the corona.

        Antonio Carlos M. de Queiroz
      • mebikerider
        Thank you. I m trying to visualize how charges distribute themselves around the shafts and spheres and points of a working Wimshurst, and I m not sure how to
        Message 3 of 3 , Nov 10, 2012
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          Thank you. I'm trying to visualize how charges distribute themselves around the shafts and spheres and points of a working Wimshurst, and I'm not sure how to do it.

          It's easy to find physics descriptions of charge distributions around
          simple shapes like conducting spheres and plates, but not so easy to
          find what happens around irregular shapes. But it's very interesting learning!

          Your web page has some very helpful tips; thank you also for providing that.

          David



          --- In VanDeGraaffGenerator@yahoogroups.com, "Antonio" <acmdq@...> wrote:
          >
          > --- In VanDeGraaffGenerator@yahoogroups.com, "mebikerider" <david_beals@> wrote:
          > >
          > > I have an inexpensive commercially made Wimshurst machine.
          > >
          > > The charge accumulator parts are built from fairly small diameter chromed steel rod, with some sections about 0.15 inches in diameter.
          > >
          > > When people estimate the voltage that Van de Graaf spheres can support before they lose charge into the surrounding air, they use an equation like 70KV per inch radius.
          > >
          > > According to that formula, those Wimshurst metal rods, with a radius of 0.075", will start to lose charge when the voltage reaches about 5 kv.
          >
          > This would be valid for spheres, or spherical ends of conductors. Straight rods support much greater voltages, as most of the charge moves to the ends.
          >
          > > But I can get 1.5" sparks fairly easily between its discharge gap, so using the voltage estimate of 30 kv per cm, it could be generating over 100 kv. How can that be?
          >
          > 30 kV/cm is approximately valid between flat surfaces (or large spheres at small distance). Between spheres the breakdown voltage is always smaller. Se here:
          > http://www.coe.ufrj.br/~acmq/hvmeasurements/
          >
          > Two 2 cm spheres separated by 4 cm spark at ~62 kV.
          >
          > > One thought is that those small diameter parts are being shielded in some way by nearby larger diameter metal parts so their field strength is less than a bare exposed sphere.
          >
          > This may happen.
          >
          > > Another thought is that there is leakage from the parts, not large visible sparks but a steady loss, a leakage of charge. Could that be? It is true that to get those large sparks, I have to crank the Wimshurst to a pretty high speed, and I'm wondering if that cranking is needed to feed the charge loss through all the small diameter parts.
          > >
          > > Is there such a thing as silent, invisible charge leakage?
          >
          > Leakage through the air is always visible, at least above a certain current. The maximum voltage that the machine can reach is limited by leakages, specially between the sectors in the disks. If you look at the machine operating in the dark you can see the corona.
          >
          > Antonio Carlos M. de Queiroz
          >
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