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*To*: tesla-at-pupman-dot-com*Subject*: Re: Geek Gap Mark-1*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Sat, 14 Apr 2001 19:48:11 -0600*Resent-Date*: Sat, 14 Apr 2001 20:12:11 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <inlsHD.A.LZB.3NQ26-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-uswest-dot-net>" <evp-at-pacbell-dot-net> Tesla list wrote: > > Original poster: "cougercat by way of Terry Fritz <twftesla-at-uswest-dot-net>" <felix1063-at-home-dot-com> > > I think but have no real proof that the larger and more rounded "load" at > the top of the coil is, the higher that voltage can be before it "leaps off" > to a ground someplace in the room. Static discharge conductors are almost > always end in a sharp point like lightning rods. If the voltage that is > supplied to a discharge point is 50kv, weather it is a sharp point or a > rounded ball, there is still 50kv and that voltage will more likely tend to > leap off a sharp point than from a rounded surface. Hence the same length > spark. > > IMHO > > --jeff The breakdown really is a function of the voltage GRADIENT, not the absolute voltage. An often-used value is about 30 kV/cm. For isolated spheres the gradient is equal to the voltage divided by the radius, so smaller spheres or sharp points will break down at much lower voltages than large terminals. Ed

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