Re: breakout voltage

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:

 > Original poster: "Luke" <Bluu-at-cox-dot-net>

 > Field strength = Voltage / Radius.
 > The break down voltage for air is approximately 76KV / inch.
 > This value varies with humidity, frequency, temperature, barometric
 > pressure etc.
 > So combining these gives:
 > Max Voltage (before air breakdown) = 76KV * Radius (in inches)
 > Is this correct for a close approximation?

This is exact for a sphere, far from other objects. Can be used
approximately also for rounded parts of other shapes.

 > When applying this to a toroid I was assuming the minor radius should be
 > considered since it is the smallest curve on the toroid so the field
 > strength would be strongest there.
 > Am I correct in assuming this formula could be applied to the minor radius
 > of a toroid to get it's approximate break out voltage?

This doesn't work for a toroid. The actual "effective radius of
curvature" is somewhere between the minimum and the maximum radii.
The exact expression is known, but is a complicated series.
Max. Voltage = 3000 x radius in meters kV:
Example: A 50x20 cm toroid should break down at a voltage between
3000*0.25 = 750 kV and 3000*0.1 = 300 kV.
An exact analysis results in 512.5 kV.
50x10 cm:
Maximum=750 kV, minimum=150 kV. Exact: 367.5 kV
50x5 cm:
Maximum=750 kV, minimum=75 kV. Exact: 249.6 kV

See the Inca program, that can calculate these things, at:
http://www.coe.ufrj.br/~acmq/programs

Antonio Carlos M. de Queiroz