E = (CV^{2}) / (2 . 10^{6}).

Since one joule is nearly equal to three-quarters of a foot-pound, the

energy storage in foot-pounds F is roughly given by the rule F =

(3/8)CV^{2}/10^{6}. For spark lengths of the order of five to fifteen

millimetres, the disruptive voltage in air of ordinary pressure is at

the rate of 3,000 volts per millimetre. Hence, if S stands for the

spark length in millimetres, and C for the aerial capacity in

microfarads, it is easy to see that the energy storage in foot-pound

is

Hertzian Wave Wireless Telegraphy · 第 6 / 10 题