Paradigm Lost: Nikola Tesla's True Wireless Page 16

What Tesla must explain is how his transmitter in the air might communicate with the receiver on the ground in spite of its not having a direct connection to the earth that would be capable of effectively launching his crucial earth currents. His explanation is that “we are merely working through a condenser.” Stating incorrectly there is a capacity that “is a function of a logarithmic ratio between the length of the conductor and the distance from the ground,” he says the receiver is affected in the same manner as with an ordinary transmitter. Evidently, we are to believe that the capacitance between earth and ground makes possible the earth currents crucial to his argument.

The formula for the capacity of a wire that he is most likely referring to would have been well known by the 1910s when it already had appeared in textbooks and handbooks:⁴⁶

\[ C = \frac{1.111 \, L}{2 \ln \left( \frac{2h}{r} \right)} \quad \text{picofarads.} \]

This expression is the capacity of a wire of length \( L \) above, and parallel to, the earth’s surface, which is assumed to be highly conducting. An airplane in flight dragging a wire antenna behind itself would create this situation. The wire is at height \( h \) above the earth, and its radius is \( r \). All dimensions are in centimeters, and the logarithm is base \( e \). Note that the capacity is proportional to the wire length \( L \), not to the logarithm of \( L \) as Tesla asserted.⁴⁷ Using the well-known formula for capacitive reactance \( X = 1/(2\pi f C) \), where \( f \) is the operating frequency, we could in principle obtain the impedance between the wire and earth. Dividing the voltage of the antenna, with respect to the earth,