DIPOLE CALCULATION

The following relationship between wavelength and frequency
is reasonably easy to remember : -

wavelength (metres) = 300
MHz

or  MHz      =      300
wavelength (m)

Now the theoretical length of a dipole is half the wavelength but
because the wave travels more slowly on a wire than in space and also because the insulators at the ends cause additional
capacitative loading ( the "end effect") a usable dipole is shorter than predicted by around five percent. In our calculation we must therefore introduce a correction factor, as follows :-

Dipole length = 0.95 x 1/2 x wavelength
= 0.95 x 150/MHz

The correction factor is even less if insulated wire is used and
also at higher frequencies; it can be as low as 0.90.

As an example we need a dipole for 7.05 MHz.

The required length is 0.95 x 150 / 7.05 which is 20.2 metres.

Most of the radiation from such an antenna comes from the
middle where the current is highest. Therefore the inverted Vee
suspended from a single mast is an economical proposition as
the performance is hardly affected even if the ends are quite
close to the ground.

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