The following relationship between wavelength and frequency
is reasonably easy to remember : -
wavelength (metres) = 300
or MHz = 300
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.