**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/MHzThe 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 for7.05 MHz.

The required length is0.95 x 150 / 7.05which is20.2metres.

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.