I believe that the traditional length in feet =3D 468/frequency (MHz) is
intentionally shorter than the free space value to get even closer to
the exact practical half-wave wire dipole length. I don't think it is
just velocity factor - the current and voltage distribution on a half
wave dipole are not exact sinusoids and I think that makes a few
percent difference in the optimal length.

Sean


On Mon, Oct 24, 2011 at 10:56 PM, RussellMc <apptechnz@gmail.com> wrote:
>> > You mean like "to get the length of your dipole antenna, divide 468 by
>> > the frequency"? (which gives you the length in feet, of course.)
>
>> ... if you use the right unit for frequency, of course :)
>
>> Exactly this. I hate it; it doesn't convey understanding as more
>> sensible formulas tend to do, and I just can't remember such stuff.
>
> I too like my formulae to make sense. This one does once you remove
> the rubbish and convert to 'propr units' [tm].
>
> As written it has a velocity factor and/or and end effect allowance,
> both of which 'muck it up' .
>
> If instead you use (in real units [tm])
>
> =A0 =A0 =A0 =A0 =A0Wavelength ~=3D 300 / f =A0 =A0 =A0 =A0 =A0 =A0 =A0 =
=A0F in Mhz, wavelength in metres
>
> You see (of course): Speed of light ~=3D 3 x 10^8 m/s poking its head up.
> Divide by E6 for Hz: Mhz and the 300 makes perfect sense.
> NOW you can remember it.
> Now we can convert to arcane units (say feet) and add whatever factors
> we see fot and divide by 2 to get =A0half-wave dipole.
>
> Sanity check.
> 1 MHz.
> 1 wavelength =3D 300/1 =3D 300 metres.
> Hlf wavelength =3D 150 metres.
> 150/0.3048 =3D 492 feet.
> The original length is short by a factor of 468/492 =3D 0.9512 -> say 0.9=
5
>
> So wavelength =3D f/300 x Kwl
>
> =A0 =A0 =A0 =A0 f in MHz
> =A0 =A0 =A0 =A0 length in metres
> =A0 =A0 =A0 =A0 Kwl =3D 0.95 =3D Murphy, end effect, velocity factor, ban=
k holidays, ...
>
> FWIW this has the (obvious but nice) reciprocal results
>
> MHz =A0=3D 300 / Wavelength
> &
> Wavelength =3D 300 / MHz
>
>
>
> =A0 =A0 Russell
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