> > 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])

          Wavelength ~=3D 300 / f                F in Mhz, wavelength in me=
tres

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  half-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 0.95

So wavelength =3D f/300 x Kwl

         f in MHz
         length in metres
         Kwl =3D 0.95 =3D Murphy, end effect, velocity factor, bank holiday=
s, ...

FWIW this has the (obvious but nice) reciprocal results

MHz  =3D 300 / Wavelength
&
Wavelength =3D 300 / MHz



     Russell
--=20
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist
.