**WAVELENGTH
CALCULATIONS **

Formulas are based on the speed of
light / electricity (approx. 186,000 miles per second). We measure frequency
by the number of cycles completed in one second. A frequency of 1 cycle per
second would travel approx. 186,000 miles from start to finish. At a frequency
of 2 cycles per second ,the first cycle would only travel 1/2 that distance
or approx. 93,000 miles from start to finish. By the time the second cycle has
completed (1 second) the full distance(186,000 miles) would have been covered.
You can see from this that as the frequency increases the distance traveled
by a single cycle decreases. A conductor long enough to accommodate 1/2 of one
cycle is an efficient radiator. With frequencies in the (millions of cycles)
megahertz range antenna lengths become quiet

manageable.

Examples:

492/1Mhz = 492 feet at 10 Mhz. only 49.2 feet and at 100Mhz only 4.92 feet

492/Fmhz = 1/2 wavelength in free space

Free space calculations
are used to:

define the height of an antenna above ground

define the spacing between antenna elements

define the length of a transmission line which may then be multiplied by it's
Velocity Factor to

determine its electrical length.

define length of antenna elements other than dipole type antennas ( inverted
v's)

Dipoles

468/Fmhz = 1/2 wave element for dipole

Dipoles exhibit a phenomenon (defined as an" observable fact") of being about 5% shorter then the 492/Fmhz formulas predicts.(492* .95 = 467.4 or approx. 468). We call this shortening " capacitive end effect". This is a fact, there are several theories as to why this is true. Since this formula (468/Fmhz) is so well known it is often used where the 492/Fmhz should be used. Often with acceptable results, the 5% being close enough .BUT NOT ALWAYS.

**
Tiny was here! **