Horizontal Element Droop
The typical wire dipole is a wire supported at each end byt a pole (tree). The wire will droop (sag) due to the weight of the wire. The amount of droop is also affected by the amount of tension holding the wire in place. The calculations below assume a single strand of wire (without a feed line) supported at each end at the same height.
The information required for various calculations is:
Fh - Hoizontal pulling force to produce tension.
H - Droop (sag) in the wire from horizontal. (feet).
L - Distance required between support poles (feet).
S - Actual length of the wire (feet).
T - Tension applied to the wire (pounds).
w - Weight of the wire per foot. Click here for copper wire properties.
The basic formula to estimate the amount of horizontal force for a specific droop is:
Fh = W *( S2 - ( 4 * H2 ) ) / ( 8 * H )
As an example, design a 20-meter dipole resonant at 14.150 MHz using 14 AWG copper wire.
How much tension will be needed to provide a 6 inch droop?
The conventional formula for a 1/2 wave dipole is 486 / MHz.
The length of the wire will be ( 486 / 14.150 ), or 34.35 feet long . ( S = 34.35 )
The weight of 14 gauge wire is 1.24 pounds per 100 feet. ( w = 0.0124 )
The expected droop is 6 inches. ( H = 0.5 )
Fh = w *( S2 - ( 4 * H2 ) ) / ( 8 * H )
= 0.0124 * ( 34.352 - ( 4 * 0.52 ) ) / ( 8 * 0.5 )
= 0.0124 * ( 1180 - ( 4 * 0.25 ) ) / ( 8 * 0.5 )
= 0.0124 * ( 1180 - 2 ) / 4
= 0.0124 * 1178 / 4
= 3.65 pounds
The distance between the poles (L) is:
L = ( 2 * Fh / W ) * arcsinh( S * W / (2 * Fh) )
Note that arcsinh is a hyberolic function, not the more common circular sin function.
= ( 2 * 3.65 / 0.0124 ) * arcsinh( 34.35 * 0.0124 / ( 2 * 3.65) )
= ( 588.7 ) * arcsinh( 0.4259 / 7.3 )
= 588.7 * arcsinh(.0583)
= 588.7 * .05827
= 34.30 feet
These formulas do not include attachments at the center of the dipole or end insulators.