I was operating under the impression that the arm length of a half dipole antenna is λ/4 while making one that operated at the 145Mhz frequency using a piece of 9.525 mm (3/8 inch) copper tubing, at a heigh of about 50 ft (15.24 meters)

However, a friend of mine informed me that the velocity factor of the material plays a role and that the real equation of Vf * λ/4, where Vf is the Velocity Factor of the material.

How does this work? Are there other variables in the equation? Does the amount of space between the arms play a role, since every NEC representation I've seen assumes there's no space in the middle of the arms?

What is the one true equation for the arm lengths of a dipole antenna?


1 Answer 1


There are several factors that come into play in determining the length of a center fed 1/2 wavelength dipole - all of which center on the notion of a near resonant feedpoint impedance.

To begin with, an infinitely thin, 1/2 wavelength center fed dipole in free space has a feedpoint impedance of 73+j42.5 ohms. By slightly shortening the antenna, the reactive component fades away and we are left with a ~73+j0 ohm feedpoint impedance.

When we construct a practical version of this dipole near the surface of the earth with real wire, things change. The proximity to the earth changes the impedance as a function of $\lambda$ above the earth and the soil conditions. The wire ends create a small self capacitance due to their real thickness. The result is reflected in an empirically derived formula for the total length (both arms) of a 1/2 wavelength dipole:

$$l=142.6 \text{ meters}/f_{\text{MHz}} \tag 1$$

For imperial units, the numerator is replaced with 468 feet.

Formula 1 yields a starting point. A good practice is to start slightly longer and then symmetrically reduce the length while monitoring the complex impedance of the feedpoint. Depending upon the antenna height, the real part of the impedance can range from ~48 to 73 ohms when optimally tuned.

Formula 1 assumes bare, copper wire. If the wire is insulated, then the velocity factor of the insulation must be considered. This will result in a reduction of the calculated length. For example, when household THHN (Thermoplastic High Heat-resistant Nylon) wire is used, the velocity factor is approximately 97% according to most field reports. The result of formula 1 is therefore multiplied by 0.97.

Finally, there are resistive losses due to the RF resistance of the conductors. While this will not alter formula 1, these losses will directly add to the real part of the idealized feedpoint impedance.

Regarding your question of the gap between the arms, this is assumed to be minimal compared to the wavelength. On HF frequencies (3-30 MHz) a gap of 5 cm is immaterial. At higher frequencies, this gap should be reduced. At 145 MHz, a gap of 2 cm or less should not pose a problem.

  • $\begingroup$ Thank you for the detailed answer! If we were to add those variables in the equation (material thickness, type, and antenna height) would it affect the equation? $\endgroup$ Mar 10, 2019 at 13:38
  • 1
    $\begingroup$ @AminShahGilani It is difficult to have an all in one formula that would reliably predict resonance (if that is the goal). Formula 1 is generally good for typical wire gauges but it does not accommodate the height and soil conditions. We typically model it with NEC or get our hands dirty in the field. You will find some experiential graphs that get you close for most common HF bands. $\endgroup$
    – Glenn W9IQ
    Mar 10, 2019 at 17:16

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