← All tools

Dipole length calculator

A half-wave dipole is a conductor cut to roughly half a wavelength and fed at its centre. It is never exactly λ/2. Capacitance at the open ends makes the wire behave electrically longer than it measures, so a resonant dipole is cut about 5% short: L = k · λ/2 with k ≈ 0.95. Each leg is half of that.

The slider covers 100 kHz to 30 GHz. 14.2 MHz sits in the 20 m amateur band.

0.95 is the classic cutting rule. Thin wire in the clear runs 0.96 to 0.97, thicker elements need a smaller k.

Total length (½λ dipole)
10.03 m
= 32.9 ft
Each leg (¼λ)
5.01 m
= 16.45 ft
Free-space ½λ
10.56 m
before end-effect shortening
Classic rule
32.96 ft
468 / f (MHz), the ham formula
I = current, peaks at the feedpointTXfeedpoint ≈ 73 ΩL/2 per leg = 5.01 mL total = 10.03 mschematic · span drawn on a log scalef = 14.2 MHz · λ = 21.11 mend effect · k = 0.95

A centre-fed half-wave dipole, seen from the side. The dashed green curve is the current along the wire: zero at the tips, largest at the feedpoint, which is why the antenna is split and fed in the middle. The amber arcs at the tips stand for the end effect. The two arrows give the total length and one leg; the drawn span follows a log scale, so the sketch keeps its proportions from 100 kHz to 30 GHz.

k = 0.95 fits thin wire in the clear. Thick tubing, loading coils, insulated wire and low mounting all move resonance, so treat the result as a cutting length and trim to measured VSWR.

How it works

  1. 01

    Start from the half wave

    Find the free-space wavelength, λ = c / f, and halve it. At 14.2 MHz λ is 21.11 m, so the starting point is 10.56 m tip to tip. The dipole is split in the middle and fed there, because that is where the current is highest and the impedance lowest.

  2. 02

    Shorten it for the end effect

    Charge piles up at the open ends and the resulting capacitance makes the antenna resonate as if it were longer than its physical length. Multiplying by k ≈ 0.95 compensates. Thin wire in the clear sits near 0.95. Thick tubing, insulated wire and end insulators pull k down to about 0.92–0.94.

  3. 03

    Cut each leg to L/2 and feed the centre

    Two equal legs of L/2 with the feedline at the gap give the classic dipole. A half-wave dipole in free space presents about 73 Ω resistive at resonance, close enough to 50 Ω coax for a match under 1.5:1, so most installations need no matching network.

  4. 04

    Cut long, then trim to resonance

    Height above ground, wire diameter, insulation, guy wires and nearby metal all shift the resonant frequency, usually downward. Add 2–3% to each leg, measure the VSWR minimum, then trim symmetrically. Roughly 1% of length moves resonance by about 1% in frequency.

Formulas

Resonant half-wave dipole
L = k · λ2 = k · c2·f
  • L — overall tip-to-tip length, m
  • k — end-effect factor, ≈0.95 for thin wire
  • λ — free-space wavelength, m
  • c — speed of light, 299 792 458 m/s
  • f — frequency, Hz
Classic ham rule, imperial
L (ft) ≈ 468f (MHz)
  • the long-standing wire-dipole rule of thumb
  • equivalent to k ≈ 0.952, so it runs marginally longer than k = 0.95
Classic ham rule, metric
L (m) ≈ 142.6f (MHz)
  • the same rule in metres: 468 · 0.3048 = 142.65, usually rounded to 142.6
  • each leg is half of that

Worked example

A 20 m band dipole at 14.2 MHz
  1. λ = 299 792 458 / 14 200 000 = 21.11 m
  2. λ/2 = 10.56 m
  3. L = 0.95 · 10.56 = 10.03 m → each leg L/2 = 5.01 m
  4. check: 468 / 14.2 = 32.96 ft = 10.05 m
  5. → cut two 5.1 m wires and trim to resonance

FAQ

Why is a real dipole shorter than half a wavelength?
Because of the end effect. The open ends store charge against their surroundings, adding capacitance that makes the antenna look electrically longer than it measures. Compensating for it shortens a thin-wire dipole by roughly 5%, and more for fat elements, since thicker conductors have a larger end capacitance.
What is the feed impedance of a half-wave dipole?
About 73 Ω resistive at resonance in free space, which is a 1.46:1 match to 50 Ω coax. Real installations differ. Half a wavelength above ground the resistance is back near 70 Ω, while at a tenth of a wavelength it falls to roughly 20–30 Ω.
Does mounting height change the length?
Barely. The resonant length shifts by only a fraction of a percent with height. What height does change is the feed impedance and the vertical pattern. Low dipoles fire straight up, while a half wavelength or more of clearance gives useful low-angle radiation.
Is an inverted-V the same length as a flat dipole?
Close, but usually a few percent shorter. Sloping the legs downward lowers both the resonant frequency and the feed impedance, typically to 50–60 Ω, so an inverted-V often matches 50 Ω coax slightly better while needing a little trimming to land on frequency.
Does the same formula work at microwave and mmWave frequencies?
Yes, the physics does not change: at 30 GHz λ/2 is about 5 mm, so a resonant dipole measures roughly 4.7 mm tip to tip. What changes is the tolerance, which shrinks with the wavelength — a 0.1 mm cutting error at 30 GHz is what 3 cm would be at 100 MHz. Above a few gigahertz dipoles are usually printed on a substrate as well, and the dielectric shortens them far beyond the 5% end-effect figure. Treat the result as the free-space starting point.

Antenna cut? Now see how far it talks from your site. Waveshed puts your height, terrain and horizon on a real map and draws the coverage, free in your browser.

Simulate your antenna site →

Related guides

Sources & further reading

Related tools

Frequency ↔ wavelength calculator

λ = c / f, with the ITU band for any frequency.

Quarter-wave vertical calculator

Radiator and radial lengths for a ground-plane vertical at any frequency.

VSWR calculator

Convert VSWR, return loss, reflection coefficient and mismatch loss.

Antenna downtilt calculator

Tilt angle and ground footprint from antenna height and target distance.