Quarter Wave Antenna Length: 234/f, 246/f and Why • The Science 360

Q: How long is a quarter wave antenna?

A free-space quarter wavelength is 74.95 divided by the frequency in MHz, in metres, or 246 divided by MHz in feet. A real resonant whip is a little shorter, typically 2 to 7% depending on how thick the conductor is relative to the wavelength. For 433.92 MHz the free-space value is 172.7 mm and a 1 mm wire resonates at about 165 mm.

Q: What is the difference between 234/f and 246/f?

246/f (feet, f in MHz) is the exact free-space quarter wavelength. 234/f is 95.1% of that, a rule of thumb for the shortening a real conductor needs to be resonant. The right factor depends on conductor thickness: about 0.97 for thin wire at HF and about 0.94 for a 1 mm wire at 2.4 GHz.

Type "quarter wave antenna length" into a search engine and you will get a dozen calculators that agree on one formula, 234 divided by the frequency in MHz gives feet, and disagree on why. Several call the 5% trim from the true quarter wavelength ( $246/f$ ) is a "velocity factor" of 0.95. It is not, and the difference matters: for a thin wire at 14 MHz the 234 rule cuts the antenna 10 cm too short and pushes it out of the band, while for a fat whip at 2.4 GHz it leaves it too long.

Below we compute the resonant length from first principles for several real cases, show why the correction depends on wire thickness, and explain why a perfectly cut quarter-wave whip still does not present 50 Ω.

Quick answerA free-space quarter wave is

74.95/f

metres (

246/f

feet), with f in MHz. A real whip resonates slightly shorter, by 2 to 7% depending on conductor thickness, which is where

234/f

(95%) comes from. At 433.92 MHz: 172.7 mm theoretical, about 165 mm for 1 mm wire. Then trim for lowest SWR.

Infographic: How long is a quarter-wave whip? A vertical whip stands on a ground plane with its mirror image below, current peaking at the feed and voltage at the tip. At 433.92 MHz the exact quarter wave is 172.7 mm, but a real whip resonates 2 to 7% shorter depending on conductor thickness. The formula is <span class= — Tap the image to view it full size. View the accessible, editable SVG.

The quarter-wave formula, done properly

A wavelength is the speed of light divided by frequency, and the speed of light is defined as exactly 299,792,458 m/s. A quarter of that, with frequency in MHz:

\begin{aligned}\lambda/4 &= \frac{c}{4f} = \frac{299{,}792{,}458}{4 \times f \times 10^6} \\ &= \mathbf{74.95/f_{MHz}\ metres}\end{aligned}

In feet:

\begin{aligned}74.95/0.3048 &= \mathbf{245.9/f_{MHz}}\end{aligned}

which rounds to $246/f$ .

\begin{aligned}234/245.9 &= \mathbf{0.951}\end{aligned}

The "234 rule" is a 4.9% shortening. Metric version of the 234 rule:

\begin{aligned}0.951 \times 74.95 &= 71.3/f_{\mathrm{MHz}}\ \text{metres}\end{aligned}

Calculators often print 71.5.

Why a quarter wave at all? A vertical element of height h over a conducting plane behaves like a dipole of length 2h, because the plane creates a mirror image of the element and of its current. A half-wave dipole is the shortest straight wire that is naturally resonant, so its half, the quarter-wave monopole, is too. The current is largest at the feed and falls to zero at the tip; the voltage does the opposite. That standing-wave pattern is the same one you see on any resonant line, and it is why the feed impedance is low.

Figure 1. A quarter-wave monopole and its image form a half-wave dipole. High current and low voltage at the feed give a low impedance, about 36 Ω. The charge piled up at the high-voltage tip is what makes the element act electrically longer than it is.

The misconception: it is not a velocity factor

Velocity factor is a property of transmission lines. In a coax cable the field travels through a plastic dielectric, so the wave is slower than light, about 66% of it for solid polyethylene cable. That is a real slowing, and it makes a quarter-wave length of coax physically shorter than a quarter wave in air.

A bare wire in air has no such dielectric, so the signal along it travels at essentially the speed of light. Even Wikipedia's dipole article frames the shortening as a "reduced speed of radio waves in the conductor", and several of the calculator pages on page 1 call it a velocity factor of 0.95. The cause is different, and it shows up in the numbers.

The real cause is the antenna's own reactance. Cut a monopole to exactly λ/4 and its feed impedance is not purely resistive. The classic induced-EMF result is:

Exact λ/4 monopole over a perfect ground plane (half of the λ/2 dipole's $73.1 + j42.5\ \Omega$ ):

\begin{aligned}Z &= \mathbf{36.5 + j21.3\ \Omega}\end{aligned}

Slightly inductive.

To cancel the +j21 Ω you shorten the element until the reactance passes through zero. How much shortening that takes depends on how fast the reactance changes with length, and that rate depends on the conductor's thickness. A thin wire has a steep reactance curve, so a small trim fixes it. A thick conductor has a shallow curve and needs a bigger trim. Physically, a fat element has more capacitance near its tip, the "end effect", which loads it like a capacitor hung on the end of a line. Wikipedia gives the dipole shortening factor as about 0.98 for thin wires down to about 0.94 for thick conductors. A true velocity factor would not care about thickness.

Figure 2. Feed impedance from the induced-EMF method. At exactly λ/4 the element is 21 Ω inductive regardless of thickness, but the thin wire's reactance falls off faster with length, so it reaches resonance with less trimming (0.240λ) than the thick one (0.233λ). In this model the resistance does not depend on thickness.

How the right shortening factor depends on wire thickness

Sweep the ratio of wavelength to conductor diameter and find where the reactance crosses zero each time, and you get a design chart (Figure 3). It is our own calculation with the induced-EMF formulas from Balanis, which assume a sinusoidal current; it is a good approximation for thin wires and becomes unreliable for very fat elements, so we stopped the curve at $\lambda/d = 30$ .

Resonant monopole height divided by a free-space quarter wavelength, plotted against wavelength over conductor diameter on a log axis from 30 to 100,000. The blue curve rises from about 0.89 at a ratio of 30 to 0.978 at 100,000. A dashed orange line marks 0.951 for the 234 over f rule. Green points: 2.45 gigahertz with 1 millimetre wire at 0.937, 433.92 megahertz with 1 millimetre wire at 0.958, and 14.2 megahertz with 2 millimetre wire at 0.972. — Figure 3. The shortening factor from our induced-EMF calculation. The $234/f$ rule matches a conductor with λ/d of a few hundred, which is roughly a 1 mm wire at UHF. Thin HF wires need less shortening; thick or high-frequency elements need more.

Source note Figures 2 to 4 come from our own induced-EMF model, not from measurement or a full method-of-moments simulation. The model reproduces the textbook

73.1 + j42.5\ \Omega

for a half-wave dipole and lands inside Wikipedia's 0.94 to 0.98 range, but expect the real resonant length to differ by a percent or so, and more if the ground plane is small.

Worked example: a 433 MHz whip

The 433.05 to 434.79 MHz ISM band, centre 433.92 MHz, is used by remote controls, weather sensors and cheap transmitter modules. Here is the length three ways for a 1 mm copper wire:

\begin{aligned}\lambda &= 299{,}792{,}458/433.92{\times}10^6 \\ &= 0.6909\ \mathrm{m}\end{aligned}

$\lambda/d = 691$ for 1 mm wire. Free-space quarter wave:

\begin{aligned}\lambda/4 &= 0.6909/4 = \mathbf{172.7\ mm}\end{aligned}

234 rule:

\begin{aligned}234/433.92 &= 0.5393\ \mathrm{ft} \\ 0.5393 \times 304.8 &= \mathbf{164.4\ mm}\end{aligned}

Our model at $\lambda/d = 691$ : factor 0.958:

\begin{aligned}0.958 \times 172.7 &= \mathbf{165.4\ mm}\end{aligned}

R = 32.3 Ω at resonance.

Here the rule of thumb and the model agree within about a millimetre, because a 1 mm wire at 433 MHz happens to sit near the thickness the rule describes. The real mistake is the one many hobby guides make: cutting the wire to 172.7 mm, the exact quarter wave, because "a quarter wave is a quarter wave". Figure 4 shows what that does.

SWR on 50 ohm coax against frequency from 380 to 480 megahertz for three 1 millimetre whips. The 172.7 millimetre whip has its SWR minimum of about 1.53 near 419 megahertz, well below the 433 megahertz band, where its SWR is about 1.8. The 165.4 and 164.3 millimetre whips have minima near 438 and 441 megahertz and SWR of about 1.55 in the band. — Figure 4. SWR of the 433 MHz whip cut three ways, from our model over a perfect ground plane. The full 172.7 mm whip resonates near 419 MHz, about 3.5% low. Note that no cut gets below about 1.53:1, because the radiation resistance is only about 32 Ω.

Worked example: 2.4 GHz and an HF wire

At 2.45 GHz the wavelength is only 122 mm, so a 1 mm wire is relatively fat:

\begin{aligned}\lambda/4 &= 74.95/2{,}450 = 0.03059\ \mathrm{m} \\ &= \mathbf{30.6\ mm}\end{aligned}

$\lambda/d = 122$ for 1 mm wire. 234 rule:

\begin{aligned}0.951 \times 30.6 &= \mathbf{29.1\ mm}\end{aligned}

Our model: factor 0.937 → 28.7 mm (0.5 mm wire: 0.947 → 29.0 mm)

234 rule error for 1 mm wire:

\begin{aligned}29.1/28.7 &\Rightarrow 1.5\%\ \text{too long} \\ &\to \text{resonance about 37 MHz low}\end{aligned}

Thirty-seven megahertz is still mostly inside a 2.4 to 2.5 GHz band, so a 2.4 GHz whip cut by the rule will work. A trace antenna on a circuit board is another matter: there the field passes partly through the board material, so a genuine dielectric slowing appears on top of the end effect, and the resonant length shrinks a good deal further. Use the chip vendor's reference layout rather than any formula here.

Source note We did not open a vendor reference design to quantify how much shorter a 2.4 GHz PCB trace monopole is than a wire whip; it depends on board material, thickness and ground clearance, so we give no number.

Now a thin wire on the 20 m amateur band, 14.0 to 14.35 MHz:

At 14.2 MHz:

\begin{aligned}\lambda/4 &= 74.95/14.2 = \mathbf{5.278\ m}\end{aligned}

2 mm wire gives $\lambda/d = 10{,}556$ . Our model: factor 0.972 → 5.131 m

234 rule:

\begin{aligned}234/14.2 &= 16.48\ \mathrm{ft} = \mathbf{5.023\ m}\end{aligned}

That is 108 mm shorter. Resonance moves up by:

\begin{aligned}5.131/5.023 &\Rightarrow 2.1\% \\ &\to \text{about } \mathbf{14.50\ MHz}\end{aligned}

That is above the band edge.

This is why experienced builders say to cut long and trim. For a thin HF wire the 234 rule overcorrects. Insulated wire complicates the picture again, since the plastic jacket does add a real dielectric effect of a few percent, but that only reinforces the point: the correction is not one universal 0.95.

Why a quarter-wave whip is about 36 Ω, not 50 Ω

The monopole sees only half the space a dipole does, the half above the ground plane, so it radiates half the power for the same current and has half the radiation resistance: $73.1/2 = 36.5\ \Omega$ at exactly λ/4, and a little less, about 32 to 34 Ω, once trimmed to resonance. Feed that from 50 Ω coax and some power reflects:

\begin{aligned}\Gamma &= \frac{50 - 32.3}{50 + 32.3} = 0.215 \\ \text{SWR} &= \frac{1 + 0.215}{1 - 0.215} = \mathbf{1.55} \\ \text{Reflected power} &= \Gamma^2 = 4.6\% \\ \text{Mismatch loss} &= -10\log_{10}(1 - 0.046) \\ &= \mathbf{0.21\ dB}\end{aligned}

A fifth of a decibel is nothing to worry about for most links, and it is why you should not chase a 1.0:1 SWR on a ground-plane antenna by trimming: no length gives it, and trimming for lowest SWR actually lands a little above the resonant frequency (Figure 4), because the resistance rises with length. The VSWR explainer covers what these numbers mean in practice.

If you do want 50 Ω, bend the radials down. The US amateur radio General exam pool states that sloping the radials of a ground-plane vertical down to about 45° raises the feed-point impedance from about 35 Ω to about 50 Ω. A tilted ground plane is less of a mirror, so the antenna starts to behave partly like a dipole, whose impedance is higher.

What counts as a ground plane?

The image trick only works if there is something to make the image. Wikipedia's monopole article puts it at a plane extending at least half a wavelength from the element, which is 35 cm at 433 MHz and 6 cm at 2.4 GHz. Three or four radials about a quarter wave long are the usual substitute. A small transmitter module on a 5 cm board has neither, so the board and its power leads become the other half of the antenna; its feed impedance drops and its resonance moves, and the only reliable length is the one you measure. Using the coax shield as the counterpoise also puts RF current on the outside of the cable, which is why such setups are sensitive to how the cable is routed.

Resistance in the element matters little for a full-size whip: a copper wire's RF resistance, set by the skin effect, is a fraction of an ohm against 32 Ω of radiation resistance. It matters a great deal for shortened, loaded antennas, where the radiation resistance falls to a few ohms.

For a sense of scale, the same formula says a quarter-wave antenna for 50 Hz mains would be 1,499 km tall (see why mains is 50 or 60 Hz), and at the 2.45 GHz of a microwave oven it is about 3 cm. An antenna is just a resonator with distributed L and C, the open-air relative of the LC tank circuit.

The short version: start from

74.95/f

metres (

246/f

feet). For a thin wire at HF shorten by about 3%, for 1 mm wire at UHF by about 4 to 5% (where

234/f

fits), for thick elements or 2.4 GHz wire by 5 to 7%. Cut a little long, trim on the real ground plane, and expect a best SWR of about 1.5 on 50 Ω coax unless you droop the radials.

FAQ

How long is a quarter wave antenna?

A free-space quarter wavelength is $74.95/f$ metres, or $246/f$ feet, with f in MHz. A real resonant whip is a little shorter, typically 2 to 7% depending on how thick the conductor is relative to the wavelength. For 433.92 MHz the free-space value is 172.7 mm and a 1 mm wire resonates at about 165 mm.

What is the difference between 234/f and 246/f?

$246/f$ (feet, f in MHz) is the exact free-space quarter wavelength. $234/f$ is 95.1% of that, a rule of thumb for the shortening a real conductor needs to be resonant. The right factor depends on conductor thickness: about 0.97 for thin wire at HF and about 0.94 for a 1 mm wire at 2.4 GHz.

What is the impedance of a quarter wave antenna?

Over a large ground plane, an exactly quarter-wave monopole has about 36.5 Ω of resistance plus about 21 Ω of inductive reactance. Trimmed slightly shorter to resonance, it is about 32 to 35 Ω resistive. Sloping the radials down at about 45° raises it to roughly 50 Ω.

Does a quarter wave antenna need a ground plane?

Yes. A quarter-wave monopole is half of a dipole; the ground plane or radials supply the other half as a mirror image. Without one, the coax shield or the circuit board becomes the other half, and the length, impedance and pattern all change.

Sources

NIST CODATA, speed of light in vacuum, 299,792,458 m/s exact: physics.nist.gov/cgi-bin/cuu/Value?c
Wikipedia, "Dipole antenna" ( $73\ \Omega + j43\ \Omega$ for a true half-wave dipole; shortening factor about 0.98 for thin wires to about 0.94 for thick conductors; $468/f$ for $k \approx 0.95$ ; attributes the shortening to reduced wave speed): en.wikipedia.org/wiki/Dipole_antenna
Wikipedia, "Monopole antenna" (36.5 Ω for λ/4; element and image act as a dipole; ground plane at least half a wavelength; resonant length slightly shorter due to end effects): en.wikipedia.org/wiki/Monopole_antenna
NCVEC General class question pool G9B02 and G9B03 via HamStudy (sloping radials raise the feed-point impedance from about 35 Ω to 50 Ω at 45°): hamstudy.org/browse/E3_2015/G9B
Wikipedia, "ISM radio band" (433.05 to 434.79 MHz, centre 433.92 MHz; 2.4 to 2.5 GHz, centre 2.45 GHz): en.wikipedia.org/wiki/ISM_radio_band
66pacific.com, quarter-wave vertical calculator (typical page-one treatment: $234/f$ and $71.5/f$ citing the ARRL Antenna Book, radials at least a quarter wave, cut long and trim for best SWR): 66pacific.com quarter-wave calculator
C. A. Balanis, Antenna Theory: Analysis and Design, Wiley, induced-EMF expressions for the radiation resistance and reactance of a finite dipole (the formulas used for Figures 2 to 4; we checked our code against the $73.1 + j42.5\ \Omega$ half-wave result rather than against the book itself).

Method note: Figures 2 to 4 were computed in Python with the induced-EMF formulas (sinusoidal current, perfect ground plane, monopole impedance = half the dipole's). Figure 1 is a schematic with the ideal sinusoidal current and voltage distributions.

Header image: NASA (image iss005e08718), regraded for this article.