A bass-reflex (ported) box is a Helmholtz resonator: the air inside the box acts as a spring and the slug of air in the port as a mass. At the tuning frequency the port air resonates and radiates strongly, extending the bass of the driver. The tuning depends on the box volume, the port’s cross-section and its length – a larger box, a longer port or a narrower port all lower the tuning.
The port also behaves as if it were a little longer than the physical tube, because air just outside each end moves with it. This end correction is k × D, with k ≈ 0.85 if both ends are flanged (flush with a baffle), 0.732 for one flanged and one free end (the usual case for a port with its inner end inside the box) and 0.614 for two free ends. Subtracting it from the acoustic length gives the tube length you cut.
Small ports cause audible “chuffing” when the air moves too fast. If you know your driver’s cone area (Sd) and maximum excursion (Xmax), the calculator estimates the peak air speed in the port; keep it below about 17 m/s (5% of the speed of sound), or use a bigger or flared port.
How to use the Port Length Calculator
- Number of ports
- Port inner diameter
- Net box volume
- Tuning frequency
- Port ends
- End correction factor k
- Speed of sound — 344 m/s at about 21 °C.
- Driver cone area Sd (for air speed) — About 480–530 cm² for a 12″ driver, 330 cm² for a 10″.
- Driver Xmax (one way)
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
The Helmholtz equation is solved for the effective port length using the total port area of N round ports, the net box volume and the tuning frequency, with the speed of sound you enter; the end correction k × D is subtracted. Port volume and the air-speed estimate use the same port area.
Worked example
A 50 L box tuned to 35 Hz with one round port of 10 cm diameter, one end flanged (k = 0.732), speed of sound 344 m/s. Port area = π × 0.1² ÷ 4 = 0.007854 m². Effective length = 344² × 0.007854 ÷ (4π² × 35² × 0.05) = 0.3844 m; subtracting 0.732 × 10 cm = 7.32 cm leaves 31.1 cm. With a 12″ driver (Sd 480 cm², Xmax 10 mm) the peak air speed is 0.048 × 0.01 × 2π × 35 ÷ 0.007854 ≈ 13.4 m/s – below the chuffing limit.
How port diameter changes length (50 L, 35 Hz, k = 0.732)
| Port diameter | Port length |
|---|---|
| 5 cm | 5.9 cm |
| 7.5 cm | 16.1 cm |
| 10 cm | 31.1 cm |
| 12.5 cm | 50.9 cm |
| 15 cm | 75.5 cm |
Inputs explained
| Field | What to enter |
|---|---|
| Number of ports | A number |
| Port inner diameter | Units: cm, in, mm, m |
| Net box volume | Units: L, ft³, in³, m³, cm³ |
| Tuning frequency | In Hz |
| Port ends | Choose one: One end flanged, one free (k = 0.732), Both ends flanged (k = 0.850), Both ends free (k = 0.614), Custom k |
| End correction factor k | A number |
| Speed of sound | Units: m/s, ft/s, km/h. 344 m/s at about 21 °C. |
| Driver cone area Sd (for air speed) | Units: cm², in², m². About 480–530 cm² for a 12″ driver, 330 cm² for a 10″. |
| Driver Xmax (one way) | Units: mm, in, cm |
What each result means
- Effective acoustic length = [Speed of sound]^2×[Total port area (m²)]÷(4×pi^2×[Tuning frequency]^2×[Net box volume])
- Port length to cut (each port) = [Effective acoustic length] – [End correction factor used]×[Port inner diameter]
- Total port area = [Total port area (m²)]
- Volume taken up by the port(s) = [Total port area (m²)]×max([Port length to cut (each port)], 0)
- Box volume needed incl. ports = [Net box volume] + [Volume taken up by the port(s)]
- Peak port air speed at tuning = [Driver cone area Sd (for air speed)]×[Driver Xmax (one way)]×2×pi×[Tuning frequency]÷[Total port area (m²)]
- Air speed as % of the speed of sound = ([Peak port air speed at tuning]÷[Speed of sound])×100
Formula
L = (c² × N × π D² ÷ 4) ÷ (4π² × F² × V) − k × D
with c = 344 m/s in cm, D in cm, V in litres: L ≈ 23,542 × D² × N ÷ (V × F²) − k × D
Peak port air speed ≈ Sd × Xmax × 2πF ÷ (N × π D² ÷ 4)
Worked example
Inputs — Number of ports: 1; Port inner diameter: 10 cm; Net box volume: 50 L; Tuning frequency: 35 Hz; Port ends: One end flanged, one free (k = 0.732); Speed of sound: 344 m/s; Driver cone area Sd (for air speed): 480 cm²; Driver Xmax (one way): 10 mm.
Results — End correction factor used = 0.732; Effective acoustic length = 38.44 cm; Port length to cut (each port) = 31.12 cm; Total port area = 78.54 cm²; Volume taken up by the port(s) = 2.444 L; Box volume needed incl. ports = 52.44 L; Peak port air speed at tuning = 13.44 m/s; Air speed as % of the speed of sound = 3.91 %.
Key terms
- Bass reflex
- A speaker enclosure with a port that reinforces low frequencies.
- Tuning frequency (Fb)
- The Helmholtz resonance frequency of box and port.
- End correction
- Extra effective length from air moving beyond the port ends.
- Sd
- Effective radiating area of the driver cone.
- Xmax
- Maximum linear one-way excursion of the cone.
Frequently asked questions
How long should a subwoofer port be?
For a 50 L box tuned to 35 Hz with one 10 cm port (k = 0.732): about 31 cm.
Does a longer port lower the tuning?
Yes – tuning is proportional to 1/√(effective length), so a 4× longer port tunes an octave lower.
What end correction should I use?
0.732 for a typical port flush with the baffle on one end and free inside the box; 0.85 if both ends are flanged; 0.614 if both are free.
What if the result is negative or tiny?
The port is too narrow for that box and tuning – choose a smaller diameter or a lower tuning, or use a passive radiator.
How fast can air move in a port before it chuffs?
Keep peak speed under about 17 m/s (5% of the speed of sound); flared ports tolerate a bit more.
Do I need to add the port volume to the box?
Yes – the port tube displaces air. The calculator shows the port volume and the gross volume needed.
Can I use two ports instead of one?
Yes; two ports of the same diameter double the area, so each must be about twice as long for the same tuning.