Module 01 · Fundamentals · about 20 minutes
How diaphragm movement creates sound
A woofer sits on your workbench. Cone, basket, magnet: it already looks like a loudspeaker. Why put a box around it?
The answer starts with the air in front of and behind the diaphragm. This module explains how a diaphragm creates sound, why a bare woofer struggles in the bass and what an enclosure changes. No prior knowledge is needed. Paper and a pen are enough.
1 · The disturbance travels; the air stays nearby
Imagine people standing in a line with small gaps between them. The first person briefly moves forwards, the next reacts, then the next. A disturbance travels down the line although nobody walks from one end to the other.
Sound in air works in a similar way. Air particles move back and forth over small distances. Regions of slightly higher and lower pressure form, and these pressure changes spread. At your ear, they make your eardrum vibrate.
A loudspeaker does not send a lump of air across the room. The analogy has limits: air is three-dimensional and its particles already move randomly. The additional back-and-forth motion represents the sound motion.
2 · What moves the diaphragm?
The diaphragm is the moving surface of the driver. In a conventional moving-coil loudspeaker, it is attached to a voice coil, a winding of wire in a magnetic field. The amplifier supplies a changing electrical signal. Current through the coil produces a force. Reversing the current reverses the force, moving the coil and diaphragm backwards and forwards.
For this first model, we treat the diaphragm as a rigid surface moving as one piece. Real drivers do not behave exactly like this at every frequency.
Moving forwards, the diaphragm displaces air in front and creates more space behind it. In a simplified picture, the air in front is compressed while the air behind expands. Moving backwards reverses the situation. Both sides produce sound contributions of opposite polarity. The rear is not silent.
Introductory schematic, not to scale. This is not a complete description of pressure and phase in a sound field.
3 · Moving faster and moving further are different things
| Term | Meaning | Example |
|---|---|---|
| Frequency f | Complete cycles per second | 100 Hz = 100 cycles per second |
| Displacement | Instantaneous distance from the resting position | The diaphragm is currently 1 mm forwards. |
| Amplitude | Maximum displacement of the oscillation considered | −1 to +1 mm: 1 mm amplitude, 2 mm peak-to-peak |
For a pure tone, frequency determines pitch. At the same frequency and under otherwise unchanged conditions, greater displacement normally produces a higher sound level, provided the driver stays linear. However, “1 mm displacement” alone does not determine the level. Frequency, diaphragm area, mounting arrangement and listening distance also matter.
4 · Why rear radiation interferes with the bass
With a bare driver, there is a short path around the edge from rear to front. The sound contributions combine. When a positive pressure fluctuation meets a negative one of similar magnitude, the resulting fluctuation is smaller. This cancellation can be particularly strong in the bass. It is sometimes called an acoustic short circuit, which has nothing to do with an electrical short circuit.
Low frequencies have long wavelengths. The path around a small driver is short by comparison. Its travel time changes the initially opposing phase relationship only slightly, allowing substantial cancellation. This does not mean silence everywhere: frequency, geometry and observation direction all affect the result.
5 · Now the equation has a purpose
Wavelength λ is the distance between successive points at the same stage of a wave, such as two pressure peaks. For air at approximately 20 °C, we use a sound speed c of 343 metres per second.
Wavelength in m = sound speed in m/s ÷ frequency in 1/s
For 100 Hz:
The seconds cancel, leaving metres. A driver only a few tens of centimetres across is small compared with this wavelength.
| Frequency | Wavelength |
|---|---|
| 50 Hz | 6.86 m |
| 100 Hz | 3.43 m |
| 1,000 Hz | 0.343 m = 34.3 cm |
6 · What an enclosure changes
A sealed enclosure separates front and rear radiation. Rear sound can no longer take the short route around the basket to the front. The trapped air also acts as an additional spring: compression or expansion creates a restoring force on the diaphragm.
The enclosure is an acoustic component. Under otherwise comparable conditions, a smaller sealed cabinet has a stiffer air spring. Its volume changes the behaviour of the vibrating system. Neither smaller nor larger is universally correct.
| Design | Principle |
|---|---|
| Sealed | Separates front and rear; the enclosed air acts as a spring. |
| Bass reflex | Air in the cabinet and port forms a resonant system. The port contributes strongly around the tuning region. |
| Open baffle | Extends the path between front and rear. Rear sound continues to radiate. |
A bass-reflex port is not an arbitrary hole: its dimensions must work with the cabinet volume. A loudspeaker does not necessarily need a closed box. It needs a mounting arrangement suited to the driver and design goal.
7 · Your transfer activity
Finish · 2 minutes: Without looking back, complete: “The diaphragm creates sound by … Behind the diaphragm … An enclosure changes …”
Model limits: small oscillations, a simplified rigid diaphragm and direct radiation without room effects. All numbers are teaching examples, not measurements from an LNGR project.
Module 02 · Enclosure design · about 20–25 minutes
From outside dimensions to actual net volume
320 × 320 × 300 millimetres sounds like a clearly defined cabinet. It still does not tell us how much air can work inside.
The walls occupy space, followed by the driver, braces and electronics. In this practical exercise, you calculate the available internal volume step by step. You need basic arithmetic, paper, a pen and a calculator. For the physical activity, add a ruler and a small box.
1 · Outside volume is not working volume
Multiplying the external dimensions gives:
0.03072 m³ × 1,000 l/m³ = 30.72 l
Those 30.72 litres include all the walls. Gross internal volume is the space inside the walls before components are deducted. Net volume remains after subtracting their displacement.
2 · The design task
| Quantity | Value |
|---|---|
| Outside width / height / depth | 320 / 320 / 300 mm |
| Material thickness on all six sides | 15 mm |
| Woofer displacement | 1.20 l |
| Brace displacement | 0.45 l |
| Electronics displacement | 0.35 l |
| Target net volume | 20.0 l |
Assume a simple rectangular cabinet with equally thick walls. The displacement figures are given teaching values, not specifications for a particular driver or construction plan.
3 · Try the calculation first
4 · The complete worked solution
Open Part 1: Internal dimensions
Left and right panels bound the width. Top and bottom bound the height; front and rear bound the depth. Subtract twice the wall thickness in every direction.
hinside = 320 − 2 × 15 = 290 mm = 0.290 m
dinside = 300 − 2 × 15 = 270 mm = 0.270 m
Open Part 2: Gross volume and cross-check
First calculate the internal cross-sectional area:
Then multiply by the internal depth:
0.022707 m³ × 1,000 l/m³ = 22.707 l ≈ 22.71 l
Check in millimetres:
84,100 mm² × 270 mm = 22,707,000 mm³
22,707,000 mm³ ÷ 1,000,000 mm³/l = 22.707 l
Both methods give the same result. Round at the end so small rounding differences do not propagate through later steps.
Open Part 3: Net volume and required depth
The components displace:
Vnet = 22.707 − 2.00 = 20.707 l ≈ 20.71 l
At least 20 litres are available, but the design does not provide exactly 20.0 litres. The excess is 20.707 − 20.0 = 0.707 l, approximately 0.71 l.
Adding the port:
20.000 − 19.707 = 0.293 l missing
The reserve is insufficient. With unchanged internal width and height, calculate the additional depth:
Δd = 293,000 mm³ ÷ (290 mm × 290 mm)
Δd ≈ 3.48 mm
The outside depth would increase from 300 to approximately 303.48 mm. A practical 305 mm depth is easier to build. It gives 290 × 290 × 275 ÷ 1,000,000 − 3.00 = 20.1275 l. This includes a small reserve, rather than giving exactly 20 litres.
Without the port, exactly 20 litres net requires 22 litres gross:
doutside ≈ 261.59 + 30 = 291.59 mm
Alternatively, retain the original depth and add a solid component displacing 0.707 litres. Whether such an adjustment is acoustically necessary depends on the specific design.
5 · Understand volume using a small box
Measure a rectangular box externally and determine its wall thickness. Calculate its internal volume. Put an eraser or wooden block inside, measure the object and calculate its volume.
6 · Avoid four common mistakes
- Subtracting only one wall
- Each direction has two opposing boundaries. For equal thicknesses: inside dimension = outside dimension − 2 × wall thickness.
- Confusing millimetres and litres
- Volume is three-dimensional. 1,000 mm equals one metre, but 1,000,000 mm³ equals one litre.
- Estimating drivers and ports too loosely
- Count only the portion inside the enclosure. The external envelope of an open driver basket is not entirely solid material. Use suitable manufacturer data when possible. An internal bass-reflex port is deducted including its walls and air passage: port air belongs to the resonator rather than the compliant air volume of the chamber. Do not subtract shared walls twice.
- Treating damping material as a wooden block
- Loose damping material consists mostly of air. Its full apparent volume is not deducted as a solid block. It may increase the effective acoustic compliance. Geometric net volume and acoustically effective volume are therefore not always identical; the effect depends on material, amount, placement and frequency.
Teaching model, not an approved LNGR construction plan. Real designs must also account for passive radiators, terminal plates, cable ducts and separate electronics compartments. A volume figure alone does not establish whether a particular driver will work well in it.
Continue learning
References
Further reading on enclosure systems and loudspeaker fundamentals:
Martin Colloms & Paul Darlington: High Performance Loudspeakers, 6. edition, Wiley, 2005. ISBN 0-470-09430-3.
The calculations on this page use the explicitly stated teaching values. The book is listed for further reading; no full text is included.