Horns, step by step

Start by observing: what moves?

Imagine a line of people passing a gentle push from shoulder to shoulder. The disturbance travels along the line without anyone walking to the other end. In sound, small regions of air oscillate around their resting positions while a pressure pattern travels.

The diaphragm first compresses the air and then retreats. Compression and rarefaction alternate. A horn guides this alternating motion. It does not work like a garden hose carrying a steady outward flow.

Watch the red dot below. It stays near its starting point while the pattern travels. Motion is greatly enlarged and slowed down; this is not a real displacement scale.

From the driver through the horn

The compression driver creates the vibration. Its diaphragm sits inside the driver body. A phase plug guides sound through several passages to the exit, helping limit path-length differences. It is not an electrical phase switch.

The throat is the small inlet. The profile describes how the horn expands. The mouth is the large opening into the room. Here, length L means the distance along the centre axis, not the curved wall length.

“One inch” describes the nominal driver exit, not its diaphragm diameter. B&C specifies a 25 mm throat for the DE250. Our teaching calculation deliberately uses 25.4 mm, the exact inch conversion. This is a learning example, not a machining dimension.

Why use a horn?

A small sound source does not couple equally well to its surroundings at every frequency. A horn changes the acoustic load seen by the driver. It also changes the directions into which sound is radiated. Keep these two effects separate.

Acoustic impedance is the ratio of sound pressure p to volume velocity U: Z_a = p/U. Pressure is the alternating deviation from ambient pressure. Volume velocity is the volume of air moving back and forth through a section per unit time. Both quantities also have a phase.

A horn creates no extra energy. Higher sound level in front may result from better energy transfer, stronger directivity, or both. One on-axis microphone reading cannot establish efficiency on its own.

Frequency → time → wavelength

Frequency f counts complete cycles per second. 1,000 Hz means 1,000 cycles each second. Period T is the time taken by one cycle. During that time the sound pattern travels one wavelength λ.

1. Use c = 343 m/s for air at roughly 20 °C.
2. At f = 1,600 Hz, find T = 1/f.
3. Substitute: T = 1/1,600 s = 0.000625 s = 0.625 ms.
4. Distance = speed × time: λ = c × T.
5. λ = 343 m/s × 0.000625 s = 0.214375 m.
6. Convert metres to centimetres: 0.214375 × 100 = 21.44 cm.

Doubling f to 3,200 Hz halves λ to 10.72 cm. The speed of sound does not double.

From diameter to area

Horn calculations need cross-sectional areas. Comparing diameters alone is insufficient: area grows with diameter squared. Start with a circular throat of d = 25.4 mm.

1. Halve the diameter: r = d/2 = 25.4/2 = 12.7 mm.
2. Square it: r² = 12.7 × 12.7 = 161.29 mm².
3. Circle area: S_t = πr² = 3.14159 × 161.29 = 506.71 mm².
4. 1 cm² = 100 mm². Thus 506.71/100 = 5.067 cm².
5. 1 m² = 1,000,000 mm². Thus S_t = 0.0005067 m².

A rectangular example mouth of 300 × 180 mm would have an area of 54,000 mm² = 540 cm². Its area ratio would be 54,000/506.71 = 106.57. Rounded corners reduce the actual open area. This ratio is neither sound-level gain nor the driver's compression ratio.

Building the profile step by step

In a simple exponential model, area grows by the same factor over equal distances. The equation is S(x) = S_t · e^mx. Here x is distance from the throat and m controls the rate of expansion. The equation specifies area, not the outline of a rectangular horn.

Our freely chosen teaching model has a 25.4 mm throat, a circular 200 mm mouth and 200 mm length. These are explicitly not verified dimensions of the LNGR V10 horn.

1. S_t = 506.71 mm²; S_m = π × 100² = 31,415.93 mm².
2. Area ratio R = S_m/S_t = 62.0001.
3. L = 200 mm = 0.2 m.
4. From R = e^mL, ln(R) = mL. The natural logarithm ln reverses exponentiation.
5. m = ln(62.0001)/0.2 = 4.12714/0.2 = 20.6357 m⁻¹.
6. At 50 mm: x = 0.05 m; mx = 1.03179.
7. e^1.03179 = 2.8061; S(50 mm) = 506.71 × 2.8061 ≈ 1,421.9 mm².
8. Back to diameter: d = 2√(S/π) ≈ 42.55 mm.

Each 50 mm multiplies area by 2.806. Diameter only grows by √2.806 ≈ 1.675. Halfway along, the diameter is therefore not the arithmetic mean of throat and mouth diameters.

Three different frequency limits

Model cutoff: An ideal infinitely long exponential horn has f_c = mc/(4π). With our definition S(x) = S_te^mx:

1. m × c = 20.6357 m⁻¹ × 343 m/s = 7,078.05 s⁻¹.
2. 4π = 12.5664.
3. f_c = 7,078.05/12.5664 ≈ 563 Hz.

Real horn: It ends after 200 mm. Mouth size, profile, reflections and driver affect the actual response. The 563 Hz value is not a guaranteed operating limit, a −3 dB point or a recommendation for the DE250.

Crossover frequency: This belongs to the complete system of driver, horn, woofer and filters. B&C lists a recommended 1.6 kHz crossover for the DE250. Clean, reliable operation in a specific combination still requires appropriate filtering and measurements. “The horn model gives 563 Hz, so I will cross there” is incorrect.

Wall angle is not coverage angle

For a straight-sided geometric example, throat radius is 12.7 mm, mouth radius 100 mm and length 200 mm. Radius increases by 87.3 mm. The half wall angle α follows tan(α) = 87.3/200 = 0.4365. Thus α = arctan(0.4365) ≈ 23.58°. The full geometric angle is 47.16°.

This calculation describes a straight wall. Our exponential horn has a curved wall with changing slope. Even a straight wall does not guarantee a 47° acoustic coverage angle.

Compare mouth diameter D = 200 mm with λ. At 500 Hz, λ = 686 mm and D/λ = 0.29. At 5,000 Hz, λ = 68.6 mm and D/λ = 2.92. At high frequencies the opening is much larger relative to a wavelength. This helps explain frequency-dependent directivity, but does not supply an exact coverage angle.

Understanding a higher sound level

Decibels express a ratio logarithmically. For sound pressures, ΔL = 20 log₁₀(p₂/p₁). Doubling pressure at the microphone gives p₂/p₁ = 2; log₁₀(2) = 0.30103; 20 × 0.30103 = 6.02 dB.

For powers use 10 log₁₀(P₂/P₁). Double power gives 10 × 0.30103 = 3.01 dB. Applying this to electrical amplifier requirements assumes, among other things, the same load and linear operation.

A 6 dB higher on-axis level does not automatically establish four times the total efficiency: the horn may concentrate energy into a narrower angle. Off-axis measurements are therefore part of the picture.

Surface: why not golf-ball dimples?

The analogy initially sounds plausible: a textured surface can change a flow. A horn has a different starting condition. In normal small-signal operation, air oscillates rather than flowing continuously at high speed around a ball. Golf-ball behaviour therefore does not demonstrate a horn improvement.

A scale comparison helps without pretending to predict an effect. At 10,000 Hz, λ = 343/10,000 m = 34.3 mm. A 0.2 mm print step is 0.2/34.3 = 0.00583, or 0.58% of a wavelength. A 2 mm depression is already 5.83%. Those percentages are not predictions of frequency-response error.

For a meaningful experiment, first make an airtight horn with a clean throat transition and no burrs. Then measure a surface variant with otherwise identical geometry and level. Do not infer “rough sounds better” or “every print line is audible” from a drawing.

From teaching model to LNGR prototype

A good-looking CAD profile establishes a geometric idea. Acoustic testing must establish the driver transition, resonances and off-axis behaviour. This learning path contains no invented measurement curves and does not validate the separate V10 design.

1. Check mechanics: actual driver exit, gasket, fasteners and a smooth transition. Support the driver independently instead of hanging its weight from a thin throat.
2. Document microphone position, distance, angle, filters and amplifier setting. Start with a suitable high-pass filter and low level.
3. Measure on axis, then horizontally and vertically at, for example, 15°, 30°, 45° and 60°. Keep microphone distance from the rotation axis constant.
4. Use a common reference for angle comparisons. Normalising every curve separately to 0 dB would hide real level differences.
5. Gate room reflections if the measurement software allows it. A short time window limits low-frequency validity.
6. Add distortion measurements at documented levels. Only then decide crossover, equalisation and integration with the woofer.

The existing LNGR measurement equipment includes a Scarlett 2i2 and Sonarworks measurement microphone. Check the calibration file and device settings. The exact DSP and Fosi boards still need confirmation before a definitive wiring guide.

Calculate and explain it yourself

Try these tasks before opening the worked answers. Use c = 343 m/s and include units on every calculation line.

1. What is the wavelength at 2,000 Hz?
2. A circular throat increases from 25 to 50 mm diameter. How does area change?
3. Our exponential model keeps its throat and mouth but becomes 400 rather than 200 mm long. What happens to m and ideal cutoff?
4. Does that allow the DE250 to operate at the new model cutoff?
5. Why is the red dot in the first animation not a parcel flying from driver to ear?

Sources and model limits

The calculations are independently worked teaching examples. Reference reading: B&C DE250 · Bjørn Kolbrek: Horn Theory.

No driver simulation, CAD approval or measured response is implied. Progress is stored only in this browser; if storage is unavailable, the page remains usable.