Hong Life CalculatorsUnderstanding the noise barrier insertion loss calculator — concept and principle
A noise barrier such as a wall or earth berm reduces sound by blocking the direct line between source and receiver, so that sound must bend (diffract) over the top. This calculator estimates that diffraction loss with Maekawa's method, used for first assessments of road, rail and industrial noise walls.
How it is calculated
- Path difference δ = a + b − d, where a is the distance from source to barrier top, b from barrier top to receiver, and d the direct source–receiver distance, all in metres.
- Wavelength λ = 343 ÷ f, using the speed of sound 343 m/s (air at about 20 °C).
- Fresnel number N = 2δ ÷ λ.
- Attenuation ΔL = 10 log₁₀(3 + 20N) dB, a widely used approximation of Maekawa's chart, applied when N > −0.2; otherwise 0.
The principle
Sound that passes over a barrier travels a longer path than the blocked straight line. What matters is how long that extra path is compared with the wavelength, which is what the Fresnel number expresses. Maekawa measured attenuation behind thin screens and showed that it depends almost only on N. When the receiver just sees the top of the barrier (N = 0), the loss is already about 5 dB; it then rises by roughly 3 dB each time N doubles. Because high frequencies have short wavelengths, they give a larger N and are blocked better than low frequencies.
Worked example
- Inputs: a = 10.2 m, b = 20.4 m, d = 30 m, f = 500 Hz.
- δ = 10.2 + 20.4 − 30 = 0.6 m.
- λ = 343 ÷ 500 = 0.686 m.
- N = 2 × 0.6 ÷ 0.686 = 1.749.
- ΔL = 10 log₁₀(3 + 20 × 1.749) = 10 log₁₀(37.99) = 15.8 dB.
| Direct distance (m) | Fresnel number N |
|---|---|
| 7.5 | 67.34693878 |
| 15 | 45.48104956 |
| 22.5 | 23.61516035 |
| 30 | 1.749271137 |
| 37.5 | -20.11661808 |
| 45 | -41.98250729 |
| 60 | -85.71428571 |
How to read the result
About 5 dB is noticeable, 10 dB is often perceived as roughly halving loudness, and 15–20 dB is a large reduction. In practice a single thin barrier rarely gives more than about 20 dB, and berms about 25 dB, because sound also leaks around the ends, through gaps and by scattering in the atmosphere. Check several frequencies: a barrier that gives 16 dB at 500 Hz will give less at 125 Hz.
Common mistakes
- Measuring a and b to the barrier base instead of its top edge.
- Using a single frequency to represent broadband traffic noise.
- Applying point-source results directly to a long road, a line source that is usually attenuated less.
Frequently asked questions
Why does the barrier work better at high frequencies?
Short wavelengths make the path difference worth more wavelengths, so N and the attenuation are larger.
Does barrier thickness or material matter?
Diffraction depends on geometry, but the panel must also be airtight and heavy enough that less sound goes through it than over it.
What if the receiver can see the source over the barrier?
Then δ is negative. For slightly negative N there is still a small effect, but once N falls below −0.2 the calculator assumes no attenuation.
Sources: Z. Maekawa, "Noise reduction by screens," Applied Acoustics 1(3), 157–173, 1968 · ISO 9613-2, Acoustics — Attenuation of sound during propagation outdoors — Part 2: General method of calculation