Home/Environmental Engineering/Air Quality/Noise Level Attenuation with Distance

Formula & Calculator

Noise Level Attenuation with Distance

Estimates how much sound pressure level decreases as distance from a point noise source increases (inverse square law for sound).

EnvironmentalAir QualityNoise Pollution

Noise Level Attenuation Calculatorwith Distance

L₂ = L₁ − 20·log₁₀(d₂ / d₁)
L₁ = sound level at d₁  ·  L₂ = sound level at d₂  ·  d₁ = reference distance  ·  d₂ = new distance
⟹ SolveL₁, L₂, d₁, d₂
dB
m
m
dB
Solve for:
Presets:
L₂
L₁: d₁: d₂: L₂:
✓ Copied!
Sound Level Gauge
Quiet (< 60 dB) Moderate (60–90 dB) Loud (> 90 dB)
L₂ = L₁ − 20·log₁₀(d₂/d₁)  ·  All distances in same units (e.g., meters)

Interpretation

L2 = L1 − 20×log₁₀(d2/d1). Sound level decrease with distance (inverse square law). Used to predict noise impact from point sources and to design noise barriers.

L2 = L1 - 20*log10(d2/d1)
Noise Level Attenuation with Distance

Variables

SymbolQuantityUnit
L2Sound level at new distancedB
L1Sound level at reference distancedB
d2New distance from sourcem
d1Reference distance from sourcem

What it means

This formula describes the geometric spreading of sound waves from a point source. The sound pressure level (L) decreases by 20 dB per decade of distance increase in a free‑field condition. It is derived from the inverse square law for sound intensity. This is used in environmental noise assessment to estimate noise levels at receptors (e.g., residences) from industrial sources, traffic, or construction. It does not account for atmospheric absorption or ground effects, so it provides a conservative estimate. Example: A compressor produces 90 dB at 1 m. The level at 10 m is L2 = 90 − 20×log₁₀(10/1) = 90 − 20 = 70 dB. At 100 m, L2 = 90 − 40 = 50 dB. This helps in determining setback distances for noise‑sensitive areas.

Worked example

Noise Attenuation with Distance – Two Examples

Real‑World
Scenario: A machine produces 90 dB at 1 m. What is the sound level at 10 m?
ParameterValue
L190 dB
d11 m
d210 m
1L2 = 90 - 20×log₁₀(10/1) = 90 - 20×1 = 70 dB
Result 70 dB ✓ 20 dB drop
Scenario: A generator measures 95 dB at 10 m. What is the level at 100 m?
ParameterValue
L195 dB
d110 m
d2100 m
1L2 = 95 - 20×log₁₀(10) = 95 - 20 = 75 dB
Result 75 dB ✓ Still loud
Environmental insight: Sound level decreases by 6 dB per doubling of distance (spherical spreading) – 20 dB drop per 10x distance.

Common mistakes

  • Distance ratio: d2/d1 must be greater than 1 for attenuation; if d2 < d1, the level increases (i.e., moving closer).
  • Units: d1 and d2 in the same units (e.g., metres).
  • Inverse square law: This formula assumes a point source in free field (no reflections). For line sources or indoor spaces, use different models.
  • Atmospheric absorption: Not included – for long distances, air absorption may be significant (add an extra term).
  • Background noise: The formula gives the level from the source alone; combine with background using logarithmic addition.

Applications

Noise level attenuation with distance follows the inverse square law for a point source, L₂ = L₁ − 20·log₁₀(d₂/d₁). This equation describes the reduction in sound pressure level as distance from the source increases. Environmental acousticians use it to assess noise impacts from transportation, industrial, and construction activities, and to design noise barriers or buffer zones. It is also applied in occupational health to determine safe distances from noisy equipment. By calculating attenuation, engineers can predict noise levels at receivers, evaluate compliance with local noise ordinances, and implement mitigation measures such as silencers, enclosures, or setback distances.

  • Noise impact assessment for highways, railways, and airports
  • Industrial noise control and compliance with occupational limits
  • Design of sound barriers and noise‑reducing enclosures
  • Urban planning and zoning to minimise noise exposure
  • Community complaint investigation and mitigation design

Frequently Asked Questions

Q01What is the noise level attenuation with distance formula?
A01

For a point source in free space, the sound pressure level decreases by 6 dB per doubling of distance (inverse square law). The formula is L2 = L1 – 20·log10(d2/d1), where L1 is the level at distance d1, and L2 is the level at distance d2.

Q02What is the common mistake when using this formula?
A02

Applying the point‑source inverse‑square formula to line sources (like highways) or in environments with significant reflections, where a different attenuation rate applies. For line sources, the level drops by 3 dB per doubling of distance (cylindrical spreading).

Q03What are the typical attenuation rates for different source types?
A03

  • Point source (spherical spreading): 6 dB per doubling of distance.
  • Line source (cylindrical spreading): 3 dB per doubling of distance.
  • Infinite plane source: no attenuation with distance.

Q04How do you calculate the noise level at 50 m if the level at 10 m is 80 dB?
A04

L2 = 80 – 20·log10(50/10) = 80 – 20·log10(5) = 80 – 20×0.699 = 80 – 13.98 ≈ 66 dB.

Q05What factors, besides distance, affect noise attenuation?
A05

  • Atmospheric absorption (depends on frequency, temperature, humidity).
  • Ground reflection and absorption.
  • Obstructions (buildings, barriers).
  • Wind and temperature gradients (refraction).

Q06How do you account for barriers in noise attenuation?
A06

Barriers (walls, berms) can reduce noise by diffraction. The attenuation depends on the path difference between the direct and diffracted paths. Simple empirical formulas or modelling software are used for accurate prediction.

Q07What is the difference between noise level and sound power level?
A07

Sound pressure level (Lp) is what we measure (dB) and depends on distance. Sound power level (Lw) is the total acoustic energy emitted by the source and is independent of distance. The formula relates Lp to Lw and distance.

Q08What is the significance of the 20·log10 term?
A08

The factor 20 comes from the definition of decibels for pressure ratios. Since sound pressure is inversely proportional to distance (for a point source), the level difference is 20·log10(d2/d1).

Q09How do you calculate the distance at which a certain noise level is reached?
A09

Rearrange the formula: d2 = d1 × 10^((L1 – L2)/20). For example, if L1=90 dB at 1 m, and you want L2=70 dB, d2 = 1 × 10^((90‑70)/20) = 1 × 10^1 = 10 m.

Q10What are the limitations of this simple attenuation formula?
A10

It assumes free‑field propagation with no reflections, no absorption, and no barriers. In real outdoor environments, more complex models (e.g., ISO 9613) are used. The formula is a good approximation for open areas at moderate distances.