Sound level vs distance for point and line sources. Forward and inverse modes. Free-field model.
Assumptions
Idealized theoretical model: point source, free field, no reflections, no ground absorption. In real environments with obstacles or reflective surfaces, observed attenuation will differ. For regulatory assessments, measure rather than calculate.
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Why doubling the distance reduces sound by 6 dB (point source)
A point source radiates sound energy uniformly in all directions, creating a spherical wavefront. The acoustic intensity — power per unit area — is distributed over the surface of that sphere. Since the surface area of a sphere grows as the square of the radius (A = 4πr²), doubling the distance quadruples the area over which the same total power is spread. Because intensity is inversely proportional to distance squared, this is the inverse-square law.
In decibels, intensity is: L = 10 × log₁₀(I/I₀). When distance doubles, intensity is divided by 4, so the change in level is: ΔL = 10 × log₁₀(1/4) = 10 × (−0.602) ≈ −6 dB. For every decade of distance (×10), the drop is 20 × log₁₀(10) = −20 dB. The general formula: L₂ = L₁ − 20 × log₁₀(d₂ / d₁).
Note: this applies only in a free field (no reflections, no boundaries). In practice, ground reflection, barriers, and room acoustics all modify the actual attenuation.
Point source vs line source — when to use each model
A point source model applies when the source is small relative to the distance — a single machine, an HVAC exhaust, a compressor, or any source you can approximate as a dot in space. Sound energy radiates spherically, and doubling the distance gives −6 dB.
A line source model applies when the source is elongated compared to the observation distance — a busy highway, a rail line, a long conveyor, or a row of machines. Sound energy radiates cylindrically, and the active area grows linearly with distance (A = 2πrL), not as a square. This gives only −3 dB per doubling: L₂ = L₁ − 10 × log₁₀(d₂ / d₁).
Rule of thumb: if your observation point is closer to the source than the source is long, use a point source. As you move farther away (distance >> source length), the line source approximation becomes valid.
Frequently Asked Questions
Does this calculator account for ground absorption?
No. This is a free-field model only. Ground absorption and ground reflection (the excess attenuation term Agr in ISO 9613-2) are not included. Soft ground (grass, soil) can add several dB of extra attenuation over hard ground (asphalt, concrete), especially at low frequencies and grazing angles. For outdoor propagation predictions that include ground, atmospheric absorption, barriers, and foliage, use the full ISO 9613-2 method.
When should I use the line source model?
Use the line source model for extended sources such as highways, rail lines, pipelines, or long rows of identical machines — when the receiver is much closer to the source than the source is long. If the source length is L and your distance is r, the line source model is appropriate when r << L. When r >> L, the source appears as a point and the point-source model applies. In the transition zone, neither model is exact.
What about indoor reverberant fields?
This tool does not apply to reverberant fields. In an enclosed space, reflected sound adds to direct sound, and beyond a certain distance (the reverberation radius or critical distance), the reverberant field dominates and level no longer drops with distance. For indoor noise assessment, use room acoustics models (Sabine equation, ISO 11690-1) and measure the reverberation time T60 of the space. The direct-field attenuation formula applies only close to the source, inside the critical distance.
Why is the calculation different from the inverse-square law for radiation?
The inverse-square law is the same physical principle — intensity drops as 1/r². The difference is in what you measure. For ionizing radiation, dose rate (in mSv/h) drops as 1/r², so doubling distance divides dose rate by 4. For sound, we express level in decibels using a logarithmic scale. The same 1/r² drop in intensity gives −10 × log₁₀(4) ≈ −6 dB when distance doubles. In both cases the physics is identical; the apparent difference is purely a unit convention (linear vs. logarithmic scale).