Distance Attenuation Calculator

Calculate how sound level (or any point/line-source radiation) falls off with distance using the inverse-square law, Lp₂ = Lp₁ − 20·log₁₀(d₂/d₁).

Quick Facts

Point source
Lp₂ = Lp₁ − 20·log₁₀(d₂/d₁)
Spherical spreading: intensity ∝ 1/d², so level drops ≈6.02 dB every time distance doubles.
Line source
Lp₂ = Lp₁ − 10·log₁₀(d₂/d₁)
Cylindrical spreading (e.g. traffic, pipelines): intensity ∝ 1/d, so level drops ≈3.01 dB per doubling.
Intensity ratio
I₂/I₁ = 10^(−ΔL/10)
A 6 dB drop leaves ~25% of the original intensity; a 20 dB drop leaves 1%.
Beyond geometry
Real-world loss = geometric + extra
Air absorption, ground effect, and barriers add on top of the inverse-square term (see ISO 9613-2).

Your Results

Calculated
Level at Target Distance
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Lp₂ = Lp₁ − total attenuation
Total Attenuation
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Geometric spreading + additional loss (dB)
Intensity Remaining
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I₂/I₁ as a percent of the reference intensity
Distance Doublings
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log₂(d₂/d₁), each worth ~6 dB (point) or ~3 dB (line)

Ready

Enter a reference level and distance, a target distance, and press Calculate.

How Distance Attenuation Works

Any wave or radiation spreading out from a source loses strength with distance simply because the same energy is spread over a larger area — no absorption required. For a small ("point") source radiating equally in all directions, that area grows with the square of the distance, so sound pressure level, light intensity, or radio-signal power all fall off following the same inverse-square law. This calculator converts a known level at a reference distance into the level at any other distance, using the standard decibel form of that law, and lets you add a source type and an extra loss term for more realistic outdoor scenarios.

The inverse-square law and the decibel formula

For a point source in a free field (no reflecting surfaces), intensity I is proportional to 1/d². Because sound pressure level is defined as Lp = 20·log₁₀(p/p₀), and pressure is proportional to √I, the level at distance d₂ relative to a known level at d₁ works out to Lp₂ = Lp₁ − 20·log₁₀(d₂/d₁). Doubling the distance (d₂ = 2d₁) gives a drop of 20·log₁₀(2) ≈ 6.02 dB — the well-known "6 dB rule." The same inverse-square math applies to any point-source quantity, including illuminance (lux) and radiated power density (W/m²).

Choosing point vs. line source

Not every source is a point. A long, continuous source — a busy highway, a pipeline, or a conveyor line — radiates more like a cylinder than a sphere, so its intensity falls off as 1/d instead of 1/d². That gives Lp₂ = Lp₁ − 10·log₁₀(d₂/d₁), a drop of only 10·log₁₀(2) ≈ 3.01 dB per doubling of distance — line sources "carry" much farther than point sources for the same source strength. Pick "line source" only when the source is genuinely long relative to your measurement distance; otherwise the point-source formula is the correct default.

Adding real-world losses

The formulas above capture geometric spreading only — the loss you'd see in an open field with no wind, humidity, or obstacles. Real outdoor propagation also loses energy to atmospheric (air) absorption, which grows with distance, frequency, and dryness; to ground effect from reflecting or absorbing terrain; and to barriers, walls, or vegetation. Standards such as ISO 9613-2 give detailed methods for estimating each term. Rather than model each one individually, this calculator's "Additional attenuation" field lets you add a single lump-sum estimate (in dB) for all of those extra losses combined, on top of the geometric-spreading result.

Frequently Asked Questions

How much does sound level drop when I double the distance from a point source?
For a point source radiating spherically in a free field (no reflections), sound pressure level drops by 20·log₁₀(2) ≈ 6.02 dB every time the distance doubles. This is the classic 6 dB rule and comes directly from the inverse-square law, since sound intensity falls off as 1/d².
What is the difference between a point source and a line source for distance attenuation?
A point source (a loudspeaker, machine, or explosion) radiates spherically, so level falls with 20·log₁₀(d₂/d₁) — about 6 dB per doubling of distance. A line source (a long line of highway traffic, a pipeline, or a conveyor) radiates cylindrically, so level falls more slowly, with 10·log₁₀(d₂/d₁) — about 3 dB per doubling.
Does this calculator include air absorption, ground effect, or barrier losses?
The base formula only models geometric spreading (the inverse-square law). Real outdoor sound propagation also loses energy to atmospheric absorption, ground effect, and barriers or screens — standards such as ISO 9613-2 detail those terms. Use the optional "Additional attenuation" field to add a lump-sum estimate of those extra losses in dB.
Can I use this calculator for light or radio-signal intensity instead of sound?
Yes — the inverse-square law governs any point-source wave or radiation, including light intensity and radio signal power. Just enter the reference level in the appropriate units (for example dB, lux, or W/m²) at your reference distance; the point-source geometric spreading math is identical. For radio links specifically, a dedicated free-space path loss calculation also factors in frequency/wavelength.