Decibel Distance Attenuation Calculator
Sound from a point source spreads out over an ever-larger area, so its level falls predictably with distance. In free field the inverse-square law gives a 6 decibel drop for every doubling of distance. This calculator takes a known sound pressure level at a reference distance and computes the level at any new distance, along with the attenuation between them. It is the standard tool for predicting how loud a machine, loudspeaker, or alarm will be further away. The result is geometric spreading only and does not include air absorption or reflections.
Inverse-square distance formula
Attenuation = 20 * log10(r2 / r1)
Level at r2 = level at r1 - attenuation
Doublings = log2(r2 / r1)
The level difference depends only on the ratio of the two distances. Each doubling of distance subtracts 20 times log10(2), about 6.02 decibels, from a point source in free field.
Worked example
A source measured at 90 decibels at 1 metre, observed at 8 metres: attenuation = 20 * log10(8 / 1) = 18.06 decibels, so the level at 8 metres is 90 - 18.06 = 71.94 decibels. That is three doublings of distance (1 to 2 to 4 to 8 metres), about 6 decibels each.
Distance attenuation: frequently asked questions
How does sound level fall with distance?
For an ideal point source radiating into free space, sound pressure level falls by 6 decibels for every doubling of distance. This is the inverse-square law: intensity is proportional to one over distance squared, and because decibels are a logarithmic measure of pressure, the drop is 20 times the base-10 logarithm of the distance ratio.
What is the distance attenuation formula?
Level at distance r2 = level at r1 minus 20 times log10(r2 divided by r1). The attenuation between the two distances is 20 times log10(r2 divided by r1) decibels. Doubling distance (ratio 2) gives 20 times log10(2), about 6.02 decibels of loss.
Does this account for air absorption or reflections?
No. This is the geometric spreading loss for a point source in free field only. Real outdoor and indoor sound also loses energy to air absorption (significant at high frequencies and long distances) and gains energy from reflections. Use this as the free-field baseline and add atmospheric and boundary effects separately.
Why 6 decibels per doubling and not 3?
A 6 decibel change corresponds to a halving or doubling of sound pressure, which is the inverse-square behaviour of a point source. A 3 decibel change corresponds to a halving or doubling of sound power or intensity, which applies to a line source. Point sources follow the 6 decibel per doubling rule.
Official sources
- International Organization for Standardization: ISO 9613-1 attenuation of sound during propagation outdoors.
- U.S. National Institute of Standards and Technology: nist.gov.
Reviewed by the CalculatorHub team, edited by James Graham, 19 June 2026. See our methodology.