How to Calculate Sound Level and Distance

Finds the sound level further from a source as the reference level − 20×log₁₀(the ratio of distances). From a point source the level falls by about 6 dB each time the distance doubles. Decibels are logarithmic, so this works by subtraction.

This finds how far the sound level falls as you move away from a point source.

L_2 = L_1 - 20\log_{10}\dfrac{r_2}{r_1}

$L_1$ is the level at the reference distance $r_1$ , and $L_2$ the level at distance $r_2$ .

Why twenty and not ten

Sound intensity falls with the square of the distance. A decibel is ten times the base-ten logarithm of intensity, so $10\log_{10}(r^2)$ becomes $20\log_{10} r$ , and the twenty appears. Each doubling of distance costs $20\log_{10} 2 = 6.02$ dB.

Example

A source reads 90 dB at 1 m. At 10 m, $20\log_{10}10 = 20$ , so the level is $90 - 20 = 70$ dB. In terms of intensity that is one hundredth. At 2 m it is $90 - 6.02 = 83.98$ dB: twice the distance for only 6 dB.

Decibels do not add

Two sources of 60 dB together do not make 120 dB. The intensity merely doubles, which adds $10\log_{10}2 = 3$ dB, giving 63 dB. Decibels are a logarithmic scale, so what can sensibly be added and subtracted are differences.

Notes

This assumes a point source in the open with nothing to reflect off. Indoors, walls send sound back and the level does not fall this far.

A source spread along a line, such as a road, loses only 3 dB per doubling of distance.