Formula
Formula used
\( L_2 = L_1 + 10 \log_{10} \left( \dfrac{4 \pi r_1^2}{Q \times 4 \pi r_2^2} \right) \)
Inputs
Result
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Understand the theory — physics concept and calculationsinteractive
The golden rule to remember: in a free field, doubling the distance loses 6 dB — no more, no less. A point source spreads its energy over an ever-larger sphere: at twice the distance, the same power covers four times the area. Drag the receiver below to watch the level drop.
The inverse-square law
A point source of acoustic power \(W\) radiates over a sphere of area \(4\pi r^2\). Intensity therefore falls off as \(1/r^2\). In decibels, the sound pressure level at distance \(r\) is:
\(L_W\) is the source sound power level and \(Q\) the directivity factor. At 1 m from an omnidirectional source (\(Q=1\)), \(L_p \approx L_W - 11\) dB. In practice, equipment usually sits on the ground (\(Q=2\), i.e. +3 dB), so a handy value is \(L_p \approx L_W - 8\) dB at 1 m.
Where do the −6 dB come from?
Between two distances \(r_1\) and \(r_2\) (same source), the difference depends only on the distance ratio:
Doubling the distance (\(r_2 = 2r_1\)) gives \(-20\log_{10}(2) = -6.02\) dB. Multiplying it by 10 gives −20 dB.
The directivity factor Q
Placing the source near a reflecting surface divides the radiation space and concentrates the energy. The gain is \(10\log_{10}(Q)\). A loudspeaker sitting on the floor is already +3 dB louder than the same speaker hanging in mid-air, for the very same power:
| Configuration | Q | Gain |
|---|---|---|
| Free space (sphere) | Q = 1 | 0 dB |
| On a plane (½ space) | Q = 2 | +3 dB |
| Edge / dihedral (¼) | Q = 4 | +6 dB |
| Corner / trihedral (⅛) | Q = 8 | +9 dB |
If the source no longer radiates into all space but against a reflecting surface, the same energy concentrates into a smaller volume: the level rises. That's the directivity factor Q. Pick the configuration:
3D unavailable here — see the table above.
Real conditions: the 1/r² law is not enough
The formula assumes an ideal free field. Outdoors and over long distances, extra terms add up (standard ISO 9613-2): atmospheric absorption (which eats the highs first), the ground effect, the weather (wind, temperature gradient) and obstacles. Indoors, beyond a certain distance the reverberant field dominates and the level stops decreasing at all. This is the critical distance: the distance at which the direct and reverberant fields are equal; beyond it, moving away from the source barely lowers the noise, and you must treat the room (absorption) rather than the distance. In short, the clean −6 dB-per-doubling rule only describes the direct field; a full prediction adds each of these correction terms one by one, which is exactly what engineering standards formalise.
Near field, far field
The 1/r² law only holds in the far field — at a distance large compared to the source size and the wavelength. Too close to a large machine, the level does not follow the −6 dB rule: you're in the near field, where pressure and particle velocity are no longer in phase. Rule of thumb: you reach the far field beyond about twice the largest source dimension, and at least one wavelength away.
Worked examples
| Situation | Result | Reading |
|---|---|---|
| 85 dB at 1 m → 10 m | 65 dB | −20·log(10) = −20 dB |
| 85 dB at 1 m → 2 m | 79 dB | one doubling = −6 dB |
| Source on the ground | +3 dB | Q = 2 vs free field |
| 90 dB at 2 m → 8 m | 78 dB | two doublings = −12 dB |
- Thinking doubling the distance halves the level: no, it's −6 dB (not −50%).
- Applying −6 dB/doubling in a reverberant room: wrong beyond the critical distance.
- Forgetting the Q factor when the source sits on the ground or against a wall.
- Using the law in the near field, too close to a large source.
Related tools
Sources : J.-C. Pascal, Vibrations et Acoustique (ENSIM, Le Mans Université); D. A. Bies & C. H. Hansen, Engineering Noise Control; ISO 9613-2 (outdoor sound propagation).