The Inverse Square Law for Sound, and Why Doubling the Distance Costs 6 dB

Walk twice as far from a loudspeaker in the open and the level drops by 6.02 dB — not by half, and not by 3 dB. This takes a level measured at one distance and gives you the level at another, the intensity and pressure ratios behind it, the number of doublings, and the distance you would have to retreat to reach a level you name. It also says plainly where the law stops being true, which is indoors.

Last updated 11 Nov 2026

Level at a new distance, and the distance for a target level

Free field, 20 log₁₀ of the distance ratio

Sound level measured at the reference distanceIn decibels. For sound pressure level the reference is 20 µPa and the definition is 20 log₁₀(p/20 µPa), as UCL’s acoustics course states it. The inverse square law does not care which dB scale you use, as long as both ends of the comparison use the same one: it is a difference, and the reference cancels. 100 dB at one metre is about what a small PA speaker or a chainsaw does, and it is chosen here because the arithmetic to 10 m is exactly 20 dB.

Distance that level was measured atManufacturers quote loudspeaker sensitivity at 1 m, which is why 1 m is the default. A level quoted without a distance is meaningless, and that is the single most useful thing on this page: “this speaker does 100 dB” is not a statement about anything until someone says where. The law is a ratio of distances, so the units cancel — metres, feet or furlongs, as long as both boxes agree.

Distance you want the level atLarger than the first distance gives a drop, smaller gives a rise, and the page is happy either way — the formula is symmetric. Moving closer is the dangerous direction: from 4 m to 0.5 m is three doublings the other way and adds 18 dB. Very small values are a fiction, because within about one source dimension of a real loudspeaker you are in its near field and the law does not hold at all.

A level you would like to get down toDoes not affect the headline. The page solves the law backwards and tells you how far from the source that level occurs. 85 dB is the default only because it is a round number people recognise; this page makes no claim about hearing damage, exposure limits or any legal standard, all of which depend on duration as well as level and are the subject of the occupational-noise pages elsewhere on this site, not of this geometry page.

The sound spreading out from the source on the left. The three grey arcs are at the reference distance, twice it and four times it — each one 6.02 dB quieter than the last. The red arrow and dot mark where YOU are, and they slide along the axis as you change the second distance; the red arc is your own wavefront while it still fits on the page. The green dot is the distance at which the level would fall to your target.

80.00dBExample

100 dB measured at 1 m, asking for the level at 10 m, with 85 dB as the target

One subtraction, and the 20 that everybody wants to make a 10

L₂ = L₁ − 20 log₁₀(r₂/r₁) · I ∝ 1/r² · p ∝ 1/r · one doubling = 20 log₁₀2 = 6.0206 dB · r = r₁ × 10^{(L₁−L)/20}

the geometry: a point source radiates equally in all directions, so the power it emits is spread over a sphere of area 4πr². Double the radius and you quadruple the area, so the intensity — power per unit area — falls to a quarter. That is the whole of the inverse square law and it is geometry, not acoustics; light, gravity and radiation do the same thing for the same reason
why 20 and not 10: intensity goes as 1/r², and 10 log₁₀ of a quarter is −6.02 dB. Pressure goes as 1/r, and 20 log₁₀ of a half is the same −6.02 dB. Both routes give 6 dB, which is the point: the factor of 20 in the pressure definition exists precisely so that the two agree. Using 10 log₁₀(r₂/r₁) gives 3 dB per doubling and is the commonest error on this topic
log₁₀: base ten, every time. This matters more than it looks: in most programming languages and in this site’s own expression engine, log means the natural logarithm, so a decibel formula written with log comes out 2.303 times too large. Every decibel expression on this page is written log10 for that reason
free field: the assumption that nothing reflects. Outdoors in the open, good. Indoors, good only out to the critical distance, after which the reverberant field takes over and the level flattens out; a hard-surfaced hall can be almost uniform in level. A half space — a source sitting on hard ground — is 3 dB louder than free field everywhere, because the same power goes into half the sphere, but the 6 dB per doubling is unchanged
what it does not include: air absorption, which removes high frequencies over long distances; ground effect; wind and temperature gradients, which bend sound rays; and barriers. This page is the geometric floor under all of those, and in practice the real level outdoors at long range is below what it predicts, never above

Worked example

100 dB measured at 1 m, asking for the level at 10 m, with 85 dB as the target
THE RATIO FIRST, AND KEEP IT A RATIO. r₂/r₁ = 10/1 = 10. Do not subtract distances, ever. The law is about the factor by which you moved away, not the number of metres, which is why going from 1 m to 2 m costs exactly as much as going from 50 m to 100 m.
TWENTY TIMES THE LOG, BASE TEN. drop = 20 log₁₀(10) = 20 × 1 = 20.00 dB. The factor of ten in distance is the tidiest case in the subject: ten times further is always exactly 20 dB quieter.
SUBTRACT. L₂ = 100 − 20 = 80.00 dB. Sanity check it the other way: 80 dB is a hundredth of the intensity of 100 dB, and a hundredth is exactly what spreading over 10² times the area gives you.
CHECK IT IN DOUBLINGS. Ten is 3.3219 doublings, because log₂10 = 3.3219. At 6.0206 dB each, that is 3.3219 × 6.0206 = 20.00 dB. Same answer by a different road, which is how you know the 6 dB rule and the 20 log rule are one rule.
NOW RUN IT BACKWARDS FOR THE TARGET. To lose 100 − 85 = 15 dB you need r = 1 × 10^15/20 = 10^0.75 = 5.623 m. Notice that 15 dB costs you less than six metres while the next 15 dB — down to 70 dB — would cost you 31.6 m. Every 20 dB is another factor of ten, so distance buys decibels at a brutally diminishing rate.
AND THE TRAP. Had you written 10 log₁₀(10) you would have got 10 dB and an answer of 90 dB. Had you written log rather than log₁₀ in a spreadsheet or in most programming languages, you would have got 20 ln(10) = 46.05 dB and an answer of 53.95 dB. Both are wrong, both look plausible, and only the second one looks obviously odd.

What distance buys you, from 100 dB at 1 m

Three patterns worth memorising from one table. **Every doubling is 6.02 dB** — look at the 1, 2, 4 rows. **Every factor of ten is 20 dB** — look at 1, 10, 100, 1000. And **the first row is the one that hurts**: moving half a metre closer to a source that is already a metre away adds as much as moving from 4 m to 2 m takes off. The last column is the physical content: at a kilometre you are receiving a millionth of the intensity you got at a metre, and you still hear it, which is a fair advertisement for the dynamic range of an ear.
Distance	Ratio to 1 m	Doublings	Drop, dB	Level, dB	Intensity fraction
0.5 m	0.5×	−1	−6.02	106.02	4
1 m	1×	0	0.00	100.00	1
2 m	2×	1	6.02	93.98	1/4
4 m	4×	2	12.04	87.96	1/16
10 m	10×	3.32	20.00	80.00	1/100
32 m	32×	5	30.10	69.90	1/1024
100 m	100×	6.64	40.00	60.00	1/10 000
1000 m	1000×	9.97	60.00	40.00	1/1 000 000

Six decibels, and the three ways people get it wrong

Stand a metre from a loudspeaker and walk away. The level you hear falls, and it falls for a reason that has nothing to do with sound: the energy the source pushes out has to spread over the surface of an ever-larger sphere, and the area of that sphere goes as the square of the radius. Double the radius, quadruple the area, quarter the intensity. That is the inverse square law, and it is geometry. Light does it. Gravity does it. Gamma rays do it. The physics in this page is entirely in the bookkeeping that turns a quarter into a number of decibels.

Why a quarter is six decibels. The decibel is a logarithmic ratio, and sound is quoted two different ways. As an intensity ratio the level is 10 log₁₀(I/I₀), and a quarter gives 10 log₁₀(0.25) = −6.02 dB. As a pressure ratio it is 20 log₁₀(p/p₀), and since pressure falls as 1/r rather than 1/r², a half gives 20 log₁₀(0.5) = −6.02 dB. The same number by two routes, which is not a coincidence — the factor of 20 in the pressure definition is chosen so the two scales agree, because intensity goes as pressure squared.

Three errors, in order of how often they appear. First, using 10 log₁₀ of the distance ratio and getting 3 dB per doubling. The 10 belongs with intensity, the 20 with pressure, and the distance ratio behaves like a pressure ratio. Second, using natural logarithms: almost every programming language and spreadsheet defines log as ln, and this site’s own expression engine does too, so every decibel expression here is written log10 deliberately. Third, subtracting distances rather than dividing them. Ten metres further away is not a fixed penalty; going from 1 m to 11 m costs 20.8 dB and going from 100 m to 110 m costs 0.83 dB.

Where the law is false, which is most of the time indoors. The derivation assumes the sound goes out and never comes back. In a room it comes back, repeatedly, off every surface. Close to the source the direct sound dominates and the law holds; past a boundary called the critical distance the reverberant field dominates and the level barely changes with distance at all. That is why the back of a hard-walled sports hall is almost as loud as the middle, and why the reverberation time of the room is the companion quantity to this one: together they tell you whether a space is governed by geometry or by its own echoes. A source sitting on hard ground radiates into a half space rather than a full sphere and is 3 dB louder everywhere, but it still loses 6 dB per doubling, because halving the solid angle is a one-off and the spreading is not.

Outdoors, over long distances, the law is optimistic. Air itself absorbs sound, and absorbs high frequencies far faster than low ones, which is why a distant thunderclap rumbles instead of cracking — the crack has been eaten. Wind gradients bend rays downwards downwind and upwards upwind, which is why you can hear a distant road at night and not in the afternoon. None of that is in the formula. Treat the number this page gives as the most sound you could possibly receive at that distance, and expect less.

Running it backwards is the useful direction. Rearranged, the law says r = r₁ × 10^{(L₁−L)/20}. Every 6 dB you want costs a factor of two in distance, and every 20 dB costs a factor of ten. Getting a 100 dB source down to 60 dB means standing a hundred times further away, and if the first distance was a metre, that is a hundred metres. This is exactly why distance is a poor noise control and why a barrier or an enclosure usually wins.

Frequently asked questions

Why is it 6 dB per doubling and not 3?

Because the distance ratio enters as a pressure ratio, and pressure ratios carry a factor of 20. Intensity falls to a quarter, and 10 log₁₀(1/4) = −6.02 dB; pressure falls to a half, and 20 log₁₀(1/2) = −6.02 dB. 3 dB is what you get from halving the power — for instance switching off one of two identical speakers — which is a different operation entirely.

Does this work indoors?

Only near the source. Indoors the reflected sound builds up a reverberant field that is roughly uniform across the room, and past the critical distance the level stops falling. In a very absorbent room — a recording studio, a room full of soft furnishing — the critical distance is large and the law holds further out. In a tiled corridor it is a metre or two. If you need the indoor answer, you need the room’s absorption, which is the reverberation-time page.

What if the source is a long road or a wall of machinery rather than a point?

Then the spreading is different and so is the rule. An infinitely long line source spreads sound over a cylinder rather than a sphere, so intensity goes as 1/r and the level falls by only 3 dB per doubling — which is why motorway noise carries so much further than you expect. A large flat source behaves like a plane wave close up and loses almost nothing with distance until you are far enough away that it looks small. This page is the point-source case, and it is the right case once you are several source dimensions away.

Two identical speakers instead of one — how much louder?

3 dB, if they are incoherent sources such as two separate noise sources: twice the power is 10 log₁₀(2) = 3.01 dB. The page shows this row. If the two are radiating the same signal in phase at your position, the pressures add rather than the powers and you get 20 log₁₀(2) = 6 dB — but only at points where they happen to be in step, and 6 dB down or worse where they are not. Real pairs of loudspeakers do both, at different frequencies and different places.

Why do the two distances not need units?

Because only their ratio appears. 10 ft from 1 ft is the same 20 dB as 10 m from 1 m. The one thing you must not do is mix them: if the reference is in feet and your distance is in metres, the ratio is wrong by 3.28 and the answer is out by 10.3 dB. The target-distance row comes back in whatever unit you used for the reference distance.

Can the level ever go up with distance?

Not from spreading, but yes in reality. Focusing by a curved surface can concentrate sound at a point further away; a temperature inversion can bend rays back down so that a ring of ground kilometres away is louder than ground closer in, which is how distant explosions are sometimes heard in a far zone and not in a near one; and indoors, a standing-wave peak can be louder than a point nearer the source. The inverse square law describes free spreading only, and all three of those are the law being overridden by something else.

Related calculators

References

University College London, Department of Speech, Hearing and Phonetic Sciences, Acoustics of Speech and Hearing, lecture 1-2 handout (read 11 November 2026). Defines the decibel sound pressure level as 20 log₁₀(p / 20 µPa), calls 20 µPa the standard reference “approximately equal to the human threshold for sound under optimum conditions”, states that in a free field “the power per unit area falls with square of the distance” and that “in decibel terms, falls by 6 dB every doubling of distance”. It notes that the dB SPL scale is strictly a measure of relative intensity. It does not write the general form L₂ = L₁ − 20 log₁₀(r₂/r₁), which this page derives from the 6 dB statement in the worked example rather than attributing to that source.
University of Illinois at Urbana-Champaign, PHYS 406 Acoustical Physics of Music, Lecture 9 notes (S. Errede, read 11 November 2026). States the Sabine equation as T₆₀ = 0.161 V/A seconds with V in m³ and A in m² (and the imperial K = 0.049 s/ft form), defines the total absorption as A = Σ a_nS_n with 0 ≤ a ≤ 1, notes that coefficients are quoted at 125, 250, 500, 1000, 2000 and 4000 Hz, and says that for lecture halls and classrooms “T₆₀ reverberation times should also be < 0.5 seconds” to avoid speech interference. It also says plainly that “optimum reverberation time is subjective”, which is why this page gives a range and not a single right answer.
The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019. The metre is fixed by defining the speed of light in vacuum as c = 299 792 458 m/s exactly, and the Boltzmann and Avogadro constants are fixed exactly too, which makes the molar gas constant R = 8.314 462 618 153 24 J mol⁻¹ K⁻¹ exact. Every speed of sound on these pages is derived from that exact R rather than copied out of a table.