Damping of sound level with distance - decibel dB damping calculation calculator distance versus sound reduction free field - decrease in sound over distance microphone distance versus dB sound at different distances

• Damping of sound level with distance •
How does the sound decrease with increasing distance?

Changing (decrease) of sound pressure level Δ L or sound pressure
with distance r in a free field (direct field), like in anechoic chambers
Conversion: Distance values → Level changing
With sound level we usually mean a logarithmic ratio of sound pressure

These calculations are meant only for engineers and the distance from a musician or a loudspeaker to a
microphone in a direct field - No air damping and frequency dependance of e.g. the thunder in a distance.

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Sound pressure level depending on the distance

Enter the three gray boxes and you get the amount of attenuation,
you can expect with a change in sound source distance, in a free field.

The sound level depends on the distance between the sound source and the
place of measurement.
The sound pressure level L_p in dB without the given distance r to the sound
source is really useless. Unfortunately that error happens quite often.

The sound pressure p changes (decreases) with 1/r over distance.
Sometimes it is said, that it goes with 1/r². That is really wrong.
But the sound intensity (energy quantity) decreases with 1/r².

Notice: Intensity is not pressure of sound.
Sound energy quantity cannot be sound field quantity.

How is the sound level dependent from the distance to the sound source?
The sound pressure level shows in the free field situation a reduction of 6 dB per
doubling of distance; that means the sound pressure value is a half and not a quarter.

A statement of the distance law for field quantities (source quantities):

Sound level difference:

or level at far distance

Δ L = L₁ − L₂.
The sound pressure p decreases really with 1/r from the sound source!

In acoustics, the sound pressure of a spherical wave front radiating from a point source
decreases by a factor of 1/2 as the distance is doubled.
The behavior is not inverse-square, but is inverse-proportional:

p ~ 1 / r.
Relation of sound intensity I, sound pressure p and the distance law:
(r is the distance from the sound source)

From this follows

Note: The often used term "intensity of sound pressure" is not correct.
Use "magnitude", "strength", "amplitude", or "level" instead.
"Sound intensity" is sound power (acoustic power) per unit area, while
"pressure" is a measure of force per unit area. Intensity (sound energy
quantity) is not equivalent to pressure (sound field quantity).

Damping of sound level in decibels with distance

Distance ratio

Conversion of sound units (levels)

For this level damping of sound with distance we have to consider the damping of air
(air damping) at larger distances. See: Absorption of sound by the atmosphere

Sound pressure level and Sound pressure

Enter a value in the left or right box, then press the TAB bar or make
a mouse click at an empty space at the side, to get the solution.
The calculator works in both directions of the ↔ sign.

Inverse distance law 1/r for sound pressure

Law for Sound Field Quantities
Distance ratio	Sound pressure p ∝ 1/r
1	1/1 = 1.0000
2	1/2 = 0.5000
3	1/3 = 0.3333
4	1/4 = 0.2500
5	1/5 = 0.2000
6	1/6 = 0.1667
7	1/7 = 0.1429
8	1/8 = 0.1250
9	1/9 = 0.1111
10	1/10 = 0.1000

Frequently used wrong statements in the context of
sound pressure and the distance of the sound source

Wrong expression	Correct version
Sound pressure falls inversely proportional to the square of the distance 1/r² from the sound source.	Sound pressure falls inversely proportional to the distance 1/r from the sound source. That is the 1/r law or distance law.
Sound pressure level decreases as the distance increases per doubling of distance from the source by (−)3 dB.	Sound pressure level decreases by (−)6 dB per doubling of distance from the source to 1/2 (50 %) of the sound pressure initial value.
Sound intensity (energy) falls inversely proportional to the distance 1/r from the sound source.	Sound intensity (energy) falls inversely proportional to the square of the distance 1/r² from the sound source.
Sound intensity level decreases inversely as the square of the distance increases per doubling of sound source with (−)3 dB per doubling.	Sound intensity level decreases by (−)6 dB per doubling of distance from the source to 1/4 (25 %) of the sound intensity initial value.

Sound pressure p and the inverse distance law 1/r

How does an acoustic sound level depend on distance from the source?

Consider a source of sound and imagine a sphere with radius r, centered on the source.
The sound source outputs a total power P, continuously. The sound intensity I is the same
everywhere on this surface of a thought sphere, by definition. The intensity I is defined as the
power P per unit area A. The surface area of the sphere is A = 4 π r², so the sound intensity
passing through each square meter of surface is, by definition:
I = P / 4 π r².
We see that sound intensity is inversely proportional to the square of the distance away
from the source:
I₂ / I₁ = r₁² / r₂².
But sound intensity is proportional to the square of the sound pressure, so we could equally
write:
p₂ / p ₁ = r₁ / r₂. The sound pressure p changes with1 / r of the distance.

So, if we double the distance, we reduce the sound pressure by a factor of 2 and
the sound intensity by a factor of 4: in other words, we reduce the sound level by
6 dB. If we increase r by a factor of 10, we decrease the level by 20 dB.

The sound intensity level and the sound pressure levels in dB have the same
value, but the quantity of sound pressure and the quantity of acoustic intensity
is different, because I = p².

How many decibels (dB) level change is twice (double, half) or three times as loud?

The beginners question is quite simple: How does the sound decrease with distance?
More specifically asked: How does the volume (loudness) decrease with distance?
How does the sound pressure decrease with distance?
How does the sound intensity (not the sound power) decrease with distance?

The decrease of sound with distance

For a spherical wave we get:
The sound pressure level (SPL) decreases with doubling of distance by (−)6 dB.
It falls on the 1/2 fold (50%) of the initial value of the sound pressure.
The sound pressure decreases with the ratio 1/r to the distance.

The sound intensity level decreases with doubling of distance also by (−)6 dB.
It falls on the 1/4 fold (25%) of the initial value of the sound intensity.
The sound intensity decreases with the ratio 1/r² to the distance.

For a cylindrical wave we get:
The sound pressure level (SPL) decreases with doubling of distance only by (−)3 dB.
It falls on the 0.707 fold (70.7%) of the initial value of the sound pressure.
The sound pressure decreases with the ratio 1/√r to the distance.

The sound intensity level decreases with doubling of distance also by (−)3 dB.
It falls on the 1/2 fold (50%) of the initial value of the sound intensity.
The sound intensity decreases with a ratio 1/r to the distance.

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Reference distance r₁ from source m or ft	Sound level L₁ at reference distance dBSPL	The 1/r law. There really is no square and no power!
New distance r₂ from source m or ft	Sound level L₂ at new distance r₂: dBSPL	Sound level difference Δ L = L₁ − L₂ dB

Sound pressure level L_p: dB-SPL	↔	Sound pressure p: Pa = N/m²

Standard reference sound pressure p₀ = 20 μPa = 2 × 10⁻⁵ Pa ≡ 0 dB