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

• 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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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 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. Intensity is not pressure.
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.

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 per unit area, while "pressure" is a
measure of force per unit area. Intensity is not equivalent to pressure.

Damping of sound level in decibels with distance dB and distance ratio - sengpielaudio

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

Inverse Distance Law

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

Wrong expressions	Correct version
Sound pressure decreases inversely as the square of the distance increases with 1/r²from the sound source.	Sound pressure decreases inversely as the distance increases with 1/r from the sound source.
Sound pressure level decreases as the distance increasesper 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 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, bydefinition. The intensity I is defined as the
power P per unitarea 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 are identical
quantities, but the quantities of sound pressure and acoustic intensity are not,
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?

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