Level changing with dB decibel distance calculator sound reduction loss free field decrease in sound over distance microphone different distances versus dB calculator drop-off distance drop ratio - sengpielaudio
 
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Damping of sound level with distance
 
Distance law - Distance damping
 
This is an approximation when the venue is a direct sound field or an anechoic room
 
How does the sound decrease with increasing distance?
How can you calculate the distance drop of the sound?


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 - The 1/r law
With sound level ususally a level of a logarithmic ratio of sound pressure is meant
 
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 dependence of e.g. the thunder in a distance.

Sound pressure level depending on the distance
for point-shaped sound sources

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.

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

dBSPL
Sound level difference  
Δ L = L2L1
dB
 
Pressure and distance

Given sound levels and calculation of the distance:  Distance and level

 
 The sound level depends on the distance between the sound source and the 
 place of measurement, possibly one ear of a subject.

 The sound pressure level Lp in dB without the given distance r to the sound
 source is really useless. Unfortunately this error (unknown distance) is quite often.

 

A doubling of distance from the sound source in the direct field will reduce the "sound level"
by (−)6 dB, no matter whether that are sound pressure levels or sound intensity levels! This
will reduce the sound pressure p (field quantity) to 1/2 (50 %) and the sound intensity I
(energy quantity) to 1/22 = 1/4 (25 %) of the initial value.

The inverse distance law 1/r shows the distance performance of field quantities
and the inverse square law 1/r2 shows the distance performance of energy
quantities. Squared field quantities are proportional to energy quantities; e.g.
p2 ~ I

The sound pressure p changes (decrease - drop - fall) with 1/r
over distance.
Sound pressure level decreases by (−)6 dB per doubling
of distance from the source to 1/2 (50 %) of the sound pressure initial value.

Sometimes it is said, that the soundpressure would change with 1/r².
That is really wrong.


Notice: Intensity and power 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. 

 

Statement of the distance law:
 
Distance law for field quantities (source quantities)

 
Sound level difference: Entfernungsgesetz               Level at far distance: Pressure and distance
 
Δ L = L1L2.
 
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 ratio of 1/2 as the distance is doubled.
The law is not inverse-square, but is inverse-proportional:

p ~ 1 / r
 
Calculating sound pressure with the inverse distance law

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

                     Intensity-distance
From this follows    sound pressure-distance
Aha!
 
Formel Pegelabnahme Schalldruck
 
Distance and pressure
 
Where:
p1  =  sound pressure 1 at reference distance r1 from the sound source
p2  =  sound pressure 2 at another distance r2 from the sound source

The sound pressure level Lp to plot against
the distance of the sound source r
Schallfeld
Kugelwelle = spherical waves
D: direct field of the spheric source
R: reflected field (diffuse field)
rH: critical distance
 
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).   
I ~ p2
 
Since the sound intensity level is difficult to measure, it is common
to use sound pressure level measured in decibels instead. Doubling the
Sound Pressure raises the Sound Pressure Level with 6 dB.
 
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
It is often asked what formula we use for the calculation of air damping:
Formula for calculating the damping of air.

Sound pressure level and Sound pressure

Enter a value in the left or right box.
The calculator works in both directions of the sign.

 
Sound pressure level Lp:
dB-SPL
 ↔  Sound pressure p:
Pa = N/m2
Formula 1   Formula 2
Standard reference sound pressure p0 = 20 μPa = 2 × 10−5 Pa ≡ 0 dB
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  

Conversions and Calculations - Sound Quantities and their Levels

Frequently used false statements in the context of
sound values 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.                                                    wrong
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.                             wrong
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.                                                               wrong
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.       wrong
Sound intensity level decreases by also (−)6 dB per
doubling of distance from the source to 1/4 (25 %)
of the sound intensity initial value.
 
The sound power level or the sound power is firmly committed to
the sound source and is really independent from the distance.

 
Notice: A 100 watt light bulb has in 1 m and in 10 m distance really
always the same 100 watt, which is emmited always from the bulb!

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, centred 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

I2 / I1 = r1² / r2² or 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 or p2 = p1 (r1 / r2)

The sound pressure p changes with 1 / r of the distance.

So, if we double the distance, we reduce the sound pressure by a ratio of 2 and
the sound intensity by a ratio of 4. In other words, we reduce the sound level by
6 dB. If we increase r by a ratio 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 power (also the acoustic power) cannot decrease with distance.

The decrease of sound with distance

For a spherical wave of a point source we get:
The sound pressure level (SPL) decreases with doubling of distance by (−)6 dB.
The sound pressure falls to 1/2 times (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.
The intensity falls to 1/4 times (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 of a line source we get:
The sound pressure level (SPL) decreases with doubling of distance only by (−)3 dB.
The sound pressure falls to 0.707 times (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.
The intensity falls to 1/2 times (50%) of the initial value of the sound intensity.
The sound intensity decreases with a ratio 1/r to the distance.

The acoustic power level does not decrease with the doubling of distance.
A lamp with the power of 100 watts has in 1 m or in 10 m distance still the same watts.

Inverse distance law
p2 / p1 = (r1 / r2)
Let us take a look at a practical application of this formula: if r1 = 1 m and at this
distance from the sound source we measure Lp1 = 100 dB at a distance of r2 = 2 m
we will have a sound pressure equal to: Lp2 = Lp1 – 20 lg (r2 / r1)
 
Application of the inverse distance formula
Lp2 = 100 − 20 lg (2 / 1) = 100 − 6.02 = 94 dBSPL. From this simple example we
obtain an easy, useful rule: every time we move away from a sound source, doubling
our distance from it, the sound pressure level decreases by 6 dBSPL. Vice versa if we
move closer to the sound source, halving our distance from it, we perceive an increase
in sound pressure of approximately 6 dBSPL.This rule is called the inverse distance law.

Question: What is the standard distance to measure sound pressure level away from equipment?
There is no standard distance. It depends on the size of the sound source and the sound pressure level.
 
Why do some engineers want to use constantly the impractical sound energy quantities,
such as sound intensity and sound power with the reference sound quantity I0 = 10−12 W/m²
or P0 = 10−12 W, when applying the human ear? For noise control calculations that is all right.
 
How does the sound decrease with distance?
 
 
 The constant unsureness is the answer of the question:
 "How many dBs are doubling a sound"?

 
 Answer: Doubling means the "factor 2". What does doubling of a"sound" mean?
 Twice the (sound) intensity we get by an increase of the (sound intensity) level of 3 dB.
 Twice the sound pressure we get by an increase of the (sound pressure) level of 6 dB.
 Twice the loudness feeling we get by an increase of the (loudness) level of about 10 dB. 

 
 
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