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| 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. |
| 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. |
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Given sound levels and calculation of the distance: 
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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)
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 |
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| 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  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 |
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| 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. |
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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: We see that sound intensity is inversely proportional to the square of the distance away But sound intensity is proportional to the square of the sound pressure, so we could equally write: |
| 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. |
| 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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