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| In the real world, the inverse distance law p ∝ 1/r is always an idealization because it assumes exactly equal sound pressure p as sound field propagation in all directions. If there are reflective surfaces in the sound field, then reflected sounds will add to the directed sound and you will get more sound at a field location than the inverse distance law predicts. If there are barriers between the source and the point of measurement, you may get less than the distance law predicts. Nevertheless, the inverse distance law is the logical first estimate of the sound pressure you would get at a distant point in a reasonably open area. The reference sound pressure level SPL = 0 dB is the sound pressure of p0 = 20 µPa = 20 × 10-5 Pa or N/m2. Field quantities, as the sound pressure, will always be shown as effective values (RMS). |
| You can explore numerically to confirm the 1/r law that doubling the distance drops the sound pressure p to a half (0.5) by a sound pressure level of about 6 dB and that 10 times the distance drops the sound pressure p to a tenth (0.1), that is a level drop by 20 dB. |
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. |
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Calculating sound pressure with the inverse distance law
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| Where: | ||
| p1 | = | sound pressure 1 at r1 |
| p2 | = | sound pressure 2 at r2 |
| r1 | = | distance 1 from source |
| r2 | = | distance 2 from source |
| 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 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 theinitial 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 propotional to energy quantities; e.g. p2 ∝ I |
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Sound Field Quantities
 Sound pressure, sound or particle velocity, particle displacement or particle amplitude, (voltage, current, electric resistance). Inverse Distance Law 1/r |
Sound Energy Quantities Sound intensity, sound energy density, sound energy, acoustic power. (electrical power). Inverse Square Law 1/r² |
| Conversions and Calculations Sound Quantities and their Levels Damping of Sound Pressure Level with Distance |
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. |
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. |
How does the 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 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, I ∝ p2, so we can 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. |
How many decibels (dB) level change is twice (double, half) or four times as loud?
| The beginners question is quite simply: How does the sound decrease with distance? More specifically questioned: 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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