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Inverse Square Law

Reference data and engineering information about inverse square law for miscellaneous applications.

inversesquarelaw

Overview

Engineering reference data for Inverse Square Law in miscellaneous.

Key Formulas

Unit Conversion

y=xky = x \cdot k

Multiply by conversion factor.

Linear Interpolation

y=y1+(xx1)(y2y1)x2x1y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}

Estimate between two known points.

Percentage

p=partwhole×100%p = \frac{\text{part}}{\text{whole}} \times 100\%

Part as fraction of whole.

Variables

SymbolDescriptionUnit
xxInput value
yyOutput value
kkConversion factor

Practical Applications & Examples

The inverse square law provides critical predictions for sound attenuation in unobstructed environments. Below are practical calculations and reference data.

Sound Propagation Data Table

The following table shows how sound pressure level (Lp) decreases with distance for a point source in a free field, consistent with a 6 dB reduction per distance doubling.

13 rows
Sound pressure level at various distances from a source, illustrating the inverse square law.
Distance(ft)
Sound Pressure Level (Lp)(dB)
1.25134
2.5128
5122
10116
20110
40104
8098
16092
32086
64078
128074
256068
512062

Source: engineeringtoolbox.com

Calculation Examples

Example 1: Rifle Shot Sound Attenuation Given a measured sound pressure level (Lp1L_{p1}) of 134 dB at 1.25 feet, find the level at 80 feet.

  1. Calculate the difference (ΔL\Delta L) using the formula: ΔL=20log(R2R1)=20log(801.25)=20log(64)36 dB\Delta L = 20 \log\left(\frac{R_2}{R_1}\right) = 20 \log\left(\frac{80}{1.25}\right) = 20 \log(64) \approx 36 \text{ dB}
  2. Apply the difference to find Lp2L_{p2}: Lp2=Lp1ΔL=134 dB36 dB=98 dBL_{p2} = L_{p1} - \Delta L = 134 \text{ dB} - 36 \text{ dB} = 98 \text{ dB}

Example 2: Industrial Noise Assessment A machine has a sound pressure level (Lp1L_{p1}) of 110 dB at 1 meter. Assess the level in a working area 5 meters away.

  1. Calculate the reduction (ΔL\Delta L): ΔL=20log(51)=20log(5)14 dB\Delta L = 20 \log\left(\frac{5}{1}\right) = 20 \log(5) \approx 14 \text{ dB}
  2. Determine the new sound level (Lp2L_{p2}): Lp2=110 dB14 dB=96 dBL_{p2} = 110 \text{ dB} - 14 \text{ dB} = 96 \text{ dB} Note: A level of 96 dB often exceeds permissible exposure limits for continuous work, necessitating engineering controls like barriers or enclosures.

Inverse Square Law Calculator

Use the formula below to calculate the sound pressure level at a new distance. Input values for the initial level (Lp1L_{p1}), initial distance (R1R_1), and the target distance (R2R_2).

The change in sound pressure level is given by: ΔL=20log10(R2R1)\Delta L = 20 \log_{10}\left(\frac{R_2}{R_1}\right) The resulting sound pressure level is: Lp2=Lp1ΔLL_{p2} = L_{p1} - \Delta L

Interactive Charts

sound pressure level vs. distance from source - inverse square law

References