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Outdoor Sound Partial Barriers

Reference data and engineering information about outdoor sound partial barriers for acoustics applications.

outdoorsoundpartialbarriers

Overview

Engineering reference data for Outdoor Sound Partial Barriers in acoustics.

Key Formulas

Speed of Sound

c=γRTc = \sqrt{\gamma R T}

Speed of sound in an ideal gas.

Sound Level

L=10log10(I/I0)L = 10 \log_{10}(I/I_0)

Decibel level.

Wavelength

λ=c/f\lambda = c / f

Wavelength = speed / frequency.

Variables

Symbol Description Unit
cc Speed of sound m/s
LL Sound level dB
λ\lambda Wavelength m
ff Frequency Hz

Sound-Barrier Interaction

When a barrier is interposed between a sound source and a receiver, the sound energy is affected in three ways:

  • Reflected — Sound bounces back from the barrier surface
  • Transmitted — Sound passes through the barrier material
  • Diffracted — Sound bends around the edges of the barrier

The attenuation achieved by the barrier primarily depends on the diffraction component, which is quantified using the Fresnel number.

Fresnel Number Behavior

The Fresnel number determines the effectiveness of barrier attenuation:

N=2δλN = \frac{2\delta}{\lambda}

where:

  • δ=A+Bd\delta = A + B - d (path length difference in m or ft)

Important characteristics:

  • High frequencies (short wavelengths): The Fresnel number increases, resulting in greater attenuation
  • Low frequencies (long wavelengths): The Fresnel number decreases toward zero, resulting in less attenuation

Note: The attenuation is reduced for moving sources (such as vehicles) compared to stationary sources.

Worked Example: Highway Noise Barrier

Parameter Value
Distance from highway to barrier top (AA) 20 m
Distance from barrier top to receiver (BB) 30 m
Direct distance source to receiver (dd) 43 m

Step 1: Calculate path length difference: δ=A+Bd=20+3043=7 m\delta = A + B - d = 20 + 30 - 43 = 7 \text{ m}

Step 2: Calculate Fresnel numbers and attenuation:

Frequency Wavelength Fresnel Number Attenuation
500 Hz 0.69 m N=2×70.69=20N = \frac{2 \times 7}{0.69} = 20 ~17.5 dB
2000 Hz 0.17 m N=2×70.17=82N = \frac{2 \times 7}{0.17} = 82 ~20 dB

Higher frequencies achieve greater attenuation due to the increased Fresnel number.

References