DIFFRACTION OF SOUND
Diffraction around Solid Objects
Preliminary note:
This calculation gives the maximum possible noise reduction by a
barrier. In practice it is less for the following reasons:
1. The ground, before the application of a
barrier, has noise reducing properties of its own. This formula assumes
there is none. Refer to elsewhere on this site for details.
2. The formula given here is for a point source. In reality with a
distributed speaker system, it approaches a line source. This reduces
calculated reductions by up to 3 dB,
3. Scattering of sound into acoustical shadow by houses and trees,
reduces the barriers effectiveness at higher frequencies.
When a tall, thin sound barrier is interposed between
a point source and a microphone, as shown in Fig., the attenuation provided
by the barrier is theoretically determined by the Fresnel Number, given
by FN = 2ft. Here 'f' is the center frequency of the frequency band used.
't' is the increase in the time that the sound takes to travel over the
barrier over the direct route with no barrier present.
Or the increase in distance the sound must travel (d), caused by
placement of the barrier, is equal to the velocity of sound multiplied
by t.
Fig.
The theoretical relation between insertion loss and FN is shown in Fig.10.
Fig.10
Some examples of barrier applications can be tabulated
Across the top row are the number of feet from the top of the barrier down to the top of the loudspeaker stack. In the first column is the distance from the loudspeaker stack to the barrier. The listen is presumed to be back in an adjacent community. In each grayed cell is the number of addition feet (d) that the sound must travel.
|
l ft \ h ft |
2 | 5 | 10 | 15 |
| 300 | 0.01 | 0.08 | 0.33 | 0.75 |
| 400 | 0.01 | 0.06 | 0.25 | 0.56 |
| 500 | 0.01 | 0.05 | 0.20 | 0.45 |
| 600 | 0.01 | 0.04 | 0.17 | 0.37 |
Using a found value of d from the table above, use it in the table
below, looking it up in the left hand column. It the first row is the
frequency of the sound. Look up the Fresnel number and refer to the
Fig. 10 above to find the dB reduction.
|
d ft \ f Hz
|
50 |
200 |
500 |
1000 |
0.1 |
0.01 |
0.04 |
0.09 |
0.18 |
0.2 |
0.02 |
0.07 |
0.18 |
0.36 |
0.4 |
0.04 |
0.15 |
0.36 |
0.73 |
0.8 |
0.07 |
0.29 |
0.73 |
1.45 |
Example: the top of the barrier is 10 feet above the top of the loudspeaker stack and is 400 feet away. Additional feet is found to be .25. The speaker is a woofer, so use 50 Hz. FN is found be be .025. From the graph the dB reduction is 6. For 500 Hz FN is about 0.22 and the dB reduction is about 9.
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