We use Optical Character Recognition (OCR) during our scanning and processing workflow to make the content of each page searchable. You can view the automatically generated text below as well as copy and paste individual pieces of text to quote in your own work.
Text recognition is never 100% accurate. Many parts of the scanned page may not be reflected in the OCR text output, including: images, page layout, certain fonts or handwriting.
ARCHITECTURAL ACOUSTICS 329
3. The Control of Reverberation. In the foregoing discussion it is shown that the proper control of reverberation in an enclosure is a paramount factor in the securing of optimal acoustic conditions. It is appropriate therefore that we outline the method of calculating the reverberation characteristics of a closed or partially closed space.
The simplest and the most useful equation for calculating the time of reverberation in an enclosed space is the following :
' t_£ (2)
CI
t represents the time of reverberation in seconds.
V represents the volume of the enclosed space in cubic feet.
k is a constant, equal to .05 for simple rectangular spaces, and varying between .04 and .05 for more irregular spaces.
a is the total absorption in the enclosed space.
a is given by the relation a = ax sx + a2 s2 + a3 s3 + . . . , where slt s2, s3, . . . represent the areas of the different materials bounding the enclosure or materials within the enclosure, and ai> <*2> a3> • • • are the corresponding coefficients of sound absorption of the different materials.
In general it is necessary to calculate the reverberation at several representative frequencies, say 128 d. v., 512 d. v. and 2048 d. v., since the coefficients of absorption are not the same for all frequencies. In order to illustrate the use of equation (2), suppose we calculate the time of reverberation for a tone of 512 d. v. in a sound stage 80' x 100' x 35'. Suppose for the sake of simplicity that the ceiling is level, and that the entire walls and ceiling are treated with a wool blanket or a wool fill 4" thick. The floor is soft wood. No sets or other equipment are in the stage. The volume of the stage is 280,000 cubic feet, the area of the floor or ceiling is 8,000 square feet, and the area of the walls is 12,600 square feet. The coefficient of absorption for the wool is about .60 and the coefficient for the wood floor is .06. Therefore the total absorption "a" in the room is 20,600 X .60 + 8,000 X .06 = 12,360 + 480= 12,840 units. The