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.
ACOUSTIC POWER LEVELS
253
But, obviously, area alone cannot be made the sole factor; the distribution must be uniform and effective. Accordingly, it would be better to speak of "effective area" rather than "area" where the term "effective" should weight the efficiency of direct covering and any aid from reflection.
15,000
AUDIENCE AREA IN SQUARE FEET
lig. 5. Relation between Area Occupied by Audience and Auditorium Volume to the Two-thirds Power.
We may assume that effective area will be a function of the seating area and hence vary in some manner as the volume of theatres. Fig. 5 represents the variation of seating area, including aisles and crosswalks, with volume to the two-thirds power. The relation is suggested by the geometrical consideration that on an average the surface of enclosures increases according to the square of the cube root of the volume. Of course, over a large number of theatres there will be considerable deviation from the values indicated by the curve, but a general average is of interest and serves our purpose. The equation connecting volume and seating area is:
S = 2.56V2/3. (5)
If we asume that aid by reflection just compensates for the increase of power that would be needed because of faulty distribution, then the effective area will be equal to the audience area. A refinement of the theory, which we have not attempted as yet, might consider some application of the inverse square law. The acoustic power in microwatts required to produce a sensation level of 80 decibels in an auditorium would be, from equation (5) and the desired flux per square foot,