Handbook for Sound Engineers

Sound System Design 1249

(34-15)

where,

and

where,

x is any distance.

Example:

Let,

LP = 90 dB,

D = 4 ft,

Dc = 125 ft,

Q = 5,

S = 28,000,

= 0.35.

then,

and

Note: Except for round-off errors, this answer and the
answer to the example for Eq. 34-14 are the same.

Eq. 34-15 is more common in the literature but Eq.
34-14 may be easier to understand and to use in a
computer program. The two equations are mathemati-
cally the same and will produce the same answers given
the same data.

34.2.3.4 The Four Questions Again

Question 2: “Can everybody hear?” is discussed in Sec-

tion 34.3.2. Questions 1, 3, and 4, however, can be

answered with the information obtained so far.

34.2.3.4.1 Question 1: Is It Loud Enough?

In the simplified (outdoor) system, the answer to this

question depended on the required LP at the farthest lis-

tener (at D 2 ), the required head room in decibels and the

sensitivity of the loudspeaker. The answer was given in

terms of the required electrical power to be supplied to

the loudspeaker (the EPR).

In the indoor system, reverberation in the room

affects the analysis. Yet, although the room adds

complexity to the answer to Question 1, it makes things

a little easier in the actual design of an indoor sound

reinforcement system, because, after the critical

distance is passed, the sound can only attenuate another

3 dB. Thus, for distances beyond Dc, no more power is

needed to maintain the same LP. Unfortunately, intelligi-

bility suffers at distances well into the reverberant field.

But that is the topic of Question 3. For now, here is the

equation for indoor electrical power required

(34-16)

where,

LP is the average LP required at distance D 2 ,

H is the head room in dB,

Ls is the sensitivity of the loudspeaker (1 W/1 m),

D 2 is the distance to the farthest listener,

For metric distances, replace the constant 3.28 with the

constant 1.00.

Example:

Let

LP = 90 dB,

H = 10 dB,

Ls = 113 dB (1 W/1 m),

Dc = 31.2 ft,

D 2 = 128 ft.

To compare with the simplified system (outdoor) EPR

equation (Eq. 34-6), simply ignore the term

Lpc LP–= 'dB

'dB 'Dc–= 'D

'x 10 Q
4 Sx^2

------------^4

Sa

–= log©¹§·--------+

a

'D 10 5

4 S 4

----------- 2 -

4

28,000 0.35

–= log©¹§·+-----------------------------------

= 16 dB

'Dc 10 5

4 S 1252

------------------^4

28,000 0.35

–= log©¹§·+-----------------------------------

=33.6 dB

' dB=33.6 dB–16 dB

=17.6 dB

LPc 90 17.6–=

=72.4 dB.

EPR 10

Lp HLS 20

D 2

–1+ log3.28---------- 0

gD (^2)

• – logg------------------ 3.28
  = ----------------------------------------------------------------------------------------------------^10
  EPR 10
  90 +10 113–1+ 20 log----------3.28^128 – 0 log17,357----------------989.4
  = ------------------------------------------------------------------------------------------------------^10
  =4.3 W