Assignment 7: Room Acoustics In last week's assignment, you calculated mean free path, total absorption, and reverberation time for a room that is 4 m long, 7 m wide, and has a ceiling height of 3 m.
Lesson Notes
Surface | S | a | S a |
Floor | 15 m2 | 0.10 | 1.5 Sab |
Ceiling | 15 m2 | 0.90 | 13.5 Sab |
Walls (total of 4) | 40 m2 | 0.45 | 18 Sab |
∑S = 70 m2 | A = 33 sabins |
Using these values, we can now calculate the room constant:
Here are the dimensions of the room:
Calculate the following:
total absorption
mean free path
reverberation time
room constant
LpD is the sound pressure level measured at an arbitrary distance directly in front of the directional source.
LpO is the sound pressure level measured at the same distance from an omnidirectional source of the same total sound power output.
Both measurements must be made in the direct field. For example, if you were to measure a sound pressure level of 50 dB at a distance of 1 m from an omnidirectional source, but 53 dB at a distance of 1 m from a directional source of the same sound power output, you could calculate the Q of the directional source
This source has a directivity factor of Q = 2. The directivity factor is unitless.
Another way to think of directivity factor is by considering what would happen if you placed an omnidirectional source up against a non-absorbing wall. Half of the power emitted from the source would be immediately reflected in the other direction. The effect of this would be that the sound power would double in the direction opposite the wall. A doubling of sound power gives us a 3 dB increase in level. We know from the example above that a 3 dB increase gives us a Q of 2.
If we put the source into a corner, all of the power would be confined to ¼ of the sphere of propagation. This gives us a Q of 4.
A person speaking has an approximate Q of 2.
Let's revisit the example that we used earlier in the lesson to calculate room constant, R.
We used the table below to find the total absorption, A, and the total surface area, ∑S.
Surface | S | a | S a |
Floor | 15 m2 | 0.10 | 1.5 Sab |
Ceiling | 15 m2 | 0.90 | 13.5 Sab |
Walls (total of 4) | 40 m2 | 0.45 | 18 Sab |
∑S = 70 m2 | A = 33 sabins |
Let's now calculate the critical distance for a person speaking in this room. We'll assume the Q = 2.
Here is the room we used in the previous "Room Values" exercise.
This time, calculate the critical distance for a source with a directivity of Q = 4.
Converting Power to Pressure
Lp is the sound pressure level
LW is the sound power level
Q is the directivity factor
r is the distance from the source
R is the room constant
If we look inside of the parentheses, we see factors: Q/4r2 and 4/R. The first, Q/4r2, represents the direct sound. The second, 4/R, represents the reverberant sound.
You'll notice that the direct component is dependent on r, the distance from the source, while the reverberant component is not. The reverberant component is dependent only on the room constant.
As r gets larger, Q/4r2 gets smaller. For a very large value of r, Q/4r2 will be inconsequential. This means that if we are a long distance form the source in a reverberant room, the sound pressure level will be entirely dependent on the reverberant contribution, 4/R.