The number of panels depends on three things: the room's volume, how much its surfaces already absorb, and how short you want the echo to be. Here is the calculation for a typical lecture hall, step by step.
Step 1: the room
A typical lecture hall of 70 × 50 ft with a 18 ft ceiling has a volume of about 63,000 ft³. With typical surfaces (drywall, a rug and some furniture), its echo time is about 2.73 seconds. The target for lecture halls is 0.90 seconds.
Step 2: how much absorption is missing
Sabine's formula says RT60 = 0.161 × V ÷ A, with V in cubic meters and A, the total absorption, in square-meter sabins. To reach 0.90 s, the room needs about 3,435 ft² sabins of absorption in total. It already has about 1,132 from its surfaces, so the panels must add the rest.
Step 3: what one panel adds
A 48 × 96 in rectangular panel, 2 in thick has about 32.0 ft² of face. At NRC 1.00, and allowing for the surface it covers, it adds about 28.8 sabins.
Step 4: the answer
About 80 panels bring the room from 2.73 s to 0.90 s.
| Thickness | NRC | Panels needed | Echo time after |
|---|---|---|---|
| 1 in | 0.8 | 103 | 0.90 s |
| 2 in | 1 | 80 | 0.90 s |
| 4 in | 1.15 | 69 | 0.90 s |
| Room | Size | Volume | Echo time now | Panels needed |
|---|---|---|---|---|
| Small | 53 × 38 × 18 ft | 36,252 ft³ | 2.44 s | 44 |
| Typical | 70 × 50 × 18 ft | 63,000 ft³ | 2.73 s | 80 |
| Large | 105 × 75 × 20 ft | 157,500 ft³ | 3.37 s | 219 |
These are estimates for speech frequencies (500–1,000 Hz). Use the room calculator with your own measurements, then load the result into the designer.