Acoustic designers are frequently asked "how much treatment do we need?" before any detailed specification work. The Sabine equation can be rearranged to answer this question directly: given a room volume and a target RT60, calculate the total absorption area required, compare it to what already exists, and the deficit is the treatment needed. This article works through a 150 m³ meeting room in full numerical detail.
The Room
A mid-size corporate meeting room:
- Length: 8.0 m
- Width: 5.5 m
- Height: 3.0 m (solid concrete slab above, 2.6 m to suspended ceiling)
- Volume: 8.0 × 5.5 × 3.0 = 132 m³ (using actual height to ceiling, 3.0 m)
- Floor area: 44.0 m²
- Total surface area: 2 × (44.0 + 24.0 + 16.5) = 169.0 m²
| Surface | Area (m²) | Material (existing) |
|---|---|---|
| Floor | 44.0 | Polished concrete |
| Ceiling | 44.0 | Painted plasterboard on battens |
| Long wall A (glazed) | 24.0 | 50% double glazing (12 m²) + 50% plasterboard (12 m²) |
| Long wall B | 24.0 | Painted plasterboard |
| Short wall A (whiteboard) | 16.5 | Whiteboard 60% (9.9 m²) + painted plasterboard 40% (6.6 m²) |
| Short wall B | 16.5 | Painted plasterboard |
Target RT60
The client specifies RT60 = 0.60 s at 500 Hz per ISO 22955:2021 (acoustic quality criteria for offices and meeting rooms). This is a reasonable target for a corporate meeting room where video-conferencing is the primary use.
Step 1 — Calculate Required Total Absorption
Rearranging Sabine: A_required = 0.161 × V / RT60_target
A_required = 0.161 × 132 / 0.60 = 21.25 / 0.60 = 35.4 m²
We need 35.4 metric sabins at 500 Hz to achieve the target.
Step 2 — Calculate Existing Absorption
Absorption Coefficients (Existing Materials)
| Material | 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|---|
| Polished concrete | 0.01 | 0.01 | 0.02 | 0.02 | 0.02 | 0.02 |
| Painted plasterboard on battens | 0.15 | 0.10 | 0.06 | 0.05 | 0.04 | 0.04 |
| Double glazing (6/12/6 mm) | 0.20 | 0.15 | 0.10 | 0.07 | 0.05 | 0.04 |
| Whiteboard (painted hard surface) | 0.05 | 0.04 | 0.03 | 0.03 | 0.02 | 0.02 |
Existing Absorption at 500 Hz
| Surface | Area (m²) | α (500 Hz) | Si × αi |
|---|---|---|---|
| Polished concrete floor | 44.0 | 0.02 | 0.88 |
| Painted plasterboard ceiling | 44.0 | 0.06 | 2.64 |
| Double glazing | 12.0 | 0.10 | 1.20 |
| Plasterboard long wall A | 12.0 | 0.06 | 0.72 |
| Plasterboard long wall B | 24.0 | 0.06 | 1.44 |
| Whiteboard | 9.9 | 0.03 | 0.30 |
| Plasterboard short wall A | 6.6 | 0.06 | 0.40 |
| Plasterboard short wall B | 16.5 | 0.06 | 0.99 |
| Total A_existing (500 Hz) | 8.57 m² |
Existing RT60 at 500 Hz
RT60_existing = 0.161 × 132 / 8.57 = 21.25 / 8.57 = 2.48 s
That is a very reverberant room — typical of an untreated concrete-and-plasterboard meeting room.
Step 3 — Calculate the Absorption Deficit
A_deficit = A_required − A_existing = 35.4 − 8.6 = 26.8 m²
We need to add 26.8 m² of absorption at 500 Hz to bring the room to the 0.60 s target. Let's evaluate three treatment options.
Option A — Suspended Acoustic Ceiling Tiles
Replace the painted plasterboard ceiling with 600×600 mm acoustic ceiling tiles in a T-bar grid, NRC 0.70 product (medium-performance ceiling tile).
Absorption coefficients for NRC 0.70 ceiling tile:
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|
| 0.18 | 0.40 | 0.65 | 0.85 | 0.80 | 0.75 |
Absorption added at 500 Hz = 44.0 × 0.65 = 28.60 m² (replacing plasterboard which contributed 2.64 m²)
Net additional absorption = 28.60 − 2.64 = 25.96 m²
This is marginally short of the 26.8 m² deficit (−0.84 m²). The actual RT60 achieved:
A_total = 8.57 − 2.64 + 28.60 = 34.53 m²
RT60 = 0.161 × 132 / 34.53 = 21.25 / 34.53 = 0.615 s
Just above the 0.60 s target — close enough for most practical purposes, and easily met with a higher-performance NRC 0.75 tile.
Option A — Full Octave-Band Check
With NRC 0.70 ceiling tiles installed:
| Band (Hz) | A_existing (floor+walls+glass+wb) | A_new ceiling | A_total | RT60 (s) |
|---|---|---|---|---|
| 125 | 5.93 + old ceiling 6.60 → 5.93 | 0.18×44 = 7.92 | 13.85 | 2.47 s |
| 250 | 5.93 − 4.40 + 4.40 | 17.60 | 23.53 | 1.45 s |
| 500 | 5.93 − 2.64 + 28.60 | — | 31.89 | 1.08 s |
Wait — let me recalculate properly. The existing floor + walls + glass + whiteboard absorption (excluding ceiling):
| Band (Hz) | A_floor | A_walls | A_glass | A_wb | A_excl_ceiling |
|---|---|---|---|---|---|
| 125 | 0.44 | 5.09 | 2.40 | 0.50 | 8.43 m² |
| 250 | 0.44 | 3.39 | 1.80 | 0.40 | 6.03 m² |
| 500 | 0.88 | 2.04 | 1.20 | 0.30 | 4.42 m² |
| 1000 | 0.88 | 1.70 | 0.84 | 0.30 | 3.72 m² |
| 2000 | 0.88 | 1.36 | 0.60 | 0.20 | 3.04 m² |
| 4000 | 0.88 | 1.36 | 0.48 | 0.20 | 2.92 m² |
Add new ceiling tile absorption:
| Band (Hz) | A_excl_ceiling | A_ceiling_new (44 × α) | A_total | RT60 (s) | Target met? |
|---|---|---|---|---|---|
| 125 | 8.43 | 7.92 | 16.35 | 2.09 s | No |
| 250 | 6.03 | 17.60 | 23.63 | 1.44 s | No |
| 500 | 4.42 | 28.60 | 33.02 | 0.65 s | Borderline |
| 1000 | 3.72 | 37.40 | 41.12 | 0.52 s | Yes |
| 2000 | 3.04 | 35.20 | 38.24 | 0.56 s | Yes |
| 4000 | 2.92 | 33.00 | 35.92 | 0.60 s | Yes |
The ceiling tiles alone control 500–4000 Hz but leave 125–250 Hz highly reverberant. This is a known characteristic of ceiling tiles, which become thin and resonant at low frequencies.
Option B — Wall Panels (100 mm Mineral Wool, Fabric-Wrapped)
Instead of or in addition to ceiling tiles, install fabric-wrapped 100 mm mineral wool panels on the walls.
Absorption coefficients for 100 mm mineral wool, fabric-wrapped:
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|
| 0.45 | 0.75 | 0.95 | 0.98 | 0.97 | 0.95 |
Required area at 500 Hz: A_deficit / α_panel = 26.8 / 0.95 = 28.2 m² of panels
Available wall space (excluding glazing, whiteboard, doors): approximately 52 m² of plasterboard. 28.2 m² represents 54% of the available wall area — achievable but visually dominant.
Panel arrangement: 600×1200 mm panels (0.72 m² each). 28.2 / 0.72 = 40 panels.
At 125 Hz, 40 panels contribute: 40 × 0.72 × 0.45 = 12.96 m² extra absorption. A_total (125 Hz) = A_existing + 12.96 − (0 — we're adding to walls, not replacing)
Actually A_existing at 125 Hz was 12.66 m² (all surfaces). Adding wall panels (net over replaced plasterboard): 28.2 × (0.45 − 0.06) = 28.2 × 0.39 = 11.0 m² additional.
A_total (125 Hz) = 12.66 + 11.0 = 23.66 m² RT60 (125 Hz) = 0.161 × 132 / 23.66 = 1.44 s — much better than Option A (2.09 s).
Option C — Suspended Ceiling Baffles
Hanging ceiling baffles are exposed on both faces, so their effective absorbing area = 2 × physical area.
Absorption coefficients for 50 mm hanging baffle (both faces counted):
| 125 Hz | 250 Hz | 500 Hz | 1000 Hz | 2000 Hz | 4000 Hz |
|---|---|---|---|---|---|
| 0.35 | 0.65 | 0.90 | 0.95 | 0.95 | 0.90 |
Note: baffle coefficients are quoted per face. For a 600×1200 mm baffle (one face = 0.72 m²), total absorption per baffle at 500 Hz = 2 × 0.72 × 0.90 = 1.30 m².
Required absorption: 26.8 m². Number of baffles = 26.8 / 1.30 = 21 baffles.
Physical ceiling area consumed: 21 baffles × 0.72 m² per baffle = 15.1 m² of ceiling. Since the ceiling is 44 m², the baffles cover 34% of the ceiling area — a reasonable coverage that leaves adequate daylighting.
Three Options Summary at 500 Hz
| Option | Treatment | Area (m²) | Additional Absorption (m²) | RT60 Achieved (s) |
|---|---|---|---|---|
| A | NRC 0.70 ceiling tiles | 44.0 (full ceiling) | 26.0 | 0.62 |
| B | 100 mm wall panels | 28.2 | 27.5 | 0.59 |
| C | 50 mm baffles | 15.1 (ceiling footprint) | 27.3 | 0.59 |
All three options achieve approximately the target. The selection depends on:
- Structural: Can the ceiling support baffles? Is there a T-bar grid for ceiling tiles?
- Visual: Wall panels are prominent; baffles add industrial character; ceiling tiles are neutral.
- Low-frequency performance: Wall panels (Option B) provide the best bass control. Ceiling tiles (Option A) are weakest at 125–250 Hz.
Step 4 — Optimised Combined Solution
For best all-round performance, combine:
- Ceiling: NRC 0.70 ceiling tiles (full 44 m²) — handles 500–4000 Hz
- Wall panels: 100 mm mineral wool, fabric-wrapped, 14.4 m² on the two solid end walls (whiteboard end gets 8 panels, opposite end gets 12 panels)
- Ceiling tiles: 7.92 m² (additional over plasterboard: 7.92 − 6.60 = 1.32 m²)
- Wall panels net: 14.4 × (0.45 − 0.06) = 14.4 × 0.39 = 5.62 m²
- Total additional: 6.94 m²
| Band (Hz) | A_existing (no ceiling, no new panels) | A_ceiling tiles | A_wall panels (net) | A_total | RT60 (s) |
|---|---|---|---|---|---|
| 125 | 8.43 | 7.92 | 5.62 | 21.97 | 1.55 s |
| 250 | 6.03 | 17.60 | 9.94 | 33.57 | 1.02 s |
| 500 | 4.42 | 28.60 | 12.81 | 45.83 | 0.47 s |
| 1000 | 3.72 | 37.40 | 13.25 | 54.37 | 0.40 s |
| 2000 | 3.04 | 35.20 | 13.11 | 51.35 | 0.42 s |
| 4000 | 2.92 | 33.00 | 12.81 | 48.73 | 0.44 s |
The 125 Hz band remains at 1.55 s. To reach the 0.60 s target at 125 Hz would require approximately 35.4 − 21.97 = 13.4 m² of additional low-frequency absorbers (bass traps, 200 mm mineral wool panels, or resonant panel absorbers tuned to 100–150 Hz). In practice, most meeting room specifications do not require RT60 control below 250 Hz.
Summary: The Inverse Sabine Workflow
- Define V and target RT60 → A_required = 0.161 × V / RT60_target
- Inventory existing surfaces → A_existing = Σ(Si × αi)
- Deficit = A_required − A_existing
- Choose treatment material → required area = deficit / α_treatment
- Check available surface area — can the deficit area physically fit in the room?
- Verify at all octave bands — does the treatment solve the problem across the full spectrum?