A structural engineer specifies a partition with STC 55 from a manufacturer's data sheet. The owner expects STC 55 in their building. The contractor installs it correctly. The acoustic consultant measures it at practical completion: FSTC 47. Everyone is angry. This scenario is preventable with a flanking transmission prediction performed at design stage, which takes less than two hours and requires no specialist software.
This article walks through a complete prediction for a meeting room partition between two open-plan offices, starting from a lab STC 55 and arriving at a predicted FSTC.
The Configuration
Partition: 90 mm steel stud, dual-layer 15 mm fire-rated gypsum board each side, resilient bar one side, acoustic insulation infill. Lab STC: 55. (Rw: 54 in European notation.)
Room geometry:
- Room A (source): 6.0 m × 4.0 m × 2.7 m (h)
- Room B (receive): 6.0 m × 4.0 m × 2.7 m (h)
- Shared partition: 4.0 m wide × 2.7 m high = 10.8 m²
- Ceiling: suspended 600 × 600 mm mineral fibre tile at 2.7 m, plenum height 0.5 m to structural soffit
- Floor: concrete slab, carpet finish
- Direct partition (the test specimen)
- Suspended ceiling tile + plenum
- Concrete floor
- Supply air duct passing through partition
- Electrical outlet back-boxes (recessed, back-to-back)
Step 1 — Direct Partition Transmission Loss Spectrum
The partition's transmission loss (TL) spectrum must be reconstructed from its STC 55 rating. STC is derived from a standard contour fit; working backwards, a typical dual-layer gypsum on steel stud with resilient bar has approximately this TL spectrum:
| Octave Band (Hz) | 125 | 250 | 500 | 1k | 2k | 4k |
|---|---|---|---|---|---|---|
| TL_direct (dB) | 36 | 42 | 52 | 57 | 60 | 62 |
Note the "mass-law dip" at 125 Hz — low-frequency performance is always significantly weaker than the mid-frequency STC value. This is characteristic of lightweight partitions and is one reason the STC single-number rating can mislead.
Convert TL to transmission coefficient τ:
τ = 10^(−TL/10)
| Band (Hz) | TL (dB) | τ |
|---|---|---|
| 125 | 36 | 2.512 × 10^−4 |
| 250 | 42 | 6.310 × 10^−5 |
| 500 | 52 | 6.310 × 10^−6 |
| 1k | 57 | 2.000 × 10^−6 |
| 2k | 60 | 1.000 × 10^−6 |
| 4k | 62 | 6.310 × 10^−7 |
Step 2 — Flanking Path 1: Ceiling Plenum
The suspended ceiling tiles provide some transmission loss, but the 0.5 m plenum above them is open to both rooms. Sound crosses the partition at the ceiling level, radiates into the plenum, travels over the partition top, and re-radiates into Room B through the ceiling tiles.
Ceiling tile TL (15 mm mineral fibre, area weight ~4 kg/m²):
| Band (Hz) | 125 | 250 | 500 | 1k | 2k | 4k |
|---|---|---|---|---|---|---|
| TL_tile (dB) | 12 | 18 | 25 | 32 | 38 | 40 |
The flanking path through the plenum involves two passes through a ceiling tile (source side and receive side) plus the plenum propagation loss. The plenum propagation loss depends on the plenum geometry and lining. For an unlined 0.5 m plenum:
Plenum propagation loss ≈ 3–6 dB (use 5 dB conservative estimate for an open plenum)
Effective TL of ceiling flanking path: TL_flank,ceiling = TL_tile(source) + plenum_loss + TL_tile(receive) = TL_tile + 5 + TL_tile = 2 × TL_tile + 5
| Band (Hz) | TL_tile × 2 | Plenum | TL_flank,ceiling (dB) |
|---|---|---|---|
| 125 | 24 | 5 | 29 |
| 250 | 36 | 5 | 41 |
| 500 | 50 | 5 | 55 |
| 1k | 64 | 5 | 69 |
| 2k | 76 | 5 | 81 |
| 4k | 80 | 5 | 85 |
Area of flanking ceiling path (ISO 15712: use area of source room floor = 24 m²):
S_flank,ceiling = 24 m²
Convert to τ_flank,ceiling = 10^(−TL_flank,ceiling/10):
| Band | TL_flank (dB) | τ_flank |
|---|---|---|
| 125 | 29 | 1.259 × 10^−3 |
| 250 | 41 | 7.943 × 10^−5 |
| 500 | 55 | 3.162 × 10^−6 |
| 1k | 69 | 1.259 × 10^−7 |
| 2k | 81 | 7.943 × 10^−9 |
| 4k | 85 | 3.162 × 10^−9 |
Step 3 — Flanking Path 2: Concrete Floor
The concrete slab transmits structure-borne sound around the partition through floor-wall junctions. For a 150 mm concrete slab with carpet finish:
Floor flanking TL (ISO 15712 junction method — simplified): Flanking TL ≈ Direct TL of floor + Junction attenuation − 5 dB (junction penalty)
For 150 mm concrete: floor TL ≈ 44 dB at 500 Hz (mass law). Junction attenuation ≈ 8 dB.
| Band (Hz) | 125 | 250 | 500 | 1k | 2k | 4k |
|---|---|---|---|---|---|---|
| TL_flank,floor (dB) | 28 | 33 | 39 | 45 | 51 | 55 |
Area: S_flank,floor = 24 m²
τ_flank,floor:
| Band | TL (dB) | τ |
|---|---|---|
| 125 | 28 | 1.585 × 10^−3 |
| 250 | 33 | 5.012 × 10^−4 |
| 500 | 39 | 1.259 × 10^−4 |
| 1k | 45 | 3.162 × 10^−5 |
| 2k | 51 | 7.943 × 10^−6 |
| 4k | 55 | 3.162 × 10^−6 |
Step 4 — Flanking Path 3: Supply Air Duct
A 300 × 150 mm sheet metal supply duct passes through the partition. Duct area = 0.3 × 0.15 = 0.045 m². The duct is unlined inside for 2 m either side of the partition.
Sound propagation loss in an unlined rectangular duct (ASHRAE method): For 300 mm wide duct: attenuation ≈ 0.6 dB/m at 125–250 Hz, rising to 2 dB/m at 1–2 kHz.
Total duct path length = 2 m × 2 sides = 4 m effective path.
| Band (Hz) | 125 | 250 | 500 | 1k | 2k | 4k |
|---|---|---|---|---|---|---|
| Duct attenuation (dB/m) | 0.6 | 0.8 | 1.2 | 2.0 | 2.0 | 2.5 |
| × 4 m path | 2.4 | 3.2 | 4.8 | 8.0 | 8.0 | 10.0 |
| TL_flank,duct (dB) | 2 | 3 | 5 | 8 | 8 | 10 |
Area: S_duct = 0.045 m²
τ_duct = 10^(−TL_duct/10):
| Band | TL (dB) | τ |
|---|---|---|
| 125 | 2 | 0.631 |
| 250 | 3 | 0.501 |
| 500 | 5 | 0.316 |
| 1k | 8 | 0.158 |
| 2k | 8 | 0.158 |
| 4k | 10 | 0.100 |
Step 5 — Flanking Path 4: Electrical Outlets (Back-to-Back)
Two duplex outlet boxes installed back-to-back through the partition, total open area behind both covers approximately 0.005 m² per pair. With an air gap connecting source and receive sides, this effectively functions as a hole with minimal TL.
Back-to-back outlet TL ≈ 5 dB (the cavity between boxes and the wall plates provide minimal attenuation).
| Band | TL_outlet (dB) | τ_outlet |
|---|---|---|
| All bands | 5 | 0.316 |
Area: S_outlet = 0.005 m²
Step 6 — Combine All Paths
ISO 15712 combined TL using parallel path summation. Reference area = S_partition = 10.8 m².
Combined τ_total = (1/S_partition) × Σ(τ_i × S_i)
For each band, calculate τ_i × S_i for each path:
At 125 Hz:
| Path | τ | Area (m²) | τ × S |
|---|---|---|---|
| Direct partition | 2.512 × 10^−4 | 10.8 | 2.713 × 10^−3 |
| Ceiling plenum | 1.259 × 10^−3 | 24.0 | 3.022 × 10^−2 |
| Floor | 1.585 × 10^−3 | 24.0 | 3.804 × 10^−2 |
| Duct | 0.631 | 0.045 | 2.840 × 10^−2 |
| Outlets | 0.316 | 0.005 | 1.580 × 10^−3 |
| Sum | 1.003 × 10^−1 |
τ_total,125 = 0.1003 / 10.8 = 9.285 × 10^−3
TL_combined,125 = −10 × log₁₀(9.285 × 10^−3) = 20.3 dB
At 500 Hz:
| Path | τ | Area (m²) | τ × S |
|---|---|---|---|
| Direct partition | 6.310 × 10^−6 | 10.8 | 6.815 × 10^−5 |
| Ceiling plenum | 3.162 × 10^−6 | 24.0 | 7.589 × 10^−5 |
| Floor | 1.259 × 10^−4 | 24.0 | 3.022 × 10^−3 |
| Duct | 0.316 | 0.045 | 1.422 × 10^−2 |
| Outlets | 0.316 | 0.005 | 1.580 × 10^−3 |
| Sum | 1.770 × 10^−2 |
τ_total,500 = 0.01770 / 10.8 = 1.639 × 10^−3
TL_combined,500 = −10 × log₁₀(1.639 × 10^−3) = 27.9 dB
Full combined TL spectrum:
| Band (Hz) | 125 | 250 | 500 | 1k | 2k | 4k |
|---|---|---|---|---|---|---|
| TL_direct | 36 | 42 | 52 | 57 | 60 | 62 |
| Dominant path | Floor | Floor | Duct | Duct | Duct | Duct |
| TL_combined (dB) | 20 | 25 | 28 | 33 | 35 | 38 |
Step 7 — Fit the STC Contour to get FSTC
Apply the STC contour to the combined field TL spectrum. The STC contour is fitted by shifting the standard contour downward in 1 dB steps until:
- No single-band deficiency exceeds 8 dB
- Sum of all deficiencies ≤ 32 dB
| Band (Hz) | 125 | 250 | 500 | 1k | 2k | 4k | STC contour at STC 32 |
|---|---|---|---|---|---|---|---|
| Combined TL | 20 | 25 | 28 | 33 | 35 | 38 | — |
| STC 32 contour | 19 | 25 | 32 | 36 | 36 | 36 | — |
| Deficiency | 0 | 0 | 4 | 3 | 1 | 0 | Sum = 8 ✓ |
Predicted FSTC = 32
This compares to the lab STC of 55 — a penalty of 23 dB. The duct is catastrophically bad (TL of only 2–5 dB), dominating every octave band. The ceiling plenum and floor flanking are secondary but significant.
Step 8 — Remediation Priority
Rank the flanking paths by their contribution to the energy sum at 500 Hz:
| Path | τ × S at 500 Hz | % of total |
|---|---|---|
| Duct | 1.422 × 10^−2 | 80.3% |
| Outlets | 1.580 × 10^−3 | 8.9% |
| Floor | 3.022 × 10^−4 | 1.7% |
| Ceiling | 7.589 × 10^−5 | 0.4% |
| Direct | 6.815 × 10^−5 | 0.4% |
The duct is causing 80% of the acoustic failure. Fitting an in-duct attenuator (silencer) with 25 dB insertion loss at all frequencies would reduce duct τ × S from 1.42 × 10^−2 to 1.42 × 10^−4 — a 100× reduction. The predicted FSTC would then rise to approximately 44–46, within 9–11 dB of the lab value.
The remaining gap is then dominated by floor flanking (now 55% of energy), which could be partially mitigated by resilient skirting at the partition base.
Predicted vs Measured: Typical Results
| Configuration | Lab STC | Predicted FSTC | Expected Measured FSTC |
|---|---|---|---|
| As designed (with duct) | 55 | 32 | 30–35 |
| Duct attenuator added | 55 | 45 | 43–47 |
| Duct attenuator + resilient skirting | 55 | 48 | 46–50 |
| Full isolation: sealed duct, plenum barrier, resilient base | 55 | 52 | 49–53 |
The full isolation package brings field performance within 2–3 dB of lab performance, which is the realistic ceiling for standard construction. Achieving FSTC 55 in the field requires extraordinary attention to all flanking paths simultaneously.
Run the same calculation for your project using AcousPlan's Sound Insulation Calculator, which models direct transmission plus up to four flanking paths and produces a predicted FSTC and STC contour comparison.