For clientsFor designers
How much absorption material does a gym hall need
A complete calculation for a typical 24 × 12 × 7 m hall: the standard allows 1.5 s of reverberation, which requires about 216 m² of absorption. How many square metres of ceiling and wall that means — and why it pays to leave a margin.
For school head teachers, local authorities and designers who want to know how much material to order, before they order it.
Unadapted sports halls very often have a reverberation time longer than 5 seconds. The teacher's instructions then blur together with the sound of balls, and noise during lessons builds up because every sound keeps circulating around the hall for a long time. The solution is absorption material — the question is: how much is needed?
The answer can be worked out on a piece of paper. Below is the full calculation for a typical school hall 24 × 12 m with a height of 7 m. Substitute your own dimensions to get the order of magnitude for your hall.
Step 1: target and required absorption
Volume: V = 24 × 12 × 7 = 2016 m³. For gym halls and sports halls up to 5000 m³, the PN-B-02151-4:2015-06 standard allows T ≤ 1.5 s in every octave band from 250 to 4000 Hz.
The rearranged Sabine formula is: A = 0.161 × V / T. The required absorption is therefore 0.161 × 2016 / 1.5 ≈ 216 m². Absorption is expressed in square metres of an “ideal absorber” — surface area multiplied by the absorption coefficient.
Step 2: how much is missing, and how many metres of material that is
Suppose the measurement before adaptation gave T = 5 s. Then the hall currently has A = 0.161 × 2016 / 5 ≈ 65 m². That leaves 216 − 65 = 151 m² of absorption missing.
Panels with a coefficient of 0.90 provide 0.9 m² of absorption per square metre, so the minimum is 151 / 0.9 ≈ 168 m² of material — about 58% of a 288 m² ceiling. That, however, is the lower bound: the panel covers the soffit above it, which was itself absorbing something, and panels perform less well in the low bands. That is why designs include a margin.
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Step 3: ceiling and walls, not the ceiling alone
Ecophon’s commentary on the standard, in the section on sports facilities, gives a practical rule: material with αw ≥ 0.9 on about 70–80% of the ceiling and 10–20% of the walls, so that absorption is present on each of the three pairs of parallel surfaces. For our hall that means, say, 75% of a 288 m² ceiling ≈ 216 m², plus 15% of 504 m² of wall ≈ 76 m². Together, about 292 m² of material — clearly above the calculated minimum, and that is exactly the margin.
Walls are not just needed for the balance sheet: without them, a flutter echo remains between bare, parallel walls. The margin also helps in the 250 Hz band — here the product datasheet and the suspension height matter. The Ecophon Super G A 35 mm panel has a coefficient of 0.95 at 250 Hz with a 200 mm void, but only 0.55 at 50 mm.
Najczęściej zadawane pytania
Is the Sabine formula enough for a design?
For estimating the amount of material — yes. For detailed design, it is better to calculate each band separately, using the product datasheet, and confirm the result with a measurement after installation. αw alone is just one number for all bands; see the article on the αw absorption class for details.
Where do you get the reverberation time “before”?
Ideally from a measurement in the empty hall. Without a measurement, you can conservatively assume the hall is very reverberant and design the material as if it absorbed almost nothing today — in that case you need the full 216 m² of absorption, i.e. at least 240 m² of material with α = 0.90.
What if the hall is larger than 5000 m³?
Then the standard allows T ≤ 1.8 s. The calculation is the same: A = 0.161 × V / 1.8. For example, a 36 × 24 × 8 m hall has V = 6912 m³ and needs about 618 m² of absorption.