Reverberation time: Sabine, Eyring, and where the formula stops working

RT60 is the time for the reverberant field to fall by 60 dB, and almost everyone computes it with one equation copied from a textbook. That equation carries an assumption — that the absorption in the room is small and evenly spread — which is true of a concert hall and false of the room you are sitting in.

The calculator gives both answers, Sabine and Eyring, so the size of the disagreement is visible rather than implied, and it works backwards from a target to the absorption the room actually needs.

The energy balance behind both formulas

Sound in a room carries a fixed amount of energy. Each reflection removes a fraction of it, and the time to lose 60 dB is set by how much is removed per reflection and how often reflections happen. Written out, the time between reflections is the mean free path divided by the speed of sound, and the loss per reflection is the absorption coefficient.

With c = 343 m/s at 20 °C that constant becomes 0.161, which is the version printed in every textbook. It is worth knowing that the 0.161 is not a physical constant: at 0 °C it is 0.171, a 6 per cent difference, and the formula is only ever as good as the absorption figure that goes into it.

Sabine's assumption, stated plainly

The step from the energy balance to that equation is a substitution: the energy lost per reflection is approximated as proportional to α. That is a first-order expansion, and it is accurate only while α is small — which is exactly the case in a hall with plaster, wood and an audience, and exactly not the case in a studio with 100 mm absorbers on two walls.

Nothing in the derivation is wrong. What fails is the linearisation, and it fails in one direction only: Sabine always over-predicts the decay time once the absorption stops being small, because α is larger than the logarithm it replaced.

Eyring, and where the 10 per cent arrives

Eyring keeps the reflection-by-reflection bookkeeping instead of linearising it. Each reflection leaves a fraction (1 − α) of the energy, so the decay depends on ln(1/(1 − α)) rather than on α directly.

The two converge as α goes to zero, because ln(1/(1−α)) ≈ α there. The divergence is worth having numbers for:

• α = 0.15 — about 4 per cent apart. Either is fine.
• α = 0.30 — about 18 per cent. Sabine is already optimistic.
• α = 0.50 — about 39 per cent. Sabine predicts half a second where the room does 0.36 s.
• α = 0.80 — a factor of two. Sabine's answer is not usable.

This matters because the rooms people actually treat are in the last two rows. A small studio with broad-band absorption averaging 0.5 is normal, and that is precisely where the textbook equation is worst.

Mean free path tells you whether the field is diffuse

The mean free path is 4V/S, and in a domestic room it comes out around 2 to 3 metres. That is the average distance between reflections, and it sets the time between them — about 6 to 9 ms at room temperature.

Both formulas assume the field is diffuse, which requires many reflections from many directions before the decay is measured. In a 5 by 4 by 2.6 m room the first reflection arrives 6 ms after the direct sound and the field is not remotely uniform until well after that. The result is that the decay you measure in such a room often has two slopes: a fast early part from the nearby surfaces and a slower tail from the rest.

Absorption is not one number

A published absorption coefficient is valid for one material, one thickness, one mounting and one frequency band. A 50 mm porous panel on a wall might be 0.15 at 125 Hz and 0.9 at 2 kHz; the same panel held 50 mm off the wall is materially better at the bottom and unchanged at the top; a thin carpet is 0.05 at 125 Hz and 0.6 at 4 kHz.

Averaging those into a single α is the single biggest source of error in a hand calculation, and it hides the thing that matters. Nearly every untreated small room has an RT60 that rises towards low frequency — a 0.25 s mid-band with a 0.6 s 125 Hz band is ordinary — and the average flattens exactly that shape away.

Work band by band, six octaves from 125 Hz to 4 kHz, and treat the mid-band figure as the one to quote. In a room this size, air absorption is worth about 0.01 per cubic metre at 4 kHz and nothing below 1 kHz, so it can be ignored here and cannot be in a hall.

The Schroeder frequency divides the room

Below roughly 2000 · √(T/V) hertz the room's response is a set of discrete modes rather than a diffuse field, and RT60 stops describing anything. For the room in the calculator that boundary lands near 200 Hz, which is well inside the range where a subwoofer operates.

The practical consequence is that RT60 and modal behaviour are not two views of one thing; they are two regimes. Treating absorption to fix a 60 Hz boom is the wrong tool, because the absorber that works at 200 Hz is 100 mm thick and the one that works at 60 Hz is a pressure-based device in a corner.

Five mistakes worth avoiding

1. Using Sabine on a treated room. Above an average α of about 0.3 the answer is optimistic, and the more successful the treatment, the worse the error.
2. One absorption number for all frequencies. It averages away the low-frequency rise that is the actual problem in most rooms.
3. Quoting RT60 below the Schroeder frequency. There is no diffuse field there for it to describe.
4. Reading a single-slope decay as gospel. In a small room the decay is often not linear in dB; T30 and T20 that disagree are telling you the field is not diffuse.
5. Designing to a target instead of a range. RT60 within ±15 per cent across 250 Hz to 4 kHz is the useful goal; a single number at 1 kHz is not.

Summary

Compute both, and read the gap between them as a confidence figure: a few per cent means the assumption holds, tens of per cent means you are outside the model's range and should be measuring instead. Then break the answer into octaves, because the average is the part that was never true.

Values shown are engineering aids rather than measurements; see the disclaimer and the tool index.