Phon.Audio
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Room & listening

Listening distance

How far you sit from the speakers — sets how much power you need.

Simple

How far your ears are from the speakers. The farther away you sit, the quieter the sound gets and the more power the amp needs to compensate. (Headphones ignore this — your "distance" is essentially zero.)

The theory

In a free field, sound from a point source obeys the inverse-square law: intensity falls with the square of distance, which is −6 dB SPL per doubling. The engine applies 20·log₁₀(distance) of attenuation relative to the 1 m reference, then asks the amp to make up the difference in power. Double your distance and you need 4× the power for the same loudness.

For experts

Inverse-square attenuation with distanceExpanding wavefront arcs from a point source at 1, 2, and 4 metres, losing 6 dB per doubling, with a dashed critical-distance arc where the room's reverberant field takes over.Source1 m0 dB ref2 m−6 dB4 m−12 dB (needs 4× power vs 2 m)Critical distance — beyond here thereverberant field dominates and the−6 dB/doubling law goes softFree-field model: intensity falls with the square of distance — the conservative end for sizing an amp

Real rooms aren't free fields, which is why this is a model, not gospel. Below the room's Schroeder frequency (~200–300 Hz domestically) you're in modal/pressure-zone behavior where the inverse-square law breaks down and standing waves dominate; above it, you transition toward a diffuse field where the falloff is gentler than 6 dB/doubling because reflected energy fills in. The full picture is the critical distance — the radius at which direct and reverberant energy are equal. Sit inside it and the direct-sound, inverse-square approximation holds well; sit beyond it and you're mostly hearing the room. The engine's 6 dB/doubling is the conservative (worst-case-for-power) end, which is the right way to size an amp.

ΔL=−20⁢log10(dd0)

Free-field attenuation: −6 dB per doubling of distance from a point source.

Check this in your chain →