Last updated: 2026-07-13

Why sound gets quieter with distance

Sound carries energy outward from its source. As the wave spreads, that fixed amount of power is smeared over a larger and larger surface, so the energy passing through any small patch — the intensity — keeps falling. The sound-level meter reads this as a steady drop in decibels the farther you stand from the source.

In a free field — open air with no reflecting surfaces and no meaningful absorption — the drop depends only on the ratio of two distances. If the level is $L_1$ at a reference distance $d_1$, then at a farther distance $d_2$ the level for a point source is

$L_2 = L_1 - 20\log_{10}\frac{d_2}{d_1}$

Only the distance ratio appears, so it does not matter whether you measure in metres or feet as long as both distances use the same unit.

The inverse-square law

A compact source that radiates equally in all directions is a point source. Its energy spreads over the surface of a sphere, and the area of a sphere grows with the square of its radius. Doubling the distance quadruples the area, so the intensity drops to one quarter:

$I \propto \frac{1}{d^2}$

Converting an intensity ratio to decibels uses $10\log_{10}$, and a quarter of the intensity is

$10\log_{10}(4) \approx 6.02\ \text{dB}$

That is the origin of the well-known rule: about 6 dB quieter for every doubling of distance. The factor of 20 in the level formula is just $10\log_{10}$ applied to a squared distance ratio, since $\log_{10}(d^2) = 2\log_{10}(d)$.

Point source vs line source

Not every source is compact. A steady stream of traffic on a highway, or a long railway line, acts like an unbroken row of sources — a line source. Its energy spreads over the curved surface of a cylinder rather than a sphere, and a cylinder's area grows in direct proportion to the radius, not its square. The intensity then falls as $1/d$, and the level formula loses a factor of two:

$L_2 = L_1 - 10\log_{10}\frac{d_2}{d_1}$

So a line source quiets by only about 3 dB per doubling of distance. This calculator switches the multiplier between 20 and 10 when you pick the source type.

Worked example

A machine reads $L_1 = 100\ \text{dB}$ at $d_1 = 1\ \text{m}$. How loud is it at $d_2 = 10\ \text{m}$ as a point source?

$L_2 = 100 - 20\log_{10}\frac{10}{1} = 100 - 20 \times 1 = 80\ \text{dB}$

The level falls by 20 dB. That is consistent with the doubling rule: going from 1 m to 10 m is about 3.3 doublings, and $3.3 \times 6 \approx 20\ \text{dB}$. If the same source were a line source, the drop would be only 10 dB, leaving 90 dB at 10 m.

What the estimate leaves out

The formula captures geometric spreading alone — the pure effect of the wavefront growing. Several real-world factors change the true level:

• Reflections. Indoors, walls, floors, and ceilings return energy to the listener, so the level drops far less than the free-field prediction. Close to a source in a room the inverse-square behaviour still holds, but beyond the "critical distance" the reverberant field dominates and the level flattens out.
• Air absorption. Over hundreds of metres the air itself absorbs sound, especially at high frequencies, adding attenuation on top of spreading.
• Ground and weather. Soft ground, barriers, wind, and temperature gradients bend and absorb sound outdoors, which is why noise maps use more elaborate models than geometric spreading alone.

Use this calculator for the spreading loss — the dominant effect outdoors near a source — and treat it as a clean baseline rather than a complete acoustic prediction.