Case 1
- Measured level
- 100dB
- Measured at distance
- 1m
- New distance
- 10m
80.00dB
Open with these valuesSound Level at Distance Calculator
80.00dB
Result: 80.00 dBDoubling your distance from a point source costs about 6 dB, ten times the distance costs 20 dB. Enter the level you measured, the distance you measured it at, and the distance you care about. This is a free-field model — outdoors, away from reflecting walls.
Worked examples
Case 2
- Measured level
- 90dB
- Measured at distance
- 2m
- New distance
- 8m
77.96dB
Open with these valuesCase 3
- Measured level
- 80dB
- Measured at distance
- 10m
- New distance
- 1m
100.00dB
Open with these valuesHow it's calculated
L₂ = L₁ − 20 × log₁₀(d₂ ÷ d₁)
- StepEnter the sound level you measured and the distance you measured it at.
- StepEnter the distance you want the level for.
- ResultRead the level; a closer distance than the reference gives a higher one.
Reference table
| Level, measured at, new distance | Change | Level (dB) |
|---|---|---|
| 100, 1, 1 | same spot | 100.00 |
| 100, 1, 2 | twice as far, −6 dB | 93.98 |
| 100, 1, 4 | four times as far, −12 dB | 87.96 |
| 100, 1, 10 | ten times as far, −20 dB | 80.00 |
| 100, 1, 100 | a hundred times as far, −40 dB | 60.00 |
| 80, 10, 1 | ten times closer, +20 dB | 100.00 |
Questions
How do I calculate sound level at a different distance?
Use L2 = L1 − 20 × log10(d2 / d1), where L1 is the level at the reference distance d1 and d2 is your new distance. For 100 dB at 1 m, the level at 10 m is 100 − 20 × log10(10) = 100 − 20 = 80 dB. Take the ratio of the distances, take its base-10 log, multiply by 20, and subtract.
Why does doubling the distance drop the level by 6 dB?
Because log10(2) ≈ 0.301, and the formula multiplies that by 20: 20 × 0.301 ≈ 6 dB. So every time you double your distance from a point source, the sound pressure level falls by about 6 decibels — quadruple it and you lose about 12 dB.
What is a point source, and why does it matter?
A point source is small compared with your distance from it, so its sound spreads out evenly over a sphere and the inverse-square law applies. Most single machines, speakers, and alarms behave like point sources once you are a few metres away. A line source like a busy road spreads cylindrically and falls only about 3 dB per doubling instead of 6.
Does this work indoors?
Not directly. Indoors, sound reflects off walls, floor, and ceiling, so the reverberant field keeps the level up and the real drop is much smaller than 20 × log10(d2/d1). This calculator models a free field — outdoors, away from reflecting surfaces.
What happens if the new distance is closer than the reference?
Then d2 is smaller than d1, log10(d2/d1) is negative and the level rises. For example, 80 dB measured at 10 m becomes 100 dB at 1 m. Subtract the measured level from the result to read the change in decibels.
Sources and last check
Information, not professional advice.