Distance Attenuation Calculator
Estimate intensity reduction and distance loss for a point source using the inverse-square law.
About distance attenuation
Distance attenuation examples
| Reference values | Target result | Distance change |
|---|---|---|
| 100 W/m² at 10 m, target 20 m | 25 W/m²; 6.0206 dB | Distance doubled |
| 90 W/m² at 5 m, target 15 m | 10 W/m²; 9.5424 dB | Distance tripled |
| 16 W/m² at 8 m, target 4 m | 64 W/m²; -6.0206 dB | Distance halved |
How to calculate attenuation
- Enter the measured or known intensity at the reference location.
- Enter the positive reference distance from the point source.
- Enter the positive target distance using the same unit.
- Select Calculate attenuation to view target intensity and decibel loss.
Frequently asked questions
Why does intensity follow an inverse-square law?
Power from an ideal point source spreads across a spherical area proportional to distance squared. The power per unit area therefore falls in inverse proportion to that square.
How much attenuation occurs when distance doubles?
Intensity falls to one quarter of its reference value. This geometric reduction corresponds to approximately 6.0206 decibels.
Why can the decibel result be negative?
A negative attenuation appears when the target is closer than the reference point. It represents an intensity gain relative to the selected reference.
Does this include air or cable absorption?
No, the calculation includes geometric spreading only. Add frequency-dependent absorption, cable loss, obstacles, and equipment loss separately.
Does the formula work close to the source?
Not always, because an extended source can have a complex near field. Use the inverse-square model only where the source behaves approximately like a point.