Distance Attenuation Calculator

Estimate intensity reduction and distance loss for a point source using the inverse-square law.

Calculate distance attenuation
Compare intensity at two distances from an unobstructed point source.

About distance attenuation

Distance attenuation describes how energy from an ideal point source spreads as it travels outward. In free space, the same emitted power is distributed over the surface of an ever-larger sphere. Because a sphere's area increases with the square of its radius, intensity decreases according to the inverse-square law. This calculator applies I2 = I1 × (d1 / d2)², where I1 is intensity measured at reference distance d1 and I2 is the predicted intensity at target distance d2. It also reports the corresponding distance ratio in decibels as 20 × log10(d2 / d1). Doubling the distance produces one quarter of the original intensity and approximately 6.0206 decibels of attenuation. Tripling distance produces one ninth of the intensity and about 9.5424 decibels of attenuation. Moving closer gives a negative loss value, representing gain relative to the reference location. The decibel expression uses a factor of twenty for distance because field amplitude falls inversely with distance while power intensity is proportional to amplitude squared. The equivalent power ratio calculation, 10 × log10(I1 / I2), gives the same result. The model is appropriate for an isotropic source radiating uniformly in unobstructed three-dimensional space. Common examples include approximate sound, light, radio, and radiation intensity away from a source when the observation distances are large compared with source dimensions. Actual transmitters and speakers are directional, and nearby measurements may occur in a near field where the simple law is not valid. Walls, terrain, reflections, absorption, scattering, atmospheric effects, and antenna gains can all change real measurements. Use the same length unit for both distance fields; meters are shown for clarity, but any matching unit gives the same ratio. Initial intensity must describe the value at the stated reference distance, not total source power. This calculator isolates geometric spreading and does not add material absorption or system losses. It is useful for quick link intuition, acoustic planning, optical estimates, and inverse-square homework checks. For compliance, exposure, communications, or safety decisions, use calibrated measurements and a domain-specific propagation model that includes the source pattern and environment.

Distance attenuation examples

Reference valuesTarget resultDistance change
100 W/m² at 10 m, target 20 m25 W/m²; 6.0206 dBDistance doubled
90 W/m² at 5 m, target 15 m10 W/m²; 9.5424 dBDistance tripled
16 W/m² at 8 m, target 4 m64 W/m²; -6.0206 dBDistance halved

How to calculate attenuation

  1. Enter the measured or known intensity at the reference location.
  2. Enter the positive reference distance from the point source.
  3. Enter the positive target distance using the same unit.
  4. 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.