Fresnel zone clearance calculator

Place an obstacle along the path and enter how far its top sits below the direct antenna-to-antenna ray. The shaded zone shows why visual line of sight is only part of a radio survey.

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Path length1,000Frequency868
AB

First zone and your 60% target. Vertical scale exaggerated; the obstruction is entered relative to the ray.

Clearance below direct ray6Required clearance3.345

2.655 m above the clearance target at this obstacle. A negative entered clearance means the obstacle crosses the direct ray.

Use this distance and frequency in the link budget

Radio path
MHz
m
Obstacle
%

Negative means the obstacle crosses the direct ray.

m
%

First Fresnel-zone radius at obstacle

5.575m

Obstacle 100 m from the first antenna on a 1,000 m path.

Required clearance at your target
3.345 m
Above clearance target
2.655 m

Geometry only. Enter clearance relative to the direct ray after accounting for terrain and Earth curvature. This is not a coverage prediction.

How it’s calculated
  1. Wavelength = speed of light ÷ frequency=299792458 ÷ (868 × 1000000)=0.3454 m
  2. First-zone radius = √(wavelength × d₁ × d₂ ÷ distance)=√(0.3454 × 100 × 900 ÷ 1,000)=5.575 m

The first-zone radius is 5.575 m; 60% clearance requires 3.345 m below the direct ray.

Tip: Measure clearance from the direct ray, not from ground level. Survey terrain and curvature separately on long paths.

Check an obstacle on the radio path

A radio path needs space around its direct antenna-to-antenna ray. An obstacle can sit below visual line of sight and still intrude into the first Fresnel zone. The zone is widest around the midpoint and narrows towards either antenna.

This tool calculates that geometry at one obstacle. It does not fetch terrain or predict diffraction loss. Use the radio survey to supply clearance relative to the direct ray, including curvature corrections where needed.

Locate the obstruction

d₁ + d₂ = total path distance

The position slider is measured from antenna A. The two distances locate the obstruction along the path; they are not the antenna heights.

Calculate the first zone

r₁ = √(λ × d₁ × d₂ / (d₁ + d₂))

Use metres for distances and wavelength. This is the standard path-planning approximation. At the midpoint the radius simplifies to √(λ × distance / 4). A higher frequency has a smaller first-zone radius for the same path.

Compare a clearance target

Required clearance = target fraction × radius

The initial 60% value is an editable design target, not a guarantee of negligible loss in every setting. Positive margin means the entered obstacle is below that target boundary. A negative clearance means it intersects the direct ray itself.

Continue with the link budget

Geometry and power margin are separate checks

Carry the path length and frequency into the link-budget tool. Terrain, reflections, foliage and interference remain separate factors. A clear Fresnel zone cannot compensate for insufficient receiver power.

Fresnel zone clearance examples

A 1 km link at 868 MHz

Wavelength is about 0.3454 m. At the midpoint, the first-zone radius is approximately 9.292 m. A 60% target requires about 5.575 m clearance. A measured 6 m clearance is approximately 0.425 m above that geometric target.

First Fresnel-zone radius at obstacle 9.292 m Open in the calculator

Move the obstacle towards one end

At 100 m from antenna A, the radius is 60% of its midpoint value, approximately 5.575 m. The selected 60% clearance target becomes approximately 3.345 m. Moving an obstacle changes the geometry even though frequency and total path length stay the same.

First Fresnel-zone radius at obstacle 5.575 m Open in the calculator

Questions about Fresnel zone clearance

Is the radius the required antenna height?

No. It is a distance perpendicular to the direct ray at a particular point. Antenna heights must also account for terrain, obstacle elevation, curvature and the chosen path geometry.

Does this include Earth curvature?

No. Clearance must already be measured relative to the direct ray with any necessary terrain and effective-Earth-radius correction. Long-path refraction varies and needs a propagation study.

Why is the endpoint radius zero?

The path-planning formula narrows to zero at either antenna. That mathematical endpoint says nothing about antenna near-field behaviour or clearance from nearby mounting structures.

Does 60% clearance guarantee reception?

No. It checks an entered geometry target. Power margin, interference, changing vegetation and multipath still affect communication.

Limits of this result

  • The 60% starting target is editable. It is not proof of a reliable link or a universal site acceptance rule.
  • No terrain database, Earth curvature, atmospheric refraction, diffraction loss, foliage or interference model is included. Enter clearance after those geometric corrections.
  • At an antenna endpoint the mathematical radius is zero; this does not establish antenna near-field suitability.

Related guides

Sources

  1. Propagation by diffraction (opens in a new tab) (PDF) ITU-R, P.526-16, 2025, Fresnel zones