Fresnel Zone Clearance Calculator

How it works

r₁ (m) = 17.32 × √(d₁ × d₂ ÷ (D × f)) with d in km, f in GHz; mid-path r₁ = 17.32 √(D ÷ 4f); clearance needed ≈ 0.6 r₁ above the obstacle + Earth bulge d₁d₂ ÷ (2 × 8,495 km)

A radio link is not a pencil beam: the energy that arrives travels through an ellipsoid around the straight line, and the first Fresnel zone is the region whose paths are within half a wavelength of the direct one, so they add constructively. Anything intruding into that zone — a rooftop, a tree line, the ground itself — diffracts and reflects energy that arrives out of phase and cancels part of the signal, even when the direct line is visually clear. The radius is largest at mid-path and grows with the square root of distance and inversely with the square root of frequency, so a 2.4 GHz link needs about 1.5 times the clearance of a 5.8 GHz one. Engineers require at least 60% of the first zone free (worth about 0 dB loss; grazing the line of sight itself costs 6 dB), and add Earth bulge because the ground rises between the ends of any long path.

Worked example

A 5.8 GHz link over 10 km with an 8 m obstacle at mid-path: the first Fresnel radius is 11.4 m, so 6.8 m of clearance is needed. Antennas at 20 m give a 20 m line of sight, minus 1.5 m of Earth bulge, leaving 10.5 m above the obstacle — comfortably clear (92% of the zone).

Frequently asked questions