What is a Fresnel Zone in Wireless Networking?
A Fresnel zone is an invisible ellipsoid of radio-frequency energy that exists between a transmitting and a receiving antenna. When a microwave or Wi-Fi signal leaves the transmitter it does not travel as a single thin ray. Instead it expands into a cigar-shaped volume whose surface is defined by the path-length difference between the direct ray and every possible reflected ray. The 1st Fresnel zone (F₁) is the most important of these ellipsoids: roughly half of the signal power is carried inside it, and the center of the zone coincides with the straight line-of-sight beam between the two antenna tips.
If any object — a hill, building, tree, or the ground itself — intrudes into this ellipsoid, it diffracts the signal and robs the receiver of energy. The rule used by wireless engineers is the 60% Clearance Rule: keep at least 60% of the 1st Fresnel zone radius (0.6 × F₁) clear of all obstructions along the entire path. At 60% clearance, diffraction loss stays under roughly 1 dB and the link behaves close to free-space. Below 60%, loss climbs quickly: an obstacle sitting exactly on the line-of-sight path can add 6 dB or more of loss, which is the difference between a working link and a dead one.
How to Calculate Fresnel Zone Clearance
The radius of the n-th Fresnel zone at any point along the link is:
Where λ is the wavelength in meters (λ = c / f, with c = 3 × 10⁴ m/s), d₁ is the distance from the transmitter to the point, d₂ is the distance from that point to the receiver, and D = d₁ + d₂ is the total link distance. At the exact midpoint of the path, where d₁ = d₂ = D/2, the formula simplifies to F₁ = 0.5 × √(λ × D). The 60% clearance boundary is simply F₆₀% = 0.6 × F₁.
Quick midpoint examples: at 2.4 GHz a 5 km link has F₁ ≈ 12.5 m and a 10 km link has F₁ ≈ 17.7 m (60% ≈ 7.5 m and 10.6 m). At 5.8 GHz the same 5 km link is F₁ ≈ 8.0 m and the 10 km link is F₁ ≈ 11.4 m. Notice that higher frequencies produce smaller Fresnel zones because the wavelength is shorter — this is why high-frequency links need less antenna height for the same clearance.
Earth Curvature & Refraction Correction (K = 4/3)
Over long paths (roughly beyond 8 km / 5 miles) the curvature of the Earth begins to matter. Even on perfectly flat terrain, the chord between two antenna bases passes above the ground at the midpoint of the link. The height of this bulge is:
Where d₁ and d₂ are in kilometers and the result is in meters. The factor K models atmospheric refraction. The standard value is K = 4/3, which assumes a mildly refractive atmosphere that bends the beam slightly toward the ground, making the effective Earth radius larger and the bulge smaller than in a vacuum (K = 1). A 30 km link with K = 4/3 has a midpoint bulge of about 30 × 30 / (12.74 × 1.333) ≈ 13.2 m — a real obstacle to clearance even on flat land.
Putting it together, the minimum antenna height needed at each end (assuming equal heights and a terrain baseline of zero) is Hmin = hobs + hearth + (0.6 × F₁), evaluated at every obstacle on the path and taking the largest result. The calculator above does exactly this: it computes F₁, the 60% boundary and the Earth bulge at every obstacle, colours the obstacle red if the boundary is breached, and reports the recommended tower height automatically.