What is a Fresnel zone and why does clearance matter?
The first Fresnel zone is the ellipsoid around a line-of-sight path within which path differences are less than λ/2. Its radius at a point d1 from one end and d2 from the other is r1 = √(λd1d2/(d1 + d2)), or in practical units r1(m) = 17.32·√(d1d2/(f·d)) with distances in km and f in GHz. Keeping at least about 60% of the first zone clear typically gives near free-space loss.[1]
What is the radius for a typical link?
At 5.8 GHz over 10 km, the midpoint radius is 17.32·√(5×5/(5.8×10)) ≈ 11.4 m; 60% clearance requires about 6.8 m above obstacles, plus Earth bulge.
What happens when the zone is obstructed?
A single sharp obstacle is modeled as a knife edge with parameter ν = h·√(2(d1 + d2)/(λd1d2)). At grazing (ν = 0) the loss is 6 dB; ITU-R P.526 gives the approximation J(ν) = 6.9 + 20·log10(√((ν − 0.1)² + 1) + ν − 0.1) dB for ν > −0.78.[1]
How does Earth curvature enter?
Earth bulge at a point is h = d1d2/(12.74·k) meters (km units), with effective Earth-radius factor k ≈ 4/3 for standard atmosphere. Planners check clearance for low k values to ensure availability during sub-refractive conditions.
Frequently asked questions
Why 60% clearance?
It keeps diffraction loss near zero relative to free space for typical obstacles.
Do Fresnel zones shrink with frequency?
Yes, as the square root of wavelength.
Is a visual line of sight enough?
Not necessarily; the Fresnel zone may still be obstructed.
See also
References
Online sources accessed October 21, 2026; printed works cited by edition. See all sources.
- “Recommendation ITU-R P.526: Propagation by diffraction”. ITU-R. current version.