Fresnel zone and radio link clearance
The first Fresnel zone radius at a point on a radio link, the 60% clearance it needs, and the Earth's bulge there, summed into the clearance above a smooth Earth.
Planning and education aid. Not for primary navigation. Full disclaimer
At that point the link needs 8.29 m of clearance above a smooth Earth.
- This assumes a smooth Earth: terrain, buildings, and trees can block the view or the signal.
- First Fresnel radius
- 60% of the Fresnel radius
- Earth bulge
Provenance
- Computed by
- navigation.los.fresnel 1.0.0, core 0.1.0
- Model
- First Fresnel zone and Earth bulge with k = 0.25 (effective radius 8494667 m)
- Accuracy
- Exact for the model; the effective Earth radius varies with the weather, so plan margin for lower k
- Notes
- 1 shown with the answer
- Cites
- International Telecommunication Union, ITU-R P.530: Propagation data and prediction methods for terrestrial line-of-sight systems
Something look off?
How we got thisFormula, worked example, sources, and proof
Model: First Fresnel zone √(λ d1 d2 / d) and Earth bulge on an effective radius R/(1 − k)
Accuracy: Exact for the model; the effective Earth radius varies with the weather, so plan margin for lower k
When to use this: Use this when planning a radio path rather than a sightline: a drone control link, a point-to-point backhaul, a repeater shot across a valley. A radio link needs more room than a bare line of sight, because the signal travels in a zone around the straight path and an obstacle intruding into that zone costs signal even when nothing blocks the view. This gives the first Fresnel zone's radius at a point along the path, the 60% of it that is the usual planning rule, the Earth's bulge there, and the two added: the height a smooth Earth path has to clear.
Limitations: This is clearance above a smooth Earth, not above the ground you are actually shooting over, so a terrain profile still has to be laid under it. It is one zone at one point on the path, not a diffraction loss: an obstacle inside the 60% figure degrades the link by an amount this does not compute. The Earth bulge depends on the refractive K factor, held at the standard 4/3; real air departs from it, and a sub-refractive day flattens the effective Earth and raises the bulge, which is why margin is planned against lower K rather than the nominal. Frequency is treated as a single wavelength, so a wideband or frequency-hopping link should be planned at its lowest frequency, where the zone is widest.
Worked example: A 5.8 GHz drone link of 10 km, at the midpoint. Source: ITU-R P.530's printed form F1 = 17.3 √(d1 d2 / (f d)) gives 11.3580 m against the tool's 11.3675 m, and the microwave-path Earth bulge d1 d2 / (12.75 K) gives 1.4706 m against 1.4715 m; each published constant is short by its own rounding, 0.084% and 0.063%, at every frequency and distance alike. It is golden test vector v001, and every build checks the tool still gives its answer within its tolerance.
You enter
- Link length
- 10 km
- Frequency
- 5.8 GHz
You get
- Required clearance
- 8.29 m
- First Fresnel radius
- 11.37 m
- 60% of the Fresnel radius
- 6.82 m
- Earth bulge
- 1.47 m
Review: Not yet independently reviewed by a geodesist.
Last verified: 2026-09-19, when a maintainer last confirmed this tool's sources at the issuer. See the sources ledger.
Status: version 1.0.0, core 0.1.0. See this tool in the verification report.
Checked against: 23 golden test vectors (download the test vectors, each with its source and tolerance). See how results are checked and every source.
Sources
- ITU-R P.530: Propagation data and prediction methods for terrestrial line-of-sight systems, International Telecommunication Union, P.530-19 (09/2025). Fresnel ellipsoid clearance and the effective Earth radius factor.