Fresnel Zone Calculator
Fresnel Zone Calculator
The first Fresnel zone radius anywhere along a radio path, the 60% of it that has to stay clear for a link to behave as if nothing were there, the earth’s own bulge under the straight line with a k-factor you can set, and the knife-edge diffraction loss when the clearance is not enough.
Fresnel zone
a 10 km link at 5.8 GHz with a ridge 4 km from the near end, 10 m below the straight line, k = 1.3333 and a 60% clearance rule
The zone, the bulge and the 60%
in ITU units: F₁ = 17.3 √(d₁d₂ / (f d)) m (f in GHz, d in km)
earth bulge: h = d₁d₂ / (2 k a) = d₁d₂ / (12.742 k) m (d in km)
k = 157 / (157 + dN/dh) ⟹ k = 4/3 at −39 N/km
clearance c below the line: ν = −c√2 / r₁ J(ν) = 0 for ν ≤ −0.78
- d1, d2
- distances from each end of the path to the obstacle; d is their sum
- lambda
- wavelength, c ÷ f. The zone gets narrower as the frequency rises — one of the few things that gets easier at microwave
- k
- effective earth radius factor. 4/3 in a standard atmosphere, because refraction bends the ray towards the ground and inflating the earth is the way to draw the ray straight
- nu
- the diffraction parameter of ITU-R P.526. Negative means clearance, positive means the obstacle is up through the line of sight
Worked example
a 10 km link at 5.8 GHz with a ridge 4 km from the near end, 10 m below the straight line, k = 1.3333 and a 60% clearance rule
λ = c ÷ f = 51.69 mm
r₁ = √(λ·d₁·d₂ ÷ d) = √(0.051688 × 4,000 × 6,000 ÷ 10,000) = 11.138 m at the ridge. At mid-path it would be 11.368 m — the zone is widest in the middle
60% of that is 6.683 m
The earth bulges up under the chord by d₁d₂ ÷ (2ka) = 1.413 m at the ridge, which eats into the clearance before any obstacle does
So the line of sight has to pass 6.683 + 1.413 = 8.095 m above the ridge top. You have 10 m, so there is 1.905 m of margin
In diffraction terms the clearance is 77.1% of the first zone, ν = -1.090. ITU-R P.526 gives zero diffraction loss for any ν below −0.78, so there is 0.00 dB to add — which is the real reason the 60% rule is 60%: ν = −0.78 corresponds to 55.2% clearance, and 60% is the small margin on top
How wide the first zone actually is at mid-path
| Path length | 868 MHz | 2.4 GHz | 5.8 GHz | 18 GHz |
|---|---|---|---|---|
| 100 m | 2.94 m | 1.77 m | 1.14 m | 0.65 m |
| 1 km | 9.29 m | 5.59 m | 3.59 m | 2.04 m |
| 5 km | 20.78 m | 12.50 m | 8.04 m | 4.56 m |
| 10 km | 29.38 m | 17.67 m | 11.37 m | 6.45 m |
| 30 km | 50.90 m | 30.61 m | 19.69 m | 11.18 m |
| 50 km | 65.71 m | 39.51 m | 25.42 m | 14.43 m |
What the clearance fraction costs you
| Clearance | ν | Diffraction loss (P.526) | Exact Fresnel integral |
|---|---|---|---|
| 100% of F₁ | -1.414 | 0.00 dB | -1.00 dB |
| 80% of F₁ | -1.131 | 0.00 dB | -1.29 dB |
| 60% of F₁ | -0.849 | 0.00 dB | -0.36 dB |
| 55% of F₁ | -0.780 | 0.00 dB | 0.01 dB |
| 40% of F₁ | -0.566 | 1.48 dB | 1.39 dB |
| 20% of F₁ | -0.283 | 3.65 dB | 3.62 dB |
| 0% of F₁ | -0.000 | 6.03 dB | 6.07 dB |
| obstacle level with the line | 0.000 | 6.03 dB | 6.07 dB |
| 0.5 F₁ above the line | 0.707 | 11.89 dB | 11.94 dB |
| 1 F₁ above the line | 1.414 | 16.34 dB | 16.12 dB |
Why a clear line of sight is not the test
Radio does not travel along a line. A signal arriving at the receiver by any route within half a wavelength of the direct path arrives in phase enough to add to it, and the set of points satisfying that is an ellipsoid with the two antennas at its foci — the first Fresnel zone. Its radius at a point d₁ from one end and d₂ from the other is √(λd₁d₂/d), fattest at the middle of the path. Everything else about clearance follows from that one shape.
Where 60% comes from. The usual rule is to keep 60% of the first zone clear, and it is normally handed over without explanation. Here is the explanation. Express the clearance as the diffraction parameter of ITU-R P.526, ν = −c√2/r₁, and the recommendation’s own formula gives exactly zero diffraction loss for any ν at or below −0.78. That is a clearance of 0.78/√2 = 55.2% of the first zone. Sixty per cent is that number rounded up with a little margin — no more mysterious than that. The table above checks it against the Fresnel-Kirchhoff integral itself rather than against the approximation, and at 60% clearance the integral gives a fraction of a decibel of gain, not loss.
The earth gets in the way too. Over any distance worth planning, the ground bulges up under the straight line between the antennas by d₁d₂/(2ka). At mid-path on a 50 km link with k = 4/3 that is 36.8 m — usually far more than the Fresnel radius, and the dominant term on a long path. The k-factor is the standard trick for handling refraction: the atmosphere’s refractive index falls with height, so a ray bends gently downward, and rather than draw a curved ray over a round earth you draw a straight ray over a bigger one. k = 1/(1 + a·dn/dh), and since 1/a is 157 N-units per kilometre, k = 157/(157 + dN/dh). The average low-level gradient of about −39 N/km gives k = 1.33, which is where 4/3 comes from. It is an average, not a constant: on a still night an inversion can push k below 1, the effective earth grows, and a path that cleared comfortably in the afternoon stops clearing. That is why ITU-R P.530 asks for two clearance tests rather than one — the full first zone at k = 4/3, and a smaller fraction at the worst-case k for your region.
What to do with the answer. If the margin is positive, the path behaves like free space and the free-space path loss page describes it. If the margin is negative but the obstacle is still below the line, the link works but is no longer free space, and the diffraction figure here is the extra loss to budget. If the obstacle is above the line the path is obstructed, and the loss above is a single knife edge — a real ridge with a rounded top, or two ridges, loses more, which is what the rest of ITU-R P.526 is about. Take the total to the link budget calculator to see whether the receiver can stand it.
Two things that surprise people. Higher frequencies need less clearance, because the zone narrows as √λ — one of very few things that get easier as you go up in frequency. And trees are not obstacles you can knife-edge: foliage both absorbs and scatters, it changes with the season and with rain, and it should be budgeted as an attenuation per metre of vegetation depth rather than as a diffraction edge.
Frequently asked questions
What is the first Fresnel zone?
The ellipsoid around the direct path within which any reflected or diffracted route arrives less than half a wavelength late, so its contribution adds to the direct signal rather than cancelling it. Its radius is √(λd₁d₂/d), largest at the middle of the path.
Why 60% clearance and not 100%?
Because 60% is where the loss has already gone to zero. Converting clearance to ITU-R P.526’s diffraction parameter, ν = −c√2/r₁, the recommendation gives no diffraction loss for ν ≤ −0.78, which is 55.2% of the first zone. Sixty per cent is that with a little margin. Clearing the whole zone buys nothing at k = 4/3 — but ITU-R P.530 still asks for it, because it is the margin that keeps the link up when the k-factor moves.
How much clearance do I need over a hill?
Your clearance fraction of the first zone radius at that hill, plus the earth bulge d₁d₂/(2ka) at the same point, measured above the top of the hill and below the straight line between the antennas. Both terms are on this page; the bulge dominates on anything longer than about 15 km.
What is the k-factor and why is it 4/3?
It is the factor by which you inflate the earth’s radius so that a refracted ray can be drawn straight. k = 157/(157 + dN/dh) with the refractivity gradient in N-units per kilometre, and the average gradient near the ground of about −39 N/km gives 1.33. It varies with weather: sub-refraction with k below 1 makes the effective bulge worse and is the case worth checking on a long link.
Does the Fresnel zone get bigger or smaller at higher frequency?
Smaller, as the square root of the wavelength. A 10 km path needs 29.4 m of mid-path radius clear at 868 MHz and 11.4 m at 5.8 GHz. It is one of the few things that gets easier as the frequency rises — everything else about the link gets harder.
My path is clear to the eye. Is that enough?
No. The zone is metres to tens of metres wide, so a ridge, a rooftop or a row of trees can sit well below the visual line of sight and still cost several decibels. That is exactly what this page is for: put in the distance to the obstacle and how far below the line it sits, and it will tell you whether the path is really clear.
Related calculators
References
- Recommendation ITU-R P.526-14 (01/2018), Propagation by diffraction. Equation (26) defines ν = h√(2(d₁+d₂)/(λd₁d₂)), equation (31) gives the single knife-edge loss J(ν) = 6.9 + 20 log₁₀(√((ν−0.1)²+1) + ν − 0.1) dB for ν greater than −0.78 (and zero below it), and §2.3 defines the diffraction zone as beginning where clearance falls to 60% of the first Fresnel zone radius. The Fresnel radius is given in practical units as Rₙ = 550√(n d₁d₂ / (f(d₁+d₂))) m with f in MHz.
- Recommendation ITU-R P.530-18 (09/2021), Propagation data and prediction methods required for the design of terrestrial line-of-sight systems. Equation (3) gives F₁ = 17.3√(d₁d₂/(fd)) m with f in GHz and d in km, and §2.2.2.1 sets the clearance procedure for a non-diversity path: 1.0 F₁ at k = 4/3, plus a second test at the worst-case k of 0.0 F₁ for a single isolated obstruction or 0.3 F₁ for an extended one in a temperate climate, and 0.6 F₁ on tropical paths over about 30 km.
- Recommendation ITU-R P.834-9 (12/2017), Effects of tropospheric refraction on radiowave propagation: the effective earth radius Re = k·a with k = 1/(1 + a·dn/dh), and the statement that for heights below 1,000 m the average refractive index profile of Recommendation ITU-R P.453 linearises to a k-factor of 4/3. The 157 in k = 157/(157 + dN/dh) is 10⁶ divided by the earth’s radius in kilometres, which this page’s build script re-derives as 156.96.
- Rappaport TS. Wireless Communications: Principles and Practice, 2nd ed., Prentice Hall, 2002, Chapter 4 — the Fresnel zone geometry, the knife-edge diffraction model and the Fresnel-Kirchhoff integral that the table above evaluates numerically.