Fresnel Zone Calculator

Introduction to Fresnel zone clearance on terrestrial radio links

When engineers say that two antennas have line of sight, the phrase sounds as if the link travels along a single infinitely thin ray. Real propagation is messier. Recommendation ITU-R P.526 describes the space between transmitter A and receiver B as being subdivided by a family of ellipsoids, all sharing A and B as focal points, such that any point M on the n-th ellipsoid satisfies AM + MB − AB = n λ/2. Those are the Fresnel ellipsoids. As a practical rule the same Recommendation assumes propagation is effectively line-of-sight, with negligible diffraction, only when there is no obstacle inside the first ellipsoid.

This calculator sizes that corridor at one specific point along a path and then does the two things most quick Fresnel tools skip. First, it works at any point, not just the midpoint, using the full d₁·d₂/(d₁+d₂) product form. Second, it adds the earth bulge implied by your effective earth-radius factor k, because on a 30 km or 50 km hop the curvature of the earth raises the terrain into your corridor far more than the Fresnel radius itself does. A Fresnel figure quoted without curvature is a flat-earth answer, and it is optimistic exactly where the money is spent: on tall towers for long links.

The third thing the page does is turn geometry into decibels. If you know how far the direct ray clears the obstacle top, the calculator reports the ITU-R P.526 diffraction parameter ν and the knife-edge loss J(ν), alongside the ITU-R P.530 average-terrain estimate. That is what makes the famous “keep 60% of the first Fresnel zone clear” rule checkable rather than folkloric: at 0.6 F₁ the parameter ν falls to about −0.85, which is past the ν = −0.78 point where the knife-edge loss curve reaches zero. The rule is the diffraction curve, restated as a distance.

Location matters more than most planners expect. The corridor is not the same size everywhere: it is widest where d₁ and d₂ are most nearly equal and it pinches to zero at each antenna. A modest rise near the centre of a hop can therefore be more damaging than a taller obstacle sitting a few hundred metres from one tower. The earth bulge peaks in the same place, which is why the midpoint is almost always the first point you should test — and why this page draws the whole profile rather than returning a single number.

How to use this Fresnel zone and earth bulge calculator

Distances d₁ and d₂ are measured along the great-circle path from the checkpoint you care about to each end of the link. Pick your unit once from the distance-unit menu — metres, kilometres, feet or statute miles — and both distances are read in that unit, so there is no chance of pairing a kilometre value with a metre value. For a midpoint check on a 3 km hop enter 1.5 and 1.5 with kilometres selected. For a tree line 400 m from the first antenna on the same hop, enter 0.4 and 2.6. The total path length d is simply d₁ + d₂; you never enter it separately, which removes another classic inconsistency.

Frequency f may be given in gigahertz or megahertz. ITU-R P.526 states its practical-unit formula in megahertz and kilometres, while ITU-R P.530 states its equivalent in gigahertz and kilometres, and field notes circulate in both, so the menu removes the factor-of-1000 slip that silently changes an answer by 31.6×. Internally everything is converted to hertz and metres before any arithmetic runs.

Zone order n defaults to 1. Set it to 2 or 3 when you are reasoning about reflection nulls rather than clearance: the even zones contribute out of phase with the direct ray, so a strong ground reflection whose extra path length places it in the second zone can cancel rather than reinforce. Clearance planning itself always uses n = 1.

Effective earth-radius factor k defaults to 1.333. ITU-R P.530 notes that variations in atmospheric refractivity change the effective earth radius, or k-factor, from its median value of approximately 4/3 for a standard atmosphere, and advises using k = 4/3 in the absence of local data. Sub-refractive conditions, the ones that actually break links, correspond to smaller k: enter 0.7 or 0.6 to see how much the bulge grows on a long hop when the atmosphere stops helping you.

Ray clearance h is optional. Leave it blank to get pure geometry. Enter the vertical distance from the top of the obstacle up to the direct ray — positive when the ray passes above the obstacle, negative when the obstacle pokes through the line of sight — and the calculator adds a diffraction-loss estimate and a pass/fail verdict against the 0.6 F₁ target. Result units lets you read every output in metres or feet without re-entering anything.

Press Compute clearance and the page returns a breakdown, a scale drawing of the path profile with the Fresnel envelope and the earth bulge, and a permalink that encodes every input so you can paste a specific scenario into a ticket or an email.

Formula set: Fresnel radius, earth bulge and diffraction loss

Everything on this page comes from four equations. The first is the radius of the n-th Fresnel ellipsoid, given in self-consistent units as equation (2) of ITU-R P.526-14. The calculator converts your frequency to a wavelength, combines it with the two distances, and reports the radius in your chosen output unit.

Here λ is the wavelength, c = 299 792 458 m/s is the speed of light in vacuum as fixed by the SI definition of the metre, and f is the frequency in hertz. Two consequences fall straight out of the algebra rather than out of habit. The radius grows with wavelength, so a 2.4 GHz link needs a corridor about √(5.8/2.4) = 1.55× wider than the same path at 5.8 GHz. And because the radius depends on the product d₁·d₂ divided by their sum, it is maximised when d₁ = d₂ — the midpoint bulge is a theorem, not a rule of thumb.

The practical-unit forms, and the 8.657 shortcut checked against them

ITU-R P.526-14 equation (3) restates the same relation in engineering units as R_n = 550 √[n d₁d₂ / ((d₁+d₂) f)] metres, with f in MHz and the distances in km. ITU-R P.530-16 equation (3) gives the first-zone form with gigahertz instead:

with f in GHz, d = d₁ + d₂ the path length in km, and the answer in metres. Those constants are not arbitrary and they are worth checking rather than trusting. Since λ (m) = 0.299 792 458 / f_GHz and one kilometre is 1000 m, the exact coefficient is √(1000 × 0.299792458) = 17.3145, which the Recommendation rounds to 17.3 — a 0.08% understatement. The megahertz version uses 550, whereas the exact coefficient is √(1000 × 299.792458) = 547.53; P.526 rounds up, so equation (3) is about 0.45% conservative. Both roundings are far smaller than any survey error, but you should know which way they lean.

The same check settles the shortcut every field engineer quotes. At the midpoint d₁ = d₂ = d/2, so d₁d₂/(d₁+d₂) = d/4 and the first-zone radius collapses to R₁ = ½ √(λd). Substituting λ = 0.299792458/f_GHz with d in kilometres gives ½ √(1000 × 0.299792458) = 8.6572, so:

The familiar 8.657 is therefore exact to four figures, but only at the midpoint. Applying it at a checkpoint that is not halfway along the hop is the single most common error in Fresnel spreadsheets, and it always errs the wrong way: it overstates the required clearance near the towers, so the sheet looks conservative while hiding the fact that the real risk is elsewhere. This calculator never uses the shortcut internally; it evaluates the product form at the point you specify, and it displays the midpoint shortcut only as a cross-check when you are actually at the midpoint.

Earth bulge and the effective earth-radius factor

Fresnel geometry alone assumes a flat earth. Over a 40 km hop that assumption is wrong by tens of metres. ITU-R P.530-16 §2.2 explains that variations in atmospheric refractive conditions change the effective earth radius, or k-factor, from its median value of approximately 4/3 for a standard atmosphere, and instructs designers to use k = 4/3 when no local refractivity data is available. ITU-R P.526-14 states the same thing as a distance: an effective earth radius of 8500 km may be taken as a basis, which is exactly 4/3 of the 6375 km true radius the Recommendations use. The rise of the earth above the chord joining the two antennas, evaluated at the checkpoint, is:

with the distances in kilometres and b in metres. The 12.75 is simply 2 × 6375 km expressed so that the answer comes out in metres, so it is the same earth model the ITU uses and not a separate approximation. Note that the bulge shares the d₁d₂ numerator with the Fresnel radius, which is why both peak together at the midpoint — but the bulge grows with the square of path length while the Fresnel radius grows only with its square root. At 5 km and 6 GHz with k = 4/3 the midpoint bulge is 0.37 m against a 7.90 m Fresnel radius and can safely be ignored; at 50 km the bulge is 36.8 m against a 24.99 m Fresnel radius and it dominates the tower height. Drop to a sub-refractive k = 0.7 on that 50 km hop and the bulge jumps to 70.0 m.

From geometry to decibels: the diffraction parameter and J(ν)

Equation (27) of ITU-R P.526-14 defines the dimensionless diffraction parameter for a knife-edge, where h is the height of the obstacle top above the straight line joining the two ends of the path (negative if the obstacle is below that line):

Comparing this with the first line of the page shows that ν = √2 · h/F₁ exactly. The diffraction parameter is nothing more than the normalised clearance rescaled, which is what lets a purely geometric ratio predict a loss. For ν greater than −0.78, P.526-14 equation (31) approximates the knife-edge loss as:

and zero otherwise. Now the 60% rule becomes arithmetic. Clearing the obstacle by exactly 0.6 F₁ makes h = −0.6 F₁ in P.526’s sign convention, so ν = −0.6√2 = −0.8485, which sits below the −0.78 cut-off where J(ν) reaches zero. That is precisely why P.526-14 §2.3 defines the diffraction zone as beginning where path clearance equals 60% of the first Fresnel zone radius, and why P.530-16 §2.2.2 says the direct path needs clearance of at least 60% of the first Fresnel zone radius to achieve free-space propagation conditions. Grazing incidence, h = 0, gives ν = 0 and J = 6.0 dB — the classic 6 dB knife-edge grazing loss.

Knife-edge loss is the optimistic bound. For a broad ridge or smooth spherical earth the loss is larger, so P.530-16 equation (2) supplies an average-terrain estimate, A_d = −20 h/F₁ + 10 dB, where h is the height difference between the most significant blockage and the path trajectory. The calculator reports both, and you should size fade margin against the pessimistic one.

Worked example: a 12 km 6 GHz hop crossing a ridge at 4 km

A licensed 6 GHz backhaul runs 12 km between site A and site B. The path profile shows a ridge 4 km from A. Enter d₁ = 4 km, d₂ = 8 km, f = 6 GHz, n = 1, k = 1.333, and read the results in metres.

Step one, the wavelength: λ = 299 792 458 / 6×10⁹ = 0.049965 m. Step two, the product term in metres: d₁d₂/(d₁+d₂) = 4000 × 8000 / 12 000 = 2666.67 m. Step three, the radius: F₁ = √(0.049965 × 2666.67) = √133.24 = 11.54 m. The P.530 practical form agrees to four figures: 17.3145 × √(4×8 / (6×12)) = 17.3145 × 0.66667 = 11.543 m. Step four, the planning target: 0.6 F₁ = 6.93 m.

Now the curvature that a flat-earth calculator would silently drop. With k = 4/3 the earth rises b = 4 × 8 / (12.75 × 1.3333) = 1.88 m above the chord at that ridge. The clearance you must find above the surveyed ridge elevation is therefore 1.88 + 6.93 = 8.81 m, which is 27% more than the Fresnel figure alone. On this hop that is a real but survivable difference; on a 40 km hop the same omission would understate the requirement by 23.5 m, which is an entire extra tower section.

Suppose the surveyed line of sight passes 5 m above the ridge top. Enter that as the ray clearance. The normalised clearance is h/F₁ = 5 / 11.54 = 0.433, below the 0.6 target, so the link is inside the diffraction zone. The P.526 parameter is ν = −√2 × 0.433 = −0.613, and equation (31) gives J(ν) = 1.14 dB of knife-edge loss. A ridge is not a knife edge, though, so the P.530 average-terrain estimate A_d = −20 × 0.433 + 10 = 1.34 dB is the number to budget against. Roughly one and a third decibels is not a catastrophe, but it is a permanent tax that vanishes entirely if the geometry is fixed.

Fixing it is cheap if you do the lever arithmetic. You need 6.93 − 5.00 = 1.93 m more clearance at a point four-twelfths of the way along the path. Raising the antenna at site A lifts the ray at that point by only d₂/d = 8/12 = 0.667 of the mast increase, so site A needs +2.89 m. Raising site B instead lifts the ray there by just d₁/d = 4/12 = 0.333, so site B would need +5.78 m for the same effect. Always raise the end nearer the obstruction; the shorter lever arm costs twice the steel.

One more sanity check the calculator performs for you: the midpoint of this same hop has F₁ = 12.24 m and a bulge of 2.12 m, so the ridge at 4 km is not the tightest geometric point on the path even though it is the tallest obstacle. That is exactly the trap the shortcut formula sets, and exactly why you should test several stations along a profile.

The table below fixes the frequency at 6 GHz and the checkpoint at the midpoint, then sweeps path length so you can see the earth bulge overtake the Fresnel requirement.

Setting the two middle columns equal and solving gives a crossover at d = (264.9 k/√f_GHz)^2/3, which is 27.5 km at 6 GHz with k = 4/3. Beyond that distance curvature, not diffraction, is what your towers are paying for. The crossover moves as f^−1/3, so it sits near 44 km at 1.5 GHz and near 22 km at 11 GHz — and it collapses dramatically in sub-refractive weather, because the bulge scales as 1/k while the Fresnel radius does not depend on k at all.

How to read the numbers when you are actually siting a mast

The result is not a verdict on whether the link works. It tells you how much geometric breathing room the wavefront wants at the point you examined, and — when you supply a clearance — roughly what it will cost you in decibels if you do not provide it. Compare the reported total against the real vertical gap between the line of sight and the obstacle at that location, taken from a surveyed profile rather than from memory.

P.530-16 §2.2.2.1 lays out the procedure the Recommendation actually advises, and it is a two-pass test rather than a single number. Pass one: size the antennas for the median k (use 4/3 absent local data) with a full 1.0 F₁ clearance over the highest obstacle. Pass two: repeat for the k_e value exceeded 99.9% of the worst month for that path length, requiring 0.0 F₁ for a single isolated obstruction, 0.3 F₁ for an obstruction extended along part of the path, or 0.6 F₁ for temperate paths longer than about 30 km. Then take whichever pass demands the taller antenna. The 60% figure this calculator highlights is the threshold at which diffraction loss disappears; the full P.530 procedure trades a little of that margin against the fact that excessive clearance worsens surface multipath fading.

When a point fails, the cheapest fixes in order are usually: raise the mast at the end nearer the obstruction (best lever arm, as the worked example shows), shift one site laterally if the obstruction is narrow, accept the computed diffraction loss and pay for it in fade margin, or move to a higher band where F₁ shrinks as 1/√f. That last option helps the geometry and hurts almost everything else — rain attenuation above roughly 10 GHz and oxygen absorption near 60 GHz both grow quickly — so it is a link-budget decision, never a Fresnel decision.

Treat a marginal pass as a fail. Foliage grows and sways, cranes appear, rooftop plant gets replaced, and lidar-derived terrain models carry their own vertical error. P.530-16 explicitly warns that above about 13 GHz the accuracy with which you can estimate obstacle height starts to approach the Fresnel radius itself, and says that estimation accuracy should be added to the clearance. If your survey is good to ±2 m and your target is 6.9 m, you should be building for 8.9 m.

Limitations and assumptions behind this ITU-R P.526 geometry

The Fresnel radius, the earth bulge and the diffraction estimate all rest on idealisations, and knowing which ones bite is what separates a screening tool from a design tool.

• A single knife edge. J(ν) assumes one thin, isolated, perfectly absorbing obstacle. P.526-14 §2.5 calls an obstacle isolated only when there is no overlap between the penumbra widths, when clearance on both sides is at least 0.6 F₁, and when there is no specular reflection either side. Two ridges, a rounded hilltop, or a forested slope all lose more than J(ν) predicts; P.526 offers the cylinder, Bullington, Epstein-Peterson and delta-Bullington methods for those cases and this page implements none of them.
• k is a fixed number here, and in reality it is a distribution. The bulge uses one k value. P.530 obtains k from refractivity-gradient statistics in the lowest 100 m of the atmosphere, as described in ITU-R P.453, and then requires a k_e exceeded 99.9% of the worst month that depends on path length. Ducting, elevated layers and coastal inversions all break the single-k picture entirely.
• Geometry only, no link budget. Free-space path loss, antenna gains and patterns, feeder loss, receiver threshold, atmospheric gas absorption (P.676), rain fading (P.838 and P.530 §2.4), multipath fading, interference and cross-polarisation discrimination are all absent. A geometrically perfect path can still fail on any of them.
• The corridor is a three-dimensional ellipsoid. A profile drawing shows only the vertical slice. A building shoulder, silo or tower leg beside the path intrudes into the same ellipsoid and this calculator will never see it.
• The bulge formula is a small-angle approximation to a spherical earth, using the 6375 km radius implied by the ITU 8500 km effective radius at k = 4/3. It is excellent for terrestrial hops and meaningless for satellite or near-horizon geometries.
• Vegetation is not terrain. Trees move, grow several percent a year, attenuate rather than diffract, and change with the season. ITU-R P.833 covers vegetation attenuation; treat a tree line as terrain plus a growth allowance plus a margin.

Two input mistakes cause most wrong answers, and both are unit or geometry errors rather than physics errors. The first is entering d₁ and d₂ measured to different references, or entering the total path length in one of the boxes; remember that this calculator derives d = d₁ + d₂ and never asks for it. The second is quoting a Fresnel radius as though it were a tower height. It is a radius about a line whose own height you must establish from a surveyed profile — the calculator gives you the size of the corridor, not the elevation of its centre.

If the link is long, expensive, or carries service-level obligations, use this page for screening, scenario comparison and sanity-checking a vendor’s numbers, then commission a proper path-profile study with measured refractivity statistics and a full budget before anyone orders steel.

Frequently asked questions about Fresnel zone clearance

Sources and further reading

Formulas, constants and clearance criteria on this page were taken from the primary Recommendations, not from secondary summaries. Fresnel ellipsoid definition and the n-th zone radius, equations (2) and (3), the diffraction-zone definition in §2.3, the isolated-obstacle conditions in §2.5, the diffraction parameter ν in equation (27) and the knife-edge loss J(ν) in equation (31) all come from ITU-R Recommendation P.526-14, “Propagation by diffraction” (January 2018). The 60% planning criterion, the k-factor median of 4/3, the average-terrain diffraction loss of equation (2), the F₁ = 17.3 √(d₁d₂/fd) form of equation (3) and the antenna-height procedure in §2.2.2 come from ITU-R Recommendation P.530-16, “Propagation data and prediction methods required for the design of terrestrial line-of-sight systems” (July 2015). Refractivity gradients and the statistics behind the k-factor are covered by ITU-R Recommendation P.453, “The radio refractive index: its formula and refractivity data”, and the reference atmosphere behind k = 4/3 by ITU-R P.310. The speed of light, 299 792 458 m/s exactly, is fixed by the SI definition of the metre published by the BIPM SI Brochure. The 550, 17.3 and 8.657 coefficients quoted above were re-derived from those definitions rather than copied, and the small rounding differences are stated explicitly in the formula section.