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Knife-edge diffraction calculator

Knife-edge diffraction is the extra loss a radio signal takes when a ridge, a rooftop or a tree line cuts across the path. The wave bends around the edge, but arrives weaker. This calculator applies the ITU-R P.526 single-edge model. Enter the frequency, the two distances and the obstacle height to read the loss in dB.

m

From the straight line A–B to the tip. Negative when the tip stays below the line.

Diffraction loss J(ν)
10.2 dB
tip on or above the sight line
Diffraction parameter ν
0.49
ν = √2 · h / F₁
F₁ at the obstacle
28.9 m
= 94.7 ft
Clearance h / F₁
0.347
−0.6 or lower satisfies the 60% rule
Free-space loss over d₁ + d₂
111.5 dB
Total path loss
121.7 dB
FSPL + J(ν)
h = 10 mside view · vertical scale exaggerated ×36– – 60% of F₁J = 10.2 dBν = 0.49obstructedABd₁ = 5 km · d₂ = 5 kmd = 10 km

Side view of the path. The green ellipse is the first Fresnel zone, the dashed amber ellipse its 60% boundary. The ridge stands at its true position along the path, and the arrow marks the obstacle height h on the same vertical scale as the zone, so where the apex falls against the two ellipses is the real clearance. The loss reaches 6 dB when the apex touches the sight line, and stays at zero only while the apex is below the amber line. As soon as the tip crosses the line, an amber V appears. That is the route the energy still takes, up to the edge and back down to B, and the reason a tall ridge costs far less than it looks.

The model idealises the obstacle as an infinitely thin, perfectly absorbing edge across the whole path, without ground reflection or vegetation. Real ridges are rounded and lose more. Heights are measured against the straight line A–B, so add the earth bulge yourself on long paths.

How it works

  1. 01

    Measure h from the sight line, not from the ground

    h is the height of the obstacle tip above the straight line joining the two antennas. It is positive when the tip pokes above that line and negative when it stays below. On long paths, add the earth bulge to every terrain height before you read h off the profile.

  2. 02

    Collapse the geometry into ν

    Height, both distances and wavelength fold into one dimensionless number: ν = h · 2·(d1 + d2) / (λ·d1·d2). Two very different paths that share the same ν suffer the same diffraction loss.

  3. 03

    Read the loss J(ν)

    ITU-R P.526 gives J(ν) = 6.9 + 20·log10((ν − 0.1)2 + 1 + ν − 0.1) dB for ν > −0.78. At ν = 0 the tip only grazes the line and the loss is already 6 dB. Below ν = −0.78 the obstacle is far enough out of the way and the model returns 0 dB.

  4. 04

    Expect slow growth once the path is blocked

    Above roughly ν = 2.4 the formula collapses to J ≈ 13 + 20·log10(ν). Since ν is proportional to the obstacle height, every doubling of h costs about 6 dB and every factor of ten costs 20 dB. Nothing is fully blocked here. The energy bends over the edge, and the deeper the obstruction the weaker that bending gets, but only logarithmically so.

  5. 05

    Cross-check with the 60% Fresnel rule

    ν and the Fresnel radius say the same thing: ν = 2·h/F1. Zero loss starts at ν = −0.78, which is a clearance of 0.55·F₁. The familiar 60% rule keeps a little margin on top of that.

  6. 06

    Add the free-space loss

    Diffraction loss is an extra on top of the free-space loss over the whole path d₁ + d₂. Add the two, then take the sum into the link budget to see what is left of the fade margin.

Formulas

Diffraction parameter (ITU-R P.526, eq. 26)
ν = h · 2 · (d1 + d2)λ · d1 · d2
  • h — obstacle tip above the straight line A–B, m (negative below it)
  • d1, d2 — distances from the two ends to the obstacle, m
  • λ — wavelength, m (λ = c / f)
Knife-edge loss (ITU-R P.526, eq. 31; valid for ν > −0.78)
J(ν) = 6.9 + 20 · log10( (ν − 0.1)2 + 1 + ν − 0.1 )
  • J(ν) — diffraction loss, dB
  • J = 0 dB for ν ≤ −0.78
  • total loss = FSPL over (d1 + d2) + J(ν)
The same geometry through the Fresnel radius
ν = 2 · hF1 , F1 = λ · d1 · d2d1 + d2
  • F1 — first Fresnel radius at the obstacle, m
  • h / F1 = −0.6 → ν = −0.85 → J = 0 dB (the 60% rule)

Worked example

A 10 km path at 900 MHz, ridge 10 m above the sight line
  1. λ = 299 792 458 / 900×106 = 0.333 m
  2. F1 = 0.333 · 5 000 · 5 000 / 10 000 = 28.9 m
  3. ν = 2 · 10 / 28.9 = 0.490
  4. J(ν) = 6.9 + 20·log10( (0.490 − 0.1)2 + 1 + 0.390 ) = 10.2 dB
  5. free-space loss over 10 km: 111.5 dB → total 121.7 dB

FAQ

Why is there 6 dB of loss when the obstacle only grazes the line of sight?
Because the edge cuts the wavefront in half. At h = 0 the parameter ν is zero, half the energy heading for the receiver is blocked and the field strength halves, which is 20·log10(2) ≈ 6 dB. Optical line of sight is not the same as a clear radio path.
A ridge 200 m above the line at 22 GHz costs only about 47 dB. Is that right?
Yes, and the number is easy to check. With d₁ = d₂ = 5 km the wavelength is 13.6 mm and the first Fresnel radius at the ridge is 5.84 m, so ν = 2 · 200 / 5.84 = 48.5 and J = 6.9 + 20·log10(96.7) = 46.6 dB. Diffraction loss follows the logarithm of ν, so a 100 m ridge on the same path already costs 40.6 dB and doubling it to 200 m adds just 6 dB more. The free-space loss over those 10 km is another 139.3 dB, so the link really sees 185.9 dB. The figure stays optimistic for two reasons. A knife edge is the lowest-loss shape a blockage can have, and terrain rarely offers only one.
How much clearance removes the diffraction loss?
The model returns 0 dB once ν ≤ −0.78, which means the tip sits at least 0.55·F₁ below the line. Planners ask for 0.6·F₁ so that survey error, vegetation growth and changing refraction do not eat the margin.
When is a real hill not a knife edge?
Whenever it is thick or rounded. A broad crest, a forested hilltop or a wide building attenuates more than an ideal edge, and ITU-R P.526 § 4.2 adds a rounded-obstacle term for exactly that case. Treat the knife-edge figure as the optimistic bound.
What if the path crosses several obstacles?
A single edge is then a first estimate, and usually an optimistic one. ITU-R P.526 covers two isolated edges, rounded obstacles and the general terrestrial path with the Bullington construction. The classic Deygout and Epstein-Peterson methods chain single-edge results in the same spirit.
Does knife-edge loss grow with frequency?
Yes, for a fixed geometry. ν grows with the square root of the frequency because the Fresnel zone shrinks. The 10 m ridge in the worked example costs 10.2 dB at 900 MHz and 12.6 dB at 2.4 GHz, on exactly the same path.

Whether a ridge really pokes into your path is a terrain question, not a formula question. Waveshed draws the sight line and the Fresnel band over real elevation data, free in your browser.

Check the profile on the map →

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