Why do leaves matter so much above 1 GHz?

Because above 1 GHz the typical loss rate rises in proportion to frequency. Figure 2 of P.833 gives 0.2 dB/m at 1 GHz, 2 dB/m at 10 GHz and about 6 dB/m at 30 GHz. Leaves, needles and twigs are then similar in size to the wavelength and scatter energy out of the path. Near 1 GHz trees in leaf attenuate about 20 % more per metre than bare trees, and the Mulhouse campaign saw seasonal swings of 2 dB at 900 MHz but 8.5 dB at 2200 MHz.

Why does the loss stop growing in deep woodland?

Because the signal finds another way in. P.833 limits the loss to a maximum Aₘ, set by a surface wave along the top of the canopy and by forward scatter inside it. Aₘ depends on the species, the density, the antenna pattern and how far below the treetops the antenna sits. In the example 100 m of woodland at 1 GHz costs 13.8 dB, not the 20 dB the linear rate suggests, and 500 m costs 24.5 dB against a maximum of 24.9 dB.

What does “depth of vegetation” mean?

It is the length of the radio path that runs inside the woodland, from the forest edge to the antenna, measured along the ray. It is neither the height of the trees nor the total path length. Eq (1) assumes one terminal inside the woodland and the other outside it, with the trees as the only obstruction besides terrain.

Does Waveshed’s tree layer apply this model?

Not yet. The simulator adds mapped forest to the terrain at a canopy height you choose, and line-of-sight runs treat it as an opaque obstacle. RF runs ignore vegetation for now, and a P.833 loss is on the roadmap. Until then, use the map to see how far your antenna sits inside the forest, enter that depth here and add Aₑᵥ to your link budget.

What about a single tree or a row of trees?

For one tree below 1 GHz, P.833 §3.1 uses A = d·γ, the dotted line in the diagram, capped by the loss of the best path around the tree, such as diffraction over it (P.526). Eq (1) does not apply because both terminals stand outside the vegetation. P.833 says this estimate tends to overestimate a wanted signal and may badly underestimate interference. Above 1 GHz it uses radiative-transfer and scattering models that need far more inputs than this page.

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Vegetation attenuation calculator (ITU-R P.833)

Vegetation attenuation is the extra loss a radio signal takes on its way through trees. This calculator applies ITU-R P.833-10 to a path that ends inside woodland: a typical loss rate per metre from the recommendation’s Figure 2, capped by a measured maximum. A second mode gives the statistical loss of a slant path towards a satellite or aircraft.

Path type

Frequency

Figure 2 of P.833 spans 30 MHz to 30 GHz.

Polarisation

Above 1 GHz Figure 2 has one line for both polarisations, so V and H give the same result here.

Depth of vegetation

d, the length of the path inside the woodland, from the forest edge to the antenna. The other end stays outside the trees.

Maximum attenuation Aₘ

These are all three eq (2) fits in P.833-10, each from one measurement campaign (read 7 November 2026). Own values start at the St Petersburg fit.

A₁

Aₘ at 1 MHz in eq (2), dB.

Frequency exponent

α in eq (2): how fast Aₘ grows with frequency. The published fits use 0.42 to 0.752.

Excess attenuation Aₑᵥ

13.8 dB

γ ±20 %: 11.8–15.4 dB

Specific attenuation

0.200

γ, a typical value from Figure 2

Maximum attenuation Aₘ

24.9 dB

1.37 · f^0.42

Without saturation

20.0 dB

d·γ, a constant rate over the whole depth

Excess attenuation against depth of vegetation, both axes linear. The green curve is the P.833 model, the shaded band shows it with γ 20 % lower and higher (roughly leafless vs in leaf), and the dot marks your depth. The dotted line is d·γ, the loss if it never saturated. The dashed line is the maximum attenuation Aₘ the curve approaches. Aₘ is a fit to one measurement campaign and scatters too (8.7 dB standard deviation at Mulhouse).

Typical values, not a site survey. Figure 2 was digitised by us, and P.833-10 itself calls it typical. The woodland mode assumes one terminal inside the trees and the other outside. Eq (6) states no frequency or elevation range and does not define p in words. Models often attributed to P.833, such as Weissberger’s, COST 235 and FITU-R, do not appear in edition 10 and are not used here.

How it works

01
Read the specific attenuation γ

Figure 2 of P.833 gives a typical loss per metre of woodland. It is 0.2 dB/m at 1 GHz and grows in proportion to frequency above that, to 2 dB/m at 10 GHz and 6 dB/m at 30 GHz. Below 1 GHz vertical polarisation loses more than horizontal. At 100 MHz it is 0.044 dB/m against 0.020 dB/m.
02
Choose the maximum attenuation Aₘ

Eq (2), A_m = A₁·f^α, comes from three measurement campaigns. At 1000 MHz they give 24.9 dB (St Petersburg), 22.4 dB (Mulhouse) and 32.5 dB (Rio de Janeiro). Each fit holds for the range it was measured over. The Rio fit already reaches 85 dB at 3.6 GHz, which shows how fast an extrapolation can run away.
03
Combine them and add the result

Eq (1) starts at the rate d·γ and bends over towards Aₘ. At d = A_m/γ, 125 m in the example, it has reached 63 % of Aₘ. The result is excess to every other mechanism, so add it to free-space, diffraction and gas loss. P.833 also notes that Aₘ is equivalent to the clutter loss of a terminal in ground cover, so do not add a clutter loss for the same trees as well.
04
Read γ as a typical value

P.833 says the Figure 2 values “should be viewed as only typical”. Trees in leaf attenuate about 20 % more per metre than bare trees near 1 GHz, and the recommendation’s own St Petersburg measurements sit below the figure: 0.17 dB/m at 949 MHz where Figure 2 gives 0.19, and 0.34 dB/m at 2117.5 MHz against 0.42. The shaded band and the ±20 % readout show that spread.
05
Slant paths use a statistical model

For a path up to a satellite or aircraft, eq (6) of P.833 turns elevation angle θ and percentage p into a loss. It estimates the path length in the canopy as d = 243·(p/100)·(θ+1)^−0.93047 + 1 m, so a steep path crosses less vegetation. At 2 GHz, 10° and p = 100 % it gives 16.1 dB for Japan cedar, at 80° only 3.9 dB.

Formulas

Excess attenuation, one terminal in woodland (eq 1)

A_ev = A_m · (1 − e^{−d·γ / A_m})

A_ev — excess attenuation due to vegetation, dB
A_m — maximum attenuation for this vegetation and antenna, dB
d — depth of vegetation (path length inside the woodland), m
γ — specific attenuation for very short vegetative paths, dB/m

Specific attenuation, closed form of Figure 2

γ = 0.2 · (f / 1000)^a

f — frequency, MHz, 30 to 30 000
a — 0.657 for vertical polarisation below 1 GHz, 1 otherwise
0.2 — γ at 1 GHz, dB/m (all three segments meet there)

Maximum attenuation (eq 2)

A_m = A₁ · f^α

f — frequency, MHz
A₁, α — fit to one campaign: 0.18 / 0.752 Rio, 1.15 / 0.43 Mulhouse, 1.37 / 0.42 St Petersburg

Slant path, site-general model (eq 6)

L = A · f^B · log₁₀(d) · (θ + E)^G − 4·p/100 + 0.4

d = 243·(p/100)·(θ + 1)^−0.93047 + 1 — path length in vegetation, m
k_h = 5.5 − 5·p/100
B = (0.30281 − 0.003624·k_h) · (f/1000)^{0.0013118 − 0.026236·k_h}
A, E, G — Table 3: 1.87, 0.01, −0.12 Japan cedar; 1.5, 0.01, −0.12 Kenya juniperus
f — MHz, θ — elevation angle, degrees, p — percentage, %

Worked example

1000 MHz, vertical, 100 m into woodland, A₁ = 1.37 dB and α = 0.42 (St Petersburg)

γ = 0.2 · (1000 / 1000)^0.657 = 0.200 dB/m
A_m = 1.37 · 1000^0.42 = 1.37 · 18.197 = 24.93 dB
d·γ / A_m = 100 · 0.200 / 24.93 = 0.802
A_ev = 24.93 · (1 − e^−0.802) = 24.93 · 0.5517 = 13.75 dB → 13.8 dB
γ ±20 % (0.160 to 0.240 dB/m): A_ev = 11.8 to 15.4 dB
without saturation the same 100 m would cost d·γ = 20.0 dB

FAQ

Why do leaves matter so much above 1 GHz?: Because above 1 GHz the typical loss rate rises in proportion to frequency. Figure 2 of P.833 gives 0.2 dB/m at 1 GHz, 2 dB/m at 10 GHz and about 6 dB/m at 30 GHz. Leaves, needles and twigs are then similar in size to the wavelength and scatter energy out of the path. Near 1 GHz trees in leaf attenuate about 20 % more per metre than bare trees, and the Mulhouse campaign saw seasonal swings of 2 dB at 900 MHz but 8.5 dB at 2200 MHz.
Why does the loss stop growing in deep woodland?: Because the signal finds another way in. P.833 limits the loss to a maximum Aₘ, set by a surface wave along the top of the canopy and by forward scatter inside it. Aₘ depends on the species, the density, the antenna pattern and how far below the treetops the antenna sits. In the example 100 m of woodland at 1 GHz costs 13.8 dB, not the 20 dB the linear rate suggests, and 500 m costs 24.5 dB against a maximum of 24.9 dB.
What does “depth of vegetation” mean?: It is the length of the radio path that runs inside the woodland, from the forest edge to the antenna, measured along the ray. It is neither the height of the trees nor the total path length. Eq (1) assumes one terminal inside the woodland and the other outside it, with the trees as the only obstruction besides terrain.
Does Waveshed’s tree layer apply this model?: Not yet. The simulator adds mapped forest to the terrain at a canopy height you choose, and line-of-sight runs treat it as an opaque obstacle. RF runs ignore vegetation for now, and a P.833 loss is on the roadmap. Until then, use the map to see how far your antenna sits inside the forest, enter that depth here and add Aₑᵥ to your link budget.
What about a single tree or a row of trees?: For one tree below 1 GHz, P.833 §3.1 uses A = d·γ, the dotted line in the diagram, capped by the loss of the best path around the tree, such as diffraction over it (P.526). Eq (1) does not apply because both terminals stand outside the vegetation. P.833 says this estimate tends to overestimate a wanted signal and may badly underestimate interference. Above 1 GHz it uses radiative-transfer and scattering models that need far more inputs than this page.

The formula needs the depth of vegetation along your path, and a guess can be off by hundreds of metres. Waveshed draws mapped forest on real terrain and can block line-of-sight with it, so you can see where the trees actually are before you add a P.833 loss by hand.

See where the trees actually are →

Related guides

Including trees — how the simulator adds mapped forest to the terrain, and what it does not do yet
LOS vs RF (FSPL & ITM) — which simulator mode uses which model, and where a vegetation loss would go

Sources & further reading

Recommendation ITU-R P.833 — Attenuation in vegetation ↗ — edition P.833-10 (09/2021), the current one: eqs (1), (2) and (6), Tables 1 and 3, Figure 2
ITU-R P.833-10 (09/2021) — PDF ↗ — we digitised Figure 2 from this file. The calculator uses the closed form of its three straight segments, within 2.3 % of the curve
Recommendation ITU-R P.2108 — Prediction of clutter loss ↗ — P.833 calls Aₘ equivalent to the clutter loss of a terminal in ground cover
Recommendation ITU-R P.526 — Propagation by diffraction ↗ — the diffraction loss that caps the single-tree estimate of P.833 §3.1
Weissberger's model — Wikipedia ↗ — an older empirical model often quoted next to P.833. It is not part of P.833-10 and this calculator does not use it. Read 7 November 2026

Vegetation attenuation calculator (ITU-R P.833)

How it works

Read the specific attenuation γ

Choose the maximum attenuation Aₘ

Combine them and add the result

Read γ as a typical value

Slant paths use a statistical model

Formulas

Worked example

FAQ

Related guides

Sources & further reading

Related tools

Clutter loss calculator (ITU-R P.2108)

Knife-edge diffraction calculator

Building entry loss calculator (ITU-R P.2109)

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