Free Space Path Loss Calculator
Free Space Path Loss Calculator
Free-space path loss in decibels from distance and frequency — or the range a given loss budget buys — with the constant derived rather than quoted, the first Fresnel zone that has to stay clear for the answer to mean anything, and the two-ray breakpoint beyond which a ground-level link stops obeying it.
free-space path loss
1 km at 2.4 GHz, antennas 10 m and 2 m above ground
Friis, and the constant that falls out of it
with d in km and f in MHz: FSPL(dB) = 20 log₁₀ d + 20 log₁₀ f + 20 log₁₀(4π × 10⁹ / c)
20 log₁₀(4π × 10⁹ ÷ 299,792,458) = 32.4478 dB
first Fresnel zone: r₁ = √(λ d₁ d₂ / d) two-ray breakpoint: db = 4 ht hr / λ
- d
- range between the two antennas, in the same units as λ before the constant is folded in
- lambda
- wavelength, c ÷ f. Everything frequency-dependent about free-space loss is here: the medium does not absorb, the receiving aperture simply shrinks
- 32.4478
- not a physical constant — it is the unit conversion that lets you put kilometres and megahertz straight into the equation. ITU-R P.525 rounds it to 32.4
- r1
- first Fresnel zone radius; d₁ and d₂ are the distances from each end to the obstacle. Keep 60% of it clear or the free-space figure is fiction
- d b
- the two-ray breakpoint. Past it a ground-level link follows a fourth-power law instead
Worked example
1 km at 2.4 GHz, antennas 10 m and 2 m above ground
λ = c ÷ f = 124.9 mm
The distance term is 20 log₁₀(1) = 0.000 dB and the frequency term is 20 log₁₀(2,400) = 67.604 dB
The constant is 20 log₁₀(4π × 10⁹ ÷ c) = 32.4478 dB, which is where 32.44 comes from — it is arithmetic, not physics
FSPL = 0.000 + 67.604 + 32.4478 = 100.05 dB, a power ratio of about 10,120,472,884 to 1
For that to mean anything, 3.35 m of clearance is needed at the midpoint — 60% of the 5.59 m first Fresnel zone
Over flat reflecting ground the two-ray model gives 95.50 dB here, 4.55 dB better than free space: the reflected ray is still partly adding at this range. The breakpoint is at 640.4 m and the fourth-power law takes over from about 2.012 km, beyond which the plane-earth estimate is the one to use
Free-space path loss for links people actually build
| Link | Distance | Frequency | Wavelength | Free-space path loss |
|---|---|---|---|---|
| Bluetooth or Wi-Fi across a room | 10 m | 2.4 GHz | 124.9 mm | 60.1 dB |
| Wi-Fi to the end of the garden | 50 m | 5.8 GHz | 51.69 mm | 81.7 dB |
| A 433 MHz remote to its receiver | 100 m | 433 MHz | 692.4 mm | 65.2 dB |
| LoRa across a town | 5 km | 868 MHz | 345.4 mm | 105.2 dB |
| VHF handheld to a repeater | 20 km | 145 MHz | 2.068 m | 101.7 dB |
| Microwave backhaul hop | 30 km | 18 GHz | 16.66 mm | 147.1 dB |
| A GPS satellite overhead | 20,200 km | 1,575.42 MHz | 190.3 mm | 182.5 dB |
| Geostationary satellite downlink | 35,786 km | 11.7 GHz | 25.62 mm | 204.9 dB |
What free space means, and when it stops being true
The Friis transmission equation says that a receiving antenna of effective aperture λ²/4π, sitting in the spherical wave from an isotropic transmitter of power Pₜ, collects Pₜ(λ/4πd)². Everything else follows. Turn that ratio upside down and take ten times its logarithm and you have free-space path loss, 20 log₁₀(4πd/λ). Express the distance in kilometres and the frequency in megahertz and the unit conversions collapse into a single number: 20 log₁₀(4π × 10⁹ / c) = 32.4478 dB. That is the whole provenance of the 32.44 people quote. It is not a property of the atmosphere or of the antennas — it is arithmetic, and ITU-R P.525 prints it rounded to 32.4.
Why higher frequencies lose more, which is not what you might think. Free space absorbs nothing. Not one decibel of the loss above is energy turned into heat. The frequency term is there entirely because the receiving antenna’s effective aperture is λ²/4π: a higher-frequency isotropic antenna is a physically smaller bucket, so it catches less of the same shower. Give both ends antennas of fixed physical aperture — two dishes, say — and the link actually gets better with frequency, because each dish’s gain rises as f². The 20 dB per decade penalty is a statement about isotropic antennas, not about the medium.
Both variables behave identically. Distance and frequency sit inside the same 20 log, so 6.02 dB is the price of doubling either one, and 20 dB the price of a decade of either. Halve the frequency and you may double the range for nothing — which is exactly the trade every sub-gigahertz IoT radio is making.
What free space excludes, and what to use instead. Free space means no ground reflection, no obstruction, no building, no foliage, no rain and no atmosphere. Remove any of those assumptions and this page is optimistic, usually by a lot. Three things in particular: Fresnel clearance — a path is only ‘unobstructed’ if about 60% of the first Fresnel zone is clear of everything, and that zone is metres wide over a long link even when the line of sight is visually clear — over a 5 km link at 868 MHz its radius at the midpoint is 20.8 m, so the 60% rule wants 12.5 m of clearance, and the radius for your own link is printed above. The ground — over real terrain a second ray arrives after bouncing, and past the breakpoint distance 4hₜhᵣ/λ it is very nearly out of phase with the direct one, so the two subtract and the loss climbs at roughly 40 dB per decade instead of 20. That fourth-power behaviour is the two-ray, or plane-earth, model, and it is plotted on the chart beside the free-space line; where they cross is the breakpoint. Everything else — rain above about 10 GHz, foliage, walls, body loss, and multipath fading, which is statistical rather than a fixed number. For real coverage prediction the propagation models to reach for are ITU-R P.1546 and P.1812 for point-to-area work over terrain, Okumura-Hata or COST-231 for urban mobile bands, and ITU-R P.838 for rain attenuation. None of them is this equation with a fudge factor, and pretending free space covers the case is the single commonest way a link budget comes out wrong.
This page is the propagation term on its own. To turn it into a received power and a link margin — with transmit power, antenna gains, cable losses and receiver sensitivity — take the number to the link budget calculator, which uses this same formula and adds an explicit margin input for everything listed above. For converting the powers themselves, the dBm to watts converter; for the antennas at each end, the antenna length calculator; and for what a mismatch at either end costs, the VSWR and return loss converter.
Frequently asked questions
What is free space path loss?
The loss between two isotropic antennas with nothing between them: 20 log₁₀(4πd/λ), or 20 log d(km) + 20 log f(MHz) + 32.4478 dB. Nothing is absorbed — the loss is entirely the spreading of the wave over a sphere and the small effective aperture of the receiving antenna.
Where does the 32.44 come from?
It is 20 log₁₀(4π × 10⁹ ÷ c) = 32.4478 dB, the constant you get when you agree to put the distance in kilometres and the frequency in megahertz. ITU-R P.525 prints 32.4; textbooks variously give 32.44 or 32.45. It is a unit conversion, not a physical constant, so use whichever rounding you like — 0.05 dB will never be the reason a link fails.
Does free space path loss really increase with frequency?
Between isotropic antennas, yes — 6 dB per octave. But no energy is lost to the medium; the frequency term is there because an isotropic antenna’s effective aperture is λ²/4π and shrinks as the frequency rises. Use fixed-size dishes at both ends instead and the link improves with frequency.
Why does my real link fall short of this?
Because free space is a floor. A ground-level link past the two-ray breakpoint loses about 40 dB per decade rather than 20; foliage, walls and bodies each cost several decibels; rain matters above 10 GHz; and multipath fading comes and goes. Budget an explicit margin for all of it rather than adjusting the free-space number.
What is the first Fresnel zone and why does it matter?
The ellipsoid around the line of sight within which a reflected path arrives less than half a wavelength late. An obstacle intruding into it interferes with the direct ray even without blocking it, so the usual rule is to keep 60% of it clear. Over a 5 km link at 868 MHz its radius at the midpoint is nearly 21 m — more than 40 m across — so a hilltop that looks clear often is not.
How far will my radio reach?
Switch this page to the second mode, put in the path loss your link budget can stand, and it gives the free-space range. Then treat that as a ceiling: at ground level a third of it is a good outcome and a tenth is common.
Related calculators
References
- Recommendation ITU-R P.525-4 (08/2019), Calculation of free-space attenuation. Gives the free-space basic transmission loss as Lbf = 20 log(4πd/λ) dB and, with f in MHz and d in km, Lbf = 32.4 + 20 log f + 20 log d dB, together with the field-strength conversions E = Pt − 20 log d + 74.8 and S = E − 145.8.
- Friis HT. A Note on a Simple Transmission Formula. Proceedings of the IRE, vol. 34 no. 5, May 1946, pp. 254–256. Pr = Pt·At·Ar/(λ²d²) in terms of effective areas, which becomes Pr = Pt Gt Gr (λ/4πd)² once the areas are written as gains.
- Rappaport TS. Wireless Communications: Principles and Practice, 2nd ed. Prentice Hall, 2002. Chapter 4 — the Friis equation, the two-ray ground reflection model, the fourth-power distance dependence beyond the breakpoint, and Fresnel zone clearance.
- Recommendation ITU-R P.1812-7 (08/2023), A path-specific propagation prediction method for point-to-area terrestrial services in the frequency range 30 MHz to 6 GHz (superseded by P.1812-8, 09/2025), and Recommendation ITU-R P.1546-6 (08/2019), Method for point-to-area predictions for terrestrial services in the frequency range 30 MHz to 4 000 MHz — the models to use when the ground and the terrain are in the way, which free-space loss cannot represent.
- Recommendation ITU-R P.838-3, Specific attenuation model for rain for use in prediction methods. The k and α coefficients that turn a rain rate into dB per kilometre, which is what the extra margin above 10 GHz is for.