Why is line of sight not enough?

Because radio does not travel as a line. Energy arrives by every route, and the routes close to in-phase with the direct one add to it. The region containing them is an ellipsoid — the first Fresnel zone — and it is far fatter than intuition suggests: about 16 metres across at mid-path on a 20 km hop at 6 GHz. An obstacle inside it removes energy that was contributing, so the link works and runs several decibels worse than the budget says.

Why 60% of the first zone?

It is the point at which diffraction loss over a smooth obstruction becomes roughly negligible, and it has become the usual planning threshold. It is a convention rather than a physical boundary, and a path that meets it is not guaranteed anything.

What is the K factor?

Air density falls with height, so radio bends slightly towards the Earth, making the Earth appear flatter than it is. K expresses that as a multiplier on the Earth's radius. K = 4/3 is the usual temperate-climate planning value.

Which K factor should I check?

Both. 4/3 for the normal case, and something around 0.6 for the sub-refractive case. Sub-refractive conditions make the Earth appear more curved, and they are what turn a link that has worked for years into one that drops out on particular nights. This shows both.

Why does raising an antenna help less than expected?

Because the line of sight at a given obstacle moves by only the fraction of the added height corresponding to how far along the obstacle sits. Raising the far antenna by ten metres lifts the line of sight at a quarter-way obstacle by two and a half. That is why the solver asks which end you can actually change.

Does this model diffraction loss?

No. It is geometry: how much of the zone is clear. Converting an intrusion into a number of decibels needs a diffraction model with its own assumptions about the obstacle's shape, and a knife-edge is not a rounded hill.

Is anything I enter sent anywhere?

No. Everything is calculated in your browser and nothing is uploaded.

Fresnel Zone & Path Clearance Calculator

Q: Why does an obstacle near the end matter less?

Because the zone pinches to nothing at each antenna and is widest in the middle. A building 500 metres from a site intrudes far less than the same building halfway along — which is why a tree line outside the compound is often survivable and a hill mid-path is not.

A radio path can have perfect line of sight and still not work. The signal travels in a fat ellipsoid rather than a line, and an obstacle that intrudes into it costs real decibels — including the ground itself, once the curvature of the Earth is accounted for.

Geometry under the assumptions you enter This calculates clearance for a chosen K factor and obstacle. It does not model diffraction loss, does not know the terrain between your sites, and the 60% figure it checks against is a widely used planning convention rather than a physical boundary.

The path

Path length (km)

Frequency (GHz)

Site A antenna height (m)

Site B antenna height (m)

K factor

or enter K

The obstacle

Obstacle height above sea level (m)

Fresnel zone to show

Obstacle position along the path: 10.0 km from site A

Antenna heights and the obstacle are all measured from the same datum — usually height above sea level. Mixing a tower height above ground with an obstacle above sea level is the commonest way this calculation goes wrong.

The path profile

Line of sightFirst Fresnel zone 60% of the first zoneEarth bulge

How much antenna would clear it?

Hold Clear

The whole path, sampled

The zone is widest at mid-path and pinches to nothing at each antenna, so an obstacle near one end intrudes far less than the same obstacle in the middle.

Why line of sight is not enough

Radio does not travel as a line. Energy arriving at the receiver has taken every possible route, and the routes that arrive close to in-phase with the direct one add to it. The region containing those routes is an ellipsoid stretched between the two antennas — the first Fresnel zone — and it is much fatter than intuition suggests. On a 20 km hop at 6 GHz it is about 16 metres across at the middle.

Put an obstacle into it and you remove some of the energy that was contributing constructively. The link does not fail cleanly; it just quietly runs several decibels worse than the budget says, which is much harder to diagnose than a link that never worked at all.

The convention of keeping 60% of the first zone clear comes from the point at which diffraction loss over a smooth obstruction becomes roughly negligible. It is a planning rule of thumb, not a law, and a path that meets it is not guaranteed anything.

The Earth is in the way too

Over any distance, the ground between two antennas bulges upwards relative to the straight line between them. On a 20 km path that is nearly six metres at mid-path under a standard atmosphere — more than enough to matter.

The atmosphere complicates it usefully: air density falls with height, so radio bends slightly towards the Earth, which makes the Earth appear flatter than it is. That is what the K factor represents, and K = 4/3 is the usual temperate-climate planning value.

The number worth checking is not 4/3. Sub-refractive conditions — K below 1 — make the Earth appear more curved and are what turn a path that has worked for years into one that drops out on particular nights. If a path is marginal, check it at K = 0.6 as well, which is what the selector here is for.

The pinch at the ends

The zone is widest at mid-path and narrows to nothing at each antenna. A building 500 metres from a site intrudes far less than the same building in the middle of the hop — which is why a tree line right outside the compound is often survivable and a hill halfway along is not.

It also means the effect of raising an antenna is not linear. Raising the near antenna lifts the line of sight at a distant obstacle by only a fraction of the height added, which is why the solver above asks which end you are able to change.