Where the square root of height comes from
Stand at height h above a sphere of radius R and the tangent line from your eye touches the surface at a distance given by Pythagoras: the distance squared is 2Rh plus h squared. For any antenna height that matters, h squared is negligible against 2Rh, so the distance is the square root of 2Rh.
Put in an earth radius of 3,958.8 miles, convert height from feet, and with no atmospheric bending at all the distance in miles is 1.2246 times the square root of the height in feet. Radio does not travel in straight lines through the atmosphere, though — the refractive index falls with altitude, so the ray bends slightly downward and reaches further than geometry allows. The standard planning dodge is to keep the ray straight and inflate the earth radius instead, by a factor k. At the usual k of four-thirds the multiplier becomes 1.4140, which is where the familiar 1.415 times the square root of height comes from.
Both figures are in the calculator because k is a field. Set it to 1 and you get the pure geometric horizon; set it to 1.33 and you get the planning figure. Neither is a constant of nature — k moves with the weather, and paths that are dead all afternoon sometimes open at dawn because of it.
The diminishing return on height
Because distance goes with the square root of height, every doubling of the mast multiplies the reach by only 1.414. Worked at k = 1.33:
| Antenna height | Radio horizon | Gain over the row above |
|---|---|---|
| 10 ft | 4.47 mi | — |
| 20 ft | 6.32 mi | 1.85 mi |
| 40 ft | 8.93 mi | 2.61 mi |
| 80 ft | 12.63 mi | 3.70 mi |
| 160 ft | 17.87 mi | 5.24 mi |
The absolute gain per doubling does grow, but the cost of doubling a mast grows much faster, and the practical reading is the other way round: the first ten feet off the ground is worth more than any ten feet above it, because it is also the ten feet that clears the roof, the fence and the trees.
Clearance is mostly Fresnel, not curvature
A line of sight that grazes an obstacle is not a working path. Radio needs an ellipsoidal volume around the direct ray to be reasonably clear, and the first Fresnel zone is the innermost part of it. Its radius at a point along the path is the square root of the wavelength times the two part-distances divided by the whole distance, and at VHF it is enormous.
Take the defaults: 146 MHz over a 12 mile path, at a point 5 miles from one end. The wavelength is 6.737 feet, the first Fresnel radius works out to 322.1 feet, and the earth bulge at that point is only 17.5 feet. Asking to keep 60 percent of the zone clear means the terrain has to sit 210.8 feet below the straight line between the antennas — 193.3 feet of Fresnel plus 17.5 feet of bulge.
That is the number people miss. Sight-line thinking says a ridge 50 feet below the line is fine. At these frequencies it is sitting well inside the first Fresnel zone and the path will not behave like free space. The zone shrinks as frequency rises, which is why microwave links can thread gaps that a VHF path cannot.
What this cannot see
Terrain. All of it. The model is a smooth sphere, and there is no such site. One hill at the wrong distance blocks a path the arithmetic calls clear, and buildings, trees in leaf and a rising ridge behind the far station are all invisible here. Use this to decide whether a path is worth investigating and then get a terrain profile.
It also ignores everything that is not line of sight. Below about 30 MHz signals routinely travel by ground wave and by reflection from the ionosphere, and none of that has anything to do with the horizon; at VHF and above, diffraction over a ridge, ducting and tropospheric scatter all put signals well past this distance some of the time. The horizon figure is a planning boundary for line of sight paths, not a wall.
And before adding height to anything: the fall radius of a mast or a tower must clear every power line around it, with margin, in every direction. A raised metal structure touching a line kills. Climbing is specialist work with fall protection, and lightning protection and bonding for a raised structure is a standards-governed and often licensed subject. This page does geometry.
Questions people ask
How far can my antenna reach at 40 feet?
To the radio horizon, about 8.93 miles at the k = 1.33 refraction assumption this page starts from, or 7.74 miles with no atmospheric bending at all. If the other station also has height, the two horizons add: 40 feet and 20 feet gives 8.93 plus 6.32, so 15.25 miles. That is smooth-earth geometry with no terrain in it, so treat it as the best a perfect site could do rather than as a prediction.
Why 1.415 times the square root of the height?
It comes from Pythagoras on a sphere. The tangent distance is the square root of twice the earth radius times the height, which with a 3,958.8 mile radius and height in feet gives 1.2246 times the square root of height in miles. Radio bends slightly downward in the atmosphere, and the standard planning trick is to keep the ray straight and multiply the earth radius by four-thirds instead, which raises the coefficient to 1.4140. Both are available here by changing the k field.
Does doubling my mast height double my range?
No, it multiplies it by about 1.41, because distance goes with the square root of height. At k = 1.33, 40 feet gives 8.93 miles and 80 feet gives 12.63 miles. The absolute gain per doubling grows while the cost grows faster, which is why the first ten feet off the ground is usually the best value — it is also the ten feet that clears the roofline and the trees.
What is the earth bulge and how much clearance does it need?
The bulge is how far the curved surface stands above the straight line between two antennas at a given point along it. On a 12 mile path at a point 5 miles from one end, with k = 1.33, it is 17.5 feet. On its own that is rarely the limiting number: at 146 MHz the first Fresnel zone at the same point has a radius of 322.1 feet, and keeping 60 percent of it clear means the ground has to be 210.8 feet below the line.
Can I work a station beyond the radio horizon?
Often, yes, and this page says so when the distance you enter exceeds the horizon sum. Signals diffract around the bulge, the refraction factor moves with the weather so the horizon is not fixed, and below about 30 MHz ground wave and ionospheric propagation ignore this geometry completely. What the horizon figure bounds is a line of sight path under an assumed atmosphere, not what is possible.