Using imperial units, the distance to the horizon is
where d is in miles and h is in feet.
These formulas may be used when h is much smaller than the radius of the Earth (6371 km), including all views from any mountaintops, aeroplanes, or high-altitude balloons. With the constants as given, both the metric and imperial formulas are precise to within 1% (see the next section for how to obtain greater precision).
[edit]Exact formula for a spherical Earth
If h is significant with respect to R, as with most satellites, then the approximation made previously is no longer valid, and the exact formula is required:
where R is the radius of the Earth (R and h must be in the same units). For example, if a satellite is at a height of 2000 km, the distance to the horizon is 5,430 kilometres (3,370 mi); neglecting the second term in parentheses would give a distance of 5,048 kilometres (3,137 mi), a 7% error.
[edit]Objects above the horizon
Geometrical horizon distance
To compute the height of an object visible above the horizon, compute the distance to the horizon for a hypothetical observer on top of that object, and add it to the real observer's distance to the horizon. For example, for an observer with a height of 1.70 m standing on the ground, the horizon is 4.65 km away. For a tower with a height of 100 m, the horizon distance is 35.7 km. Thus an observer on a beach can see the tower as long as it is not more than 40.35 km away. Conversely, if an observer on a boat (h = 1.7 m) can just see the tops of trees on a nearby shore (h = 10 m), the trees are probably about 16 km away.
Referring to the figure at the right, the lighthouse will be visible from the boat if
where DBL is in kilometres and hB and hL are in metres. If atmospheric refraction is considered, the visibility condition becomes
[edit]Effect of atmospheric refraction
Because of atmospheric refraction of light rays, the actual distance to the horizon is slightly greater than the distance calculated with geometrical formulas. With standard atmospheric conditions, the difference is about 8%; however, refraction is strongly affected by temperature gradients, which can vary considerably from day to day, especially over water, so calculated values for refraction are only approximate.[5]
Rigorous method—Sweer
The distance d to the horizon is given by[7]
where RE is the radius of the Earth, ψ is the dip of the horizon and δ is the refraction of the horizon. The dip is determined fairly simply from
where h is the observer's height above the Earth, μ is the index of refraction of air at the observer's height, and μ0 is the index of refraction of air at Earth's surface.
The refraction must be found by integration of
where is the angle between the ray and a line through the center of the Earth. The angles ψ and are related by
Simple method—Young
A much simpler approach uses the geometrical model but uses a radius R′ = 7/6 RE. The distance to the horizon is then[5]
Taking the radius of the Earth as 6371 km, with d in km and h in m,
with d in mi and h in ft,
Results from Young's method are quite close to those from Sweer's method, and are sufficiently accurate for many purposes.
[edit]Curvature of the horizon
From a point above the surface the horizon appears slightly bent (it is a circle, after all). There is a basic geometrical relationship between this visual curvature κ, the altitude and the Earth's radius. It is
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