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Correct me if I'm wrong, but there's no way in hell you can see a rig that's 50 miles off shore. I think the horizon line is only like 26 miles of visibility or something like that.
Correct: Or much less if you're into math.
http://en.wikipedia.org/wiki/Horizon
| quote: | Distance to the horizon
Three types of horizon.
Three types of horizon.
[edit] Approximate formulas
For SI units, the straight line of sight distance d in kilometers to the true horizon on earth is approximately
d = 1000\sqrt{13h}
where h is the height above ground or sea level (in meters) of the eye of the observer. Examples:
* For an observer standing on the ground with h = 1.70 m (average eye-level height), the horizon appears at a distance of 4.7 km.
* For an observer standing on a hill or tower of 100 m in height, the horizon appears at a distance of 36 km.
In the Imperial version of the formula, 13 is replaced by 1.5, h is in feet and d is in miles. Thus:
d = \sqrt{1.5h}
Examples:
* For observers on the ground with eye-level at h = 5 ft 7 in (5.583 ft), the horizon appears at a distance of 2.89 miles.
* For observers standing on a hill or tower 100 ft in height, the horizon appears at a distance of 12.25 miles.
These formulas may be used when h is much smaller than the radius of the Earth (6371 km), including all views from any mountaintops, airplanes, or even high-altitude balloons. The metric formula is accurate to about 1%; the imperial one is more accurate still.
[edit] Exact formula
The exact formula for distance from the viewpoint to the horizon, applicable even for satellites, is
d = \sqrt{2Rh + h^2},
where R is the radius of the Earth (note: both R and h in this equation must be given in the same units, e.g. kilometers, but any consistent units will work).
Another relationship involves the arc length distance s along the curved surface of the Earth to the bottom of object:
\cos\frac{s}{R}=\frac{R}{R+h}.
Solving for s gives the formula
s=R\cos^{-1}\frac{R}{R+h}.
The distances d and s are nearly the same when the height of the object is negligible compared to the radius (that is, h<
[edit] Optical adjustments and objects above the horizon
To compute the height of a tower, the mast of a ship or a hilltop visible above the horizon, compute the distance-to-horizon for a hypothetical observer on top of that object, and add it to the real observer's own distance-to-horizon. For example, standing on the ground with h = 1.70 m, 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), they are probably about 16 km away.
Note that the actual visual horizon is slightly farther away than the calculated visual horizon, due to the slight refraction of light rays due to the atmospheric density gradient. This effect can be taken into account by using a "virtual radius" that is typically about 20% larger than the true radius of the Earth.
[edit] Curvature of the horizon
From a point above the surface the horizon appears slightly bent. There is a basic geometrical relationship between this visual curvature κ, the altitude and the Earth's radius. It is
\kappa=\sqrt{\left(\frac{R+h}{R}\right)^2-1}\ .
The curvature is the reciprocal of the curvature angular radius in radians. A curvature of 1 appears as a circle of an angular radius of 45° corresponding to an altitude of approximately 2640 km above the Earth's surface. At an altitude of 10 km (33,000 ft, the typical cruising altitude of an airliner) the mathematical curvature of the horizon is about 0.056, the same curvature of the rim of circle with a radius of 10 metres that is viewed from 56 centimetres. However, the apparent curvature is less than that due to refraction of light in the atmosphere and because the horizon is often masked by high cloud layers that reduce the altitude above the visual surface. |
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