Editing Horizon (section)

== Objects above the horizon ==
[[File:HorizonDistance.png|thumb|right|300px|Geometrical horizon distance]]

To compute the greatest distance ''D''<sub>BL</sub> at which an observer B can see the top of an object L above the horizon, simply add the distances to the horizon from each of the two points:
:''D''<sub>BL</sub> = ''D''<sub>B</sub> + ''D''<sub>L</sub>
For example, for an observer B with a height of ''h''<sub>B</sub>{{=}}1.70&nbsp;m standing on the ground, the horizon is ''D''<sub>B</sub>{{=}}4.65&nbsp;km away. For a tower with a height of ''h''<sub>L</sub>{{=}}100&nbsp;m, the horizon distance is ''D''<sub>L</sub>{{=}}35.7&nbsp;km. Thus an observer on a beach can see the top of the tower as long as it is not more than ''D''<sub>BL</sub>{{=}}40.35&nbsp;km away. Conversely, if an observer on a boat (''h''<sub>B</sub>{{=}}1.7{{nbsp}}m) can just see the tops of trees on a nearby shore (''h''<sub>L</sub>{{=}}10{{nbsp}}m), the trees are probably about ''D''<sub>BL</sub>{{=}}16&nbsp;km away.

Referring to the figure at the right, and using the [[#Approximation|approximation above]], the top of the lighthouse will be visible to a lookout in a [[crow's nest]] at the top of a mast of the boat if
:<math>D_\mathrm{BL} < 3.57\,(\sqrt{h_\mathrm{B}} + \sqrt{h_\mathrm{L}}) \,,</math>
where ''D''<sub>BL</sub> is in kilometres and ''h''<sub>B</sub> and ''h''<sub>L</sub> are in metres.

[[File:Horizon, Valencia (Spain).JPG|thumb|A view across a {{convert|20|km|mi}} wide bay in the coast of Spain. Note the curvature of the Earth hiding the base of the buildings on the far shore.]]
[[File:Curvatura 2.gif|thumb|A ship moving away, beyond the horizon]]
As another example, suppose an observer, whose eyes are two metres above the level ground, uses binoculars to look at a distant building which he knows to consist of thirty [[storey]]s, each 3.5 metres high. He counts the stories he can see and finds there are only ten. So twenty stories or 70 metres of the building are hidden from him by the curvature of the Earth. From this, he can calculate his distance from the building:

:<math>D \approx 3.57(\sqrt{2}+\sqrt{70})</math>

which comes to about 35 kilometres.

It is similarly possible to calculate how much of a distant object is visible above the horizon. Suppose an observer's eye is 10 metres above sea level, and he is watching a ship that is 20&nbsp;km away. His horizon is:

:<math> 3.57 \sqrt{10} </math>

kilometres from him, which comes to about 11.3 kilometres away. The ship is a further 8.7&nbsp;km away. The height of a point on the ship that is just visible to the observer is given by:

:<math>h\approx\left(\frac{8.7}{3.57}\right)^2</math>

which comes to almost exactly six metres. The observer can therefore see that part of the ship that is more than six metres above the level of the water. The part of the ship that is below this height is hidden from him by the curvature of the Earth. In this situation, the ship is said to be [[hull-down]].