# Mathematics: Distance of horizon

## Distance of horizon

AD = h is the height of eye above the earth.
DO = BO = CO = r (radius of the earth).

Fact: any angle between a tangent line to a circle and the radius of the circle is a right angle.

Since we have a right triangle ABO where AB = d,
AO = h+r and BO = r,
we can find a formula for d in terms of h:
(AO)2 = AB2 + BO2
(h+r)2 = d2 + r2
d = sqrt[(h + r)2r2)],
where r is approx. 3440.1 NM

An example: Let the eye height (h) be 4 meters (= 0.0022 NM); find the distance in NM of the geometrical horizon.
d = sqrt[(0.0022 + 3440.1)2 − 3440.12)]
d = sqrt[11834303 - 11834288]
d = sqrt[15.146]
d = 3.89 NM (geometrical)

The distance of the visible horizon as found in the table is greater (4.2 NM) due to atmospheric refraction.
The semi-empirical function used is:
d = sqrt[ (2 × 3440.1 × h) / (1852 × ρo) ], where ρo accounts for refraction (0.8279).

Next math chapter: Sextant angles,
or back to chapter 5 of the navigation course,
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