Distance of horizon

AD = h is the height of eye above the earth.
DO = BO = CO = r (radius of the earth).
Factum: 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)²-r²)],
where r is approx. 3.440.1 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 + 3.440.1)² - 3.440.1²)] ; 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[ (2×3440.1xh) / (1852xρ_o) ], where ρ_o accounts for refraction (0.8279).
Next math chapter: Sextant angles

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