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# 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).

*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} = **AB**^{2}+**BO**^{2}

(**h**+**r**)^{2} = **d**^{2}+**r**^{2}

**d** = sqrt[(**h**+**r**)^{2}-**r**^{2})],

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} - 3.440.1^{2})] ;
**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[ (2x3440.1x**h**) / (1852x**ρ**_{o}) ], where **ρ**_{o} accounts for refraction (0.8279).

Next math chapter: **Sextant angles**

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