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

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

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

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} − 3440.1^{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[ (2 × 3440.1 × **h**) / (1852 × **ρ _{o}**) ], where

**ρ**accounts for refraction (0.8279).

_{o}
Next math chapter: **Sextant angles**,

or back to **chapter 5 of the navigation course**,

or **learn to sail bareboat**, **skippered instruction** or at RYA ASA sailing schools in **Greece** and beyond…