Vertical sextant angle

The triangle OBL (see fig. below) can be described in terms of H, α and Distance: Distance = H/tan(α)
The angle in rad. (0-2π) and both height and distance in metres.

• From rad. to degrees: α = A * π/180, ‘A’ being the same angle in degrees.
• To describe angle A in minutes total, then A*60 = a, thus α = (a/60) * (π/180). So, α = a/3438, ‘a’ being the angle in arc minutes.
• FACTUM: tan(x) = x, if angle x is small.

Resulting in (with π = 3.14): Distance (m)= H * 3438/a

• Furthermore, distance in nm. = distance in meters/1852.

Voilà, la very practical equation:

It contains just two approximations, both of neglitible influence. First, we left out the tan function and second we used 3.14 for π.
Please realize that a smaller angle improves the approximation of the tan. Yet, as an opposing effect the instrument error of a smaller sextant angle increases. All in all, the factor 1.856 is not a typo, and just by chance near to the nautical mile: 1.852 kilometres. If you are still reading, you are very brave person and might perhaps agree that it originates from: (60 * 180)/(π * 1852).

So far we considered a perfect triangle (OBL) and forgot that life isn’t always perfect. Height h is usually quite small, but distance SB sometimes is not. This leads to an extra premise, which is seldom mentioned by other navigation textbooks:
Angle OLS should be bigger than 15°.

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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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The som of angles in a triangle is 180°

Let line DAE be parallel to line BC, then the angles α and α equal angles DAB and EAC, respectively. Therefore, the sum of angles in the triangle is 180° : a straight line.

“Doubling the angle” yields two equal angles

α = 30° , β = 60° thus γ = 30°

So, α + δ + γ = 180°
α + 180 - β + γ = 180°
2α = β
α + 180 - 2α + γ = 180°
180° - α + γ = 180°
-α + γ = 0
γ = α

Two equal angles render an triangle isosceles

In the triangle on the right, α = γ and β = 2α.
By constructing the bisector h of angle β we create two little triangles in which x=y.
Therefore, d₁=d₂. Next math chapter: Distance of horizon

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