This is an advanced online course on marine navigation, providing you with the “conditio sine qua non” of offshore sailing. Nowadays most sailors tend to rely on modern equipment like differential GPS or Radar to navigate them through hazardous waters. Not only is such reliance unwanted and possibly dangerous, also the act of navigating by yourself is actually a lot of fun, especially when sailing in Greece or Turkey, which are the perfect to learn how to sail.

What is navigation?

“Navigare necesse est, vivere non est necesse” is latin for: to sail is vital, to live is not. This phrase tells us that both sailing and the “conditio” of positioning are highly intertwined. Indeed, the art of navigation enables you to set a course and sail to your destination by using only nautical charts, a compass and your common sense. The aim of this course is to teach you how to navigate safely while using the minimum of resources: methods that have been in use since the Middle Ages, and are still applied by the professionals. This course greatly extends on - for instance - the ASA courses and gives you the insight and feel of a seasoned navigator.

Enjoy!

This is chapter 0: Use the moving anchor logo on the right of this page to navigate through the course. Alternatively, click on the links below to study the chapters.
Chapter 1 - Positions
Chapter 2 - Nautical chart
Chapter 3 - Compass
Chapter 4 - Plotting
Chapter 5 - Piloting
Chapter 6 - Tides
Chapter 7 - Tide prediction
Chapter 8 - Currents
Chapter 9 - Navigation aids

Math - Running fixes
Math - Distance of horizon
Math - Vertical sextant angles

Compass deviation table
Lunation, phases of the moon

The wind chill forecast
Beaufort wind scale

UK based sailing schools

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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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