Lesson 4: Applications of Trigonometry for Ship Stability

In this lesson we will take a look at some examples of the application of trigonometry in ship stability.

Please note that the examples developed for this section were inspired by those created by Martin A. Rhodes, in his text: Ship Stability for Mates/Masters, Seamanship International Ltd., 2003.

Example 1: A ship heels at an angle of 3°. If its metacentric height (GM) is 1.76 m, find the righting lever (GZ).

Solution:

Let us assume that θ represents the angle of heel. The ΔGZM is a right triangle, with the right angle at Z, and a hypotenuse GM. Figure 9.4.1 shows the given values.
For θ, GZ is the opposite side and we are given the length of the hypotenuse. From our t-ratios we know that: $\sin θ = \frac{Opposite side}{Hypotenuse}$ , therefore in this question: $\begin{matrix} \sin θ = \frac{G Z}{G M} \\ o r \\ G Z = G M \times \sin θ \end{matrix}$ Substituting the given values: $G Z = 1.76 \times \sin (3^{o}) = 1.76 \times 0.05234 = 0.0921 m$

Fig. 9.4.1

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Example 2: On a ship with a metacentric height (GM) 0.7 m, a load is shifted horizontally causing the centre of gravity to move a distance of 0.1 m starboard. Calculate the angle of list.

Solution:

Recall from module 7:
Shifting cargo moves the centre of gravity in the same direction as the cargo $\begin{array}{l} \tan θ = \frac{G G_{1}}{G M} \\ = \frac{0.1}{0.7} \\ \tan θ = 0.14286 \\ therefore, \\ θ = \tan^{- 1} (0.14286) = {8.1}^{o} \end{array}$

Fig.9.4.2

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Example 3: Find the angle of list of a ship with metacentric height (GM) 2 m and a righting lever of 0.16 metres.

Solution:

Fig. 9.4.3

Let us assume that θ represents the angle of list. Figure 9.4.3 shows the given values.
The ΔGZM is a right triangle, with hypotenuse GM.
For θ, GZ is the opposite side. From our t-ratios we know that $\sin θ = \frac{Opposite side}{Hypotenuse}$ , therefore in this question we can calculate: $\begin{array}{l} \sin θ = \frac{G Z}{G M} = \frac{0.16}{2} = 0.08 \\ θ = \sin^{- 1} (0.08) = {4.58}^{o} \approx {4.6}^{o} \end{array}$

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Example 4: : A ship has a displacement of 8000 t. The corresponding KM and KG are 9 m and 7.5 m respectively. Calculate the righting moment when the ship is heeled to 7°

Solution:

Fig. 9.4.4

To determine the righting moment we need to know the righting lever. The righting lever (GZ) can be found from the ΔGZM. Let us find the length of the hypotenuse (GM) of the triangle first. $\begin{array}{l} G M = K M - K G \\ = 9 m - 7.5 m \\ = 1.5 m \end{array}$ Now we can calculate GZ for the ΔGZM, since the angle M and the hypotenuse are given. We will use the sine ratio: $\begin{array}{l} \sin θ = \frac{Opposite side}{Hypotenuse} = \frac{G Z}{G M} \\ Therefore, \\ Righting lever G Z = G M \times \sin θ \\ G Z = 1.5 \times \sin 7^{o} = 0.1828 m \end{array}$ Finally, $\begin{array}{l} R i g h t i n g m o m e n t = G Z \times D i s p l a c e m e n t \\ R M = 0.1828 m \times 8000 t \\ = 1462.4 t m \end{array}$

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Example 5: A box-shaped vessel heels 5° as it makes a turn. Calculate the maximum draught when heeled, if the upright draught and length of beam are 8 m and 14 m respectively. (When a ship makes a turn it heels due to centrifugal and centripetal forces acting upon it; you will learn more about these forces in your stability course.)

Solution:

Fig. 9.4.5

\begin{array}{l} I n △ O A B, θ = 5^{o}, A B = ? \\ O B = \frac{1}{2} \times B e a m = \frac{1}{2} \times 14 m = 7 m \\ \sin θ = \frac{O p p o s i t e s i d e}{H y p o t e n u s e} \\ \sin θ = \frac{A B}{O B} \\ o r \\ A B = O B \times \sin θ \\ A B = 7 \times s i n 5^{o} = 0.610 m \end{array}

\begin{array}{l} I n △ B C F, θ = 5^{o}, B F = U p r i g h t d r a u g h t = 8 m, B C = ? \\ \cos θ = \frac{A d j a c e n t s i d e}{H y p o t e n u s e} \\ \cos θ = \frac{B C}{B F} \\ o r \\ B C = B F \times \cos θ \\ = 8 \times \cos 5^{o} \\ = 7.970 m \end{array}

New draught = A C = A B + B C = 0.610 + 7.970 = 8.580 m

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Example 6: A weight of 100 t is loaded on deck at a distance of 10 m above the keel and 6 m off the line of centre to port on a ship that initially displaces 5000t. The initial values of KG and KM are 4 m and 6 m respectively. Calculate the list.
Fig. 9.4.6_1

Solution:

Recall from module 7:
The centre of gravity moves in the same direction as the loaded cargo. You may wish to review the formulae used to calculate the new centre of gravity when cargo is loaded, discharged, or shifted. (Module 7, Lesson 2)

For this example let us divide the calculation into two steps:

The cargo is loaded at a distance of 10 m above the keel:
This will cause the centre of gravity to move up vertically. Let this distance be called G_V
The cargo is also shifted 6 m to port:
This will cause the new centre of gravity from Step 1 to move horizontally, in the same direction as that of the cargo. Let us denote that distance as G_H

Step 1: Find G_V :

\begin{array}{l} G G_{V} = \frac{w d}{W + w} \\ = \frac{100 \times (10 - 4)}{5000 + 100} \\ = \frac{100 \times 6}{5100} = \frac{600}{5100} = 0.12 m \end{array}

Therefore,

K G_{V} = K G + G G_{V} = 4 m + 0.12 m = 4.12 m

And

M G_{V} = K M - K G_{V} = 6 m - 4.12 m = 1.88 m

Step 2: Find G_H:

$\begin{array}{l} G_{V} G_{H} = \frac{w d}{W + w} \\ = \frac{100 \times 6}{5000 + 100} \\ = \frac{600}{5100} = 0.12 m \end{array}$ Consider figure 9.4.6_2. Since we have the adjacent and the opposite side we will use tangent ratio to find the list angle:
$\begin{array}{l} \tan θ_{L i s t} = \frac{G_{V} G_{H}}{M G_{V}} = \frac{0.12}{1.88} = 0.06383 \\ Therefore, \\ θ_{_{L i s t}} = \tan^{- 1} (0.06383) = {3.65}^{o} = {3.7}^{o} \end{array}$

Fig. 9.4.6_2

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Example 7: What is the loss in righting lever, GZ, when the centre of gravity is raised by loading cargo, if the ship is heeled at an angle of θ?

Solution:

Consider the following figure:

Recall from the readings in Lesson 1:
When two parallel lines are intersected by a transversal, the corresponding angles are equal.
Since MZ is parallel to G₁L, therefore ∠GG₁L is equal to ∠G₁MZ.
We are required to find the loss in righting lever - the value 'GL'.
Consider the Δ GG₁L,

\begin{array}{l} \sin θ = \frac{O p p o s i t e}{H y p o t e n u s e} \\ \sin θ = \frac{G L}{G G_{1}} \\ o r \\ G L = G G_{1} \times \sin θ \end{array}

Hence, the loss or decrease in GZ is equal to GG₁sinθ.

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Example 8: The righting lever 'GZ' of a ship heeled at a large angle is given by the following formula: $G Z = (G M + \frac{1}{2} B M \tan^{2} θ) \sin θ$ where GM and BM are the values for the ship in an upright position. Calculate GZ for a ship heeled at an angle of 45° when is it known that GM and BM in the upright position are 1.0 and 3.5 metres respectively. Watch your order of operations (BEDMAS) when using this formula!

Solution:

Substituting the given values in the formula we get: $\begin{array}{l} G Z = (G M + \frac{1}{2} B M \tan^{2} θ) \sin θ \\ = (1 + \frac{1}{2} \times 3.5 \times \tan^{2} 45^{o}) \sin 45^{o} \\ = (1 + \frac{1}{2} \times 3.5 \times 1^{2}) \times 0.7071 \\ = (1 + 1.75) \times 0.7071 \\ = 2.75 \times 0.7071 \\ = 1.945 m \end{array}$

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Lesson 4: Applications of Trigonometry for Ship Stability

Example 1: A ship heels at an angle of 3°. If its metacentric height (GM) is 1.76 m, find the righting lever (GZ).

Example 2: On a ship with a metacentric height (GM) 0.7 m, a load is shifted horizontally causing the centre of gravity to move a distance of 0.1 m starboard. Calculate the angle of list.

Example 3: Find the angle of list of a ship with metacentric height (GM) 2 m and a righting lever of 0.16 metres.

Example 4: : A ship has a displacement of 8000 t. The corresponding KM and KG are 9 m and 7.5 m respectively. Calculate the righting moment when the ship is heeled to 7°

Example 6: A weight of 100 t is loaded on deck at a distance of 10 m above the keel and 6 m off the line of centre to port on a ship that initially displaces 5000t. The initial values of KG and KM are 4 m and 6 m respectively. Calculate the list. Fig. 9.4.6_1

Example 7: What is the loss in righting lever, GZ, when the centre of gravity is raised by loading cargo, if the ship is heeled at an angle of θ?

Example 6: A weight of 100 t is loaded on deck at a distance of 10 m above the keel and 6 m off the line of centre to port on a ship that initially displaces 5000t. The initial values of KG and KM are 4 m and 6 m respectively. Calculate the list.
Fig. 9.4.6_1