Question

A ship is undergoing a steady starboard turn. Assume that the total hydrodynamic forces \(Y\) including the rudder forces act at the centre of buoyancy \(B\).
If \(W\) is the weight of the ship, \(G\) is the centre of gravity and \(M\) is the transverse metacentre, then the magnitude of the heel angle \(\phi\) is given by ______.
Assume that \(\phi\) is small.

• A. \( \left| \dfrac{Y \times BG}{W \times GM} \right| \)
• B. \( \left| \dfrac{W \times BG}{Y \times GM} \right| \)
• C. \( \left| \dfrac{W \times GM}{Y \times BG} \right| \)
• D. \( \left| \dfrac{Y \times GM}{W \times BG} \right| \)

Hint:

Balance the heeling moment created by the offset horizontal force Y against the usual righting moment W times GM times the heel angle.

Answer 1

The Correct Option is A

Step 1: Find the heeling moment.
During the steady turn, the hydrodynamic side force \(Y\) acts horizontally at the centre of buoyancy \(B\). Weight \(W\) effectively acts at the centre of gravity \(G\), which sits a vertical distance \(BG\) above \(B\) along the ship centreline shown in the figure. Because \(Y\) is horizontal and its line of action is offset vertically from \(G\) by \(BG\), it creates a heeling moment of magnitude \(Y \times BG\) about \(G\).

Step 2: Find the righting moment.
For a small heel angle \(\phi\), the righting arm is \(GZ = GM \sin\phi \approx GM\,\phi\), so the righting moment that resists the heel is \(W \times GM \times \phi\).

Step 3: Balance the two moments.
A steady turn means the heel angle has settled and stopped changing, so the heeling moment must equal the righting moment:
\[ Y \times BG = W \times GM \times \phi \]

Final Answer:
Solving for \(\phi\) and keeping only the magnitude gives the heel angle in a steady turn. \[ \boxed{\phi = \left| \dfrac{Y \times BG}{W \times GM} \right|} \]