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.
Show Hint
Balance the heeling moment created by the offset horizontal force Y against the usual righting moment W times GM times the heel angle.
Step 1: Set up the couple using the figure axes.
In the figure, $y$ points athwartship and $z$ points down, with $M$ above $G$ and $G$ above $B$ before the ship heels. The force $Y$ acts at $B$ along the $y$ direction and stays constant through the steady turn.
Step 2: Write the restoring couple that appears once the ship heels.
Once the ship heels by a small angle $\phi$, the line of action of the buoyancy force shifts sideways from $G$ by the righting arm $GZ$, and for a small angle $GZ = GM \sin\phi \approx GM\,\phi$. The weight $W$ acting through $G$ and the shifted buoyancy force together form a restoring couple of size $W \times GM \times \phi$ that always pushes the ship back upright.
Step 3: Balance the disturbing couple from $Y$ against this restoring couple.
The disturbing couple comes from $Y$ acting at $B$ while the mass of the ship effectively resists at $G$, a vertical distance $BG$ away, giving a disturbing moment $Y \times BG$. In steady turning the ship heels to a fixed angle where these two couples cancel: $Y \times BG = W \times GM \times \phi$.
Final Answer:
Rearranging for the heel angle and reporting it as a magnitude gives the same expression.
\[ \boxed{\phi = \left| \dfrac{Y \times BG}{W \times GM} \right|} \]
