Question:medium
The sway-yaw coupled maneuvering equation of motion for a ship is
\[ \begin{bmatrix} M - Y_{\dot{v}} & -Y_{\dot{r}} \\ -N_{\dot{v}} & I_z - N_{\dot{r}} \end{bmatrix} \begin{bmatrix} \dot{v} \\ \dot{r} \end{bmatrix} = \begin{bmatrix} Y_v & Y_r \\ N_v & N_r \end{bmatrix} \begin{bmatrix} v \\ r \end{bmatrix} + \begin{bmatrix} Y_{\delta} \\ N_{\delta} \end{bmatrix} \delta_R \]For a control-fixed straight-line stability, the eigenvalues of the function should be ______.
The sway-yaw coupled maneuvering equation of motion for a ship is
\[ \begin{bmatrix} M - Y_{\dot{v}} & -Y_{\dot{r}} \\ -N_{\dot{v}} & I_z - N_{\dot{r}} \end{bmatrix} \begin{bmatrix} \dot{v} \\ \dot{r} \end{bmatrix} = \begin{bmatrix} Y_v & Y_r \\ N_v & N_r \end{bmatrix} \begin{bmatrix} v \\ r \end{bmatrix} + \begin{bmatrix} Y_{\delta} \\ N_{\delta} \end{bmatrix} \delta_R \]For a control-fixed straight-line stability, the eigenvalues of the function should be ______.
\[ \begin{bmatrix} M - Y_{\dot{v}} & -Y_{\dot{r}} \\ -N_{\dot{v}} & I_z - N_{\dot{r}} \end{bmatrix} \begin{bmatrix} \dot{v} \\ \dot{r} \end{bmatrix} = \begin{bmatrix} Y_v & Y_r \\ N_v & N_r \end{bmatrix} \begin{bmatrix} v \\ r \end{bmatrix} + \begin{bmatrix} Y_{\delta} \\ N_{\delta} \end{bmatrix} \delta_R \]For a control-fixed straight-line stability, the eigenvalues of the function should be ______.
Show Hint
Think about what root type of the linearised sway-yaw system gives a smooth, non-oscillatory return to the straight course.
Updated On: Jul 28, 2026
- real and negative
- real and positive
- complex numbers with positive real part and negative imaginary part
- zero
Show Solution
The Correct Option is A
Solution and Explanation
Course (straight-line) stability of a ship is a classic linear-systems question hiding inside a naval architecture problem: does the free response of the sway-yaw system decay back to zero disturbance, and does it do so without swinging past the target heading?
- real and negative: correct. A negative real root gives pure exponential decay $e^{-|\lambda| t}$, so $v$ and $r$ die out monotonically and the ship settles back onto a straight course without hunting.
- real and positive: wrong. A positive real root grows without bound as $e^{|\lambda| t}$, so any small disturbance (a gust, a wave) would keep pulling the bow further off course, which is the definition of a directionally unstable hull.
- complex with positive real part and negative imaginary part: wrong. Complex roots always occur in conjugate pairs for a real system, so this description is not even self-consistent, and a positive real part still means the oscillation amplitude grows.
- zero: wrong. A zero eigenvalue means the system stays wherever it is disturbed to (neutral, marginal stability), it never actively returns to the original straight course, so this is not stability.
Only option A, real and negative eigenvalues of the sway-yaw system, describes a hull that is inherently directionally stable when the rudder is fixed amidships, so A is the correct choice.
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