Question:
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
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The Correct Option is A
Solution and Explanation
Step 1: Write the free (control-fixed) system.
Set \(\delta_R = 0\) since we want control-fixed stability. The equation becomes
\[ \begin{bmatrix} M - Y_{\dot{v}} & -Y_{\dot{r}} \\ -N_{\dot{v}} & I_z - N_{\dot{r}} \end{bmatrix} \begin{bmatrix} \dot{v} \\ \dot{r} \end{bmatrix} = \begin{bmatrix} Y_v & Y_r \\ N_v & N_r \end{bmatrix} \begin{bmatrix} v \\ r \end{bmatrix} \]
Step 2: Assume an exponential disturbance and form the characteristic equation.
A small disturbance from the straight course behaves like \(v, r \propto e^{\lambda t}\). Putting this into the matrix equation and collecting terms gives a real quadratic in \(\lambda\) of the form
\[ a\lambda^2 + b\lambda + c = 0 \]
where \(a = I_zM - I_zY_{\dot{v}} - MN_{\dot{r}} + N_{\dot{r}}Y_{\dot{v}} - N_{\dot{v}}Y_{\dot{r}}\), and \(c = N_rY_v - N_vY_r\), with \(b\) built from the mixed added-mass and hydrodynamic derivative terms.
Step 3: Interpret what each root type means physically.
If a root is complex, the disturbed motion oscillates (hunts) about the straight course as it grows or decays, so a positive-real-part complex root can still diverge. Only a purely real, negative root guarantees the sway and yaw disturbance dies away smoothly, without the ship snaking, and returns the vessel to its original heading.
Final Answer:
Control-fixed straight-line (course) stability needs both eigenvalues of this system to be real and negative, matching option A.
\[ \boxed{\lambda_1, \lambda_2 \ \text{real and negative}} \]
Set \(\delta_R = 0\) since we want control-fixed stability. The equation becomes
\[ \begin{bmatrix} M - Y_{\dot{v}} & -Y_{\dot{r}} \\ -N_{\dot{v}} & I_z - N_{\dot{r}} \end{bmatrix} \begin{bmatrix} \dot{v} \\ \dot{r} \end{bmatrix} = \begin{bmatrix} Y_v & Y_r \\ N_v & N_r \end{bmatrix} \begin{bmatrix} v \\ r \end{bmatrix} \]
Step 2: Assume an exponential disturbance and form the characteristic equation.
A small disturbance from the straight course behaves like \(v, r \propto e^{\lambda t}\). Putting this into the matrix equation and collecting terms gives a real quadratic in \(\lambda\) of the form
\[ a\lambda^2 + b\lambda + c = 0 \]
where \(a = I_zM - I_zY_{\dot{v}} - MN_{\dot{r}} + N_{\dot{r}}Y_{\dot{v}} - N_{\dot{v}}Y_{\dot{r}}\), and \(c = N_rY_v - N_vY_r\), with \(b\) built from the mixed added-mass and hydrodynamic derivative terms.
Step 3: Interpret what each root type means physically.
If a root is complex, the disturbed motion oscillates (hunts) about the straight course as it grows or decays, so a positive-real-part complex root can still diverge. Only a purely real, negative root guarantees the sway and yaw disturbance dies away smoothly, without the ship snaking, and returns the vessel to its original heading.
Final Answer:
Control-fixed straight-line (course) stability needs both eigenvalues of this system to be real and negative, matching option A.
\[ \boxed{\lambda_1, \lambda_2 \ \text{real and negative}} \]
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