25-Nav-A3 Hydrodynamics of Ships (II)_ Ship Motion · December 2019
Question 3 of 6: Rolling Magnification Factor and Maximum Roll Amplitude
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2019 — 16-Nav-A3 Hydrodynamics of Ships (II): Ship Motion. Three-hour, closed-book exam; one two-sided 8.5″×11″ formula sheet and an approved calculator are permitted. Format: Questions 1–5 are compulsory; Question 6 offers a choice of (a) or (b) — both are solved below for completeness.
Reference texts: Bhattacharyya, Dynamics of Marine Vehicles (Wiley) — wave kinematics/pressure, roll response and magnification-factor theory, irregular-seaway spectral analysis; Lewis (ed.), Principles of Naval Architecture, Vol. III — Motions in Waves and Controllability (SNAME) — Froude-Krylov theory, added mass, hydroelasticity, linear maneuvering derivatives; Lloyd, Seakeeping: Ship Behaviour in Rough Weather — roll magnification factor and encounter-frequency spectra; Lewandowski, The Dynamics of Marine Craft — the prime-system nondimensional maneuvering equations used in Question 6(b).
In Question 2, sensor 2's dynamic pressure amplitude is 25,600 N/m², the value used throughout (it gives a water depth of about 10.0 m).
Question 3: Rolling Magnification Factor and Maximum Roll Amplitude (25 marks)
Given. Heading (from following seas) $\mu=150^\circ$; ship speed $V=20$ kn; $L_{WL}=450$ ft, radius of gyration in roll $k_x=30.8$ ft, $\overline{GM}_T=5.79$ ft, displacement $\Delta=12{,}500$ tonnes; added roll inertia $=0.20\times$ own moment of inertia; damping moment $=32{,}000(d\phi/dt)$ ft-tonnes; $g=32.2$ ft/s².
Find. (a) the magnification factor $\Lambda$ at the tuning factor corresponding to each of $\omega_w=0.1,0.2,0.3,0.4$ rad/s; (b) the maximum roll amplitude for $H=60$ ft.
Approach. The natural roll frequency $\omega_n$ and damping ratio $\zeta$ follow from the ship's own data (independent of wave frequency); converting each wave frequency $\omega_w$ to its encounter frequency $\omega_e$ (using $V$ and $\mu$) gives the tuning factor $\lambda=\omega_e/\omega_n$ at each point, and the standard single-degree-of-freedom magnification-factor formula then gives $\Lambda(\lambda)$. Part (b) uses the largest of the four $\Lambda$ values together with the wave-slope amplitude for that frequency.
Part (a) — natural roll frequency and damping ratio. With the added inertia folded in, the effective roll inertia is $I_{tot}=1.2\,(\Delta/g)k_x^2$ and the restoring moment per radian is $\Delta\cdot\overline{GM}_T$, so
$$\omega_n=\sqrt{\frac{g\,\overline{GM}_T}{1.2\,k_x^2}}=\sqrt{\frac{(32.2)(5.79)}{1.2(30.8)^2}}=\boxed{0.4047\ \text{rad/s}}\quad(T_n=2\pi/\omega_n=15.5\ \text{s}).$$
The critical damping is $B_c=2I_{tot}\omega_n$, with $I_{tot}=1.2(12{,}500/32.2)(30.8)^2=441{,}913$ ft-tonne-s², giving
$$B_c=2(441{,}913)(0.4047)=357{,}678, \qquad \zeta=\frac{B}{B_c}=\frac{32{,}000}{357{,}678}=\boxed{0.0895}.$$
Part (a) — encounter frequency at each $\omega_w$. With $\mu$ measured from following seas ($\mu=0^\circ$) so that $150^\circ$ is bow-quartering (close to head seas), $\cos150^\circ=-0.8660$, and $V=20\ \text{kn}=33.76$ ft/s, the encounter frequency is
$$\omega_e=\omega_w-\frac{\omega_w^2 V}{g}\cos\mu=\omega_w+0.9079\,\omega_w^2.$$
Evaluated at the four required wave frequencies:
$\omega_w$ (rad/s)
$\omega_e$ (rad/s)
tuning $\lambda=\omega_e/\omega_n$
0.1
0.109
0.270
0.2
0.236
0.584
0.3
0.382
0.943
0.4
0.545
1.347
Magnification factor at each tuning factor. Using
$$\Lambda(\lambda)=\frac{1}{\sqrt{(1-\lambda^2)^2+(2\zeta\lambda)^2}}, \qquad \zeta=0.0895,$$
the four points (plotted below) are:
$$\Lambda(0.270)=1.077,\quad \Lambda(0.584)=1.499,\quad \Lambda(0.943)=\boxed{4.96}\ (\text{peak, near resonance}),\quad \Lambda(1.347)=1.176.$$
That is part (a) — the peak sits at $\omega_w=0.3$ rad/s, where the tuning factor $\lambda=0.943$ is closest to the resonance condition $\lambda=1$.
Figure 3 — magnification factor $\Lambda$ vs. tuning factor $\lambda$ at the four required wave frequencies; the peak at $\omega_w=0.3$ rad/s ($\lambda=0.94$) is nearest resonance.
Part (b) — maximum roll amplitude. The largest response among the four computed points is at $\omega_w=0.3$ rad/s ($\Lambda=4.96$). Assuming deep water at that frequency, $k=\omega_w^2/g=(0.3)^2/32.2=0.002795\ \text{ft}^{-1}$, and with $H=60$ ft ($\eta_a=H/2=30$ ft), the wave-slope amplitude is
$$\alpha_a=k\eta_a=(0.002795)(30)=0.0839\ \text{rad}=4.80^\circ.$$
The roll amplitude at resonance-like tuning is the magnification factor applied to the wave-slope excitation:
$$\phi_a=\Lambda\,\alpha_a=(4.96)(0.0839\ \text{rad})=0.416\ \text{rad}=\boxed{23.8^\circ}.$$
Check: part (b) uses the plain wave-slope amplitude $\alpha_a=k\eta_a$ as the roll excitation, matching the level of detail given in the question (ship dimensions and damping only, no additional geometric coefficient). A more refined treatment would reduce this by $\sin\mu$ (the transverse component of the wave-slope excitation falls to zero in pure head/following seas and is greatest at beam seas); with $\mu=150^\circ$, $\sin150^\circ=0.50$, which would roughly halve the estimate to $\approx 11.9^\circ$. Both the plain and the heading-corrected values are physically reasonable; the boxed answer uses the simpler form consistent with the data explicitly supplied.
