1. **Problem 1:** Calculate the moment of inertia of a rectangular waterplane 42m long and 9m wide about axes parallel to the beam. - The beam is the width direction, so the axis is parallel to the 9m side. - Moment of inertia about an axis parallel to the beam (lengthwise axis) through the centroid is given by: $$I = \frac{bL^3}{12}$$ where $b=9$ m (beam), $L=42$ m (length). - Calculate $I$ about the centroid: $$I_c = \frac{9 \times 42^3}{12} = \frac{9 \times 74088}{12} = \frac{666792}{12} = 55566 \text{ m}^4$$ - For the other axes: ii. Axis halfway between centroid and end: distance $d = \frac{42}{4} = 10.5$ m from centroid. Use parallel axis theorem: $$I = I_c + Ad^2$$ Area $A = 42 \times 9 = 378$ m$^2$ $$I = 55566 + 378 \times 10.5^2 = 55566 + 378 \times 110.25 = 55566 + 41674.5 = 97240.5 \text{ m}^4$$ iii. Axis at the end: distance $d = \frac{42}{2} = 21$ m from centroid. $$I = 55566 + 378 \times 21^2 = 55566 + 378 \times 441 = 55566 + 166698 = 222264 \text{ m}^4$$ 2. **Problem 2:** Calculate approximate $MCT1cm$ for a ship with beam $B=18$ m and waterplane area $A=2011$ m$^2$. - Approximate formula: $$MCT1cm = \frac{\rho g I}{100 \times V}$$ where $I$ is the moment of inertia of waterplane about longitudinal axis, $V$ is volume displacement. - Since volume $V$ is not given, approximate $I$ for rectangular waterplane: $$I = \frac{B L^3}{12}$$ But length $L$ unknown, so approximate $I$ using: $$I = \frac{B A^2}{12}$$ - Calculate: $$I = \frac{18 \times 2011^2}{12} = \frac{18 \times 4044121}{12} = \frac{72794178}{12} = 6066181.5 \text{ m}^4$$ - Without volume $V$, $MCT1cm$ cannot be precisely calculated here. 3. **Problem 3:** Calculate $MCT1cm$ for a ship 96m long with given half-ordinates and displacement 5200t. - Calculate waterplane area $A$ using trapezoidal rule: Half-ordinates: 2.10, 5.54, 6.37, 6.85, 6.85, 6.77, 5.43, 2.83, 0 Stations = 9, spacing $\Delta x = \frac{96}{8} = 12$ m - Waterplane area: $$A = 2 \times \Delta x \times \left( \frac{y_1 + y_9}{2} + y_2 + y_3 + y_4 + y_5 + y_6 + y_7 + y_8 \right)$$ $$= 2 \times 12 \times \left( \frac{2.10 + 0}{2} + 5.54 + 6.37 + 6.85 + 6.85 + 6.77 + 5.43 + 2.83 \right)$$ $$= 24 \times (1.05 + 5.54 + 6.37 + 6.85 + 6.85 + 6.77 + 5.43 + 2.83)$$ $$= 24 \times 41.69 = 1000.56 \text{ m}^2$$ - Moment of inertia $I$ about longitudinal axis: $$I = \frac{B L^3}{12}$$ Beam $B = 2 \times$ max half-ordinate $= 2 \times 6.85 = 13.7$ m - Calculate $I$: $$I = \frac{13.7 \times 96^3}{12} = \frac{13.7 \times 884736}{12} = \frac{12119863.2}{12} = 1009988.6 \text{ m}^4$$ - Displacement volume $V = 5200$ t $= 5200$ m$^3$ (assuming salt water density 1 t/m$^3$) - Calculate $MCT1cm$: $$MCT1cm = \frac{\rho g I}{100 V}$$ Using $\rho g = 9.81$ kN/m$^3$: $$MCT1cm = \frac{9.81 \times 1009988.6}{100 \times 5200} = \frac{9918773.7}{520000} = 19.07 \text{ kN/cm}$$ 4. **Problem 4:** Calculate new drafts at FP and AP after moving 80t load 60m forward. - Given: $$L=120m, B=15m, C_{WP}=0.76, T=7.5m, LCF=1.8m \text{ aft of amidships}$$ Longitudinal moment of inertia $I_x = 1,350,800$ m$^4$ - Load moved $\Delta W = 80$ t, distance $d = 60$ m - Change in trim moment: $$M = \Delta W \times d = 80 \times 60 = 4800 \text{ t-m}$$ - Change in trim angle $\theta$: $$\theta = \frac{M}{\rho g V GM_T}$$ Without $GM_T$ and $V$, approximate change in draft at FP and AP: - Change in draft at FP: $$\Delta T_{FP} = - \frac{M \times (L - LCF)}{I_x} = - \frac{4800 \times (120 - 1.8)}{1,350,800} = - \frac{4800 \times 118.2}{1,350,800} = -0.42 \text{ m}$$ - Change in draft at AP: $$\Delta T_{AP} = \frac{M \times LCF}{I_x} = \frac{4800 \times 1.8}{1,350,800} = 0.0064 \text{ m}$$ - New drafts: $$T_{FP,new} = 7.5 - 0.42 = 7.08 \text{ m}$$ $$T_{AP,new} = 7.5 + 0.0064 = 7.51 \text{ m}$$ **Final answers:** 1. i. $I_c = 55566$ m$^4$ ii. $I = 97240.5$ m$^4$ iii. $I = 222264$ m$^4$ 2. Insufficient data to calculate exact $MCT1cm$. 3. $MCT1cm \approx 19.07$ kN/cm 4. New drafts: $T_{FP} = 7.08$ m, $T_{AP} = 7.51$ m