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HW Solutions 1. 4-2 A ship has a submerged volume of 112;000 ft3 and a righting arm of 2 ft when heeling to 15◦. Calculate its righting moment when heeling at this angle. RM = GZ(φ) · ∆ = GZ(φ) · ρ · g · r assuming seawater: lb 1 LT RM = 2:0 ft · 64 · 112;000 ft3 · = 6;400 ft · LT ft3 2240 lb 1 2. 4-4 (a) Use the following data to plot the Curve of Intact Statical Stability of a ship for starboard heels only. The data is taken from a ship displacing 3;600 LT with a KG of 18:0 ft. Remember to title your plot and label the axis correctly. Angle of Heel, φ (◦) 0 20 40 60 80 100 Righting Arm, GZ (ft) 0.0 1.2 2.8 4.1 2.7 0.0 (b) Use your plot to sketch a diagram of the ship heeling to 30 degrees to starboard. Calculate the righting moment being developed at this angle. RM = GZ(φ) · ∆ = 2 ft · 3;600 LT = 7;200 ft · LT (c) By observation of your sketch, what would happen to the magnitude of the righting moment calculated in (b), if the center of gravity was raised so that KG increased to 18:5 ft. 2 If KG increases by δKG = 0:5 ft, the righting arm GZ will decrease by δKG · sin(φ) = 0:5 ft · sin(30◦) ◦ ! RM will decrease to: RMf = RM0 −∆·δKG·sin(φ) = 7;200ft·LT −3;600LT ·0:5ft·sin(30 ) RMf = 7;200 ft · LT − 3;600 LT · 0:5 ft · 0:5 = 6;300 ft · LT 3 3. 4-5 Using the Cross Curves provided for the FFG-7 in the notes, graph the Curve of Intact Statical Stability for FFG-7 at a displacement of 3500 LT with KG = 0 ft. φ [◦] GZ [ft] 0 0.0 5 2.0 10 4.0 15 5.9 20 7.8 25 9.6 30 11.5 35 13.1 40 14.8 45 16.3 50 17.5 55 18.4 60 19.2 65 19.7 70 20.1 80 20.3 4 4. 4-6 (a) Plot a curve of intact statical stability for starboard heels only for a ship with the following overall stability characteristics. Overall Stability Characteristic Value Range of Stability 0 − 90◦ Maximum Righting Arm 3:8 ft Angle of Maximum Righting Arm 50◦ Righting Arm at 30◦ of heel 2 ft (b) On your plot in part (a) sketch the curve of intact statical stability for a ship with a stiffer righting arm. Which ship is more stable? The stiffer ship in (b) is more stable. (c) How would you calculate the dynamic stability from your plot in part (a)? Find dynamic stability by integrating the righting arm plot (area under the curve) and multiplying through by the displacement ∆. Z ]Range of Stability Dynamic Stability = ∆s GZ dφ 0 5 5. 4-7 A DDG-51 has a displacement of 8;350 LT and KG = 21:5 ft. In this condition it develops a righting arm of 2:1 ft when heeling to 20◦. (a) Use a suitable diagram to derive an equation for the magnitude of the new righting arm if the center of gravity shifted so that KG increased. Gf Zf = G0Z0 − δKG · sin(φ) (b) Use the equation you derived and the data above to calculate the magnitude of the new righting arm at 20◦ of heel if the KG of the DDG-51 increased to 22:6 ft. ◦ Gf Zf = G0Z0 − δKG · sin(φ) = 2:1 ft − [22:6 ft − 21:5 ft] sin(20 ) = 1:72 ft 6 6. 4-9 A ship has a displacement of 6;250 LT and KG = 17:6 ft. In this condition the ship develops a righting arm of 5:5 ft when heeling to 30◦. (a) Draw a diagram showing the effect that lowering the center of gravity has on the righting arm. Include on your diagram the old and new locations of G, old and new locations of the center of buoyancy, the metacenter, angle of heel, initial and final righting arms, and displacement and buoyant force vectors. (b) A weight shift causes the ship's center of gravity to be lowered by 1:5 ft. Calculate the ship's righting arm at a heeling angle of 30◦ after the weight shift. ◦ Gf Zf = G0Z0 − δKG · sin(φ) = 5:5 ft − [−1:5 ft] sin(30 ) = 6:25 ft 7 7. 4-10 USS SUPPLY (AOE-6) is preparing to UNREP its Battle Group. Prior to UNREP the ship was steaming on an even keel at a draft of 38:5 ft. The center of gravity was located 37 ft above the keel. Lpp = 717ft. When UNREP is complete, AOE-6 has discharged 10;000LT (3.2 million gallons) of F76 and JP-5 to the Battle Group. The fuel is assumed to have had a center of gravity on the centerline, 25 ft above the keel of AOE-6. Using the curves of form and cross curves of stability, determine the following: (a) Displacement and draft of SUPPLY after UNREP. 300 LT From COF: ∆ = 163:75 units · = 49;125 LT 0 unit Assuming that the UNREP causes no δTrim, ∆f = ∆0 − wUNREP = 49;125 LT − 10;000 LT = 39;125 LT 39;125 LT On COF: ∆f; COF = 300 LT = 130:42 units unit ! From COF: Tf = 32:3 ft 8 (b) Ship's KG and TCG following UNREP Pn KG0 · ∆0 + i=1 ±wi · kgi ∆0 · KG0 − wfuel · kgfuel KGf = Pn = ∆0 + i=1 ±wi ∆0 − wfuel 49;125 LT · 37 ft − 10;000 LT · 25 ft KG = = 40:07 ft f 39;125 LT Pn TCG0 · ∆0 + i=1 ±wi · ±tcgi ∆0 · TCG0 − wfuel · tcgfuel TCGf = Pn = ∆0 + i=1 ±wi ∆0 − wfuel 49;125 LT · 0 ft − 10;000 LT · 0 ft TCG = = 0 ft f 39;125 LT (c) Compute and plot the righting arm curves for the initial and final conditions of the ship. Do this for starboard heeling angles only. Note: use of a spreadsheet program is encouraged. ◦ ◦ φ0 [ ] GZ0 [ft] φf [ ] GZf [ft] 0 0.0 0 0.0 5 4.0 5 4.0 10 8.0 10 8.0 20 16.0 20 16.1 30 24.5 30 24.5 40 31.7 40 32.1 50 36.4 50 37.3 60 39.0 60 40.0 70 40.0 70 41.2 80 39.6 70 40.5 90 38.4 80 38.5 9 (d) From your results in part (c), complete the following table and comment on the UNREP's effect on stability. Parameter Before UNREP After UNREP Displacement, ∆ (LT ) 49;125 LT 39;125 LT Center of Gravity, KG (ft) 37 ft 40:07 ft Maximum Righting Arm, GZmax (ft) 8:06 ft 6:61 ft ◦ ◦ ◦ Angle of GZmax ( ) 50 50 Range of Stability (◦) 0◦− ≈100◦ 0◦ − 84◦ UNREP has made the AOE less stable (lower righting arm for all angles, lower righting moment, less dynamic stability (area under the curve), and a smaller range of stability. 10 8. 4-11 A ship has a displacement of 7;250 LT and KG = 23:5 ft on the centerline. At this condition the ship has the following stability characteristics: Range of stability: 0◦ − 85◦ Maximum righting arm: 5:2 ft at a heeling angle of 50◦ (a) What happens to the ship's stability characteristics if the center of gravity is raised? Range of Stability, Dynamic Stability, , righting moment, and GZ all decrease due to ]GZmax max increase in KG. (b) What happens to the ship's stability characteristics if the center of gravity is lowered? Range of Stability, Dynamic Stability, , righting moment, and GZ all increase due to ]GZmax max decrease in KG. (c) What happens to the ship's stability characteristics if there is a change in the transverse location of G with no vertical change in G? Range of Stability, Dynamic Stability, righting moment, and GZmax all decrease due to a lateral shift in G (TCG 6= 0 ft). should increase due to the static list angle. ]GZmax 11 9. 4-13 An LCS-1 has a displacement of 3;000 LT and KG = 14:0 ft. Use Cross Curves of Stability found in the Ship's Data section. (a) Plot the Curve of Intact Statical Stability for starboard heels for the ship at TCG = 0 ft. On the same axes, plot a second curve for the LCS-1 in the same condition but with TCG = 1 ft. ◦ φ0 [ ] GZ0 [ft] 0 0.0 5 1.4 10 3.0 15 4.5 20 6.1 30 9.0 40 11.8 50 14.2 60 16.0 70 16.8 80 17.2 90 17.4 12 (b) Compare the two curves. In which condition is the ship more stable? For all angles, the LCS with TCG = 1:0 ft has a lower righting arm and is, therefore, less stable (at least for rolling to starboard). The ship is more stable with G on the centerline. (c) What is the permanent angle of list when TCG = 1:0 ft? This is seen as the angle where the GZ curve crosses zero for TCG = 0 ft, or ≈ 16◦. 13 10. 4-14 An FFG-7 class ship is underway at a displacement of 3;990 LT . KM = 22:8 ft, KG = 19:3 ft, Lpp = 408ft. 9;000 gallons (28LT ) of F-76 are transferred from a storage tank located on the centerline, 6 ft above the keel, to a service tank located 6 ft above the keel and 15 ft to port of centerline.
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