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STABILITY AND TRIM CALCULATIONS Rolling Period

STABILITY AND TRIM CALCULATIONS Rolling Period. RP = .44 B √GM. 1. You are on a vessel that has metacentric height of 4 feet and a beam of 50 feet. What can you expect the rolling period of the vessel to be? A. 10.0 s B. 10.5 s C. 11.0 s D. 11.5 s

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STABILITY AND TRIM CALCULATIONS Rolling Period

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  1. STABILITY AND TRIM CALCULATIONS Rolling Period

  2. RP = .44 B √GM 1. You are on a vessel that has metacentric height of 4 feet and a beam of 50 feet. What can you expect the rolling period of the vessel to be? A. 10.0 s B. 10.5 s C. 11.0 s D. 11.5 s 2. You are at the sea on a vessel that has a beam of 60 feet, and you calculate the period of roll to be 25 seconds. What is the vessel’s metacentric height? A. 0.8 ft. B. 1.1 ft. C. 1.4 ft. D. 1.6 ft. GM = .44 B2 RP

  3. GM = .44 B2 RP 3. Your vessel has a displacement of 10,000 tons. It is 350 ft. long and has a beam of 55 ft. You have timed its full rolling period to be 15.0 seconds. What is your vessel’s approximate GM? A. 1.18 ft. B. 1.83 ft. C. 2.60 ft. D. 3.36 ft.

  4. GM = .797 B2 RP 4. Your vessel has a measures 35.0 meters long by 5.0 meters in 6 seconds. What is the GM? A. 0.44 B. 0.54 C. 0.33 D. 0.48 5. You are on a vessel that has a metacentric height of 0.68 meters and a beam of 5.2 meters. What can you expect the rolling period the vessel to be? A. 7 seconds B. 6 seconds C. 4 seconds D. 5 seconds RP = .797 B √GM

  5. VERTICAL SHIFT OF CENTER OF GRAVITY

  6. 1. A vessel’s light draft displacement is 7400 tons. The center of gravity at this draft is 21.5 ft. above the keel. The following weights are loaded: 450 tons VCG 17 ft.; 220 tons VCG 11.6 ft., 65 tons VCG 7.0 ft. The new CG above the keel is: A. 14.7 ft. B. 17.8 ft. C. 18.7 ft. D. 20.9 ft. NEW KG = MOMENT WT = 169757 8135 = 20.867 FT.

  7. 2. Your present displacement is 15,000 short tons. The KG at this displacement is 60 ft. 100 short tons of casings are added at a VCG of 75 ft. for an ocean tow. What is the new KG? A. 59.09 ft. B. 62.09 ft. C. 60.09 ft. D. 61.09 ft. NEW KG = MOMENT WT = 907,500 15,100 = 60.099 FT.

  8. 3. A vessel with displacement of 19,700 LT and a KG of 50.96 ft. loads 300 LT of barite into P-tanks located 120 ft. above the keel. What is the change in KG? A. 1.00 ft. upward B. 0.79 ft. upward C. 1.04 ft. upward D. 1.83 ft. upward NEW KG = MOMENT WT = 1,039,912 20,000 = 52.00 OLD KG = 50.96 (-) 1.04 UPWARD

  9. GG’ = W X D Δ = 15 X 8 300 = 0.4 FT. 4. You have approximately 15 tons of fish on deck. What will the shift in the center of gravity after you shift the fish to the fish hold a vertical distance of 8 feet? Total displacement 300 tons. A. 0.4 ft. B. 0.1 ft. C. 0.2 ft. D. 0.3 ft.

  10. OLD DISPL. = 12,000 X 21.5 = 258,000 GROUNDED WT=(-) 2,500 NEW DISPL. = 9,500 NEW KG = 258,000 9,500 NEW KG = 27.16 OLD KG = 21.50 VIRTUAL RISE OF CG = 5.66 UPWARD 5. Your ship of 12,000 tons displacement has a center of gravity of 21.5 ft. above the keel. You run aground and estimate the weight aground is 2500 tons. The virtual rise in the center of gravity is: A. 1.26 ft. B. 5.66 ft. C. 3.80 ft. D. 4.80 ft. GG’ = W X D Δ = 2,500 X 21.5 9,500 = 5.66 UPWARD

  11. INCLINING EXPERIMENT

  12. GM = W X D Δ TAN LIST = 60 X 45 16,400 TAN 1.5˚ = 6.29 FT 1. Sixty tons of cargo are raised with a boom 45 feet from the centerline. The vessel’s displacement including the weight lifted is 16,400 tons.The angle of list caused by the suspended weight is 1.5˚. KM is 28.75 ft. and BM is 17.25 FT. What is the KG? A. 11.65 ft. B. 22.46 ft. C. 23.15 ft. D. 23.82 ft. KM = 28.75 FT. GM = 6.29 FT. KG = 22.46 FT.

  13. 2. A ship is inclined by moving a weight of 30 tons a distance of 30 ft. from the center line. A 28 feet pendulum shows a deflection of 12 inches. Displacement including weight moved is 4,000 tons. KM is 27.64 ft. What is the KG? A. 21.34 ft. B. 22.06 ft. C. 22.76 ft. D. 23.21 ft.

  14. INCLINING EXPERIMENT: GM = W X D Δ Tan LIST = 30 TONS X 30 FT 4,000 TONS X Tan 2.05˚ GM = 6.29 FT. KM = 27.64 FT. GM = - 6.29 FT. KG = 21.35 FT. SOHCAHTOA 28’ WL1 WL2 1’ SIN Ø = OPP HYPO = 1 28 = 2.05˚

  15. GM = W X D Δ TAN LIST TAN LIST = W X D Δ X GM = 75 X 50 6,000 X 6 = 5.946˚ 3. A cargo of 75 tons is to be lifted with a boom located 50 ft. from the ship centerline. The ship’s displacement including the suspended cargo is 6,000 tons and GM is 6 ft. The list of the ship with cargo suspended from the boom will be: A. 5.00O B. 5.40O C. 5.94O D. 6.50O

  16. TAN LIST = W X D Δ X GM = 60 X 20 6,000 X 2 = 5.7˚ 4. A cargo of 60 tons to be loaded on deck 20 feet from the ship’s centerline. The ship’s displacement including the suspended cargo is 6,000 tons and GM is 2 ft. The list of the ship with the cargo suspended from the boom will be: A. 5.4O B. 5.7O C. 6.1O D. 6.4O

  17. GM = W X D Δ TAN LIST = 40 X 40 16,000 TAN 6.5˚ = 0.88 FT. 5. To check stability, a weight of 40 tons is lifted with the jumbo boom, whose head is 40 ft. from the ship’s centerline. The clinometer shows a list of 6.5O with the weight suspended. Displacement including the weight is 16,000 tons. The GM while in this condition is: A. 0.21 ft. B. 0.43 ft. C. 0.88 ft. D. 1.02 ft.

  18. GG’ = W X D Δ = 50 X 56 18050 = 0.156 (+)3.397 3.553 6. You are making a heavy lift with the jumbo boom. Your vessel displaces 18,000 tons. The 50-tons weight is on the pier and its center is 75 feet to starboard of the centerline. The head of the boom is 112 feet above the baseline and the center of gravity of the lift when stowed on the deck will be 56 ft. above the baseline. As the jumbo takes the strain the ship lists 3.5O. What is the GM when the cargo is stowed? A. 3.19 ft. B. 3.40 ft. C. 3.56 ft. D. 3.24 ft. GM = W X D Δ TAN LIST = 50 X 75 18050 TAN3.5˚ = 3.397

  19. TAN LIST = W X D Δ X GM = 60 X 12 6,000 X .6 = 11. 309 11˚ 18’ 36” 7. A ship of 6,000 tonnes displacement has KM=7.3 meters and KG=6.7 meters and is floating upright. A weight of 60 tonnes already on board is shifted 12 meters transversely. Find the resultant list. A. 10° 00’ B. 11°15’ C. 9° 15’ D. 11°18’ 30” KM = 7.3 KG = 6.7 (-) GM = 0.6

  20. FREE SURFACE CORRECTION

  21. FREE SURFACE CORRN. • FOR SALT WATER: • F.S.CONST. = l b3 420 l = LENGTH OF THE TANK b= BREADTH OF THE TANK FOR LIQUIDS OTHER THAN SALT WATER: F.S.CONST. = r l b3 420 r = S.G OF LIQUID IN TANK S.G WHERE VSL. FLOATS F.S.CORRN. = F.S CONSTANT Δ F.S.CORRN. = r l b3 420 Δ 432

  22. FSC = r l b3 420Δ = (1)(45)(45)3 (420)(8,000) = 1.22 ft. 1.On the vessel displacing 8,000 tons, what is the reduction in metacentric height due to free surface when a tank 45 ft. long and 45 ft. wide is partially filled with saltwater? A. 1.22 ft. B. 1.16 ft. C. 1.13 ft. D. 1.10 ft.

  23. 2. On a vessel of 10,000 tons displacement, compute the reduction in metacentric height due to free surface in a hold having fresh water on the tank top. The hold is 40 ft. long and 50 ft. wide. The reduction in GM is: • A. 1.1 ft. • B. 1.2 ft. • C. 1.3 ft. • D. 1.5 ft. FSC = r l b3 420Δ

  24. FSC IS REDUCED BY 75%. 3. What is the reduction in GM due to free surface when a tank 60 ft. long and 60 ft. wide is partially filled with salt water and is fitted with a centerline bulkhead? The vessel has a displacement of 10,000 tons. A. 0.5 ft. B. 0.8 ft. C. 1.0 ft. D. 1.2 ft. FSC = r l b3X .25 420Δ

  25. 4. A shaft alley divides a vessel’s cargo hold into two tanks, each 25 ft. wide by 50 ft. long. Each tank is filled with sea water below the level of the shaft alley. The vessel’s displacement is 6,000 tons. What is the reduction in Gm due to free surface effect? • A. 0.56 ft. • B. 0.58 ft. • C. 0.62 ft. • D. 0.66 ft. FSC = r l b3X 2 420Δ

  26. FSC = r l b3 X 2 420Δ 5. A 7,000 ton displacement tankship carries two slack tanks of alcohol with a S.G. of 0.8 Each tank is 50 long and 30 ft. wide. What is the reduction in GM due to free surface with the vessel floating in sea water (SG 1.026)? A. 0.36 ft. B. 0.46 ft. C. 0.72 ft. D. 0.82 ft.

  27. F.S.Const. = r l b3 420 6. Determine the free surface constant for fuel oil tank 30 ft. long by 40 ft. wide by 15 ft. deep. The S.G. of the fuel oil is 0.85 and the ship will float in salt water (SG 1.026) A. 0.83 B. 42.7 C. 3787 D. 4571 7. Determine the free surface correction for fuel oil tank 30 ft. long by 40 ft. wide by 15 ft. deep with free surface constant of 3794. The vessel is displacing 7,000 tons in saltwater. A. 0.35 ft. B. 0.54 ft. C. 0.65 ft. D. 1.38 ft. F.S.Corrn. = F.S.Const Δ

  28. 8. A ship of 8,153.75 tonnes displacement has KM=8 m, KG=7.5 and a double bottom tank 15 m x 10 m x 2 m which is full of salt water ballast. Find the new GM if this tank is now pumped out until half empty. A. 0.225 m B. 0.825 m C. 1.75 m D. 1.45 m CHS - VIRTUAL RISE GG1.ppt

  29. LET GG1 REPRESENT THE ACTUAL RISE OF “G” DUE TO MASS DISCH. THE MASS OF WATER DISCH.(W) = L X B X D X DENS. = 15 X 10 X 1 X 1.025 = 153.75 TONS INITIAL KG = 7.5 RISE OF g = 1.5 = 6.0 INITIAL Δ = 8,153.75 DISCH.= (-) 153.75 FINAL Δ = 8,000.00 GG1 = W X D Δ = 153.75 X 6 8,000 GG1 = 0.115 M NKG = T.MOM. = 60,922 T.WT. 8,000 NKG = 7.615 OKG = 7.500 GG1 = 0.115

  30. LET G1GV REPRESENTS THE VIRTUAL LOSS OF GM DUE TO FREE SURFACE. G1GV = l b3 12V = 15(10)3(1.025) (12)(8,000) = 0.160 M VIRTUAL LOSS OF GM (G1GV) = i X d1 X 1 V d2 n2 i = SECOND MOMENT OF FREE SURFACE ABOUT THE CENTER LINE. V = SHIPS VOL. OF DISPL. d1 = DENS.OF LIQUID IN TANK. d2 = DENS.OF WATER WHERE VSL. IS FLOATING. n = NO. OF LONG’L. COMPARTMENT INTO WHICH IS PARTIALLY FILLED WITH LIQUID. BUT : d1 = d2 n = 1 G1GV = i V FOR RECTANGULAR WATERPLANE: i = l b3 12

  31. OLD KM = 8.000 M OLD KG = 7.500 M OLD GM = 0.500 M RISE OF G DUE DISCH. = 0.115 M ( - ) 0.385 M VIRTUAL RISE OF “G” = 0.160 M ( - ) NEW GM = 0.225 M

  32. FREEBOARD ALLOWANCE

  33. 1. You are loading in a port subject to the tropical load line mark and bound for a port subject to the summer load line mark. You will enter the summer zone after steaming four days. You will consume 40 tons of fuel, water, and stores per day. The hydrometer reading at the loading pier is 1.025 and the average TPI is 53. The following data is extracted from the load line Certificate: Freeboard from Deck line Load Line Tropical 67” (T) 5”above (s) Summer 72” (S) * Winter 77” (W) 5” below (s) Allowance for fresh water for freeboards 4” *Upper the edge of line at level of center of ring.

  34. What is the minimum freeboard required at the start of the voyage? • A. 65 inches • B. 69 inches • C. 72 inches • D. 75 inches CONSUMPTION = 40 TONS/DAY STEAMING TIME = (X)4 DAYS CONS.FOR 4 DAYS = 160 TONS SUMMER F.B. = 72” (-) 3” MIN. F.B.ON DEP = 69” CONS. = 160 TPI 53 = 3”

  35. 2. You are loading in a port subject to the tropical loadline mark and bound for a port subject to the summer loadline mark. You will enter the summer zone after steaming two days. You will consume 28 tons of fuel, water and stores per day. The hydrometer reading at the loading pier is 1.020 and the average TPI is 55. The following data is extracted from the load line certificate:

  36. Freeboard from Deckline Load Line • Tropical 69” (T) 7” above (s) • Summer 76” (S) * • Winter 83” (W) 7” below (s) • Allowance for fresh water all freeboards 6” • *Upper edge of line at level of center ring. • What is the minimum freeboard at the start of the voyage? • A. 62 inches • B. 66 inches • C. 70 inches • D. 74 inches

  37. CONSUMPTION = 28 TONS/DAY STEAMING TIME = (X)2 DAYS CONS.FOR 2 DAYS = 56 TONS CONS. = 56 TPI = 55 = 1.0” A.I = FWA X CH.OF DENS 25 = 6 X 5 25 = 1.2” BURN OFF/CONS.= 1.0” A.I. = 1.2” 2.2” SUMMER F.B.= 76.0” F.B.ON DEP.= 73.8”

  38. 69”(T) 76”(S) 83”(W) TF F T FWA(6”) S SUMMER ARR. W WNA TROPICAL 2 DAYS DEP.

  39. 3. You are loading in a port subject to the summer loadline mark and bound for a port subject to the winter loadline mark. You will enter the winter zone after steaming four days. You will consume 35 tons of fuel, water and stores per day. The hydrometer reading at the loading pier is 1.0083 and the average TPI is 65. The following data is extracted from the loadline Certificate: • Freeboard from Deckline Load Line • Tropical 68” (T) 6” above (s) • Summer 74” (S) * • Winter 80” (W) 6” below (S) • Allowance for fresh water all freeboards 5” • *Upper edge of line at level of center ring.

  40. A.I. = FWA X CH.OF DENS. 25 = 5 X 16.7 25 = 3.34” • What is the minimum freeboard required at he start of the voyage? • A. 74 inches • B. 78 inches • C. 80 inches • D. 86 inches CONSUMPTION = 35 TONS/DAY STEAMING TIME = (X) 4 DAYS CONS.FOR 4 DAYS = 140 TONS BURN OFF/CONS = 2.15” A.I. = 3.34” 5.49” WINTER F.B. = 80.00” DEP. F.B. = 74.51” CONS. = 140 TPI = 65 = 2.15”

  41. 4. You are loading in a port subject to the tropical loadline and bound for a port subject to the winter loadline mark. You will enter the summer zone after steaming eight days and you will enter the winter zone after ten days. You will consume 31 tons of fuel, water and stores per day. The hydrometer reading at the loading pier is 1.016 and the average TPI is 41. The following data is extracted from the loadline certificate: • Freeboard from deckline Loadline • Tropical 42” (T) 6” above (s) • Summer 48” (S) * • Winter 54” (W) 6” below (s) • Allowance for fresh water all freeboards 5” • *Upper edge of the line at the level of center ring.

  42. CONSUMPTION = 31 TONS/DAY STEAMING TIME = (X) 10 DAYS CONS.FOR 10 DAYS = 310 TONS • What is the minimum freeboard required at the start of the voyage? • A. 55.0 inches • B. 49.5 inches • C. 44.5 inches • D. 41.0 inches CONS. = 310 TPI 41 = 7.56” A.I. = FWA X CH.OF DENS. 25 = 5 x 9 25 = 1.80” CONSUMPTION = 31 TONS/DAY STEAMING TIME = (X) 8 DAYS CONS.FOR 8 DAYS = 248 TONS 41(TPI) = 6.05” CONS./BURN OFF = 6.05” A.I. = 1.80” 7.85” SUMMER F.B. = 48.00” F.B. FOR SUMMER= 40.15” CONS/BURN OFF = 7.56” A.I. = 1.80” 9.36” WINTER F.B. = 54.00” F.B.ON DEP. = 44.64”

  43. 5. You are loading in a port subject to the tropical loadline mark and bound for a port subject to the winter loadline mark. You will enter the summer zone after steaming four days and you will enter the winter zone after nine days. You will consume 29 tons of fuel, water and stores per day. The hydrometer reading at the loading pier is 1.008 and the average TPI is 53. The following data is extracted from the Load Line Certificate”. • Freeboard from Deckline Load Line • Tropical 75” (T) 8” above (s) • Summer 83” (S) * • Winter 91” (W) 8” below (S) • Allowance for fresh water all freeboards 9” • *Upper edge of line at the level center ring.

  44. CONSUMPTION = 29 TONS/DAY STEAMING TIME = (X) 9 DAYS CONS.FOR 9 DAYS = 261 TONS • 5. What is the minimum freeboard required at the start of the voyage • A. 72.5 inches • B. 75.0 inches • C. 77.0 inches • D. 80.0 inches CONS. = 261 TPI 53 = 4.92” CONSUMPTION = 29 TONS/DAY STEAMING TIME = (X) 4 DAYS CONS.FOR 4 DAYS = 116 TONS A.I. = FWA X CH.0F DENS. 25 = 9 X 17 25 = 6.12” CONS. = 116 TPI 53 = 2.19” A.I. = 6.12” CONS/BURN OFF = 4.92” 11.04” WINTER F.B. = 91.00” DEP.F.B. = 79.96” A.I. = 6.12” CONS/BURN OFF = 2.19” 8.31” SUMMER F.B = 83.00” DEP. F.B./ SUMMER ZONE = 74.7”

  45. ZONE ALLOWANCE • 1. A tanker loads at a terminal within the tropical zone. She will enter the summer zone six days after departing the loading port. She will burn off 45 tons/day and daily water consumption is eight tons. How many tons may she load above her summer loadline? • A. 270 tons • B. 278 tons • C. 291 tons • D. 318 tons BURN OFF = 45 TONS WATER CONS.= 8 TONS 53 TONS/DAY STEAMING TIME = (X) 6 DAYS 318 TONS

  46. 2. A vessel’s tropical loadline is 6 inches above her summer loadline. Her TPI is 127 tons. She will arrive in the summer zone 8 days after departure. She will burn off about 47 tons/day and water consumption is 12 tons/day.How many tons may she load above her summer loadline if she loads in the tropical zone? • A. 376 tons • B. 472 tons • C. 762 tons • D. 1016 tons BURN OFF = 47 TONS WATER CONS = 12 TONS 59 TONS/DAY STEAMING TIME = (X) 8 DAYS 472 TONS

  47. THE LOAD LINE MARK Deck Line 25 mm 1 in. 300 mm 12 in. 540 mm 21 in. 230 mm 9 in. TF F T L R S 230 mm W WNA 300 mm 450 mm 18 in.

  48. 67”(T) 72”(S) 77”(W) SUMMER TROPICAL

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