1. you are on a vessel that has metacentric height of 4 feet and a beam of 50 feet. what can you...

• 1. You are on a vessel that has metacentric height of 1. You are on a vessel that has metacentric height of 4 feet and a beam of 50 feet. What can you expect 4 feet and a beam of 50 feet. What can you expect the rolling period of the vessel to be?the rolling period of the vessel to be?

• A. 10.0 sA. 10.0 s• B. 10.5 sB. 10.5 s• C. 11.0 sC. 11.0 s• D. 11.5 sD. 11.5 s 2. You are at the sea on a vessel that has a beam of 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 60 feet, and you calculate the period of roll to be 25 seconds. What is the vessel’s metacentric height?seconds. What is the vessel’s metacentric height?

A. 0.8 ft.A. 0.8 ft. B. 1.1 ft.B. 1.1 ft. C. 1.4 ft.C. 1.4 ft. D. 1.6 ft.D. 1.6 ft.•

RP = .44 B √GM

GM = .44 B 2

RP

3. Your vessel has a displacement of 10,000 3. Your vessel has a displacement of 10,000 tons. It is 350 ft. long and has a beam of 55 ft. tons. It is 350 ft. long and has a beam of 55 ft. You have timed its full rolling period to be 15.0 You have timed its full rolling period to be 15.0 seconds. What is your vessel’s approximate seconds. What is your vessel’s approximate GM?GM?

A. 1.18 ft.A. 1.18 ft. B. 1.83 ft.B. 1.83 ft. C. 2.60 ft.C. 2.60 ft. D. 3.36 ft.D. 3.36 ft.

GM = .44 B 2

RP

4. Your vessel has a measures 35.0 meters long by 4. Your vessel has a measures 35.0 meters long by 5.0 meters in 6 seconds. What is the GM?5.0 meters in 6 seconds. What is the GM?

A. 0.44A. 0.44 B. 0.54B. 0.54 C. 0.33C. 0.33 D. 0.48D. 0.48 5. You are on a vessel that has a metacentric height 5. You are on a vessel that has a metacentric height

of 0.68 meters and a beam of 5.2 meters. What can of 0.68 meters and a beam of 5.2 meters. What can you expect the rolling period the vessel to be?you expect the rolling period the vessel to be?

A. 7 secondsA. 7 seconds B. 6 secondsB. 6 seconds C. 4 secondsC. 4 seconds D. 5 secondsD. 5 seconds

GM = .797 B 2

RP

RP = .797 B √GM

• 1. A vessel’s light draft displacement is 7400 tons. 1. A vessel’s light draft displacement is 7400 tons. The center of gravity at this draft is 21.5 ft. above the The center of gravity at this draft is 21.5 ft. above the keel. The following weights are loaded: 450 tons keel. The following weights are loaded: 450 tons VCG 17 ft.; 220 tons VCG 11.6 ft., 65 tons VCG 7.0 VCG 17 ft.; 220 tons VCG 11.6 ft., 65 tons VCG 7.0 ft. The new CG above the keel is:ft. The new CG above the keel is:

• A. 14.7 ft.A. 14.7 ft.• B. 17.8 ft.B. 17.8 ft.• C. 18.7 ft.C. 18.7 ft.• D. 20.9 ft.D. 20.9 ft.

WEIGHTWEIGHT C. GC. G MOMENTMOMENT

74007400 21.521.5 159100159100

(+) 450(+) 450 17.017.0 76507650

(+) 220(+) 220 11.611.6 25522552

(+) 65(+) 65 7.07.0 455455

81358135 169757169757NEW KG = MOMENT WT = 169757 8135 = 20.867 FT.

2. Your present displacement is 15,000 short 2. Your present displacement is 15,000 short tons. The KG at this displacement is 60 ft. 100 tons. The KG at this displacement is 60 ft. 100 short tons of casings are added at a VCG of short tons of casings are added at a VCG of 75 ft. for an ocean tow. What is the new KG?75 ft. for an ocean tow. What is the new KG?


WEIGHTWEIGHT KGKG MOMENTMOMENT

15,00015,000 6060 900,000900,000

(+) 100(+) 100 7575 (+) 7,500(+) 7,500

15,10015,100 907,500907,500

NEW KG = MOMENT WT = 907,500 15,100 = 60.099 FT.

3. A vessel with displacement of 19,700 LT and a KG 3. A vessel with displacement of 19,700 LT and a KG of 50.96 ft. loads 300 LT of barite into P-tanks of 50.96 ft. loads 300 LT of barite into P-tanks located 120 ft. above the keel. What is the change in located 120 ft. above the keel. What is the change in KG?KG?

A. 1.00 ft. upwardA. 1.00 ft. upward B. 0.79 ft. upwardB. 0.79 ft. upward C. 1.04 ft. upwardC. 1.04 ft. upward D. 1.83 ft. upwardD. 1.83 ft. upward

WEIGHTWEIGHT KGKG MOMENTMOMENT

19,70019,700 50.9650.96 1,003,9121,003,912

(+) 300(+) 300 120120 (+)36,000(+)36,000

20,00020,000 1,039,9121,039,912

NEW KG = MOMENT WT = 1,039,912 20,000 = 52.00OLD KG = 50.96 (-) 1.04 UPWARD

4. You have approximately 15 tons of fish on 4. You have approximately 15 tons of fish on deck. What will the shift in the center of gravity deck. What will the shift in the center of gravity after you shift the fish to the fish hold a vertical after you shift the fish to the fish hold a vertical distance of 8 feet? Total displacement 300 distance of 8 feet? Total displacement 300 tons.tons.


GG’ = GG’ = W X DW X D ΔΔ = = 15 X 815 X 8 300300 = 0.4 FT.= 0.4 FT.

5. Your ship of 12,000 tons displacement has a 5. Your ship of 12,000 tons displacement has a center of gravity of 21.5 ft. above the keel. You run center of gravity of 21.5 ft. above the keel. You run aground and estimate the weight aground is 2500 aground and estimate the weight aground is 2500 tons. The virtual rise in the center of gravity is:tons. The virtual rise in the center of gravity is:


OLD DISPL. = 12,000 X 21.5 = 258,000GROUNDED WT =(-) 2,500NEW DISPL. = 9,500 NEW KG = 258,000 9,500 NEW KG = 27.16 OLD KG = 21.50VIRTUAL RISE OF CG = 5.66 UPWARD

GG’ = W X D Δ = 2,500 X 21.5 9,500 = 5.66 UPWARD

• 1. Sixty tons of cargo are raised with a boom 1. Sixty tons of cargo are raised with a boom 45 feet from the centerline. The vessel’s 45 feet from the centerline. The vessel’s displacement including the weight lifted is displacement including the weight lifted is 16,400 tons.The angle of list caused by the 16,400 tons.The angle of list caused by the suspended weight is 1.5˚. KM is 28.75 ft. and suspended weight is 1.5˚. KM is 28.75 ft. and BM is 17.25 FT. What is the KG?BM is 17.25 FT. What is the KG?

• A. 11.65 ft.A. 11.65 ft.• B. 22.46 ft.B. 22.46 ft.• C. 23.15 ft.C. 23.15 ft.• D. 23.82 ft.D. 23.82 ft.

GM = W X D Δ TAN LIST = 60 X 45 16,400 TAN 1.5˚ = 6.29 FT KM = 28.75 FT.

GM = 6.29 FT.KG = 22.46 FT.

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


INCLINING EXPERIMENT:INCLINING EXPERIMENT:GM = GM = W X DW X D

ΔΔ Tan LIST Tan LIST = = 30 TONS X 30 FT30 TONS X 30 FT 4,000 TONS X Tan 2.05˚4,000 TONS X Tan 2.05˚GM = 6.29 FT.GM = 6.29 FT.

KM = 27.64 FT.KM = 27.64 FT.GM = GM = - 6.29 FT.- 6.29 FT.KG = 21.35 FT.KG = 21.35 FT.

28’

1’

SOHCAHTOA

SIN Ø = OPP HYPO = 1 28 = 2.05˚

WL1WL2

3. A cargo of 75 tons is to be lifted with a 3. A cargo of 75 tons is to be lifted with a boom located 50 ft. from the ship centerline. boom located 50 ft. from the ship centerline. The ship’s displacement including the The ship’s displacement including the suspended cargo is 6,000 tons and GM is 6 ft. suspended cargo is 6,000 tons and GM is 6 ft. The list of the ship with cargo suspended from The list of the ship with cargo suspended from the boom will be:the boom will be:

A. 5.00A. 5.00OO

B. 5.40B. 5.40OO

C. 5.94C. 5.94OO

D. 6.50D. 6.50OO

GM = W X D Δ TAN LIST

TAN LIST = W X D Δ X GM = 75 X 50 6,000 X 6 = 5.946˚

4. A cargo of 60 tons to be loaded on deck 20 4. A cargo of 60 tons to be loaded on deck 20 feet from the ship’s centerline. The ship’s feet from the ship’s centerline. The ship’s displacement including the suspended cargo displacement including the suspended cargo is 6,000 tons and GM is 2 ft. The list of the is 6,000 tons and GM is 2 ft. The list of the ship with the cargo suspended from the boom ship with the cargo suspended from the boom will be:will be:

A. 5.4A. 5.4OO

B. 5.7B. 5.7OO

C. 6.1C. 6.1OO

D. 6.4D. 6.4OO

TAN LIST = W X D Δ X GM = 60 X 20 6,000 X 2 = 5.7˚

5. To check stability, a weight of 40 tons is 5. To check stability, a weight of 40 tons is lifted with the jumbo boom, whose head is 40 lifted with the jumbo boom, whose head is 40 ft. from the ship’s centerline. The clinometer ft. from the ship’s centerline. The clinometer shows a list of 6.5shows a list of 6.5OO with the weight with the weight suspended. Displacement including the weight suspended. Displacement including the weight is 16,000 tons. The GM while in this condition is 16,000 tons. The GM while in this condition is:is:


GM = W X D Δ TAN LIST = 40 X 40 16,000 TAN 6.5˚ = 0.88 FT.

6. You are making a heavy lift with the jumbo boom. 6. You are making a heavy lift with the jumbo boom. Your vessel Your vessel displaces 18,000 tonsdisplaces 18,000 tons. The . The 50-tons50-tons weight is on the pier and its center is weight is on the pier and its center is 75 feet to 75 feet to starboard of the centerlinestarboard of the centerline. The head of the boom is . The head of the boom is 112 feet above the baseline and the 112 feet above the baseline and the center of gravitycenter of gravity of the lift of the lift when stowed on the deck will be 56 ft.when stowed on the deck will be 56 ft. above the baseline. As the jumbo takes the strain above the baseline. As the jumbo takes the strain the ship the ship lists 3.5lists 3.5OO. What is the . What is the GM when the cargo is GM when the cargo is stowedstowed??


GM = W X D Δ TAN LIST = 50 X 75 18050 TAN3.5˚ = 3.397

GG’ = W X D Δ = 50 X 56 18050 = 0.156 (+)3.397 3.553

7. A ship of 6,000 tonnes displacement has 7. A ship of 6,000 tonnes displacement has KM=7.3 meters and KG=6.7 meters and is KM=7.3 meters and KG=6.7 meters and is floating upright. A weight of 60 tonnes already floating upright. A weight of 60 tonnes already on board is shifted 12 meters transversely. on board is shifted 12 meters transversely. Find the resultant list.Find the resultant list.

A. 10° 00’ A. 10° 00’ B. 11°15’B. 11°15’ C. 9° 15’C. 9° 15’ D. 11°18’ 30”D. 11°18’ 30”

TAN LIST = W X D Δ X GM = 60 X 12 6,000 X .6 = 11. 309 11˚ 18’ 36”KM = 7.3

KG = 6.7 (-)GM = 0.6

FREE SURFACE CORRN.FREE SURFACE CORRN. FOR SALT WATER:FOR SALT WATER: F.S.CONST. = F.S.CONST. = l bl b33

420420FOR LIQUIDS OTHER THAN SALT WATER:F.S.CONST. = r l b3

420

l = LENGTH OF THE TANKb= BREADTH OF THE TANK

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 Δ432

• 1.On the vessel displacing 8,000 tons, what is 1.On the vessel displacing 8,000 tons, what is the reduction in metacentric height due to free the reduction in metacentric height due to free surface when a tank 45 ft. long and 45 ft. wide surface when a tank 45 ft. long and 45 ft. wide is partially filled with saltwater?is partially filled with saltwater?

• A. 1.22 ft.A. 1.22 ft.• B. 1.16 ft.B. 1.16 ft.• C. 1.13 ft.C. 1.13 ft.• D. 1.10 ft. D. 1.10 ft.

FSC = r l b3

420Δ = (1)(45)(45)3

(420)(8,000) = 1.22 ft.

2. On a vessel of 10,000 tons displacement, 2. On a vessel of 10,000 tons displacement, compute the reduction in metacentric height compute the reduction in metacentric height due to free surface in a hold having fresh due to free surface in a hold having fresh water on the tank top. The hold is 40 ft. long water on the tank top. The hold is 40 ft. long and 50 ft. wide. The reduction in GM is:and 50 ft. wide. The reduction in GM is:

A. 1.1 ft.A. 1.1 ft.B. 1.2 ft.B. 1.2 ft.C. 1.3 ft.C. 1.3 ft.D. 1.5 ft.D. 1.5 ft.

FSC = r l b3

420Δ

3. What is the reduction in GM due to free 3. What is the reduction in GM due to free surface when a tank 60 ft. long and 60 ft. wide surface when a tank 60 ft. long and 60 ft. wide is partially filled with salt water and is fitted is partially filled with salt water and is fitted with a with a centerline bulkheadcenterline bulkhead? The vessel has a ? The vessel has a displacement of 10,000 tons.displacement of 10,000 tons.

A. 0.5 ft.A. 0.5 ft. B. 0.8 ft.B. 0.8 ft. C. 1.0 ft.C. 1.0 ft. D. 1.2 ft. D. 1.2 ft.

FSC = r l b3 X .25 420Δ

FSC IS REDUCEDBY 75%.

4. A shaft alley divides a vessel’s cargo hold 4. A shaft alley divides a vessel’s cargo hold into two tanks, each 25 ft. wide by 50 ft. long. into two tanks, each 25 ft. wide by 50 ft. long. Each tank is filled with sea water below the Each tank is filled with sea water below the level of the shaft alley. The vessel’s level of the shaft alley. The vessel’s displacement is 6,000 tons. What is the displacement is 6,000 tons. What is the reduction in Gm due to free surface effect?reduction in Gm due to free surface effect?

A. 0.56 ft.A. 0.56 ft.B. 0.58 ft.B. 0.58 ft.C. 0.62 ft.C. 0.62 ft.D. 0.66 ft.D. 0.66 ft.

FSC = r l b3 X 2 420Δ

5. A 7,000 ton displacement tankship carries 5. A 7,000 ton displacement tankship carries two slack tankstwo slack tanks of alcohol with a S.G. of 0.8 of alcohol with a S.G. of 0.8 Each tank is 50 long and 30 ft. wide. What is Each tank is 50 long and 30 ft. wide. What is the reduction in GM due to free surface with the reduction in GM due to free surface with the vessel floating in sea water (SG 1.026)?the vessel floating in sea water (SG 1.026)?


FSC = r l b3 X 2 420Δ

6. Determine the free surface constant for fuel oil tank 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 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 the fuel oil is 0.85 and the ship will float in salt water (SG 1.026)(SG 1.026)

A. 0.83A. 0.83 B. 42.7B. 42.7 C. 3787C. 3787 D. 4571D. 4571 7. Determine the free surface correction for fuel oil 7. Determine the free surface correction for fuel oil

tank 30 ft. long by 40 ft. wide by 15 ft. deep with free tank 30 ft. long by 40 ft. wide by 15 ft. deep with free surface constant of 3794. The vessel is displacing surface constant of 3794. The vessel is displacing 7,000 tons in saltwater.7,000 tons in saltwater.


F.S.Const. = r l b3

420

F.S.Corrn. = F.S.Const Δ

8. A ship of 8,153.75 tonnes displacement has 8. A ship of 8,153.75 tonnes displacement has KM=8 m, KG=7.5 and a double bottom tank 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 15 m x 10 m x 2 m which is full of salt water ballast. Find the new GM if this tank is now ballast. Find the new GM if this tank is now pumped out until half empty.pumped out until half empty.

A. 0.225 mA. 0.225 m B. 0.825 mB. 0.825 m C. 1.75 mC. 1.75 m D. 1.45 mD. 1.45 m

CHS - VIRTUAL RISE GG1.pptCHS - VIRTUAL RISE GG1.ppt

GG1 = W X D Δ = 153.75 X 6 8,000 GG1 = 0.115 M

INITIAL Δ = 8,153.75 DISCH.= (-) 153.75 FINAL Δ = 8,000.00

INITIAL KG = 7.5RISE OF g = 1.5 = 6.0

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

WTWT CGCG MOMMOM

8,153.758,153.75 7.57.5 61,15361,153

(-)153.75(-)153.75 1.51.5 (-) 231(-) 231

8,000.008,000.00 60,92260,922

NKG = T.MOM. = 60,922 T.WT. 8,000NKG = 7.615OKG = 7.500 GG1 = 0.115

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 LOSSOF 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 = 1G1GV = i VFOR RECTANGULARWATERPLANE: i = l b3

12

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

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

• What is the minimum freeboard required at the What is the minimum freeboard required at the start of the voyage?start of the voyage?

• A. 65 inchesA. 65 inches• B. 69 inchesB. 69 inches• C. 72 inchesC. 72 inches• D. 75 inchesD. 75 inches CONSUMPTION = 40 TONS/DAY

STEAMING TIME = (X) 4 DAYSCONS.FOR 4 DAYS = 160 TONS

CONS. = 160 TPI 53 = 3”

SUMMER F.B. = 72” (-) 3” MIN. F.B.ON DEP = 69”

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

• Freeboard from DecklineFreeboard from Deckline Load LineLoad Line• Tropical Tropical 69”69” (T)(T) 7” above 7” above

(s)(s)• Summer Summer 76”76” (S)(S) * *• WinterWinter 83”83” (W) 7” below (W) 7” below

(s)(s)• Allowance for fresh water all freeboards 6”Allowance for fresh water all freeboards 6”• *Upper edge of line at level of center ring.*Upper edge of line at level of center ring.• What is the minimum freeboard at the start of the What is the minimum freeboard at the start of the

voyage?voyage?• A. 62 inchesA. 62 inches• B. 66 inchesB. 66 inches• C. 70 inchesC. 70 inches• D. 74 inchesD. 74 inches

CONSUMPTION = 28 TONS/DAY STEAMING TIME = (X) 2 DAYSCONS.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”

TROPICAL

SUMMER

69”(T)69”(T)

76”(S)76”(S)83”(W)83”(W)

TFF T

SWWNA

FWA(6”)

DEP.

ARR.

2 DAYS

3. You are loading in a port subject to the summer 3. You are loading in a port subject to the summer loadline mark and bound for a port subject to the loadline mark and bound for a port subject to the winter loadline mark. You will enter the winter zone winter loadline mark. You will enter the winter zone after steaming four days. You will consume 35 tons of after steaming four days. You will consume 35 tons of fuel, water and stores per day. The hydrometer fuel, water and stores per day. The hydrometer reading at the loading pier is 1.0083 and the average reading at the loading pier is 1.0083 and the average TPI is 65. The following data is extracted from the TPI is 65. The following data is extracted from the loadline Certificate:loadline Certificate:

Freeboard from DecklineFreeboard from Deckline Load LineLoad Line TropicalTropical 68”68” (T)(T) 6” above (s) 6” above (s) SummerSummer 74”74” (S)(S) * * WinterWinter 80”80” (W)(W) 6” below (S) 6” below (S) Allowance for fresh water all freeboards 5”Allowance for fresh water all freeboards 5” *Upper edge of line at level of center ring.*Upper edge of line at level of center ring.

What is the minimum freeboard required at he What is the minimum freeboard required at he start of the voyage?start of the voyage?

• A. 74 inchesA. 74 inches• B. 78 inchesB. 78 inches• C. 80 inchesC. 80 inches• D. 86 inchesD. 86 inches

A.I. = FWA X CH.OF DENS. 25 = 5 X 16.7 25 = 3.34”


CONS. = 140 TPI = 65 = 2.15”

BURN OFF/CONS = 2.15” A.I. = 3.34” 5.49” WINTER F.B. = 80.00” DEP. F.B. = 74.51”

4. You are loading in a port subject to the tropical 4. You are loading in a port subject to the tropical loadline and bound for a port subject to the winter loadline and bound for a port subject to the winter loadline mark. You will enter the summer zone after loadline mark. You will enter the summer zone after steaming eight days and you will enter the winter zone steaming eight days and you will enter the winter zone after ten days. You will consume 31 tons of fuel, water after ten days. You will consume 31 tons of fuel, water and stores per day. The hydrometer reading at the and stores per day. The hydrometer reading at the loading pier is 1.016 and the average TPI is 41. The loading pier is 1.016 and the average TPI is 41. The following data is extracted from the loadline certificate:following data is extracted from the loadline certificate:

Freeboard from decklineFreeboard from deckline Loadline Loadline Tropical Tropical 42”42” (T)(T) 6” above (s) 6” above (s) Summer Summer 48”48” (S)(S) * * WinterWinter 54”54” (W)(W) 6” below (s) 6” below (s) Allowance for fresh water all freeboards 5”Allowance for fresh water all freeboards 5” *Upper edge of the line at the level of center ring.*Upper edge of the line at the level of center ring.

What is the minimum freeboard required at the start of What is the minimum freeboard required at the start of the voyage?the voyage?

• A. 55.0 inchesA. 55.0 inches• B. 49.5 inchesB. 49.5 inches• C. 44.5 inchesC. 44.5 inches• D. 41.0 inchesD. 41.0 inches


CONS. = 310 TPI 41 = 7.56”

CONS/BURN OFF = 7.56” A.I. = 1.80” 9.36” WINTER F.B. = 54.00” F.B.ON DEP. = 44.64”

A.I. = FWA X CH.OF DENS. 25 = 5 x 9 25 = 1.80”

CONSUMPTION = 31 TONS/DAY STEAMING TIME = (X) 8 DAYSCONS.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”

5. You are loading in a port subject to the tropical 5. You are loading in a port subject to the tropical loadline mark and bound for a port subject to the winter loadline mark and bound for a port subject to the winter loadline mark. You will enter the summer zone after loadline mark. You will enter the summer zone after steaming four days and you will enter the winter zone steaming four days and you will enter the winter zone after nine days. You will consume 29 tons of fuel, water after nine days. You will consume 29 tons of fuel, water and stores per day. The hydrometer reading at the and stores per day. The hydrometer reading at the loading pier is 1.008 and the average TPI is 53. The loading pier is 1.008 and the average TPI is 53. The following data is extracted from the Load Line following data is extracted from the Load Line Certificate”.Certificate”.

Freeboard from DecklineFreeboard from Deckline Load Line Load Line Tropical Tropical 75”75” (T)(T) 8” above 8” above

(s)(s) SummerSummer 83”83” (S)(S) * * Winter Winter 91”91” (W)(W) 8” below 8” below

(S)(S) Allowance for fresh water all freeboards 9”Allowance for fresh water all freeboards 9” *Upper edge of line at the level center ring.*Upper edge of line at the level center ring.

5. What is the minimum freeboard required at the start 5. What is the minimum freeboard required at the start of the voyageof the voyage

• A. 72.5 inchesA. 72.5 inches• B. 75.0 inchesB. 75.0 inches• C. 77.0 inchesC. 77.0 inches• D. 80.0 inchesD. 80.0 inches CONSUMPTION = 29 TONS/DAY STEAMING TIME = (X) 4 DAYSCONS.FOR 4 DAYS = 116 TONS

A.I. = FWA X CH.0F DENS. 25 = 9 X 17 25 = 6.12”

A.I. = 6.12” CONS/BURN OFF = 2.19” 8.31” SUMMER F.B = 83.00”DEP. F.B./ SUMMER ZONE = 74.7”

CONS. = 116 TPI 53 = 2.19”

CONS. = 261 TPI 53 = 4.92”

A.I. = 6.12”CONS/BURN OFF = 4.92” 11.04” WINTER F.B. = 91.00” DEP.F.B. = 79.96”


ZONE ALLOWANCEZONE ALLOWANCE• 1. A tanker loads at a terminal within the 1. A tanker loads at a terminal within the

tropical zone. She will enter the summer zone tropical zone. She will enter the summer zone six days after departing the loading port. She six days after departing the loading port. She will burn off 45 tons/day and daily water will burn off 45 tons/day and daily water consumption is eight tons. How many tons consumption is eight tons. How many tons may she load above her summer loadline?may she load above her summer loadline?

• A. 270 tonsA. 270 tons• B. 278 tonsB. 278 tons• C. 291 tonsC. 291 tons• D. 318 tonsD. 318 tons

BURN OFF = 45 TONS WATER CONS.= 8 TONS 53 TONS/DAYSTEAMING TIME = (X) 6 DAYS 318 TONS

2. A vessel’s tropical loadline is 6 inches above 2. A vessel’s tropical loadline is 6 inches above her summer loadline. Her TPI is 127 tons. She her summer loadline. Her TPI is 127 tons. She will arrive in the summer zone 8 days after will arrive in the summer zone 8 days after departure. She will burn off about 47 tons/day departure. She will burn off about 47 tons/day and water consumption is 12 tons/day.How and water consumption is 12 tons/day.How many tons may she load above her summer many tons may she load above her summer loadline if she loads in the tropical zone?loadline if she loads in the tropical zone?

• A. 376 tonsA. 376 tons• B. 472 tonsB. 472 tons• C. 762 tonsC. 762 tons• D. 1016 tonsD. 1016 tons

BURN OFF = 47 TONS WATER CONS = 12 TONS 59 TONS/DAYSTEAMING TIME = (X) 8 DAYS 472 TONS

THE LOAD LINE MARKTHE LOAD LINE MARK

300 mm

450 mm 18 in.

300 mm 12 in.

Deck Line

WNA

W

S

TF

TF

L R230 mm

540 mm 21 in. 230 mm

9 in.

25 mm 1 in.

TROPICAL

SUMMER

67”(T)67”(T)72”(S)72”(S)

77”(W)77”(W)

1. you are on a vessel that has metacentric height of 4 feet and a beam of 50 feet. what can you...

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