•
•
•
•
•
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•
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
RP = .44 B
√GM
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.
2
GM
=
.44
B
C. 1.4 ft.
RP
D. 1.6 ft.
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
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.
GM = .44 B 2
RP
C. 2.60 ft.
D. 3.36 ft.


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
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
4. Your vessel has a measures 35.0 meters long by
5.0 meters in 6 seconds. What is the GM?
A. 0.44
GM = .797 B 2
B. 0.54
RP
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
RP = .797 B
C. 4 seconds
√GM
D. 5 seconds
•
•
•
•
•
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.
WEIGHT
C. G
MOMENT
B. 17.8 ft.
7400
21.5
159100
C. 18.7 ft.
(+) 450
17.0
7650
(+) 220
11.6
2552
D. 20.9 ft.
NEW KG = MOMENT
WT
= 169757
8135
= 20.867 FT.
(+) 65
8135
7.0
455
169757
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



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.
WEIGHT
KG
MOMENT
B. 62.09 ft.
15,000
60
900,000
C. 60.09 ft.
(+) 100
75
(+) 7,500
D. 61.09 ft.
15,100
907,500
NEW KG = MOMENT
WT
= 907,500
15,100
= 60.099 FT.
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

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?
WEIGHT
KG
MOMENT
A. 1.00 ft. upward
19,700
50.96
1,003,912
B. 0.79 ft. upward
C. 1.04 ft. upward
(+) 300
120
(+)36,000
D. 1.83 ft. upward
20,000
1,039,912
NEW KG = MOMENT
WT
= 1,039,912
20,000
= 52.00
OLD KG = 50.96 (-)
1.04 UPWARD
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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.
GG’ = W X D
A. 0.4 ft.
Δ
B. 0.1 ft.
= 15 X 8
C. 0.2 ft.
300
D. 0.3 ft.
= 0.4 FT.
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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.
OLD DISPL. = 12,000 X 21.5 = 258,000
GROUNDED WT =(-) 2,500
B. 5.66 ft.
NEW DISPL. = 9,500
C. 3.80 ft.
D. 4.80 ft.
NEW KG = 258,000
9,500
NEW KG = 27.16
OLD KG = 21.50
VIRTUAL 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
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?
GM = W X D
A. 11.65 ft.
Δ TAN LIST
B. 22.46 ft.
=
60 X 45
C. 23.15 ft.
16,400 TAN 1.5˚
= 6.29 FT
D. 23.82 ft.
KM = 28.75 FT.
GM = 6.29 FT.
KG = 22.46 FT.
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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.
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
WL2
SIN Ø = OPP
HYPO
= 1
28
= 2.05˚
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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:
GM = W X D
A. 5.00O
Δ TAN LIST
O
B. 5.40
C. 5.94O
TAN LIST = W X D
D. 6.50O
Δ X GM
= 75 X 50
6,000 X 6
= 5.946˚
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
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
TAN LIST = W X D
Δ X GM
B. 5.7O
=
60
X
20
O
C. 6.1
6,000 X 2
O
D. 6.4
= 5.7˚
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
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.
GM = W X D
B. 0.43 ft.
Δ TAN LIST
=
40 X 40
C. 0.88 ft.
16,000 TAN 6.5˚
D. 1.02 ft.
= 0.88 FT.
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
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?
GG’ = W X D
A. 3.19 ft.
Δ
GM =
WXD
= 50 X 56
B. 3.40 ft.
Δ TAN LIST
18050
C. 3.56 ft.
=
50 X 75
=
0.156
D. 3.24 ft.
18050 TAN3.5˚
(+)3.397
= 3.397
3.553
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
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’
TAN LIST = W X D
Δ X GM
B. 11°15’
= 60 X 12
C. 9° 15’
6,000 X .6
D. 11°18’ 30”
= 11. 309
KM = 7.3
11˚ 18’ 36”
KG = 6.7 (-)
GM = 0.6
FREE SURFACE CORRN.
FOR SALT WATER:
l
=
LENGTH
OF
THE
TANK
 F.S.CONST. = l b3
b= BREADTH OF THE TANK
420
FOR LIQUIDS OTHER THAN SALT WATER:
F.S.CONST. = r l b3
r = S.G OF LIQUID IN TANK
420
S.G WHERE VSL. FLOATS

F.S.CORRN. = F.S CONSTANT
Δ
F.S.CORRN. = r l b3
420 Δ
432
•
•
•
•
•
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.
FSC = r l b3
C. 1.13 ft.
420Δ
D. 1.10 ft.
= (1)(45)(45)3
(420)(8,000)
= 1.22 ft.
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.
FSC = r l b3
 C. 1.3 ft.
420Δ
 D. 1.5 ft.


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
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.
FSC IS REDUCED
BY 75%.
D. 1.2 ft.
FSC = r l b3 X .25
420Δ
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.
3 X 2
FSC
=
r
l
b
 C. 0.62 ft.
420Δ
 D. 0.66 ft.

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
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.
FSC = r l b3 X 2
B. 0.46 ft.
420Δ
C. 0.72 ft.
D. 0.82 ft.
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
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
F.S.Const. = r l b3
C. 3787
420
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.
F.S.Corrn. = F.S.Const
C. 0.65 ft.
Δ
D. 1.38 ft.
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

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
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
INITIAL KG = 7.5
= 153.75 TONS
RISE OF g = 1.5
= 6.0
WT
CG
MOM
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
8,153.75 7.5
61,153
(-)153.75 1.5
8,000.00
(-) 231
60,922
NKG = T.MOM. = 60,922
T.WT.
8,000
NKG = 7.615
OKG = 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
BUT : d1 = d2
n = 1
G1GV = i
V
FOR RECTANGULAR
WATERPLANE:
i = l b3
12
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.
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
•
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.
•
•
•
•
•
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”
•
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:
•
•
•
•
•
•
•
•
•
•
•
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
CONSUMPTION = 28 TONS/DAY
STEAMING TIME = (X) 2 DAYS
CONS.FOR 2 DAYS = 56 TONS
CONS.
TPI
=
=
=
56
55
1.0”
A.I = FWA X CH.OF DENS
25
=6X5
BURN OFF/CONS.= 1.0”
25
A.I. = 1.2”
= 1.2”
2.2”
SUMMER F.B.= 76.0”
F.B.ON DEP.= 73.8”
TF
F
69”(T)
76”(S)
83”(W)
T
FWA(6”)
S
W
WNA
SUMMER
TROPICAL
DEP.
ARR.







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.

•
•
•
•
What is the minimum freeboard required at he
start of the voyage?
A.I. = FWA X CH.OF DENS.
A. 74 inches
25
B. 78 inches
= 5 X 16.7
C. 80 inches
25
= 3.34”
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”
CONS. = 140
5.49”
TPI
= 65
WINTER F.B. = 80.00”
= 2.15”
DEP. F.B. = 74.51”







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.

•
•
•
•
What is the minimum freeboard required at the start of
the voyage?
CONSUMPTION
= 31 TONS/DAY
STEAMING TIME = (X) 10 DAYS
A. 55.0 inches
CONS.FOR 10 DAYS = 310 TONS
B. 49.5 inches
CONS. = 310
C. 44.5 inches
TPI
41
D. 41.0 inches
= 7.56”
A.I. = FWA X CH.OF DENS.
CONSUMPTION = 31 TONS/DAY
25
STEAMING TIME = (X) 8 DAYS
= 5x9
CONS.FOR 8 DAYS = 248 TONS
25
41(TPI) = 6.05”
= 1.80”
CONS./BURN OFF = 6.05”
CONS/BURN OFF = 7.56”
A.I. = 1.80”
A.I. = 1.80”
7.85”
9.36”
SUMMER F.B. = 48.00”
WINTER F.B. = 54.00”
F.B. FOR SUMMER= 40.15”
F.B.ON DEP. = 44.64”







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.

•
•
•
•
5. What is the minimum freeboard required at the start
CONSUMPTION = 29 TONS/DAY
of the voyage
STEAMING TIME = (X) 9 DAYS
A. 72.5 inches
CONS.FOR 9 DAYS = 261 TONS
B. 75.0 inches
CONS. = 261
C. 77.0 inches
TPI
53
= 4.92”
D. 80.0 inches
CONSUMPTION = 29 TONS/DAY A.I. = FWA X CH.0F DENS.
STEAMING TIME = (X) 4 DAYS
25
CONS.FOR 4 DAYS = 116 TONS
= 9 X 17
25
CONS. = 116
= 6.12”
TPI
53
A.I. = 6.12”
= 2.19”
A.I. = 6.12” CONS/BURN OFF = 4.92”
CONS/BURN OFF = 2.19”
11.04”
8.31”
WINTER F.B. = 91.00”
SUMMER F.B = 83.00”
DEP.F.B. = 79.96”
DEP. F.B./ SUMMER ZONE = 74.7”
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
BURN OFF = 45 TONS
B. 278 tons
WATER CONS.= 8 TONS
C. 291 tons
53 TONS/DAY
D. 318 tons STEAMING TIME = (X) 6 DAYS
318 TONS

•
•
•
•
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
BURN OFF = 47 TONS
WATER CONS = 12 TONS
C. 762 tons
59 TONS/DAY
D. 1016 tons
STEAMING TIME = (X) 8 DAYS
472 TONS
THE LOAD LINE MARK
Deck Line
300 mm
12 in.
25 mm
1 in.
540 mm
21 in. TF
230 mm
9 in.
F
T
L
R
230 mm
S
W
WNA
300 mm
450 mm
18 in.
67”(T)
72”(S)
77”(W)
SUMMER
TROPICAL