• 1. A ship displaces 8,800 tons at a
summer draught of 7.28m. Its Fresh
Water Allowance is 315 mms. Find its
change of draught when proceeding to a
port of UNIT DENSITY.
Answer = 31.5 cms
• 1. Solution:
• A.I. = FWA x ( Change in Density )
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25
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= 315 mms. x ( 1,025 – 1,000 )
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25
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= 315 mms. x 25
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25
• A. I. = 315 mms. or 31.5 cms.
• 2. A box has a dimension of 2ft x 2.5ft x
3ft and weighs 95 lbs. What is its
stowage factor?
• Answer = 353.7
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2. Solution:
Volume = 2 ft. x 2.5 ft. x 3 ft.
Volume = 15 ft³
Stowage Factor =
Volume
Weight / 2240 lbs.
=
15 ft³
95 lbs/ 2240 lbs
= 15 ft³ / 0.0424 lbs
Stowage Factor = 353.7
• 3. You are going to load bales of abaca
with SF 65 and lead with SF 18. The
Remaining space is 257,000 cu ft. and the
total weight to be loaded is 5,400 tons.
How much of each cargo should be
loaded to make the vessel FULL AND
DOWN?
• Answer = 2,000 t lead / 3,400 t abaca
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3. Solution:
WLF = Weight of the cargo having the large SF
WLF = Volume – ( Wt. of Cargo x Small SF )
Difference in SF
= 257,000 ft³ - ( 5,400t x 18 )
65 – 18
= 257,000 – 97,200
47
= 159,800 / 47
WLF = 3,400 t ( Abaca )
Wt. of Lead = 5,400 t – 2,400 t = 2,000 t
• 4. A vessel with a light displacement of
6,500t loaded cargoes of 1,300t, 1,400t
and 1,200t with a KG of 6.9m, 6.4m and
8.4m respectively. After loading, the new
KG was then found to be 7.07 m. What
was the KG prior to loading?
• Answer = 7.00 m
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WT
LOADED
=
1,300
LOADED
=
1,400
LOADED
=
1,200
INIT.DISPL = (+) 6,500
TOTAL DISPL= 10,400
KG = MOMENT
WEIGHT
= 45,518
6,500
OLD KG = 7.00 m
DIST
x 6.9
x 6.4
x 8.4
MOMENT
= 8,970
= 8,960
= 10,080
x 7.07 =73,528
(-) 8,970
(-) 8,960
(-)10,080
45,518
• 5. A bulk carrier has a displacement of
40,000 tons and the KG is 9 m .The KN at
10 degrees heel is 1.87 m. What is the
righting arm at this heel?
• Answer = 0.307 m
• 5. Solution:
• GZ = KN – ( KG x Sin θ )
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= 1.87 m – ( 9 m x Sin 10º )
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= 1.87 – 1.5628
• GZ = 0.307 m
• 6. A bulk carrier has a displacement of
25,000 tons. KG is 8.6 m. At an angle of
heel of 15 degrees, KN is 2.98. Solve for
the righting moment.
• Answer = 18,854 mtr-tons
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6. Solution:
GZ = KN – ( KG x Sin θ )
2.98 – ( 8.6 x Sin 15º )
2.98 – 2.2258
GZ = 0.754 m
MSS = W x GZ
= 25,000 x 0.754
MSS = 18,854 ton - meter
• 7. While inspecting the log pond, you
noticed that the average length of the logs
to be loaded is 15 m and the average
diameter is 70 cms. While floating in pond
of density 1,012 kgs./m³, 90 percent of
their volume is submerged. What is the
average weight of each log?
• Answer = 5.26 tons
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7.Solution;
Vol. of Cylinder = π x r² x h
= 3.1416 x 0.35²m x 15m
= 5.77m³
x 0.9 (90%)
Volume
= 5.2m³
Weight = Volume x Density
= 5.2m³ x 1.012ton/m³
Weight = 5.26 tons
• 8. A box–shaped vessel with 12m beam is
floating upright at a draught of 6.7m even
keel. Find the draft if the vessel is now
listed 18 degrees.
• Answer = 8.226 mtrs
• 8. Solution:
New Draft = ½ x B x Sin List + ODr. x Cos
List
= ½ x 12m x Sin 18º + 6.7m x Cos 18º
= 1.854 + 6.372
New Draft = 8.226 m
• 9. On completion of loading, a vessel of
8,800 tons displacement and a GM of
0.45 m was listing 3 degrees to stbd. How
many tons of ballast should be pumped
into the port wing tank, 6 m from the
centerline to make the vessel upright?
• Answer = 34.6 tons
• 9. Solution:
• Tan List = Wt x Dist
∆ x GM
• Weight = Tan List x ∆ x GM
Distance
• Weight = Tan 3º x 8,800 tons x 0.45m
6m
• Weight = 34.6 tons
• 10. The speed made good was 16 knots
and the engine speed 15.4 knots. What is
the percentage of slip?
• Answer = - 3.9%
• 10. Solution:
• % Slip = Eng. Spd – Obs. Spd
X 100
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Eng. Spd
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= 15.4 – 16.0 X 100
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15.4
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- 0.6 X 100
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15.4
• % Slip = - 3.9 %
• 11. A cylindrical drum, 1.4m high has a
diameter of 80 cms. It contains oil of
density 0.78 ton/m³ Find the weight of oil
if filled to capacity.
• Answer = 549 kgs
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11. Solution:
Volume = π x r² x l
= 3.1416 x 0.40² x 1.4m
Volume = 0.70 m³
Weight = Volume x Density
= 0.70 m³ x 0.78 ton/m³
Weight = 0.5489 ton x 1,000 kgs.
Weight = 549 kgs.
• 12. If the circumference of a mooring
rope is 250 mms, what is its diameter?
• Answer = 80 mms
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12. Solution:
Circumference = 3.1416 x Diameter
Diameter = C / 3.1416
= 250 mms / 3.1416
Diameter = 79.58 mms
• 13. A log is floating in a pond of relative
density 1.008. Its length is 12 m and its
diameter is 50 cms. If 95% of the log is
submerged, what is its weight?
• Answer = 2.26 tons
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13. Solution:
Vol. of Cylinder = π x r² x h
= 3.1416 x 0.25²m x 12m
Volume
= 2.36m³
x 0.95 (95%)
Volume
= 2.24m³
Weight = Volume x Density
= 2.24m³ x 1.008m³/ton
Weight = 2.26 tons
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14. On arrival at the discharging port,
the displacement was 7,800 t. After
Discharging 3,200 t of cargo with an
average KG of 5.8 m the new KG was
found to be 6.14 m. What was the
vessel’s KG prior to discharge?
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Answer = 6.00 m
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WT DIST MOMENT
7800 – 3,200 = 4600
DISCH
= 3,200 x 5.8 = 18,560
FINAL DISPL= 4,600 x 6.14 = 28,244 (+)
INITIAL DISPL = 7,800
46,804
KG = MOMENT
WEIGHT
= 46,804
7,800
OLD KG = 6.00 m
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15. A box-shaped vessel with 14 m
beam is floating upright at a draught of
5.4 m even keel. Find the draught if the
vessel is now listed 15º.
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Answer = 7.028 m
• 15. Solution:
New Draft = ½ x B x Sin List + ODr. x Cos List
= ½ x 14m x Sin 15º + 5.4m x Cos 15º
= 1.8133 + 5.22
New Draft = 7.028 m