DRAFT AND STABILITY PROBLEMS AND ANSWERS QUESTION 1 : AT A DRAFT OF 3,25 M, AN ADDITIONAL 25 TONNES OF FRESH WATER IS TAKEN ABOARD. FIND NEW DRAFT. ANSWER : FIND APPROPRIATE VALUE TPC FOR DRAFT = 3,25M

Draft and stability problems and answers

DRAFT AND STABILITY PROBLEMS AND ANSWERS

QUESTION 1 :

AT A DRAFT OF 3,25 M, AN ADDITIONAL 25 TONNES OF FRESH WATER IS TAKEN ABOARD.
FIND NEW DRAFT.

ANSWER :

FIND APPROPRIATE VALUE TPC FOR DRAFT = 3,25M. IT'S 8,61 TONNES. SINKAGE EQUALS TONNAGE ADDED THEN DIVIDED BY TPC. NEW DRAFT EQUALS OLD DRAFT PLUS SINKAGE.

SINKAGE = LOADED WEIGHT = 25,00 = 2,90 CM = 0,290 M
T P C 8,61

NEW DRAFT = OLD DRAFT + SINKAGE = 3,25 + 0,29 = 3,54 M


QUESTION 2 :

A SHIP OF 1000 TONNES DISPLACEMENT IS FLOATING IN SEA WATER. WHAT WILL THE
CHANGE OF DRAFT BE WHEN SHE SAILS INTO RIVER WATER ? T P C AT LOAD LINE IS
23,5 TONNES.

ANSWER :

1. APPARENT CHANGE IN TONNAGE = Sg FW = 1,00 = 0,98 OR 98 %
Sg SW 1,03

2. DISPL. -- ( DISPL. x Sg FW ) = 1000 -- ( 1000 x 0,98 ) = 20
Sg SW

3. INCREASE IN DRAFT = 2. = 20 = 0,85 CM ( + )
T P C 23,5


QUESTION 3 :

A SHIP HAS A FRESH WATER ALLOWANCE OF 2 CM AND THE HARBOUR SPECIFIC GRAVITY
IS 1,010.
HOW MUCH CAN THE LOADLINE BE SUBMERGED BEFORE PROCEEDING TO SEA ?

ANSWER :

VALUE OF SUBMERGE = ( 1,025 -- MEASURED Sg ) * FWA =
( 1,025 -- Sg FRESH WATER )

= ( 1,025 -- 1,010 ) * 2 = 0,015 * 2 = 0,03 = 1,3 CM
( 1,025 -- 1,000 ) 0,025 0,025


QUESTION 4 :

A SHIP IS 65 M LONG; 10 M IN BREADTH AND HAS A DRAFT OF 5 M TO THE TOP OF THE
WEATHER DECK MIDSHIP. THE FREEBOARD MIDSHIP IS 0,8 M AND THE COEFFICIENT OF DISPLACEMENT IS 0,65. WHAT IS HER DISPLACEMENT IN SALT WATER ?

ANSWER :

1. DRAFT -- FREEBOARD = " XXX " = 5 M -- 0,8 M = 4,2 M

2. DISPL. IN SALT WATER = " XXX " * L * B * COEFF. OF DISPL. = 4,2 * 65 * 10 * 0,65
1,025 1,025
= 1731,219 MT


QUESTION 5 :

A SHIP DISPLACED 6342 cu.m WHEN Sg IS MEASURED AT 1,018. DISPLACEMENT CHANGES TO 6358 cu.m . WHAT IS THE NEW Sg ?

ANSWER :

NEW Sg = OLD DISPL. * MEASURED Sg = 6342 * 1,018 = 1,015
NEW DISPL. 6358


QUESTION 6 :

A SHIP IS 65 M LONG; 10 M IN BREADTH AND IS FLOATING IN SALT WATER AT DRAFT OF
5,48 M. COEFFICIENT OF FINESSE ( SOMETIMES REFFERED TO AS A BUCK COEFFICIENT ) OF
HER WATERPLANE AT THAT DRAFT IS 0,758.
a ) FIND HER TPC
b ) WHAT WILL THE DRAFT BE IN Sg. OF 1,018 ?

ANSWER :

a ) TPC = L * B * COEFF.OF FINESSE = AREA = 138 *17 *0,758 = 17,78
100 100 100
b ) NEW DRAFT = ( OLD DRAFT * OLD Sg ) = 5,48 * 1,025 = 5,52 M
NEW Sg 1.018


QUESTION 7 :

A SHIP FLOATS AT A DRAFT OF 6,88 FWD AND 6,93 AFT. HER MTC IS 105 TM AND THE
CENTRE OF FLOTATION IS ON THE LONDITUDINAL CENTRE LINE. FIND THE CHANGE IN
TRIM AND THE NEW DRAFT IF A WEIGHT OF 40 TONNES IS SHIFTED 60 M FORWARD.

ANSWER :

1. CHANGE OF TRIM = MOMENT TO CHANGE TRIM = 40 * 60 = 22,85 CM
MTC 105,00

2. 22,85 = 11,425 CM = 0,114 M
2,00

3. FWD DRAFT = 6,88 + 0,114 = 6,994 M
AFT DRAFT = 6,93 - 0,114 = 6,816 M


QUESTION 8 :

A SHIP HAS SUMMER DRAFT OF 8,15 M AND FRESH WATER ALLOWANCE OF 17 CM. TO
WHAT DRAFT MAY SHE LOAD IN DOCK WATER OF A DENSITY OF 1,007 ?

ANSWER :

1. ( 1,025 -- Sg OF DOCK WATER ) * FWA = ( 1,025 -- 1,007 ) * 17 = 0,018 *17 = 12,24 CM
( 1,025 -- Sg FRESH WATER ) 0,025 = 0,1214 M

2. ALLOWABLE DRAFT AT DOCK WATER = SHIP'S DRAFT + 1. = 8,15 + 0,1224 = 8,2724 M


QUESTION 9 :

A SHIP HAS DISPLACEMENT OF 2000 TONNES. FIND THE CHANGE OF HER CENTRE OF
GRAVITY IF A WEGHT OF 100 TONNES SHIFTED ACROSS HER HOLD 12 METRES ?

ANSWER :

1. WEIGHT * SHIFTING = 100 * 12 = 0,60 M
DISPLACEMENT 2000


QUESTION 10 :

A WEIGHT OF 550 TONNES IS LOADED ON A SHIP. THE CG OF THE WEIGHT IS LOCATED
3 M FROM THAT OF THE SHIP. IF THE SHIP'S ORIGINAL DISPLACEMENT WAS 3250 TONNES,
WHAT WILL BE THE CHANGE OF THE CG ?

ANSWER :

550 * 3 = 1650 = 0,434 M
3250 + 550 3800


QUESTION 11 :

A SHIP AND HER CARGO DISPLACE 8400 TONNES. IF A WEIGHT OF 120 TONNES IS REMOVED
FROM A POINT 35 M FROM THE ORIGINAL CG, WHAT WILL BE THE CHANGE OF THE CG ?

ANSWER :

120 * 35 = 4200 = 0,51 M
8400 -- 120 8280


QUESTION 12 :

A WEIGHT OF 35 TONNES IS LOADED INTO A VESSEL. THE NEW DICPLACEMENT IS 2550 T.
THE OLD KG WAS 4,52 M.
WHAT IS THE NEW KG UNDER THE FOLLOWING CONDITIONS ?
( a ) LOADED 6,2 M ABOVE THE CG ?
( b ) LOADED 6,2 M BELOW THE CG ?

ANSWER :

KG = 35 * 6,2 = 0,085 M
2550

( a ) LOADED ABOVE CG =4,52 + 0,08 = 4,60 M ( NEW KG )

( b ) LOADED BELOW CG = 4,52 -- 0,08 = 4,44 M ( NEW KG )


QUESTION 13 :

A VESSEL DISPLACED 3250 TONNES, AND HER KG IS 6,8 M. 150 TONNES ARE LOADED WITH
THE CG 2,4 M ABOVE THE KEEL. WHAT IS THE NEW KG ?

ANSWER :

1. (6,8 -- 2,4 ) * 1150 = 4,4 * 1150 = 1,15 M
( 3250 + 1150 ) 4400

2. NEW KG = 6,8 -- 1,15 = 5,65 M


QUESTION 14 :

A DERRICK IS USED TO LIFT A 12 TONNES WEIGHT FROM THE HOLD. THE HEADS OF THE
DERRICK IS 15 M ABOVE THE ORIGINAL CG OF THE WEIGHT. DISPLACEMENT IS 3200
TONNES. WHAT IS THE SHIFT OF CG ?

ANSWER :

12 * 15 = 0,056 M UPWARDS
3200


QUESTION 15 :

A VESSEL DISPLACED 4600 TONNES, AND HER KG IS 4,6. A WEIGHT OF 25 TONNES IS
LIFTED FROM THE LOWER HOLD AND STOWED ON THE DECK WITH THE VESSEL'S DERRICK.
THE HEAD OF THE DERRICK IS 21,5 M ABOVE THE KEEL. THE CG OF THE WIGHT WAS 1,4 M
ABOVE THE KEEL WHEN IT WAS IN THE HOLD, AND IS 10,4 M ABOVE THE KEEL WHEN ON
DECK. FIND THE FOLLOWING :

( a ) WHAT IS THE NEW KG WHEN THE WEIGHT IS SUSPENDED FROM THE DERRICK ?
( b ) WHAT IS THE NEW KG WHEN THE WEIGHT IS LANDED ON THE DECK ?

ANSWER :

( a ) ( 21,5 -- 1,4 ) * 25 = 20,1 * 25 = 0,109
4600 4600

NEW KG = 4,6 + 0,11 = 4,71 M

( b ) ( 10,4 -- 1,4 ) * 25 = 9,0 * 25 = 0,049
4600 4600

NEW KG = 4,6 + 0,05 = 4,65 M


QUESTION 16 :

20 TONNES ARE LIFTED BY A DERRICK, THE DERRICK HEAD IS 15 M ABOVE THE ORIGINAL
POSITION OF THE WEIGHT. VESSEL'S DISPLACEMENT IS 1750 TONNES. KG IS 3,2 M. FIND
THE FOLLOWING :

( a ) WHAT IS THE NEW KG WHEN THE WEIGHRT IS 1,3 M ABOVE THE KEEL ?
( b ) 10,4 M ABOVE THE KEEL ?

ANSWER :

( a ) 20 * 15 = 0,171 ( b ) THE SAME. WEIGHT SUSPENDED FROM
1750 A DERRICK ACTS AT THE DERRICK HEAD.

3,2 + 0,17 = 3,37 KG


QUESTION 17 :

A VESSEL DISPLACED 2635 TONNES & A KG OF 4,28; LOADS A WEIGHT OF 35 TONNES; THE
DERRICK WITH IT'S HEAD ABOVE THE KEEL IS USED. WHEN THE WEIGHT IS STOWED IN THE
HOLD, THE FINAL POSITION OF IT'S CG IS 2,8 M ABOVE THE KEEL. FIND THE FOLLOWING :

( a ) WHAT IS THE KG WHEN THE WEIGHT IS SUSPENDED FROM THE DERRICK ?
( b ) WHAT IS THE KG WHEN THE WEIGHT IS LOADED IN THE HOLD ?

ANSWER :

( a ) NEW DISP. = OLD DISP. + WEIGHT = 2635 + 35 = 2670

SHIFT OF CG = WEIGHT * HEIGHT DERRICK ABOVE THE KEEL = 35 *17 = 0,22
DISPLACEMENT 2670

NEW KG = OLD KG + SHIFT OF CG = 4,28 + 0,22 = 4,50 M

( b ) NEW DISP. = OLD DISP. + WEIGHT = 2635 + 35 = 2670

? = 35 * 2,8 = 0,04
2670

? = 4,28 -- 0,04 = 4,24 M


QUESTION 18 :

DOUBLE BOTTOM TANK CAN HOLD 420 TONNES OF SEA-WATER. WHEN FULL IT'S CG IS 0,8 M
ABOVE THE KEEL. WITH THE TANK EMPTY, DISPLACEMENT IS 3700 TONNES & KG IS 8,78 M.
WHAT IS THE KG WHEN THE TANK IS FILLED ?

ANSWER :

d = OLD KG -- CG OF FULL TANK = 8,78 -- 0,8 = 7,98
NEW DISP. = OLD DISPL. + WEIGHT = 3700 + 420 = 4120

420 * 7,98 = 0,813
4120

NEW KG = 8,78 -- 0,813 = 7,97 M


QUESTION 19 :

VESSEL'S TANK HOLDS 192 TONNES OF WATER; IT'S CG IS 41,3 M FROM THAT OF THE SHIP.
IF THE DISPLACEMENT IS 2974 TONNES WHEN THE TANK IS FULL, WHAT IS THE SHIFT IN G
WHEN IT IS PUMP OUT ?

ANSWER :

NEW DISPL. = OLD DISPL. -- WEIGHT = 2974 -- 192 = 2782 TONNES

SHIFT OF G = WEIGHT * CG OF TANK = 192 * 41,3
NEW DISPLACEMENT 2782



QUESTION 20 :

THE VESSEL TRANSFERRED 135 TONNES OF FUEL FROM THE FOREPEAK TO AN AFTER TANK.
THE DISTANCE BETWWEN THE TANK'S CG IS 125 M; DISPLACEMENT IS 7620 TONNES. WHAT
IS THE SHIFT IN G ?

ANSWER :

SHIFTING = WEIGHT * DISTANCE = 135 * 125 = 2,21 M AFT
DISPLACEMENT 7620


QUESTION 21 :

THE SHIP'S LCF IS 4 M AFT OF MIDSHIPS. WEIGHTS ARE SHIFTED AFT TO CHANGE THE
TRIM 0,4 M. LBP IS 130 M. FORWARD DRAFT WAS 5,97 M. AFT DRAFT WAS 6,13 M. FIND
NEW DRAFT.

ANSWER :

FORWARD DRAFT = 5,97 -- 0,4 = 5,77 M
2

AFTER DRAFT = 6,13 + 0,4 = 6,33 M
2

MEAN DRAFT = 5,77 + 6,33 = 6,05 M
2

BY PROPORTIONAL METHODS

CHANGE FORWARD = 69 * 0,4 = 0,21
130

CHANGE AFT = 61 * 0,4 = 0,19
130

FORWARD DRAFT = 5,97 -- 0,21 = 5,76 M

AFTER DRAFT = 6,31 + 0,19 = 6,32 M

MEAN DRAFT = 5,76 + 6,32 = 6,04 M
2



QUESTION 22 :

SHIFTING

A PLANK WEIGHT 25,09 KG. WHAT WILL BE THE SHIFT OF IT'S CENTRE OF GRAVITY IF A
WEIGHT OF 17,24 KG IS PLACED A DISTANCE OF 3,66 M FROM IT'S ORIGINAL CG ?

ANSWER :

( a ) THE CG OF A BODY MOVES PARALLEL TO THE SHIFT OF THE CG OF ANY WEIGHT
SHIFTED WITHIN IT.

( b ) THE DISTANCE IT MOVES IS EQUAL TO THE WEIGHT SHIFTED, MULTIPLIED BY THE
SHIFT OF IT'S CG; ALL DIVIDED BY THE TOTAL WEIGHT OF THE BODY.

17,24 * 3,66 = 63,09 = 1,49 M
25,09 + 17,24 42,33


QUESTION 23 :

A BEAM CARRIES 2438 KG AT A DISTANCE OF 1,83 M FROM ONE END & THE CENTRE OF
GRAVITY OF THE WHOLE MASS IS 6,09 M FROM THE END. IF THE WEIGHT OF THE BEAM
ALONE IS 4267,47 KG, WHAT IS THE DISTANCE OF THE CG OF THE ORIGINAL BEAM FROM
THE END ?

ANSWER :

GG = w * d = 2438,55 * 4,26 = 2,43
W 4267,47

2,434 + 6,09 = 8,524


QUESTION 24 :

T P C

FIND THE TONNES PER CENTIMETER IMMERSION OF A BOX SHAPED VESSEL 64 M LONG BY
10,668 M BREADTH.

ANSWER :

T P C = 1,025 * AREA OF THE WATERPLANE ( A ) = 1,025 * 10,668 * 64,0 = 6,99
100 100


QUESTION 25 : ( METRIC )

WEIGHT OFF THE CENTRE OF FLOTATION

VESSEL'S LENGHT IS 115,82 M, T P C IS 14,96 TONNES, M T C IS 330,7 TONNES. CENTRE
OF FLOTATION ON THE CENTRE LINE.THERE IS 57,9 TONNES REMOVED 39,62 M AFT OF THE
STERN. ORIGINAL DRAFTS ARE 5,23 M FWD & 5,33 M AFT. FIND NEW DRAFTS.

ANSWER :

LENGTH = 115,82 = 5791 -- 39,62 = 18,29 M
2

MOMENT TO CHANGE TRIM = W * D = 57,9 * 18,29 = 1058,99

CHANGE OF TRIM = 1059,99 = 3,20 CM
330,7

REMOVAL OF WEIGHT THE VESSEL WILL RISE = W = 57,9 = 3,6 CM
T P C 14,96

ORIGINAL DRAFT RISE FWD = 0,23 M AFT = 5,33 M

5,30 5,298
TRIM -0.036 + 0,036
5.162 M 5,334 M

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