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Longitudinal Stability Yuvaraj Adithya
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Transcript of Longitudinal Stability Yuvaraj Adithya
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8/9/2019 Longitudinal Stability Yuvaraj Adithya
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By
Yuvaraj R
AdithyaManda
LONGITUDINAL STABILITY
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CONTENTSLongitudinal ta!ilityLongitudinal "#ta$#ntr#
MCTC %
Tri"Lot !uoyan$y "#thod
Add#d ight "#thod
'looda!l# l#ngth
'a$tor o( u!diviion
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Longitudinal StabilityLongitudinal Metacenter: Similar to the definition of the
transverse meta center, when a ship is inclined longitudinallyat a small angle, A vertical line through the center of
buoyancy intersects the vertical line through (before the
ship is inclined) at .
1B
0B
LM
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The Location of the LongitudinalMetacenter
For a small angle inclination, volumes of forward wedge immersedin water and bacward wedge emerged out of water are!
10
" 1 "0
0 0
(" ) ( tan ) where is the half breadth.
(" ) ( tan ) , .
#hus, " " , which indicates!
moment of area forward of $ moment of area after .
is the cente
l
L l
l L l
v y x dx y
v y x dx v v v
yxdx yxdx
F F
F
=
= = ==
0 0
0 0
r of (mass) gravity of waterline , %
is called of . #herefore, for e&ual
volume longitudinal inclination the new waterline always
passes through the ('.F
W L
W Lcenter of flotation
center of flotation ).
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Location of the LongitudinalMetacenter
sing the same argument used in obtaining transverse metacenter.0 1 1 "
0
1 "0
0" "
0
0 1 0
,
(" ) ( tan ) " tan
tan " " tan
is the moment of inertia with respect to the
transverse a*is passing the center of flotation.
t
l
L l
l
FCL l
FC
B B vg g
vg g y x xdx yx xdx
yx dx yx dx I
I
B B B M
=
= +
= + =
=
0
0
an , .
.
FC
FC FC
ML B B L B g
IB M
I IH B M Z Z GM Z Z
=
= + = + = +
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Location of the LongitudinalMetacenter
"
0,
0,
sually Floatation 'enter ('.F) of a waterplane is not at the
midship,
,
where is the moment of inertia w.r.t. the transverse a*is
at midship (or station +) and is the distance from F
FC T
T
I I Ax
I
x
=
.'. to
the midship.
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Moment to Change Trim One cm(MCTC 1)
#-! (moment to alter (change) the ships trim per inch) at
each waterline (or draft) is an important &uantity. /e may
use the longitudinal metacenter to predict #-
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MCTC 1 ( a function of draft)ue to the movement of a weight, assume that the ship as 1 trim,
and floats at waterline /.2., also e*pressed as MTIin inches
( )
( )
0 1
0 1
0 1 0
13 1tan , where is in feet.
1"
ue to the movement of the weight, moves to ,,
tan tan
tan tan
1
1"
w
L L G
FCw L G w B G
FCw B G
LL L
G GM w h G G
G G G M HM Z
IM HM Z Z Z
IZ Z
L
= =
= =
= =
= = + = +
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MCTC 1 ( a function of draft)
-f the longitudinal inclination is small, MCTC 1can be used to
find out the longitudinalposition of gravity center ( ).
4
1 ,
1" 1"
56 lbft , 2ong #on $ ""60 lb, #- (ton7ft)6"0
FC FC w FCG B w
FCw
I I IZ Z M
L LI
L
>> = =
= =
1GX
0 1
0 0 0
1 0 1
#rimtan
tan ,
Since is in the same vertical line as . under ,
F A
FC FC FCB G G B
FC
G B B
T T T
L L L
I I ITG G Z Z Z Z L
G C B W L
I TX X G G X
L
= = =
= +
= + +
)
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when a compartment is open to S
0 0
1 1
/.2. before the damage
/.2. after the damage
W L
W L
-f W1L1 is higher at any point than the main deckat which the
bulheads stop (the bulkhead deck) it is usually considered
that the ship will belost (sink)because the pressure of water
in the damaged compartments can force off the hatches and
unrestricted flooding will occur all fore and aft.
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(1) Lost buoyancy method
0
0 0
(
1) 'omputing the( )
7 volume of lost buoyancy up to
7 intact . . area8 7 area of the damage
") Find the (e*cluding damage
aera ) area , '.F.,m y m
vyA a
v W L
A W L a
a A I
=
parallel sinkage
midway !L!
) 0at draft 0.+
( 7 draft before the damage).
cf d y
d
+
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"
"
14) pdate the sinage .
6) Find the trim , #rim (in)6"0
distance between the '.F. of mean .
% centroid of lost buoyancy of .
(0.+ )+) raft aft #
y
y
m
mfc w
d
d
m
vyA
I v x
L
x W L
v
v Ld
A L
+
+
=
= =
= +
MTI MTI
rim
(0.+ ) raft forward #rim
is the distance between '.F. % midship
Aassuming that '.F. is aft of midship % the
damage taes place of forward.
m
v Ld
A L
+= + +
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! update the midway . .
0.+ , then find new , and
repeat the procedures given in the previous
page.
#he iteration may be stopped if the results are
convergent, e.g. the mid
m m
W L
d v A A+"econd iteration
L
way . . of the
previous iteration % present iteration are
different by an amount smaller than aprescribed tolerance error.
W L
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(0) (0), , wv x A a
(())
(0) (0)
w
vy
A a=
(0)
1"
(0) "yd d= +(0)
mA(0)
ma
(1)(0) (0)
m m
vyA a
=
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(0) (1)y y
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(2) dded !eight Method(considering the loss of buoyancyas added weight)
also a #rial 9 error (iterative) method
1) Find added weight vunder /020. #otal weight $ W + v
".) According to hydrostatic curve , determine W1L1 (or T) % trim (moment
caused by the added weight % MCTC 1).
4.) Since we have a largerT, and vwill be larger, go bac to step 1) re7compute
v.
#he iterative computation continues until the difference
between two added weights vobtained from the two
consecuti#e computationis smaller than a prescribed error
tolerance.
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$loodable length and its computation
$loodable Length! #he $!L. at any point within the length ofthe ship is the ma*imum portion of the length, having its center
at the point which can be symmetricallyflooded at the
prescribed permeability, without immersing the margin line.
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%ulkhead deck! #he dec tops the watertight bulhead
Margin line! is a line :+ mm (or 4) below the bulheadat the side of a ship
ithout loss of the ship: /hen the /.2. is tangent to
the margin line.
$loodable length(in short) #he length of (part of) the
ship could be flooded without loss of the ship.
&etermine $loodable length is essential to determine
1. ;ow many watertight compartments (bulheads) needed
'! $actor of subdi#ision(;ow many water compartments
flooded without lost ship)
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0 0 0 0
1 1
1 1 1 1
1 0
Steps for computing the F.2. given , , or
1)
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T*AN+
YOU,,,