4.9 Stability at Large Angles of Inclination
The transverse metacenter height is a measure of the
stability under ‘initial stability’ (aka small angle
stability).
When the angle of inclination exceeds 5 degrees, the
metacenter can be no longer regarded as a fixed point
relative to the ship. Hence, the transverse metacenter
height (GM) is no longer a suitable criterion for measuring
the stability of the ship and it is usual to use the value of the
righting arm GZ for this purpose.
•The Derivation of Atwood’s Formula
W0 L0 : W.L. when the ship is at upright position.
W1 L1 : W.L. when the ship is inclined at an angle θ.
If the ship section is not vertically sided, the two W.L.,
underneath which there must be the same volume, do not
intersect on the center line (as in the initial stability) but at S.
v  he hi
B0 R 

GZ  B0 R  GB0  sin 
v  he hi
 GB0  sin 

Atwood Formula

B0 R 
v  he hi

Atwood Formula
GZ  B0 R  GB0  sin  
v  he hi
 GB0  sin 

• GZ vs.

For each angle of θ, we compute GZ, the righting arm.
The ship is unstable beyond B. (even if the upsetting moment
is removed, the ship will not return to its upright position).
From 0 to B, the range of angles represents the range of
stabilities.
Ex. Righting arm of a ship vertically sided (A special
example to compute GZ at large angle inclinations)
Transverse moment of volume shifted =

L
0
L2
1
 4 
3
y

y
tan


y
dx

tan

y
dx  I x tan 

 


0 3
2
3

 

Volume
arm
Transverse shift of C.B.
I x tan 

 B0 M tan 

2 L 3
I x   y dx
3 0
Ex. Righting arm of a ship vertically sided (A special
example to compute GZ at large angle inclinations)
Similarly, vertical moment of volume shifted =

L
0
L1
1
1
 2
2
3
2
y

y
tan


y
tan

dx

tan

y
dx

I
tan

x



0
3
2
2
 3
Volume
Arm
Vertical shift of C.B.

1
I x tan 2  / 
2
Ex. Righting arm of a ship vertically sided (A special
example to compute GZ at large angle inclinations)
Ix
1 Ix
B0 R   cos    sin   sin  
tan 2  sin 

2
Ix  1

 sin  1  tan 2  
 2

when 
1 small angle inclination, B0 R  sin  I x /     I x /     B0 M
1


GZ  B0 R  B0G  sin   sin   B0 M  B0G  B0 M tan 2  
2


Ix
1

2 
 sin  GM  B0 M tan   ,
where B0 M  .
2



This formula is called 'walled sided formula' or the 'box formula'.
when 
1 small angle inclination,
GZ    GM
• Cross Curves of Stability
It is difficult to ascertain the exact W.L. at which a ship would
float in the large angle inclined condition for the same
displacement as in the upright condition. The difficulty can be
avoided by obtaining the cross curves of stability (see p44).
How to Computing
them
•Assume the position
of C.G. (not known
exactly)
•W.L. I - V should
cover the range of
various displacements
which a ship may
have.
• Cross Curves of Stability
Computation Procedures
1. The transverse section area under waterline I, II, III, IV, V
2. The moment about the vertical y-axis (passing through C.G)
3. By longitudinal integration along the length, we obtain the
displacement volume, the distances from the B.C. to y-axis
(i.e. the righting arm GZ) under the every W.L.
4. For every   5 ,10 ,
, we obtain  and GZ for W.L. I - V
5. Plot the cross curves of stability.
Cross Curves of Stability
These curves show that the righting arm (GZ) changes with the
change of displacement given the inclination angle of the ship.
For the sake of understanding ‘cross curves of stability’ clearly,
here is a 3-D plot of ‘cross curves of stability.’
The curved
surface is
GZ  f ( , )
• Curve of Static Stability
‘Curve of static stability’ is a curve of righting arm GZ as a
function of angle of inclination for a fixed displacement.
Computing it based on cross curves of stability.
1. How to determine a curve of statical stability from a 3-D of
‘cross curves of stability.’ (C.C.S.), e.g., the curve of static
stability is the intersection of the curved surface and the plane
of a given displacement.
2. Determining a C.S.S. from 2-D ‘C.C.S.’ is to let displcement
= const., which intersects those cross curves at point A, B,…,
see the figure.
GZ
• Influences of movement of G.C on ‘curve of static
stability’
1. Vertical movement (usually due to the correction of G.C
position after inclining experiment.)
G1Z1  GZ  GG1 sin 
• Influences of movement of G.C on ‘curve of static
stability’
2. Transverse movement (due to the transverse movement of
some loose weight)
G1Z1  GZ  GG1 cos 
wh
GG1 

Weight moving from the
left to the right
• Features of A Curve of Static Stability
1. Rises steadily from the origin and for the first few degrees is
practically a straight line.
Near the origin GZ = θ * slope & slope = ?, why?
2. Usually have a point of inflexion, concave upwards and
concave downwards, then reaches maximum, and afterwards,
declines and eventually crosses the base (horizontal axis).
1 radian
The maximum righting arm & the range of stability are to a
large extent a function of the freeboard.
(the definition of freeboard)
Larger freeboard
stability
Larger GZmax & the range of
Using the watertight superstructures
the range of stability
Larger GZmax &
4.10 Dynamic Stability
Static stability: we only compute the righting arm (or
moment) given the angle of inclination. A true measure of
stability should considered dynamically.
• Dynamic Stability: Calculating the amount of work done by
the righting moment given the inclination of the ship.

 max
0
0
W    w  GZ  d Wmax  
 w  GZ  d
• Influence of Wind on Stability (p70-72)
Upsetting moment due to beam wind
M  kAhV 2 cos 2 
k - an empirical coeff.
A - projected area of the
ship above the W.L
V - nominal wind velocity
h - distance from the half
draft to wind pressure
center.
 - the angle of inclination
w.r.t. the beam
When the ship is in upright position, the steady beam wind starts to
blow and the ship begins to incline. At point A, the M(wind) =
M(righting), do you think the ship will stop inclining at A? Why?
The inclination will usually not stop at A. Because the rolling
velocity of a ship is not equal to zero at A, the ship will continue to
incline. To understand this, let’s review a simple mechanical
problem
The external force
F = constant
The work done by it
WF  Fx1
F
If at x  x1 , R  kx1  F
the work done by the
spring force R,
x1
WR    Rdx   
0
x1
0
R
F
m
No Friction
X=
0
X = X1
1 2
kxdx   kx1
2
1 2 1 2
The total work WF  WR  Fx1  kx1  mv  ER
2
2
Hence, the block will continue to move to the right. It will not
stop until
ER  WF  WR  0, at x  x2
X
In a ship-rolling case:
Work done by the upright moment

Wup    w   GM  d
0
Work done by the wind force

Wwind   M wind  d
0
It will stop rolling (at E)
Wup  Wwind
In a static stability curve
AreaOAEGCO  AreaDAFGCOD
or simply,
AreaODA  Area AEF
• Consideration in Design (The most sever case
concerning the ship stability)
Suppose that the ship is inclining at angle  0 and begins to
roll back to its upright position. Meanwhile, the steady
beam wind is flowing in the same direction as the ship is
θ
going to roll.
0
Wind
Standards for USN warships:
1)  =30 ,
AGZ  0.055m  rad
 =45 or  max , AGZ  0.09m  rad
30    45 , AGZ  0.03m  rad
2) GZ  0.2m, at  =30
3) GZ MAX occurs at   30
4) GM  0.15m
Standards of Stability: ships can withstand
1. winds up to 100 knots;
2. rolling caused by sever waves;
3. heel generated in a high speed turn;
M turning  MV 2 h / r , h - distance between C.G to the half draft
r - radius of turning circle, M - mass, V - velocity of ship.
4. lifting weights over one side (the C.G. of the
weight is acting at the point of suspension);
5. the crowding of passengers to one side.
4.11 Flooding & Damaged Stability
So far we consider the stability of an intact ship. In the event of
collision or grounding, water may enter the ship. If flooding is
not restricted, the ship will eventually sink. To prevent this, the
hull is divided into a number of watertight compartments by
watertight bulkheads. (see the figure)
Transverse (or longitudinal) watertight bulkheads can
• Minimize the loss of buoyancy
• Minimize the damage to the cargo
• Minimize the loss of stability
Too many watertight bulkheads will increase cost & weight of the
ship. It is attempted to use the fewest watertight bulkheads to
obtain the largest possible safety (or to satisfy the
requirement of rule).
Forward peak bulkhead
(0.05 L from the bow)
After peak bulkhead
Engine room: double bottom
Tanker: (US Coast Guard) Double Hull (anti pollution)
This section studies the effects of flooding on the
1) hydrostatic properties
2) and stability
• Trim when a compartment is open to Sea
W0 L0  W.L. before the damage
W1 L1  W.L. after the damage
If W1L1 is higher at any point than the main deck at which the
bulkheads stop (the bulkhead deck) it is usually considered
that the ship will be lost (sink) because the pressure of water
in the damaged compartments can force off the hatches and
unrestricted flooding will occur all fore and aft.
• (1) Lost buoyancy method
1) Computing the parallel sinkage
y0  v
( A  a)
v - volume of lost buoyancy up to W0 L0
A - intact W .L. area; a - area of the damage
2) Find the midway W.L. (excluding damage
aera a) area Am , C.F., I y ( mcf ) at draft d  0.5 y0
(d - draft before the damage).
3) Update the sinkage
y1  v
4) Find the trim MTI d  y 
2
Am
I mfc
420 L
.
, Trim 
 w v x
MTI d  y
x  distance between the C.F. of mean W .L
& centroid of lost buoyancy of v.
v (0.5 L   )
5) Draft aft  d 

 Trim
Am
L
v (0.5 L   )
Draft forward  d 

 Trim
Am
L
 is the distance between C.F. & midship
Aassuming that C.F. is aft of midship & the
damage takes place of forward.
2
(in)
Second iteration: update the midway W .L.
d  0.5v / Am
, then find new Am , and
repeat the procedures given in the previous
page.
The iteration may be stopped if the results are
convergent, e.g. the midway W .L. of the
previous iteration & present iteration are
different by an amount smaller than a
prescribed tolerance error.
Ex. p121-123
A vessel of constant rectangular cross-section L = 60 m, B = 10 m,
T = 3 m. ZG = 2.5 m l0 = 8 m.
w1
L1
w0
1)v  T  l0  B  3  8 10
 240m
3
Aw  L  B  60 10  600m 2
a  l0  B  8 10  80m 2
A  Aw  a  520m 2
L0
l0 = 8 m
2) Parallel sinkage
y0 
v 240

 0.46m
A 520
3) Draft at midway between W0L0 – W1L1 :
A  520m2 ,
T
y1  0.46m, same as y0 ,
y
 3.23m
2
no iteration is necessary in this case.
4) Computing Aw  a,  , I MFC .
A  Aw  a  520m 2 ,
New (damaged) W .L's. C.F.   4m & x  30m
1
(52)3 10 m 4 longitudinal moment of inertia
12
I MFC
I MFC
GM L 
 Z B  ZG 


I MFC
Because
 Z G  Z B

I MFC 
Moment for Trim per meter:
I MFC   w
1m
MTI(m)  GM L    w

L(m)
L
1 (52)3 10 3
kN

m 1.025  9.81 3
12
60
m
 19, 637 kN-m
TF
L 


 wvx
2
T  y

L
34  w 30  240  60
 3  0.46  
MTI
60  w 121  52 3 10
 3  0.46  2.09  5.55m
TA
L 


 wvx
2
T  y

L
MCT
 1.86m
v, x , Aw(0) a (0)
y (()) 
v
Aw(0)  a (0)
d 12  d 
(0)
y( 0 )
2
(0)
Am(0) am
y (1) 
v
Am(0)  am(0)
if y
(0)
y (1)
 y (1)  
Find trim.
MTI ( at
MTI 
TF  d  y
(1)
d
y (1)
2
)
I mfc
420 L
L 

   xv
2

L
MTI
L 

   xv
2
(1)
TA  d  y 
L
MTI
• (2) Added Weight Method (considering the loss
of buoyancy as added weight)
also a Trial – error (iterative) method
1) Find added weight v under W0L0. Total weight = W + v
2.) According to hydrostatic curve , determine W1L1 (or T) &
trim (moment caused by the added weight & MTI).
3.) Since we have a larger T, and v will be larger, go back to
step 1) re-compute v.
The iterative computation continues until the difference
between two added weights v obtained from the two
consecutive computation is smaller than a prescribed error
tolerance.
• Stability in damaged condition
Increase in draft increases Z B ,
 ZB 
 vbb1

where b is the C.G of the lost buoyancy,
b1 is the C.G of the buoyancy recovered on
above the original W.L,  the porosity coeff.
of the damaged compartment.
The loss of the moment of inertia w.r.t to the
longitudinal axis I d , thus
I  Id
 KM 

GM  GM (intact) 
 vbb1


Id
I
 GM (intact)  d


• Asymmetric flooding
1. If the inclination angle is large,
then the captain should let the
corresponding tank flooding.
Then the flooding is
symmetric.
2. If the inclination angle is small,
Lost buoyancy : v w
Heeling moment: v w y
 w GM sin   v w y,
vy
sin    
,
GM
 is the inclination angle.
•Floodable length and its computation
Floodable Length: The F.L. at any point within the length of
the ship is the maximum portion of the length, having its center
at the point which can be symmetrically flooded at the
prescribed permeability, without immersing the margin line.
Bulkhead deck: The deck tops the watertight bulkhead
Margin line: is a line 75 mm (or 3”) below the bulkhead
at the side of a ship
Without loss of the ship: When the W.L. is tangent to
the margin line.
Floodable length (in short) The length of (part of) the
ship could be flooded without loss of the ship.
• Determine Floodable length is essential to determine
1. How many watertight compartments (bulkheads) needed
2. Factor of subdivision (How many water compartments
flooded without lost ship)
Steps for computing the F.L. given  0 , B0 , or W0 L0
1) Obtain a limit W.L.: W1 L1 ,
2) using the Bonjean Curves to obtain 1 and B1 under W1 L1.
3) 1   0  v, v is the loss of buoyancy due to the flooding.
4) 1  B0 B1  v  x   1   0   x ,
x
1  B0 B1
1   0
x is the distance of the center of the lost buoyancy from B0 .
5) 1   0  v  A0l ,
A0 is the area of the cross section, l is the F.L.
Setting a half of l on either side
of the center of lost Buoyancy.
Near the ends of a ship, A0 is
changing rapidly then using the
iterative process to determine l.
6) Repeating (1) to (5) for
a series of W .L.s tangent to
the margin line at different
positions in the length of the
ship. Then a series of values
of the F.L. can be obtained
for different positions along
the ship.
7) Considering the different
permeability coefficients 
at different positions along
the ship.
8) Factor of Subdivision F
Factor of subdivision is the ratio of a permissible length to
the F.L.
For example, if F is 0.5, the ship will still float at a W.L. under
the margin line when any two adjacent compartments of the
ship are flooded. If F is 1.0, the ship will still float at a
W.L. under the margin line when any one compartment of
the ship is flooded.
Rules and regulations about the determination of F are set by
many different bureaus all over the world (p126-127)