Chapter IV
(Ship Hydro-Statics & Dynamics)
Floatation & Stability
4.1 Important Hydro-Static Curves or
Relations (see Fig. 4.11 at p44 & handout)
• Displacement Curves (displacement [molded, total]
vs. draft, weight [SW, FW] vs. draft (T))
• Coefficients Curves (CB , CM , CP , CWL, vs. T)
• VCB (KB, ZB): Vertical distance of Center of
Buoyancy (C.B) to the baseline vs. T
•LCB (LCF, XB): Longitudinal Distance of C.B or
floatation center (C.F) to the midship vs. T
4.1 Important Hydro-Static Curves or
Relations (Continue)
• TPI: Tons per inch vs. T (increase in buoyancy due
to per inch increase in draft)
• Bonbjean Curves (p63-66)
a) Outline profile of a hull
b) Curves of areas of transverse sections (stations)
c) Drafts scales
d) Purpose: compute disp. & C.B., when the vessel
has 1) a large trim, or 2)is poised on a big wave crest or
trough.
How to use Bonjean Curves
• Draw the given W.L.
• Find the intersection of the W.L. & each station
• Find the immersed area of each station
• Use numerical integration to find the disp. and C.B.
4.2 How to Compute these curves
• Formulas for Area, Moments & Moments of Inertia
L
A   ydx,
a) Area d A  ydx
0
L
M   xydx,
b) Moments d M  xydx
0
Center of Floatation x  M / A
c) Moments of Inertia d I  x 2 d A  x 2 ydx
L
I 0   x ydx,
2
0
2
I C .F  I 0  x A
Examples of Hand Computation of Displacement
Sheet (Foundation for Numerical Programming)
• Area, floatation, etc of 24’ WL (Waterplane)
•Displacement (molded) up to 8’ WL
•Displacement (molded) up to 24’ and 40 ‘WL (vertical
summation of waterplanes)
•Displacement (molded) up to 24’ and 40 ‘WL (Longitudinal
summation of stations)
•Wetted surface
•Summary of results of Calculations
4’wl
area
32’wl
area
8’wl
area
40’wl
area
16’wl
area
Up to
4’wl
24’wl
area
Up to
8’wl
MTI
Disp. Up
to 16’wl
MTI
Wetted
surface
Disp. Up
to 40’wl
Disp. Up
to 32’wl
•1-6 Areas & properties (F.C.,
Ic, etc) of W.L
•7-11 Displacement, ZB , and
XB up W.L., vertical
integration.
Up to
24’ & 40
wl
Disp. Up
to 24’wl
Red sheet will be studied in
detail
Summary
•12-15 Transverse station area,
longitudinal integration for
displacement, ZB , and XB
•16-18 Specific Feature (wetted
surface, MTI, etc.
•19 Summary
2
S 2
3
S  The distance
m1 
between the
two stations
2
Simpson's 1st
3
2 1
  y0  2 y1 
3 2
y 2  2 y3 

2 1
  y0  y 1 
2
3 4
3

y1  2 y2  .. 
4

2 Symmetric
Formulas for the remaining coefficents mi
2 2
m2   S  2
3
S  the distance between the two stations
2
Simpson's 1st rule coeff.; 2 - Symmetric
3
2 3
m3   S  2
3
1 2
1
y3
m4    S  2 ( from
)
3 3
3
3
2
h
2 2 h
m5   S   2,
m6   S   2
3
3
3
3
Illustration of Table 4:
C1
Station
C2
Half Ordinate copy from line drawing table ( 24’ WL).
(notice at FP. Modification of half ordinate)
C3
Simpson coefficient (Simpson rule 1) (1/2 because of half station)
C4 = C3 x C2
FP-0
AP-10 (half station)
(area function)
displacement
C5 = Arm (The distance between a station and station of 5 (Midship)
C6 = C5 x Function of Longitudinal Moment with respect to Midship (or station 5)
C7 = Arm (same as C5)
C8 = C6 x C7 Function of Longitudinal moment of inertia with respect to Midship.
C9.= [C2]3
C10. Same as C3. (Simpson Coeff.)
C11. = C9 x C10.
Transverse moment of inertia of WL about its centerline
Table 5 is similar to Table 4, except the additional computation of appendage.
Illustration of Table 8
For low WLs, their change is large. Therefore, it is first to use planimeter or other
means to compute the half-areas of each stations up to No. 1 WL (8’ WL).
C1.
Station
C2.
Half area (ft2) of the given station
C3.C3/(h/3) ( divided by h/3 is not meaningful, because it later multiplying by
h/3) (h = 8’ the distance between the two neighboring WLs)
C4.½ Simpson’s Coeff.
C5.
C4 x C3
C6.Arm distance between this station and station 5 (midship)
C7 C5 x C6
f(M)
Illustration of Table 9
C1.
WL No.
C2.
f(V) Notice first row up to 8’. f(v)
C3.
Simpson’s coeff.
C4.
C2 x C3
C5.
Vertical Arm above the base
C6.
C4 x C5.
f(m) vertical moment w.r.t. the Baseline.
* Notice up the data in the first row is related to displacement up to 8’
WL. The Table just adding V)
Illustration of Table 12
C1.
Station No.
C2.
under 8’ WL.
(From Table 8)
C3.
8’ WL
x1
C4.
16’ WL x ¼
C5.
24’ WL x 1
C6.
(C2 + C3 + C4 + C5)
C7.
Arm
C8.
C7 x C6 (Simpson rule)
C9.
C6*h/3
(SM 1 + 4 + 1)
Function of Area of Stations
(Distance between this station to midship)
4.3 Stability
A floating body reaches to an equilibrium state, if
1) its weight = the buoyancy
2) the line of action of these two forces become collinear.
The equilibrium: stable, or unstable or neutrally stable.
•
Stable equilibrium: if it is slightly displaced from its
equilibrium position and will return to that position.
•
Unstable equilibrium: if it is slightly displaced form its
equilibrium position and tends to move farther away from
this position.
• Neutral equilibrium: if it is displaced slightly from this
position and will remain in the new position.
• Motion of a Ship:
6 degrees of freedom
- Surge
- Sway
- Heave
- Roll
- Pitch
- Yaw
Translation
Rotation
Axis
x Longitudinal
Surge
Neutral S.
Roll
S. NS. US
y Transverse
Sway
Neutral S.
Pitch
S.
z Vertical
Heave
S. (for sub, N.S.)
Yaw
NS
Righting & Heeling Moments
A ship or a submarine is designed to float in the upright
position.
• Righting Moment: exists at any angle of inclination where
the forces of weight and buoyancy act to move the ship
toward the upright position.
• Heeling Moment: exists at any angle of inclination where
the forces of weight and buoyancy act to move the ship away
from the upright position.
For a displacement ship,
W.L
G---Center of Gravity, B---Center of Buoyancy
M--- Transverse Metacenter, to be defined later.
If M is above G, we will have a righting moment, and
if M is below G, then we have a heeling moment.
For submarines (immersed in water)
B
G
G
If B is above G, we have righting moment
If B is below G, we have heeling moment
Upsetting Forces (overturning moments)
•
Beam wind, wave & current pressure
•
Lifting a weight (when the ship is loading or unloading in
the harbor.)
•
Offside weight (C.G is no longer at the center line)
•
The loss of part of buoyancy due to damage (partially
flooded, C.B. is no longer at the center line)
•
Turning
•
Grounding
Longitudinal Equilibrium
For an undamaged (intact) ship, we are usually only
interested in determining the ship’s draft and trim regarding
the longitudinal equilibrium because the ship capsizing in the
longitudinal direction is almost impossible. We only study
the initial stability for the longitudinal equilibrium.
Static Stability & Dynamical Stability
Static Stability: Studying the magnitude of the
righting moment given the inclination (angle) of the
ship*.
Dynamic Stability: Calculating the amount of work
done by the righting moment given the inclination of
the ship.
The study of dynamic Stability is based on the study of
static stability.
• Static Stability
1) The initial stability (aka stability at small inclination) and,
2) the stability at large inclinations.
• The initial (or small angle) stability: studies the right
moments or right arm at small inclination angles.
• The stability at large inclination (angle): computes the right
moments (or right arms) as function of the inclination angle, up
to a limit angle at which the ship may lose its stability
(capsizes).
Hence, the initial stability can be viewed as a special case of the
latter.
Initial stability
• Righting Arm: A symmetric ship is inclined at a small angle
dΦ. C.B has moved off the ship’s centerline as the result of the
inclination. The distance between the action of buoyancy and
weight, GZ, is called righting arm.
• Transverse Metacenter: A vertical line through the C.B
intersects the original vertical centerline at point, M.
GZ  GM sin d 
 GMd  if d 
1
Location of the Transverse Metacenter
Transverse metacentric height : the distance between the
C.G. and M (GM). It is important as an index of
transverse stability at small angles of inclination. GZ is
positive, if the moment is righting moment. M should be
above C.G, if GZ >0.
If we know the location of M, we may find GM, and thus the
righting arm GZ or righting moment can be determined
given a small angle dΦ.
How to determine the location of M?
When a ship is inclined
at small angle dΦ
WoLo – Waterline (W.L) at upright position
W1L1 – Inclined W.L
Bo – C.B. at upright position, B1 – C.B. at inclined position
- The displacement (volume) of the ship
v1, v2 – The volume of the emerged and immersed
g1, g2 – C.G. of the emerged and immersed wedge, respectively
Equivolume Inclination
(v1 =v2 )
If the ship is wall-sided with the range of inclinations of a small
angle dΦ, then the volume v1 and v2 , of the two wedges between
the two waterlines will be same. Thus, the displacements under
the waterlines WoLo and W1L 1 will be same. This inclination is
called equivolume inclination. Thus, the intersection of WoLo,
and W1 L1 is at the longitudinal midsection.
For most ships, while they may be wall-sided in the vicinity of
WL near their midship section, they are not wall-sided near
their sterns and bows. However, at a small angle of
inclination, we may still approximately treat them as
equivolume inclination.
When a ship is at equivolume inclination,
vg1 g 2
B0 B1 
,

v1  v2  v
According to a theorem from mechanics, if one of the bodies
constituting a system moves in a direction, the C.G. of the
whole system moves in the same direction parallel to the shift of
the C.G. of that body. The shift of the C.G. of the system and
the shift of the C.G of the shifted body are in the inverse ratio of
their weights.
2 L 3
y dx

B0 B1
Ix
vg1 g 2
0
3
B0 M 


 ,
tan( d )  tan( d )


L1
L
2
2
2 L 3
3
vg1 g 2   y  ( y tan d )  (2  y )dx  tan d  y dx, I x   y dx
0 2
0
3
3
3 0
I x  the moment of inertia of W0 L 0 w.r.t. the longitudinal axis
For a ship inclined at a small angle d , the location of
its transverse metacenter is approximately above its
Ix
C.B. by
, which is independent of d .

K .M . (Metacenter measured from keel ), or
H M is the height of metacenter above the baseline.
Ix
K.M.= H M = + Z B ,

where Z B is the vertical coordinates of the C.B.
The vertical distance between the metacenter & C.G,
GM  H M  Z G 
Ix
+ Z B  ZG.

If we know the vertical position of the C.G., Z G and the
C.B., Z B the righting arm at small angles of inclination, d ,
 Ix

GZ  GM  d     Z B  Z G   d 


and the righting moment is
 Ix

M   w   Z B  Z g  d .


Examples of
computing KM
d
B
B
d
a ) Rectangular cross section
d
1
, I x  LB 3 ,   LBd
2
12
Ix
B2
BM  
 12d
B2 d
KM  BM  Z B 

12d 2
ZB 
b) Triangular cross section
2d
1
1
3
ZB 
, I x  LB ,   LBd
3
12
2
I x B2
BM  
 6d
B 2 2d
KM  BM  Z B 

6d 3
Natural frequency of Rolling of A Ship
Free vibration
 2
M X 2   w    GM    0
t
 w    GM
2
 
MX
where M X is the inertia moment of the ship w.r.t. C.G.
A large GM leads to a higher natural freq.
4.4Effects of free surfaces of
liquids on the righting arm
pp81-83
• When a liquid tank in a ship is not full,
there is a free surface in this tank.
• The effect of the free surface of liquids
on the initial stability of the ship is to
decrease the righting arm.
For a small parallel angle inclination,
the movement of C.G of liquid is
G0G1  d
I OL
 tan k
•The increase in the heeling moment due to the movement
of C.G. of liquid
M heeling   F  tan k  G0G1   F I OL d
If there is no influence of free-surface liquids, the righting
moment of the ship at a small angle dΦ is:
 I ox

M  GM    w  d  
 Z B  Z g     w  d


In the presence of a free-surface liquid, the righting moment
is decreased due to a heeling moment of free-surface liquid.
The reduced righting moment M’ is
M   M  M heeling
 I ox
 F I ol 
    w  d   Z B  Z g 

w  

The reduced metacentric height GM’:
GM  
I OX
 F I OL
 ZB  Zg 

w 
Comparing with the original GM, it is decreased by an
amount,
 I OL
F
w 
.
The decrease can also be viewed as an increase in height
of C.G. w.r.t. the baseline.
 I OL
Z g  Z g 
F
w 
How to decrease IOL:
•Longitudinal subdivision: reduce the width b, and thus reduces
I OL  b3l
Anti rolling tank
4.5 Effects of a suspended
weight on the righting
arm
• When a ship inclines at a small
angle dΦ, the suspended object
moves transversely
• Transverse movement of the weight
= h dΦ , where h is the distance
between the suspended weight and
the hanging point
• The increase in the heeling moment
due to the transverse movement
M heeling  w  h  d
In the presence of a suspended object, the righting moment &
righting arm are decreased due to a heeling moment of the
suspended object. The reduced righting moment M’ &
metacentric height GM’ are:
 I ox
w 
M   M  M heeling     w  d   Z B  Z g 
h
 w 

I ox
w
w
GM   GM 
h
 ZB  Zg 
h
 w

 w
In other words, the C.G of a suspended object is actually at its
suspended point
Because the suspension weights & liquid with free
surface tend to decrease the righting arm, or
decrease the initial stability, we should avoid them.
1. Filling the liquid tank (in full) to get rid of the
free surface. (creating a expandable volume)
2. Make the inertial moment of the free surface as
small as possible by adding the separation
longitudinal plates (bulkhead).
3. Fasten the weights to prevent them from moving
transversely.
4.6 The Inclining Experiment (Test)
Purpose
1. To obtain the vertical position of C.G
(Center of Gravity) of the ship.
2. It is required by “International convention
on Safety of Life at Sea.” (Every
passenger or cargo vessel newly built or
rebuilt)
4.6 The Inclining Experiment (Continue)
Basic Principle
M: Transverse
Metacenter (A vertical
line through the C.B
intersects the original
vertical centerline at
point, M)
Due to the movement of
weights, the heeling
moment is
M heeling  wh
where w is the total weight of the moving objects and h is the
moving distance.
4.6 The Inclining Experiment (Continue)
The shift of the center of gravity is GG  wh
1
W
where W is the total weight of the ship.
The righting moment = The heeling moment
GM W tan     wh
wh
GM 
 GG1 cot( )
W tan( )
1. w and h are recorded and hence known.
2.  is measured by a pendulum known as stabilograph.
3. The total weight W can be determined given the draft T. (at
FP, AP & midship, usually only a very small trim is allowed.)
4. Thus GM can be calculated,
4.6 The Inclining Experiment (Continue)
GM  H M  Z g ,
Ix
H M   ZB

The metacenter height and vertical coordinate of C.B have
been calculated. Thus, C.G. can be obtained.
Z g  H M  GM
Obtaining the longitudinal position of the gravity center of
a ship will be explained in section 4.8.
4.6 The Inclining Experiment (Continue)
1. The experiment should be carried out in calm water & nice weather. No
wind, no heavy rain, no tides.
2. It is essential that the ship be free to incline (mooring ropes should be as
slack as possible, but be careful.)
.
3. All weights capable of moving transversely should be locked in position
and there should be no loose fluids in tanks.
4. The ship in inclining test should be as near completion as possible.
5. Keep as few people on board as possible.
6. The angle of inclination should be small enough with the range of validity
of the theory.
7. The ship in experiment should not have a large trim.
4.7 Effect of Ship’s Geometry on Stability
Transverse metacenter height GM = BM – (ZG –ZB)
Ix
GM   ( Z g  Z B )

I x C1LB 3
C1 B 2
B2


C
 CB LBT CB T
T
I 
d x 
    2 dB  dT
B
T
 Ix 
 

where C1 dpends on waterplane.
4.7 (Continue)
 Ix 
d 
dB


a) Increase B only:
2
(  CB LBT increases)
B
 Ix 
 

I 
d x 
dT


b) Decrease T only:

( decreases)
T
 Ix 
 

dB
dT
c) Change B & T but keep  fixed:

B
T
 Ix 
d 
    3 dB
B
 Ix 
 

• Conclusion: to increase GM ( Transverse metacenter
height)
1. increasing the beam, B
2. decreasing the draft, T
3. lowering C.G (ZG)
4. increasing the freeboard will increase the ZG, but will
improve the stability at large inclination angle.
5. Tumble home or flare will have effects on the stability at
large inclination angle.
6. Bilge keels, fin stabilizers, gyroscopic stabilizers, antirolling tank also improve the stability (at pp248-252).
4.7 (Continue)
Suitable metacenter height
• It should be large enough to satisfy the requirement of
rules.
• Usually under full load condition, GM~0.04B.
• However, too large GM will result in a very small rolling
period. Higher rolling frequency will cause the crew or
passenger uncomfortable. This also should be avoided.
(see page 37 of this notes)
4.8 Longitudinal Inclination
Longitudinal Metacenter: Similar to the definition of the
transverse meta center, when a ship is inclined longitudinally
at a small angle, A vertical line through the center of B1
buoyancy intersects the vertical line through B0 (before the
ship is inclined) at M L.
• The Location of the Longitudinal Metacenter
For a small angle inclination, volumes of forward wedge immersed
in water and backward wedge emerged out of water are:
l
v1   (2 y )  ( x tan  )dx where y is the half breadth.
0
v2  
L l
0
Thus,
(2 y )  ( x tan  )dx,
l
L l
0
0
 2 yxdx  
v1  v2  v.
2 yxdx, which indicates:
moment of area forward of F = moment of area after F .
F is the center of (mass) gravity of waterline W0 L0 , &
is called center of flotation of W0 L0 . Therefore, for equal
volume longitudinal inclination the new waterline always
passes through the center of flotation (C.F).
• Location of the Longitudinal Metacenter
Using the same argument used in obtaining transverse metacenter.
B0 B1  vg1 g 2 / ,
l
0
0
L l
vg1 g 2   (2 y )  ( x tan  )  xdx   2 yx tan  xdx
l
0
2

 tan   2 yx dx   2 yx 2 dx   tan  I FC
 0

L l
I FC is the moment of inertia with respect to the
transverse axis passing the center of flotation.
I FC
B0 B1  B0 M tan  , B0 M 
.

I FC
I FC
H ML  B0 M  Z B 
 Z B . GM L 
 ZB  Zg


• Location of the Longitudinal Metacenter
Usually Floatation Center (C.F) of a waterplane is not at the
midship,
I FC  I 0,T  Ax 2 ,
where I 0,T is the moment of inertia w.r.t. the transverse axis
at midship (or station 5) and x is the distance from F.C. to
the midship.
• Moment to Alter Trim One Inch (MTI)
MTI: (moment to alter (change) the ship’s trim per inch) at
each waterline (or draft) is an important quantity. We may
use the longitudinal metacenter to predict MTI
• MTI ( a function of draft)
Due to the movement of a weight, assume that the ship as 1” trim,
and floats at waterline W.L.,
1"
1
tan   
,
where L is in feet.
L 12  L
Due to the movement of the weight, G0 moves to G1 ,
M  w  h  G0G1   w ,
G0G1  G0 M L tan    HM L  Z G  tan 
 I FC

M   w  HM L  Z G  tan    w 
 Z B  Z G  tan 
 

 I FC
 1
  w 
 Z B  ZG 
 
 12  L
• MTI ( a function of draft)
 w I FC
I FC 1

M   w
 12  L 12  L
I FC
3
(ton-ft)
 w  64 lb/ft , Long Ton = 2240 lb, MTI 
420 L
I FC
 Z G  Z B ,

• If the longitudinal inclination is small, MTI can be used to find
out the longitudinal position of gravity center ( X G1 ).
Trim TF  TA T
tan  


L
L
L
I FC T I FC
 I FC

G0G1  
 Z B  Z G  tan  
,
ZG  Z B
 L

 

Since G0 is in the same vertical line as C.B under W0 L0 ,
X G1  X B  G0G1  X B 
I FC T
 L