Now, there are a couple of other coefficients,
which are also important, though the main
one is a block coefficient. One of them 
is the midship coefficient. It is defined
as the ratio of the sectional area under water
to the breadth and drafts. It is defined to
be C M. midship sectional area and C M is
equal to A M by B into T. As I told you before,
we divide the ship in two ways. If you define
this to be the ship, this direction is known
as the longitudinal direction. When you see
from here, what you see in that area? It will
look like this and the area is known as the
sectional area. The direction is known as
the traverse direction and it is known as
the sectional area. Now, this area is what
we are talking about here, so it is this area
divided by B into this draft. The midship
sectional area will look something like this;
this area sectional area, which is written
here as A M divided by B into T. So, you will
have the sectional area at the midship section.
It is known as the midship coefficient and
it is designated as C M. So, this is the second
coefficient. I already told you the first
one is the block 
coefficient and then you have the midship
sectional area coefficient.
We call something as the prismatic coefficient;
it is defined as C P is the ratio of the molded
displacement volume del. Now, it is little
more difficult to draw this figure, so I leave
it. Prismatic coefficient is defined as the
ratio of the displacement of the ship del;
it is the volume displacement of the ship.
Delta is always the volume displacement and
we call it as rho into delta. It is known
as the displacement weight of the ship. So,
you have C P, the prismatic coefficient as
delta divided by A M into L, where A M is
the midship coefficient. If 
you consider this to be the length of the
ship, then midship sectional area into L will
give you, this is length L of the ship, midship
sectional area into L is this and it is the
denominator. Delta operator takes up the numerator.
So, this is the prismatic coefficient.
There is a water plane area coefficient. It
is denoted as C WL, it is equal to A W divided
by L into B. If you have a shape like this,
A W is the water plane area and let us say
that this is the water line, this is the ship,
this is the water line and then A W is this
area. This water plane area divided by L into
B, where L is this distance and B is this
breadth. This is known as the water plane
area.
There is a vertical prismatic coefficient.
Vertical prismatic coefficient is defined
as C VP. It is equal to delta divided by A
W into T. So, I leave it, as the definition
itself says what it is. C VP is equal to delta
divided by A W into T, where T is the draft;
A W is the water plane area. You should know,
what is the water plane area? for sure. Water
plane area means, if you have the ship like
this, there is a water line. At the water
line, there is a horizontal area of ship and
that area is known as the water plane area
told as A W into T.
So, we defined a couple of things in this
chapter. We defined a block coefficient and
we defined what is known as the displacement,
displacement volume, displacement mass or
weight, midship coefficient, then midship
sectional area, prismatic coefficient, vertical
prismatic coefficient, water plane area. Among
these, I think, what is water plane area is
very important. You have to know what is a
water plane area? then the water plane area
coefficient. These are some of the terms we
learnt in this chapter.
This C P becomes C B by C M. The prismatic
coefficient is equal to the block coefficient
divided by the midship sectional area coefficient.
So, this is where we leave chapter 1. Next
is the second part, which deals with Archimedes
principle. This is first principle of this
topic. The most basic principle with which,
all rest of the study proceeds. As you know
the famous story - Archimedes is supposed
to have discovered this. When he discovered
it, he was so excited that he ran naked on
the streets shouting ‘eureka’. That is
the famous story of the Archimedes principle.
Let us take a look at what is Archimedes principle.
Before we do it, we make some assumptions:
water is incompressible; incompressible is
a word used in fluid mechanics, which implies
that there are no density changes in the medium.
So, the water has a constant density. It does
not have any variability’s in density. Water
is incompressible and then water is an ideal
fluid. It is an ideal fluid; it is again an
important term in fluid mechanics.
You have two types of fluids: ideal fluids
and viscous fluids. Now, all the liquids that
we consider, including water are all viscous
fluids. In many cases, we can assume that
the fluid is ideal without any viscosity and
that assumption is made in this case. Then,
we say that there is no surface tension and
that is an assumption made here because it
makes changes to the water line. It would
not be a straight line and that is last assumption
that is water surface is plane, so the surface
of the water is plane. So, these are some
assumptions we make.
We now go into what is known as the Archimedes
principle. The principle itself states that
if a body is immersed in a liquid, it will
experience an upward force and the magnitude
of that force is equal to the weight of the
liquid displaced. When the body is immersed,
it displaces some liquid. The volume of the
liquid displaced will be equal to the volume
of the solid immersed. The part of the body
immersed has a particular volume and that
volume will be equal to the volume of the
liquid displaced. In addition, if rho is the
density of that medium; rho into that volume
will give you the weight of the liquid displaced
and that is the amount of weight that is displaced.
This figure tells that any object wholly or
partly immersed in a fluid is buoyed up by
a force equal to the weight of the fluid displaced
by the object. So, the object displaces weight
of the fluid. Here, you have seen this weight
is put into the water and it has displaced
the 3 pounds of water. So, this is experiencing
an upward force of 3 pounds of water. This
upward force is known 
as the buoyancy force and this buoyancy force
is equal to the weight of the liquid displaced.
As this figure clearly states, there are many
derivations for the Archimedes principle.
We will take a look at some of them.
Suppose there is a solid, which looks like
this; face 1 is this, face 2 is below; it
is not this one, it is the face below, this
is 3, this is 4, this is 5, this is 6. So,
these are the six faces of this solid. Let
us assume that the water plane is here. So,
this is the water plane 
and it is touching the surface 6 and surface
3 at this point. Now, you know that as per
the physics, all these faces will experience
a pressure due to the liquid, but they will
also experience a force.
Now, for example, this is the fourth side;
this one, the curve. I mean the face opposite
to 6 is 4. So, if you are familiar little
bit with the hydrostatics of fluids, you will
see that if this is the body here, you have
the face 1 and if this is T, then you will
have the pressure distribution. Pressure distribution
is given by rho gz; this is the pressure at
any depth. It is given by rho gz, where z
is the distance from the free surface. Now,
in this book, rho g is always written as gamma.
So, we have rho g as gamma and gamma z will
give you pressure at any point.
Now, the pressure gamma z into dz will give
you the force per unit length of the side.
If you multiplied it with L, you will get
the force on the side 4. The pressure distribution
will look like this and it will look like
a triangle because it is rho gz. As z increases;
the values keeps increasing, so the whole
picture will look like a triangle between
pressure and depth. Here, p 0 L into T, where
p 0 is the atmospheric pressure. The side
4 will experience it; it is experienced on
the whole area. It is the atmospheric pressure
plus the pressure due to the liquid. This
is due to the atmospheric pressure and this
is due to the liquid. So, F 4 is known 
and then you need to find out F 6.
If you see, F 6 will be exactly opposite.
Now, pressure into area p dA will give you
dF. The direction of dA defines the direction
of the force. Pressure has no direction and
force has this direction. In this case, since
the area vectors diverge in the opposite directions,
one goes like this and another goes like this.
It will be of a different sign from the previous
one on F 6 that is minus L gamma z dz plus
minus p 0 L T and both will be negative. When
you add these two together, it implies F 4
plus F 6 is equal to 0. Both of them are opposite
of each other and so when you do F 4 plus
F 6, you get 0. It shows that when a body
is put into the water, it does not experience
any horizontal forces. This is known or otherwise,
it would be a kind of free propulsion. Without
applying any power, you get free propulsion
and it does not happen.
Now, let us look at the vertical direction.
This one is F 1 and as you can see, F 1 will
be equal to minus p 0 into L into B L is this
and B is this. So, L into B gives the area
p 0 L into B and this gives you the force
on the side 1, force on the side 2, which
is at the bottom. It will be plus p 0 into
LB. If you take this dA as negative, then
this dA you have to take it as positive.
So, it is plus p 0 LB plus gamma into LBT.
LBT is our volume under water, this is T.
Gamma into T will give you the pressure at
that depth. It is just gamma z and so gamma
into T gives you the pressure at this depth
of 2. What is the pressure? It is given by
this gamma into T. L into B will give is a
multiplication with area. Therefore, you will
get the total force. So, F 1 plus F 2 is therefore
equal to gamma into LBT. LBT is the underwater
volume and multiplied with gamma, which is
the density of water. It gives you the underwater
weight and what it really means is that it
is the weight of the liquid displaced. So,
you get that F 1 plus F 2; the net force on
the body, which is a net upward force.
It is directed in the upward direction, pushing
the body up. As you can see, it is given by
3 pounds. As you see here, 3 pound weight
of the water is directed upwards. It lifts
the body up and this is the weight of the
liquid displaced. It is equal to the force
of the body and this is the Archimedes principle.
It can be done in some other ways. In the
same figure, for instance 
that the area under a pressure length curve
would give you the force on a body.
If you take that half base into altitude and
if you consider one side as pressure, one
side as the breadth, one side as the length,
then you put it on a pressure length curve.
The area under the curve will be half base.
Base is equal to gamma T, the pressure at
the base is gamma T and half base into altitude.
Altitude is T and so you will get half gamma
T square. This will be the area of the triangle
and the force on fourth side. This will be
equal to L into half gamma T square plus p
0 LT. It comes from the atmospheric pressure
and similarly, it is the same as what we got
before. So, this is another way in which you
can do it. If you do not like integrals, you
can just take it as the area under a curve
and do it.
We can look at the general case. Let us assume
we have a coordinate system like this – y,
z, x. Suppose you have a body that is immersed
in the liquid at the center and let us say
that the surface is given by z equal to f
1 of x, y. The bottom is given by z equal
to f 2 of x, y and the surface is defined
in this fashion. The top half of the surface
or top part of the surface is defined as z
equal to the upper part of the surface. It
is defined as z equal to f 1 of x and y. The
bottom half is defined by z equal to f 2 of
x into y. Therefore, z equal to f 1 of x into
y, which is the top surface in S 1. In S 2,
z equal to f 2 of x y, this is the basic definition
of the surface.
We can see that 
the pressure force is given on the surface.
Let us take the pressure force on the upper
surface. Here, p dA equals to gamma f 1 of
x y into dA. Now, the vertical component of
this force - if you look at this figure, we
now consider a small element d, like this.
This will be in direction normal to the surface;
it is dA and it is directed normal to the
surface dA. It will have an angle with the
vertical, whatever be that angle. Gamma f
1 of the vertical component 
of this pressure force is given by gamma f
1 of x y into dx dy into cos of the angle
between n and z and so it is a cos of angle
between n and z.
This will give you the vertical component
of the force. All I have done here is, if
n is a normal like this, then the vertical
component of the force that is the component
of this force in this direction is this force
into cos of the angle between them. It will
give you the component of the force in the
vertical direction and that is what is given
here. Vertical component of the force is equal
to gamma 1 into dA into cos n z. If you put
that, it will be an element dy dx. This is
basically the projection of dA cos n z. It
is actually the projection of dA on to this
plane. It is given by gamma 1 f of this. It
becomes dx dy and similarly, the same reasoning
we will get from the bottom side of the vertical
component of the force.
The vertical component on side S 2 will be
gamma 1 f 2 and this is f 1 f 2 of x y into
dx dy. Therefore, you get the total force
F equal to the sum of these two gamma 1 into
a double integral or I will write as dF equal
to gamma 1 into f 1 of x y dx dy and the top
surface. If that direction is taken as positive,
the direction of the bottom surface has to
be taken in negative as minus f 2 of x y dx
dy. So, dF is equal to gamma 1 into double
integral F is equal to surface area f 1 of
x y minus f 2 of x y dx dy. Now, what is this?
This is basically z dx dy. It is equal to
the volume of the body and this volume is
the submerged volume. So, gamma 1 into submerged
volume.
The force on the body, the upward force or
the vertical force is equal to gamma into
volume submerged 
that is double integral over S. So, it is
equal to the volume submerged. This gives
you the upward force on the body and similarly,
if you try to look at the horizontal component,
it will be like this. In this figure, you
will have to 
get the horizontal force. You have to take
the projection on to this side, y z. It will
give you p into cos of the angle between n
and x. It will give you gamma into z dy dz
and this will give you the horizontal force
on one side. On the other side, the horizontal
force will be minus gamma z dy dz, which will
imply that when you add the two forces together,
it will sum up to 0. Therefore, the body does
not experience any horizontal movement. It
does not experience any horizontal force as
it is submerged in the liquid. So, there is
no horizontal force, but the vertical force
is given is equal to the weight of the liquid
displaced.
Now, for a body that is freely submerged in
water that is for a body that floats on water
remains at equilibrium. It is in a stable
state without any support, it remains floating
on the surface of water. You can give this
formula delta, which is equal to the weight
of liquid displaced. It is given by gamma
into delta, where delta is the underwater
volume.
If you have a body like this, it is very important
to note that we are talking about the underwater
volume and we are not talking about the total
volume of the ship. This is the total volume
of ship and we are talking about underwater
volume of the ship. It is delta into gamma,
where gamma is the density of water and not
the density of the material of ship or anything.
These are the two confusions that come up
in many places that is gamma is density. This
is actual the weight of the liquid and the
displaced volume of liquid is equal to the
volume submerged because, what is gone in
has to go out as it has no other way.
If x amount or V amount of volume goes into
the water then V amount of liquid will be
displaced that is delta. It is multiplied
by the density of water, you will get the
weight of the liquid displaced. If the body
is floating freely, then this gamma into delta
is equal to W and that is the weight of the
body itself. If density of the material of
the body is rho, it is rho into V, which is
the total volume of the body. So, rho is the
density of the material, it means rho into
total volume of the body and it will 
give you the weight of the body.
In this figure, If we had some case like that
we could say that the weight of the body in
this case, there is a tension acting upwards,
there is a weight acting downwards and there
is a 3 pound force acting downwards. In this
case, I would say that t, the tension plus
3 is equal to w, the weight of the body. So,
t is the tension in the rho plus 3, which
is the upward force equal to w, which is the
weight of the body. In all other cases, you
will have delta equal to the weight of the
liquid displaced. We can write this in many
ways, for example, gamma into C B into LBT,
where C B is a block coefficient. I have already
told you that C B is equal to delta divided
by LBT, so C B into LBT is equal to delta.
Gamma into delta is given by gamma into C
B T. This is equal to the total weight on
the ship and this is equal to the sum of all
the weights on to n W i. It is equal to sum
of all the weights in the ship, where W i
is the weight of the i th item of ship. There
are different items on the ship and that is
given here. This is something about the forces.
In this subject, we will deal with two things.
We will deal with forces and we will deal
with moments that occur in ship itself.
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For example, a body is floating like this
and the weight of the body acts down at G,
which is at the center of gravity of the body.
A weight acts down at the center of buoyancy
and a weight or a force acts upward. Now,
the center of buoyancy is defined as the centroid
of the underwater volume. If a body is floating
in water, a part of it will be in underwater.
Centroid of that underwater part is known
as the center of buoyancy. It is a centroid
of the volume; the weight is actually the
centroid of the underwater volume of the ship.
We call that as the center of buoyancy as
B and G is the center of gravity of the ship.
Now, the center of gravity of a ship depends
upon the weights on the body. It does not
have to do anything with water or the upward
force from water. It has to do with the weights
that are placed on the body. When you place
weights at different places, you get the center
of gravity. For instance, in this case, you
see center of gravity and center of buoyancy
in the same straight line. If they are in
the same straight line, you can see that they
do not produce any moment. So, this system
does not feel any moment, but let us say that
I removed some weight from here and pushed
it here. Some of the weights on the ship on
this body is taken from here and put on this
side. So, there is a shift in the center of
gravity.
You will have the system like this. Now, center
of gravity is here and the center of buoyancy
has not changed. Center of buoyancy is still
at the same place. At this point, a weight
delta acts at this point and the weight W
acts at that point. The weight W acts downwards
and at this point, delta acts upwards. We
have the center of gravity and center of buoyancy
at different points. They are not on the same
straight line. In this line, it acts here
and this acts in this line. We have two forces
acting in the opposite direction, separated
from each other. As you know, they will produce
a moment. In this case, there is a tendency
to trim like this. It has a tendency to heal;
the word is not trim, sorry it is heal. It
has a tendency to heal like this and then
the body will become like this.
In this case, what will happen is that the
body will tilt, such that G and B again come
in a same straight line. The moment it comes
in a same straight line, the moment stops
and the body becomes stable. In this case,
the body remains in inclined position. So,
this is what we mean by studying about the
moments. There comes a very important concept
that is many cases are not studied properly.
This should be looked about the water plane
of the ship.
Now, this is the initial water plane of the
ship, W 0 L 0. Suppose, the 
body tilts like this and it is very important
that you understand this figure properly.
What has happened here? This body was initially
in horizontal state. The body has tilted in
this direction and therefore, some part of
the body has gone at this site and another
part of the body has come out at this side
and this is still the water line. Instead
of trying to draw the ship like this and then
tilting it. It is easier to tilt the water
line. Initially, the water line is like this,
if the body has titled in this fashion and
something has gone under here and came out
here.
If you draw the water line in this fashion,
this is what it is. What this figure shows
that the body, which was initially at W 0
L 0, when it was in a horizontal position
is tilted in this direction to become w phi
l phi like this. We write the water line as
tilting and this is done. This is called as
W phi L phi and this is the new water line.
This figure indicates that this body is not
in a straight position, but it is in an inclined
state. There are some other formulas that
need to be satisfied. In case of floating
bodies that is for the body float upright.
It is just not enough that the body just floats,
it is also important that the body or the
ship float upright. It does not heal or trim
too much, it floats upright.
There are some other formulas that needs to
be satisfied and that is what we are going
to do next. Initially, we have G here and
this is known as G 0. At this point, you have
B 0. Here, you have W - the weight of the
ship acting. You have delta - the ship the
buoyancy force acting upwards. The body is
tilted and B 0 shifted to this point B phi.
It is the new point, this line gives the vertical
at that point 
and here, it hits the vertical. Let us say
that G 0 is here and B phi goes up and hits
the vertical. Here, this point of intersection
is known as M. Since the figure is a little
confusing, let me do it again.
This is the body and the initial water line
is here as W 0 L 0. Here, you had G 0 and
here you had B 0. Something shifted and as
a result, the body had to tilt this much.
This is the new water line W phi L phi and
it has shifted here as B phi because of this,
some wedge has gone out here and some wedge
is coming here. Therefore, B will move in
the direction of increased volume. So, B 0
to B phi and here, if you draw a vertical,
this hits the original vertical. Remember,
this line is the vertical, because this water
line is always vertical W phi L phi. It is
the horizontal line and not vertical. W phi
L phi is a horizontal line, so perpendicular
to that is always the vertical line. So, B
phi into M is a line and this M is known as
the metacenter. A very important concept in
naval architecture is known as metacenter
point M. Before I go into that we should know,
This is how it will look. In this case, the
ship has tilted in this direction. Some part
has gone in here. In this case, it is different;
the initial is WL 1. When the ship was stationary,
the ship was horizontal. When it tilted, the
water line came to WL, like this and B shifted
from here to here. Initially, you have B and
this is B prime. So, there is a transfer of
wedge to one side and this is the figure.
When you look at different points, these are
the different points that you need to know.
In this case, for example, M is a metacenter;
it will be at the top mostly. G is the center
of gravity coming below it. B is the center
of buoyancy coming here, then K is the keel.
Keel is the bottom most hull part of the ship.
It is known as the keel and k is the basic
reference point, from which all other points
are measured. So, K is defined to be z equal
0 and all these have particular values. You
have KB, KG, KM, GM and BM. We have different
things and we call differently. For example
, there are two things: one is called as GM
- this is known as 
metacentric height and then BM, which is known
as metacentric radius.
Here, you see that BM is known as a metacentric
radius and GM is the metacentric height. You
have KM that is another parameter.
Now, couple of small points here, first of
all, initially the center of buoyance water
line was WL. Suppose, WL was the initial water
line, B is the centroid of the underwater
volume. I have already told you, it is the
centroid of the underwater volume. So, you
have B. Now, due to some reason, the ship
sunk a bit more, such that the water line
has raised above and it has now gone to WL
1. If that happens, then the center of buoyancy
B 1 as the water line goes up. So that is
how the ships center of buoyancy behaves.
The next one is about G. In case you add weights,
G shifts towards that point. So, whichever
point, you add some weight, G shifts towards
that point. If you remove weight from some
place, G moves away from that point. So, G
moves away from weight removal, G moves towards
weight addition and G moves in the same direction
as the weight shift. So, if there is a shift
of weight, body will move in that direction.
In this figure, we have seen this M is above
G and that is the condition possible. Now,
another condition is also possible. As you
can see, when it reaches this condition; here,
you have a delta weight – W. You have this
weight delta and displacement delta and you
can see that. In this case, there is a tendency
for the body to shift back to its original
position of equilibrium. Before we do that
we have to define what equilibrium is.
In general, we say that there are three types
of equilibrium. We call them as stable equilibrium,
neutral equilibrium 
and unstable equilibrium. The meaning of these
terms are: a body is said to be in stable
equilibrium, if it is in some position and
some force acted on it or a moment acted on
it to change its position or its orientation,
the moment the force is removed, the body
returns back to its original position. That
kind of equilibrium is known as a stable equilibrium.
Now, on the other hand, if a body is in some
position and suppose a force acts on. It continues
as long as and it tilts, when the force is
removed, the body does not come back to its
original position, but remains as such without
changing its position in a tilted position
or in some other form. It is mainly in the
tilted form with the angle of heal. If that
happens, then it is said to be in neutral
equilibrium.
The third one is unstable equilibrium. If
the body is in a particular position and if
a force acts on it or a moment acts on it
and it tilts. If the body does not return
to its original position, but it continues
in its original tilting and increases the
tilting, such that it capsizes. Such a situation
is known as an unstable equilibrium. These
are three forms of equilibrium: stable, neutral
and unstable equilibrium. Unstable equilibrium
is - once the force is removed also, it capsizes.
In stable equilibrium - the body returns to
its original position, after the removal of
the force or moment. In neutral, it remains
at the same position. So, these are some forms
of equilibrium.
Now, we need to figure out, how this applies
to the case of ships. I mean, how this stability
parameterization that is the next step. We
will continue in the next lecture. So, today,
we will stop here. Thank you.
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