This is the third lecture. In the last class,
we talked about the Archimedes principle and
we showed some slides showing the effect of
Archimedes principle on floating bodies.
The principle states that the weight of the
volume of water displaced by the floating
body is equal to the floating body in the
case of a freely floating body.
If you have 
a freely floating body, then principle says
that the weight of the volume of water displaced
equals the weight of 
the body. Now the volume of the liquid displaced
is known as the displacement of the ship.
What this says is that displacement delta
which is the weight of the volume of water
displaced is equal to W, the weight of the
ship. Then we also know that or you can say
that gamma is equal to rho g. So, gamma into
del equals W, where del is the del is this.
It is the volume of water displaced by the
floating body; that is del.
So, gamma into del is equal to W, which you
can write as gamma into C B into LBT equals
sigma weight i, where the right hand side
sigma W i represents the net volume and it
represents the sum total of the weights and
each of them represent the different weights
that are acting on the ships.
This will include the dead weight, include
the light weight, the hall weight and the
dead weight etcetera. So, this represents
the total weight of the ship. This is the
weight of the ship and this represents gamma
C B LBT; we have already seen C B is the block
coefficient; block coefficient into LBT gives
you del which is the volume.
This is one first principle that we have to
study in order to understand the principles
of floating bodies. This is about the forces.
Then another thing that exists along with
this is the moments.
We have already seen, for equilibrium, the
net forces and net moments should be equal
to 0. This is the condition for equilibrium
of a floating body and if we consider the
moment,
Suppose, we have initially a body like this,
here you have the weight acting; this is G
and the weight acting delta. This is B; B
is the center of buoyancy; this is the centroid
of the immersed volume of the liquid. From
there, upward acts the delta and this is the
weight. So, this is weight W of the body and
from below the center of buoyancy upward acts
the displacement of 
the ship.
As you can see, there are two forces; these
two forces should be equal. W should be equal
to the delta in the case of a freely floating
body. This is the implication that the net
forces should be 0 on a ship on a freely floating
body.
The next one is that the net moment should
be 0. How will the net moment be 0? Net moment
can be 0, only if this weight W and delta
which we are given going up, and W coming
down - they act on a same straight line.
Like this, if they are on the same straight
line then you do not have a net moment. Then
net moment is equal to 0, in that case. Now,
suppose that the same body is by some method,
for example - by the shifting of weights - G
initial has moved from G 0 to this new G 1
point. The center buoyancy is here at B; initially
it was in the same straight line
But now, G 1 which is the new center of gravity
has moved to G 1 and so here the weight acts
W and here just as before, the displacement
acts upwards. So, the buoyancy force and the
weight W are separated by some distance.
Now, if this happens then there is a net moment
acting on the ship. There is a weight acting
downward here and there is a weight acting
upward which means there are two forces and
there is a distance between them; it creates
a moment. There is a tendency to turn the
ship in this fashion like this. So, when this
turns you will have,
This is the water line again and then new
G has moved here and because of this tilting
B has also moved. As a result of it, you will
get this delta that is acting upwards and
W that is acting downwards again coming on
a same straight line. This is the condition
for equilibrium. So, in this inclined position,
you will see that the net moment is 0 and
the net force is also 0.
Then, in the case of naval architecture - in
this case - instead of trying to draw all
the figures - instead of drawing the inclined
body bringing the inclined body into the picture
more than once it is easier to draw the waterline
itself
For example, instead of the body tilting,
if this is the 
initial state and this is the final state
in which case, the body has rotated by an
amount. Some more part has been submerged
here, so in this new water line, some more
part is submerged on the right side; some
more part is submerged on the left side. So,
in that case, instead of drawing the body
itself again in an inclined position, it is
better, if we change the water line to meet
this. For example, like this.
If I put a water line like this, the meaning
of this is that, this is the new waterline
this much has immersed a lot of region like
this much region has immersed and this much
region has emerged. So, this is the new waterline.
We call the initial waterline as W L and then
the new waterline as W phi and L phi. So,
you have W L and W phi L phi. This is how
you draw the inclined waterlines.
This is how you treat the inclined waterlines,
in the case of the naval architecture. This
is how we do it. Initially, you have the G
here and the B here; then here, G has shifted
here to G 1 and this is B 1 or B phi. We usually
write it as B phi. The book of Adrian Biran
writes it as always B phi. So, B 0 and B phi
and they will be in the same straight line.
Now, we already talked about the different
kinds of equilibrium stable equilibrium, neutral
equilibrium and unstable equilibrium. This
is something, which we have already discussed.
Suppose, the 
body has inclined with no change in G as such,
but the body has inclined due to some other
factor. For example, it could be a wind force;
a wind force acting on the ship will produce
an inclination heeling. So, the body is heeled.
This is B 0 and this is B phi. This is the
initial vertical and if I draw a vertical
from here. You have to remember, this is the
waterline and this is horizontal. Even though,
it looks inclined in this figure, the line
is actually horizontal. It is the horizontal
waterline that is drawn here and a perpendicular
to it, gets is known as it is always a vertical
line. So, this is vertical 
and where this line meets this point, we call
it as metacenter M and of this. GM is known
as 
the metacentric height. As we showed in the
last time, we have K, B, G and M. So, you
will see that the GM is equal to KB plus BM
minus 
KG and if this is M, another possibility exists.
That is, let us suppose the center of gravity
has shifted here and this is the new waterline.
Suppose, the vertical meets it here M, G,
B 0, M, G. In this case, you see that M is
below G and if M is below G, you can see that
the moment is again like this.
There is no restoring moment. Net moment is
to the net tendency is to increase the tilting
and this tilting will keep on increasing,
till the body capsizes. That is this figure.
So, what we see is that as long as M is above
G, as you see here, as long as M is above
G in this figure, the body is stable and if
M goes below G then the body becomes unstable.
Therefore the 
condition for stability is that GM should
be greater than 0.
Now, we will have a small recap of something
we already know. It is about the center of
mass or even center of volume. Now, we know
that the center of gravity of a body is given
by: if there are two bodies, x1 m 1 plus x
2 m 2 divided by m 1 plus m 2. This is the
position of the center of gravity of the body
x G.
If we move the mass m 2, a distance d like
this horizontally - m 2 has been shifted a
distance d horizontally - then x G will be
x 1 m 1 plus x 2 plus d m 2 by m 1 plus m
2 or the change in the center of gravity is
equal to d m 2 by m 1 plus m 2 or it is equal
to the change in moment divided by total mass.
If you are talking about a volume then the
change in the centroid of that volume will
be given by change in the moment of volume
divided by the total volume. This gives you
lemma in center of gravity, a established
result that you would all be familiar with.
Just to remind, the change in moment divided
by the total volume will give you the centroid
of the volume. That is something that we will
be using in many places.
Now, we have a problem here. Let us suppose
that we have a coordinate system like this
and we have a body; this is z, this is y and
this is x. Suppose, there is a body, let us
Suppose, this is the body here; this is the
water plane W 0 L 0 and then it tilts and
it gets a new water plane area, water plane
this is W phi L phi. Initially it is here
and then it is tilted into W phi L phi about
the axis of inclination x. The body has tilted
about x. Now, since there is no change in
the volume of the body, we can say that the
amount of volume that sinks in will be equal
to the amount of volume that emerges out.
This is true for, generally what we call as
wall sided bodies. So, W phi L phi comes out
and W 0 L 0 goes in. So, the volume that comes
out is equal to integral, let us say that
side is S 1, y tan phi that is because if
this is y, this is phi and this is y tan phi.
Here, this is y and this is y tan phi. This
is y; it is the y direction. If you put this
and if it is heeled by an angle phi, this
is y tan phi. So, the volume y tan phi into
d x d y. This is the general expression for
volume of any body. In this case, you have
y tan phi d x d y.
The volume that has come up is equal to, this
side is S 1 and this side is S 2. So, S 2
is volume that has emerged; S 1 is the volume
that has submerged. So, it is the volume that
goes in and S 2 is the volume that comes out.
S 2 is equal to minus y tan phi d x d y.
Now, from our definition of the type of body,
we have seen that V 1 is equal to V 2. That
is, the amount of volume that has submerged
is equal to the amount of volume that has
emerged.
So V 1 is equal to V 2. Therefore, double
integral y tan phi d x d y is equal to minus
double integral y tan phi d x d y. This is
s 1; this is s 2. If you bring this to the
left hand side, you will get 
double integral y tan phi d x d y over S1
plus S 2 is equal to 0. Now, tan phi is constant
because phi is a constant. So, that does not
hold. So, S 1 plus S 2 is the total surface
and let us call it as S.
y d x d y is equal to 0. Now, what is y d
x d y? y d x d y represents the centroid of
the surface S. 
For example, x c will be x d x d y divided
by double integral d x d y. What we see here
is that over the total surface S, y d x d
y is equal to 0. It just implies that this
represents the centroid.
So y c, so from this what we get 
is that, if there is a.. Let me read this.
What it says is that if you assume the initial
water line to be W 0 L 0, now after the inclination
along the axis of inclination and if this
inclination is very small then this inclination
will be a straight line that passes through
the centroid of the body.
So, this inclination will be, thus axis of
the inclination will pass through the centroid
of the water plane area. Centroid of the water
plane area in the case of a ship is known
as center of floatation and therefore, we
can say that the body inclines about the center
of floatation or the axis of inclination.
If you assume it to be the longitudinal axis,
it will be passing through the center of floatation.
This makes an assumption that the angle of
inclination is very small. So, if you have
a small angle of inclination, the body inclines
such that the axis of inclination passes through
the centroid of the water plane area.
This is a very important result and you should
know that always inclination whether it is
heeling or trimming occurs through the center
of floatation or the centroid of the water
plane area.
.
That is an important point in naval architecture.
Then, let us consider something else - that
is, BM. BM is known as the metacentric radius.
Now, let us assume because G remains the same
and because of some force acting or some wind
acting, there is a heel and there is an angle
of inclination and the center of buoyancy
shifts from B 0 to B phi. If that has happened
then the change in coordinate, how do you
find the change in coordinate? It is given
by the change in the moment of volume 
divided by the total volume. This, I write
so confidently because of the lemma on the
masses that I wrote in the previous paper.
We saw that the change in coordinate is the
change in the moment of volume divided by
the total volume.
Therefore, x B the change in this is equal
to 
the volume itself is double integral y tan
phi d x d y and that, we saw in the previous
section. y tan phi is the vertical distance
and double integral over the whole surface,
y tan phi d x d y will give you the volume
of the body. If you are trying to get the
moment then x into that y tan phi d x d y
divided by double integral y tan phi d x d
y which is equal to del. This is equal to
del. So, this becomes double integral x y
tan phi d x d y divided by del and this is
equal to I x y tan phi by del, where I x y
is the second moment of area or it is known
as the moment of inertia; the second moment
is known as moment of inertia.
So, the I x y, since it is x here, it is I
x y about the x axis. I x y tan phi d x d
y is I x y by del. So, this gives you the
expression for x B. Then the expression for
y B, the transverse moment is equal to double
integral over S again y into y tan phi d x
d y divided by del.
This gives you the shift in the centroid of
the body in the transverse direction because
of the heeling. Because of the heeling, there
is a shift in the centroid of the water plane
area and that shift is given by y into y tan
phi.
So, this is equal to double integral over
S, y squared tan phi d x d y by del. That
is equal to I by del into tan phi, where double
integral of y squared d x d y is the second
moment of area about the longitudinal axis,
this is about the x axis. About the x axis,
I is the moment of inertia about the x axis
and I by del into tan phi. This gives you
the shift in the centroid.
Then, let us consider this. Initially, this
is your metacenter; this is B 0. It has shifted
to a new point B phi 
and this is y B; this is z B.
This is W 0 L 0, W phi L phi. We can see that
the distance y B is really this distance,
but since we have already said that our angle
of inclination phi is 0, we can see that y
B is almost approximately equal to the length
of the arc B 0 B phi. This B 0 B phi is an
arc and the length of that arc is almost equal
to y B, if the angle of inclination phi is
very small.
In that case, you get B 0 B phi equals y B.
y B is equal to the length of the arc. Length
of the arc as you know is given by r theta.
So, B 0 M into phi and here we have that y
B is equal to I by del tan phi. Now, if phi
is very small, we can write this as I by del
phi because tan phi approximates to phi in
the case, phi is very small. So, I by del
phi and these two are same.
Therefore, you get I by del phi equals B 0
M phi. So, cancelling out phi, you get B 0
M equals I by del. Therefore, the Metacentric
radius is given by 
I by del; that is a very important formula.
That is always that Meta centric radius. You
can see that in a special case. For instance,
when there is no water plane area. For example,
in the case of a submarine, which is completely
submerged? If the submarine is completely
submerged then, you know that I as you can
imagine there is no I, there is no water plane
area.
Water plane area, we have already said, is
that area where the water line exists - the
area at the point of water line. If there
is no water plane area then there is no I
and therefore, B M is equal to 0. B and M
coincide in case of a submarine in a submerged
condition. So, such different kinds of ideas
you can get.
Then 
it is probably wise to remember the moments
of inertia of various bodies, various types
of water plane area. For instance, if you
have a rectangular water plane area and suppose
that we are taking this is L, this length
is L and this length is breadth.
If you are taking moment about this, then
the moment of inertia of this system is given
by l b cubed by 12 about this longitudinal
axis, the length of the water plane. The moment
of inertia is given by l b cubed by 12 and
in case, you are taking the moment of inertia
about this axis, then moment of inertia is
given by b l cubed by 12. Moment of inertia
about a transverse axis is b l cubed by 12.
Similarly, if you have a circle, the moment
of inertia of this kind of area is equal to
d power 4 by 64. It gives you the moment of
inertia of a circle about an axis that is
passing through its center. So, if any axis
passing through the center is considered,
you will get the moment of inertia as d power
4 by 64. So, these are some important formulas
and now, let us do some problems regarding
this that comes in this chapter.
For instance, suppose, there is a buoy. The
problem is as follows. There is a buoy 
and this buoy is of radius something, let
us say. It is of 
mass M and it is a hollow sphere. So, there
is an inside volume and an outside sphere
and small amount of material of some thickness
and then inside radius is a smaller radius.
So, it is a steel plate of 3 mm thickness.
So, the thickness of the steel plate is 3
mm and the body itself has a mass capital
M.
Let us suppose that the density of the material
of the buoy is rho s. Your question is, under
what condition will the body float?
So, the problem is you have a buoy which is
actually a hollow spherical shell. There is
an outside surface and there is an inside
surface; the thickness of the plate is 3 millimeter
and the rest of it is hollow and this is placed
on water. Under what conditions will the body
float? That is your question.
First of all, we bring in the Archimedes principle.
When you have the Archimedes principle, you
say that gamma del is equal to W. This is
the first principle or rho of water into this.
It is not the rho of the steel plate, this
is the density of water into del, which equals
the mass of the body M in this case. We have
removed the acceleration due to gravity - the
g term has been removed and therefore, you
get the formula that rho W into del is equal
to M. M is the mass of the floating body.
So this is the first rule, Let us assume that
the volume of the submerged sphere and this
is equal to 4 by 3 pi d cubed, 4 by 3 pi d
0 cubed; this is the volume of the submerged
half sphere. Half of it.
So, it is floating like this; with the center,
that has to be mentioned. The body is floating
with the center at the water line. So, half
of it is inside and half of it is outside.
The volume submerged is given by 4 by 3 pi
d 0 cubed into half. This is the 
volume of submerged half sphere. Volume of
the submerged half sphere L is equal to half
4 by 3 pi d 0 cubed.
Then the mass of the spherical shell 
is given by M of steel is equal to rho s into
4 by 3 pi. Remember, this is a hollow sphere;
it has an inside and outside and the mass
comes only from the steel. The mass does not
come from the hollow part inside; that does
not add to the mass. When you are finding
the mass of the body, you need to take only
the steel volume into consideration. So, 4
by 3 pi into d 0 cubed minus d 0 minus 0.003
cubed.
Now, putting this formula rho W into del is
equal to M, you get rho W into half into 4
by 3 pi d 0 cubed is equal to rho s into 4
by 3 pi into d 0 cubed minus d 0 minus 0.03
cubed. So, this gives you the formula and
from this, you will get when you do this,
you will get a cubic equation and when you
solve that, you will get the value of d 0.
So, that will give you the condition that
is required for the body to float in this
fashion, as we have just said.
Then there is the second problem. Another
problem states that there is a cone floating
upside down in water and this is a spherical
cone; this is a circle and it is a cone. This
has a draft T and it has a depth H.
So, we are told these things; this is T and
this is H and this distance, this diameter
is capital D; this diameter is small d. So,
small d and capital D. These two are given.
and then you are told also about the cone.
The specific gravity is gamma c for the cone
and specific gravity is density into g. Specific
gravity of the cone is given as gamma c, specific
gravity of the water is given as gamma W.
So, gamma c and gamma W and now, the question
is under what condition will the cone float
in this fashion.
For this kind of problem, we need to have
two conditions. The first of it is the Archimedes
principle: The weight of the volume of water
displaced is equal to the weight of the body
itself.
That is number one principle. The second principle
which we saw today is that GM should be greater
than 0. If both these conditions are satisfied
then the body will float up right as we have
said in this problem.
First Archimedes 
principle says that gamma c into volume of
the cone and this is equal to the gamma pi,
this is the weight of the cone itself which
is equal to the specific gravity of the cone
into the total volume of the cone. So, pi
D squared by 4 H into 1 by 3; the volume of
a cone is equal to 1 by 3 pi r squared H - 1
by 3 into pi D square by 4 into H; this is
the volume of any cone.
These are some common geometric terms that
you should be familiar with; that is the volume
of a cone, volume of a parallelepiped or a
volume of a rectangular barge, volume of a
sphere and similarly, the moment of inertia
of a couple of things like moment of inertia
of a spherical surface, moment of inertia
of a rectangular surface, moment of inertia
of square and parallelogram. These kinds of
things, it is better to remember when you
are studying this section.
So, Archimedes principle says that the mass
of the cone that is gamma c into 1 by 3 pi
D squared by 4 H is equal to the weight of
water displaced r w into pi, this much volume
of water has been displaced, pi d squared
by 4, 1 by 3 again, 1 by 3 pi d squared by
4 into T; in this case, the height of the
cone is T.
So, the volume of water displaced is this.
It is 1 by 3 pi d squared by 4 T. This is
the volume of water displaced; into gamma
w will give you the weight of the water displaced.
So, the weight of the water displaced is equal
to the weight of the cone here.
This just represents the weight of the cone
which is equal to specific gravity of the
cone into the volume of the whole cone. Just
to stress on this again, there are two volumes:
here, one is the volume of the whole cone
and other is the volume of the area submerged.
When you are doing the mass of the cone, you
take the whole volume of the cone and when
you take the weight of liquid displaced, you
take only the volume that is submerged. So,
this gives you gamma W, This is the first
principle.
We can put a kind of geometrical similarity
because it is the same cone. Geometrical similarity
says that d by D is equal to T by H; d by
D is equal to T by this is not D, sorry this
is H. I wrote wrongly; this is H. So, d by
D is equal.. This is D; this is small d and
this is H. So, d by D is equal to T by H is
a geometrical similarity and let us say that
alpha is equal to gamma c by gamma w power
1 by 3. So, gamma of the cone by gamma of
water power 1 by 3 and we will write it as
alpha.
So, we obtain T is equal to therefore, alpha
H from the previous principle. Now, we need
to get GM. GM is our final goal and we need
to say that GM of this system will be greater
than 0 and that will imply stability. GM is
greater than 0 will imply stability.
To get GM, we need to get KB plus BM minus
KG. This is the way to get GM. This is the
only way to get GM. KB plus BM minus KG will
give you GM, the metacentric height. The metacentric
height will be given by KB plus BM minus KG.
So, we need to find for this cone, what is
KB, what is BM and what is KG. We have formulas
for everything.
For instance, if you look at this, KB is the
centroid of this region of the cone - the
cone inside. It is given by KB. So, KB is
equal to 2 by 3 of T; this is another thing.
If you have a right angled triangle, the centroid
of this triangle is at 2 by 3 from here - 2
by 3 from this side or 1 by 3 from this side,
of the total height H.
So, 1 by 3 H or 2 by 3 H. KB is equal to 2
by 3 T and KG is equal to 2 by 3 H. KG is
for the whole cone. Remember, we are talking
about the center of gravity of the whole cone.
So, that will be at 2 by 3 of the H. 2 by
3 of this whole distance. The distance from
here will be 2 by 3; the distance from here
is 1 by 3. So, 2 by 3 H.
KG is equal to 2 by 3 H and BM we have seen
is equal to I by del. I is equal to pi d power
4 by 64. Therefore, BM is equal to pi d power
4 by 64 divided by del, where del is equal
to 1 by 3 pi d squared by 4 into T; this is
the underwater volume. I do not have to repeat
it. This is the underwater volume, divided
by del. When you do that there, you will get
BM for the whole thing.
Once you have, all those things, you can get
GM. You do KB plus BM minus KG and just putting
all these things together, we give the condition
that GM should be greater than 0; this is
the condition for stability.
So, GM is greater than 0 and when this condition
is met then you have the case of a cone exactly
sitting in an upright condition like this.
So, this is the problem. We will work on similar
problems in the next class and we will stop
here, today.
Thank you.
