Tip:
Highlight text to annotate it
X
We are moving into the lecture 32 and we continue with the bilging. We will most probably finish
up the bilging section today. As we have explained, there are couple of different ways in which
you can have the bilging that occurs in different parts of the ship in all our simulation. So
far we have considered bilging to be occurring in the midship compartment of the ship that
is definitely one possibility. You have the bilging occurring in one compartment
which is a distance that is smaller than the length. If the total length of the ship as
length between perpendiculars you have a small length between two fixed stations. You have
a compartment which has the flooding in it and bilging occurs only in that compartment.
Then, another possibility is that you can have the bilging at different other sections
of the ship for instance, you can have bilging in the front part of the ship that is, the
bore region of the ship, you can have bilging occurring at the aft region of the ship. As
this happens when you have the flooding occurring in different other regions of the ship. In
case, the flooding occurs at the front part of the ship that is in bow the ship if some
compartment there gets flooded then, the ship will trim forward. Of course, what we mean
by trim forward is, the draft in the forward section of the ship will become more than
the draft in the aft section of the ship. So, if you consider this to be the ship, if
this is the front and this is aft of the ship, the ship will trim like this .
So, this is what happens if this compartment is flooded you can look at it - in any ways
you can look at it - one possibility is, in increase of weight as we have already explained
the analysis for flooding is usually done using two methods, one is using the method
of added weights and one is using the method of loss buoyancy. Both the methods are exactly
identical in the sense that not identical in the method of calculations they are different,
but they are identical in the result. For example, if you are considering the righting
arm finally, you will get the same result whether you use the method of added weight
or the method of loss buoyancy. If you consider this problem of flooding using
the method of added weight, you can see what happens; the ship is initially on even keel,
I have already described that when you say something on even keel it means that the water
line is horizontal. Initially, if the ship is on even keel or any other trimmed condition,
suppose that one compartment in the bow region of the ship gets flooded. There is an increase
in weight in that section of the ship and the ship is trimmed by forward. So, the front
part of the ship is actually going down as you can see it will be like this now .
That is another kind of problem and that is one of the problem that we will consider today.
As we have already described there is another possibility is that the ship can be flooded
not exactly midships, means not exactly along the center line. If you consider it symmetrically
flooded means, the port side and the star board side are evenly flooded or we say that
the amount of water that is entered on the port side is equivalent to the amount of water
that is entered on the star board side. If it is like that then, it is one class of problem
but, a second class of problem occurs when the ship gets flooded on one side completely.
That is, if you have the center line suppose, the flooding occurs only on one side of the
ship, may be the star board side of the ship. Now, if you have a situation like this, first
of all, as the ship gets flooded the draft will increase, so that means the ship will
sink first of all. Another thing that happens is - you can now see that this is not exactly
the centroid of this water plane it is not on this, because if you consider the method
of loss buoyancy some volume is lost here but, you have this volume.
Therefore, the centroid shifts to here; shift to somewhere away from the center line at
some distance etched from this end which is not equal to this B by 2. If you consider
the total breadth as B, this is B by 2. Now, this distance is f, distance h from this edge
which is greater than B by 2. When this happens you will have heeling the weight of the ship
is unsymmetrical about the center line that is there is more weight on the right side
of the ship compared to the left side or if other way around depending on which side is
flooded, when this happens therefore, that side of the ship will heel.
If there is more water coming here if that is side gets flooded the ship will heel like
this. Let us consider this problem first, then we will go to the problem of trimming.
They are just different ways of looking at the problem - different possibilities that
can exist.
Now, if you consider flooding this is what I talked about in case of if you consider
this is the whole ship and as you can see one compartment has got flooded here, this
blue color is indicating water, so this gets flooded.
As you can see it keeps getting flooded and of course, if there is damage this depends
upon the ships damage condition, because of this whole region is getting flooded, so one
by one different compartment are getting flooded. As this keeps continuously happening, what
will happen is that the ship will start trimming about the forward. So, ship will trim about
the forward and that is the secondary condition that we will simulate today.
Let us consider a box shaped
vessel of existing at a water line W 0 L 0. It is at a draft initially of d I; d initial
it represent the initial draft. Now, this is the elevation view or what we call as the
profile view. This length becomes the length of the ship which is capital L and this is
the draft. Now, let us consider that a small region gets flooded in the center. If you
consider this to be the center midship, I have already told you midship is represented
like this. If you consider the midship and consider the flooding to occur symmetrically
about the midship some region has got flooded to some height greater than the draft, so
it is flooded initially. Then, let us look at it is planned view, so this is again the
length of the ship and this represents B, this distance is B by 2.
Now, this problem is designed in such a fashion that it allows for that particular case that
we talked about, so it is flooded with midship alright but, it is flooded at one side means
this region is flooded. If you have a ship, if you consider this to be the center line,
so you have the ship like this; this is the center line, it is flooded with the midship
alright but, it is flooded at one side if this is the ship, this one corner here is
flooded. So, extending over the full depth of the ship but, it is flooded in one side,
so like this it is flooded. Now, let us consider it is dimensions to be
b; this is the breadth to which it is flooded. Let us consider it is length to be l - this
length l. Now, let us say that as I have already explained initially the ship is at the upright
condition and in that case, this is the center line and this will be G the position of the
center of gravity. The position of the center of gravity does
not change, because from now all the calculations will be based on the method of loss buoyancy.
Remember, the method of loss buoyancy implies that volume is lost from the ship. Automatically
it assumes that area is also lost from the ship therefore, the centroid of this water
plane area is not going to be at the center line there at G but, it is going here, at
a distance may be, let us call it y at this point. This height is let us call it as h,
it is the distance between the one side of the ship and the position where the ship has
its centroid, so this is the final condition.
Now, how will you get h? It is very easy to get h; h can be defined as the moment of area.
We can do many things, we can calculate the moment of area about some center line, it
can be about that B by 2 and it can be about the line where that G acts initially, so along
that line is a possibility but easier because of some reasons. First of all we are calculating
edge from one side, let us calculate all the moments use about one edge of the - this is
the rectangle - so one edge; let us take the moments about the edge.
Therefore, h is defined as the moment of area about the edge divided by the total area.
Now, the moment of area, what we have here first of all, we had complete rectangle. So,
this l into b we had a complete rectangle, because of flooding some volume is lost. Some
volume is lost directly imply that some area is lost in that water area plane. So, if you
take that water plane area initially you have l into b and finally, you have l into b minus
some area here. This area is again small l into small b, so that much area is lost that
we need to figure out. Based on that we get the moment of area about the edge divided
by area will get h.
You will get h to be L into B into B by 2 and moment is means here, you have a L into
B this is the whole area and into the position of it is center of gravity which is B by 2.
So, L into B into B by 2 minus - a small area is gone here, this is of small l, it is of
breadth small b and it is centroid is at a distance small b by 2.
Therefore, its moment is l into b into b by 2 divided by the total area which is L into
B minus small l into b, so this will give you your h. This gives you this distance h
which is the position of the new centroid, this is where the new centroid occurs. The
initial centroid is here and that is when the ship was in the alter right condition
without the flooding when the flooding occur, note again that the G which is the center
of gravity has not shifted, but it is only the centroid of the area has shifted to a
new point. Now, why do we need the centroid of area? We will see in the next figure, we
will draw it. So, these become h then, let us draw the other one.
We have always drawn sectional figures, cross sections means, if you have the ship like
this, you make a section like this. We have already done these sections which the lines
plan, we actually call it as the body plan. The body plan section will give you the different
stations and all that.
Now, just like that if you draw, this is the box shaped vessel, now for this we are drawing
the. So, this is the section, let us consider this to be the initial water
line and the final water line. Initially, the ship is at G, it is at midship section.
So, this is G; somewhere you have G, where the weight of the ship is acting.
Now, what has happened is in this case, as you see from this side some area or volume
has gone, so what will happen? The ship will tilt in this side and as the result the water
line will tilt up. The ship will tilt in this fashion up that means the water will come
down like this. Initially, the water line is here, W 0 L 0 and the ship is tilted in
this direction, so water line is tilted into W 1 L 1, so this is tilted by an angle phi.
Now, first of all the position of B will shift here in some new point that means some volume
is lost from here, which implies that it is equivalent to saying that some volume is added
on this side or what it means is if some volume is lost here, we know that the center of buoyancy
has to shift here, so B shifts to this point B 1. If you draw a vertical
at G you get the metacenter which is M b. So, this will give you the position of the
metacenter. The distance is let us draw a perpendicular
from G to this point Y, let us call this distance as X or at least the distance GY - as you
can see - is the distance between G and the new position of the centroid. This is what
we did in the last time just previously, we saw that we need to find the position of the
centroid. The position of the centroid is required for this purpose that is to find
the distance between G and the new position of the centroid. So, that distance is measured
as GY, we get here GY is actually equal to h minus B by 2.
You can see here, how do I get this, just look at this, this is Y coming here. This
distance that I am talking about GY is h minus B by 2, this is h and this is B by 2. So,
h minus B by 2 is the distance between G and Y it is one thing we need.
Then other things we need are one KB, we have to find the final KB of the ship. We always
say that KB is the vertical distance of the center of the buoyancy from the keel, now
that distance is always measured as half the draft, whatever is the draft. So, it is d
b divided by 2 this will give you the KB. Another thing we need to calculate is BM,
in this case BM which is equal to defined as I by v or I by del - I by del that is the
metacentric radius is defined as v 1 M b is equal to I by del.
Now, only thing you have to know here or slight catch here is in the calculation of I note
that I finally - so what did we had? What did we have initially? We had a whole box.
Now, what happen is, it got flooded on one side as a result of which one small volume
got lost here. What are we finding? We are finding the I which is the moment of inertia
of the water plane area. When you are finding the moment of inertia of the water plane area
then, in this case this much area here is lost. So, you actually have to find the moment
of inertia of the rest of the area which is equal to the moment of inertia of the whole
area minus the moment of inertia of this small region. That is very easy but it is still
important it can be calculated very simply as LB cube by 12 minus lb cube by 12. So,
this is about the B by 2 lines. Now, in this case it is just easier to do
all the moment of inertias about one side of the box therefore, we say that I - if you
remember the formula the moment of inertia of a rectangle about a edge of the rectangle
is represented as LB cube by 3. It is equal to LB cube by 12, if it is about the center
line it is equal to LB cube by 3 if it is about an edge.
So, LB cube by 3 minus lb cube by 3, this is the moment of inertia of the whole rectangle
now about an edge, means which edge? This edge, so it is a moment of inertia about this
edge. Now, note that we always need the moment of inertia. Now, in the derivation of BM equal
to I by del that BM metacentric radius was derived using the assumptions, we have done
all this derivations that derivation of BM del assume that I is always the I about the
centroid of the water plane area. That means in this case, you need to calculate
this I about the centroid as well. Now, you know by the parallel axis theorem that once,
you have the moment of inertia about one edge, if you can just do I about that point minus
A y square will give you that moment of inertia of that rectangle about that new position,
so that we can calculate using the parallel axis theorem which proceeds as this. So, I
about the h point or the centroid will be equal to this I which I have just written
here, I minus A into h square. This I is actually I about the side, so this will give you I
about the h or I about that line through h, the position which is at a distance h from
the side.
From that you will get BM equal to I by del then once, you have that let us look at this
figure, you see that if it is tilted through an angle phi, from this you can directly read
that tan phi is equal to GY divided by YM. In this case, GY we have already calculated
it is equal to h minus B by 2, which is the distance between the G and that vertical position
of M or the position of the projection of M, so that gives you GY divided by YM.
Once, you have that you get KM is equal to KB plus BM, YM is equal to KM minus KG. Just
look at this figure YM - why is it so, because remember this will be the keel. So, if you
are looking at the vertical distances KM; this vertical distance is the same as this
vertical distance. So, this is KG, this is also equal to KG, KG YG or KY is equal to
KG. This KG is actually equal to Y KY which is equal to KG because of this you can get
this expression. We know all these things, we have to know
KG that is now it calculated. KG is the center of gravity of the ship or the box shape vessel
the vertical position that needs to be known and this is a box shaped vessel anyway. So,
you can use the formula that KM is equal to d by 2 plus b square by 12d that is a known
formula for box shaped vessels, where d is the draft of the vessel. When you have that
you can get it as b square by 12d that will give you the KM or in case of if you are dealing
with a large scale ship the value of KM will be given from the hydrostatic data.
From the hydrostatic data or from the hydrostatic particulars, you get hydrostatic curves, you
get the KM value, KG value has to be known, it has to be given by in the problem itself.
Once you have that you find YM as we have seen YM comes as KG itself then, KY is equal
to KG. As a result you get YM the distance between Y and the metacenter as KM minus KY.
Once you get that then you can just use the formula that tan phi is equal to GY divided
by YM, now that will give you phi which is the angle of heel.
Therefore, in this particular case where we had a ship which is flooded in the midship
section but close to one side, it is not flooded symmetrical to the center line, but it is
flooded on one side. When this happens the ship initially sinks
as a result of which its water line goes up and the draft increases and then the ship
heels depending on which side gets flooded, depending upon that side will go down. Either
way, if you look at it in case of weight you see that weight has increased there, so it
has to heel there. It is comes as one of the many problems that we have done in the shifting
of weights in fact the inclining experiment itself was device using this. That is, along
the deck you shift a weight from one side to another you see that the ship heels in
that direction and then you calculate the heel and the horizontal distance of the center
of gravity moved and the GM, that is the inclining experiment.
Now, just like that in this case ship gets flooded, so additional weight comes on the
right side, because of that the ship heels to that side or if you look at it the other
way round, ship loses its volume there, it is still the same thing ship heels again to
the right side either way. So, this explains one type of problem dealing with the ships
flooding.
Now, we will look at one problem; suppose, you are told that there is a box shaped vessel
of length 60 meters, breadth 9 meters and it is floating at a draft of 5 meters and
it has a K G of 3 meters. You are asked to find the list if a midships compartment of
length 6 meters and breadth 6 meters is bilged. It is a midship compartment therefore, it
is occurs in the midship but the figure is given along with a problem we say the problem
is something like this.
Just like I explain the previous problem it is come something like this. So, you have
the center line and in this case a midships compartment somewhere here a region is flooded.
The breadth of the ship initially is 9 meter - this is 9 meter - and his breadth of 6 meters.
One compartment like this of 6 meters, so this is of breadth 6 meters, so out of the
9 meters 6 meters length, this region is flooded. Now, this is the same problem that we have
done first of all, we need to find the position of centroid here somewhere else, as you can
see more volume is lost here and directly more area is lost here. If area is lost here,
area is still available here, we see that the centroid should shift here - will shift
like this. From this points it shifts here and this distance we have called as h. As
in the previous derivation this is h and this is B by 2, so this is B by 2 and that is h.
Now, the question is to find the list? Because of the ship flooding like this, it is seen
that the ship first sinks and then heels just like as I said.
Let us see how this problem has to be done? First of all we use the formula intact volume
before flooding equals intact volume after flooding. This is the most general formula intact volume
before flooding is equal to the intact volume after flooding. Intact volume before flooding
is LBd i length, breadth into the draft initially equals LBd b minus lbd b. This tells you the
intact volume after flooding which is capital LB into d b, which is the bilged draft; the
draft after the ship has bilged minus lbd b small l small b is the compartment volume,
small l represents the length of the compartment b represents the breadth of the compartment
- I mean - that small region of the compartment which is flooded. So, l into b will give you
the area into draft bilged. So, that will give you the volume of the compartment that
is flooded. As I said before, this equation says - I mean
- what it says? That the intact volume before flooding is equal to intact volume after flooding
is same as saying, the total weight after flooding is equal to the initial weight of
the ship. This equation is actually the same as if the total weight of the ship after flooding
is equal to the initial weight of the ship plus the weight of water added. So, it is
just saying a mass conservation, it or shifted it becomes the volume conservation. So, that
is this equation LBd i is equal to LBd b minus small lbd b.
From this you get d b in this problem since, you have L B, you are given the initial draft,
you are told l b which is the dimensions of the compartment that is being flooded. Once,
you have all this you can get d b which is the bilged draft, first step. Now, we need
to find the position of the h which is the centroid of the area. So, you need to find
the centroid of the area.
That is based on what we have done so far, it is very simple. What you have here? You
have area and have its distance of centroid - you make a table like this - that area is
there distance of the centroid and moment. Initial area is L into B is area of the total
box that water plane. Distance of the centroid is B by 2 you find it is moment then you have
an area of l into b in this case, its breadth is given as 6 meters and length is given as
some 6 meters.
Based on that 6 by 6 area like as we have discussed here in this figure, this 6 by 6
region is flooded, so this area is lost and therefore, it is minus lb into distance of
this is B by 2 and you calculate moment. As you know moment is just this first column
into the second column, multiplied by the second column, so it gives you this. Finally,
you need to find the final h which is equal to the final moment divided by the final area.
Therefore, this will give you the final moment and this will give you the final area.
Just doing this you will give you the h, this is the distance of the position of centroid
from this edge. This is the side, so distance from this side is here, so this distance will
give you the position of h which is the final centroid of the area.
Remember, the main purpose of our work is to find tan phi which is equal to given by
GY by YM, this explain to be the equation for tan phi. First thing we need to calculate
is GY which is equal to h minus B by 2, I have already described it I need to describe
it again h minus B by 2. For this problem just we have already calculated
h, which is the distance from the edge, from the side we have seen the distance of h we
have seen the distance of h from one edge we have seen the distance of the position
of the centroid from one edge that is what we have measured as h. So, h minus B by 2
then we need to find I this is one important thing. Remember as I have already mentioned
BM, this is to calculate BM using this formula, our purpose is to calculate BM using this
formula I by del, del is the underwater volume. You need to calculate I, in this I is the
moment of inertia of the final water plane which is actually the whole water plane minus
the small area that is lost due to the flooding - that whole area minus the small area. So,
that much remaining area moment of inertia of that about the centroid; so it is the moment
of inertia about the centroid and not about an edge or not about the symmetrical center
line, it is about that centroid. Now, to get that as we have already seen,
it is best to calculate I about one side use the parallel axis theorem and calculate I
about that edge or the centroid. First, we calculate I about this side which is given
as LB cube by 3 minus l b cube by 3, this will give you I about the side. Then, the
h point is given by I about the side minus square. This will give you the formula, then
you do BM equals I by del and this will give you an answer of 1.28 meters.
You are given in this problem KB is 2.68 therefore, KB plus BM will give you KM, is equal to 3.98
meters KM. Then, KG is given for the ship it is 3 meters - it is already given in the
problem 3 meters. We have already seen that KY is equal to KG from the figure. Therefore,
YM is equal to KM minus KY which is equal to 3.98 minus 3, so this is about 0.98 meters.
This gives you the value of YM and remember, what we were supposed do is to calculate tan
phi given by GY by YM. Now, it becomes straight forward, we have
already calculated GY which is equal to h minus B by 2; h minus B by 2 gives you GY
the distance through which the centre liner shifted. YM gives you the distance between
that Y and the metacenter. Using this equation, we can get the value of GY by YM which is
equal to the tan of the angle of phi. So, you get the angle of phi in these problems
to be about 6.38 degrees. This is one sort of problem which we work on.
Now, we move into the second type of problem that I mentioned at the beginning of this
lecture. That is, we have seen in this figure that instead of having the bilging occurring
at the midship compartment like here. The problem that we have concerned so far deals
with that, we consider that bilging occurred in the midship compartment here, instead of
that happening. We are going to consider bilging to be occurring at 1n compartment most likely
in the forward side, here it floods in the forward side. We are not considering progressive
flooding, one sides get flooded and that is the problem that we are considering next.
In this case, let us consider the ship in the box, so you have the box shape vessel,
so one compartment like this in the front part of the ship gets flooded. Let us say
that there is a hole here, this compartment gets flooded here, this is of length small
l, the total length of the ship is capital L, so this compartment gets flooded.
Let us assume this is the water line initially therefore, the initial draft is d initial
- d i - is an initial draft. Finally, because of this bilging just like I explain the previous
problem, in this case also something happens that is the ship is like this initially at
some draft. In the forward side of the ship one compartment has got flooded as the result
of which this compartment gets flooded and initially the ship sinks because of the added
weight the ship sinks. There is an increase in draft and in this
case, since this side has no more weight than this side, because of this unevenness in weight
this will trim like this. The ship is going to go down at the forward side, so the front
part of the ship will trim now. So, this is the new process that we are trying to mathematically
study here. Initially, the draft increases to d b and
then, the ship trims and it goes up the draft increases to d f, so the final draft is d
f that water line is - let us call to be W 1 L 1 this is the final water line - where
finally the ship has come to this position. We will have here G, this is the position
of G - the centroid initially - then, at G let us assume that the ship initially it is
B 0. Because of its trimming the B shifts to - it
is a loss of buoyancy, so the B is now here. So, this is B 1 or B 2. The final position
of the centre of buoyancy is B 2 and M is the position of the metacenter, this is the
same explanation as in the previous figure, weight of the ship acts here W, so B 2 in
this is the position of the centre of buoyancy, the same method as we did.
If you take that the distance through which the same concept that is, in this problem
the only difference that is, in the last problem we consider that one area is lost on one side
as the result of which that centerline shifted to slightly higher - slightly the other side
to the star board side. In this problem instead of the centerline
shifted like this, we are talking about G shifting here, it is this shift, it is no
longer this shift but it is this shift. Similarly, this area here this part of the ship is flooded,
so out of this some area will be lost in the front part of the ship.
So, how much distance does it shift? Same way that is you take the moment of the area
divided by the total area. If you take the water plane area remember water plain area
will be like this - water plane area will be section like this. Now at that point you
will have some area lost here, by doing that we can find it is very easy if you take the
moment you will see that if this is shifted by l and if you assume it is centroid to be
at l by 2 you will see this will be shifted by l by 2.
Now, let us call this to be the aft perpendicular, so the distance of centroid from aft perpendicular
is given to be L by 2 which is this - this is L by 2 - distance as we know this whole
distance is L, this is L by 2, this distance from here. Therefore, it is L by 2 minus small
l by 2. So, this will give you the distance of the centroid from the aft perpendicular.
As you can see in this figure the distance of G from the aft perpendicular is again equal
to L by 2 always, because this is again the method of loss buoyancy and the centre of
gravity remains at G. The only thing we need to do here is, let us consider the weight
acting now. So, what we have? About this new position of the centroid that is what we call
as centre of flotation - the centre of flotation is now here - centre of flotation is defined
as a centroid of the water plane area. In this case, some water plane area is lost in
the front because this side is bilged and because of this bilging, some area is lost
here, because this area is lost here the centroid of the water plane area shifted here, from
here it is shifted here. This is the centre of flotation and if you
take moments about the centre of flotation, we see that at G which is the position of
the centre of gravity, there is a weight W acting downwards. The position of the centre
of buoyancy is here, so there is no weight acting about it is along that same axis.
So, that net moment acting which is trying to cause the trim is this W, the weight of
the ship into small l by 2 which is this distance, so W the weight here, W into small l by 2
this much is the moment causing trim. We write the trimming moment equals weight into l by
2, this is your trimming moment. Now, once you have the trimming moment how do you calculate
the change in trim that is very easy, they have already done many times.
The change in trim equals the trimming moment divided by the MCTC moment to change the trim
by 1 centimeter. Now, the trimming moment is the W into l by 2 divided by MCTC will
give you the change in trim which we usually write as trim t or change in trim. Change
in trim will be given by W into l by 2 by MCTC, this will give you the change in trim.
Once you are able to find the total change in trim, you need to find the change in trim
aft change in trim forward the same way, same equation that is - I mean - in the previous
chapter not here. Here, we have used different notations small l in our previous chapters
that is the chapter dealing with trim were we were doing.
The small l actually represented the distance between the aft perpendicular and the centre
of flotation. If that is so then, the change of trim aft is given by small l which is the
distance between the aft perpendicular and the centre of flotation divided by capital
L which is the total length of the ship distance between perpendiculars multiplied by the total
change of trim. So, that will give you the change of trim aft, so small l by capital
L.
In this case, small l represents not the distance between the aft perpendicular and the centre
of flotation, but l by 2 represents this distance, small l represents this distance, so this
is l by 2, but the concept remains the same to get the change of aft draft. I need to
find the distance between aft perpendicular and the centre of flotation, centre of flotation
is this, so it will be small l by 2 or l minus small l by 2 into l into change of trim, so
l minus l by 2 l into W into l by 2 divided by MCTC.
This will give you the total change in trim in the aft side and if you want to find the change in the trim
in the forward side, instead of l minus l using the previous notations if small l is
the distance between the aft perpendicular and center of flotation. What you do is, you
do capital L minus small l divided by capital L that will give you the distance from the
- instead of taking this distance you take this distance - the front distance that will
give you the change in trim in the forward side of the ship.
Once, you find the change in trim forward change in aft, what you generally have to
do is whatever is the initial trim you add this change in trim aft to the initial trim
aft and you will get the final trim aft. Change in trim forward to the initial trim forward
you will get the final trim forward and that will give you the final drafts also that will
give you the final draft in the forward and the aft section.
This we have explain two types of dealing with the bilging problems, any kind of bilging
problem will be having - now that we have done two extreme types of problems means,
we have studied the case when the ship can heel and we have study the case ship an trim
due to bilging. These are mainly the two types of process and you can just extrapolate this
to say that if you have a time dependent bilging means, as the time goes on the flooding keeps
continuous happening. If you are trying to study that problem then it does not become
heeling as such it becomes a case of rolling, it becomes dynamic process and not a static
study and since this course is on hydrostatics. We are not dealing with the dynamic process
of flooding thought it is definitely a very important point and a very important topic
very active area of research. So, that gives to different ways of calculating the trim
or the heel as a result of bilging. These are two extreme cases when you have bilging
the front part; in the front part of the section this gives that how to the find the trim and
in case you have bilging occurring on one side of the ship and not the other side, that
gives you the another weight of calculating the heel.
You can combine the two to calculate very complicated cases, where you have some bilging
not at the centre not at the end, somewhere in between. The whole process becomes a combination
of these two, you can get the trim and the heel due to this process and combine it and
get the final value of heel and trim, it might be a combination of two and you get the value
of heel and trim, with this I will stop here today, thank you.