INFINITY MARINER: Ship Stability - I Control trim, stability and stress

Approximate calculations of Area and volume

Learning Objectives

After completing this topic you will be able to:

State Simpsons Rules 1, 2 & 3
Calculate using Simpsons Rules

Areas & Volumes

Geometric centers of Areas & Volumes

TPC & FWA

Simpson Rules 1,2 and 3

Simpsons Rules

While it is easy to calculate the area/ volume of geometric figures, the calculation of areas and volumes of irregular shapes is not so straight forward. By irregular it is meant that, the borders of the shapes should confirm to curves which follow mathematical laws i.e. equations.

Such irregular shapes with curved borders following mathematical laws are very common in shipbuilding. E.g. Water plane

Simpsons Rules are a set of three rules which allow us to calculate the areas, Volumes and centroids of such figures with relative ease and acceptable degrees of accuracy. For linear, quadratic or cubic curves the accuracy is exact.

Simpsons 1^st Rule

This rule assumes that the curved border is a parabola of the $2^{nd}$ order.

The area of the elementary strip is ydx and the area enclosed by the curve and the $\frac{X}{Y}$ axis is given by , Area = $\int_0^{2h} y\ dx$ , from which we derive Area = $\frac{h}{3}(y_1+4y_2+y_3)$ .

‘h’ is called the common interval. y_1, y_2, y_3

are the ordinates, 1,4,1 are the Simpsons multipliers. The constant is $\frac{1}{3}$ .

If there are more than 3 ordinates eg 1,4,1 and 1,4,1 then the two joining ones are combined to make 2 and the multipliers become 1,4,2,4,2,4,……..1

Usage for computing Area/ Volume:

Ordinate	Simpsons Multiplier	Product of Area/ Volume
A	1	1A
B	4	4B
C	2	2C
D	4	4D
E	1	1E

${\Sigma \ \text{Product\ of\ Area\ or\ Volume}} = (1A+4B+2C+4D+1E)$

${\text{Area\ or\ Volume}}=\frac{h}{3}(1A+4B+2C+4D+1E)$

Simpsons $2^{nd}$ Rule

This rule assumes that the curved border is a parabola of the $3^{rd}$ order.

The area of the elementary strip is ydx and the area enclosed by the curve and the $\frac{X}{Y}$ axis is given by Area = $\int_0^{3h} y\ dx$ , from which we derive Area = $\frac{3}{8}(y_1+3y_2+3y_3+y_4)$

‘h’ is called the common interval. y_1, y_2, y_3,y_4

are the ordinates, and 1,3,3,1 are the Simpsons multipliers. The constant is $\frac{3}{8}$ .

If there are more than 4 ordinates eg 1,3,3,1 and 1,3,3,1 then the two joining ordinates are combined to make 2 and the multipliers become 1,3,3,2,3,3,2,……..1

Usage for computing Area/ Volume:

Ordinate	Simpsons Multiplier	Product of Area/ Volume
A	1	1A
B	3	3B
C	3	3C
D	2	2D
E	3	3E
F	3	3F
G	1	1G

${\Sigma \ \text{Product\ of\ Area\ or\ Volume}} = (1A+3B+3C+2D+3E+3F+1G)$

${\text{Area\ or\ Volume}}=\frac{3h}{8}(1A+3B+3C+2D+3E+3F+1G)$

Simpsons $3^{rd}$ Rule

Deviating from the usage of the $1^{st}$ and $2^{nd}$ Rules the $3^{rd}$ Rule allows us to manipulate any three ordinates, to obtain the area/ volume between two consecutive ordinates amongst three. In this case $y_1 \& \ y_2,$ between which the elementary strip falls.

Area = $\int_0^n y\ dx$ from which we derive Area = $\frac{h}{12}(5y_1+8y-y_3)$

‘h’ is called the common interval. $y_1,\ y_2,\ y_3,$ are the ordinates, with the target area falling between $y_1 \& \ y_2,$ and 5,8,-1 are the Simpsons multipliers. The constant is $\frac{1}{12}$ .

Usage for computing Area/ Volume:

Ordinate	Simpsons Multiplier	Product of Area/ Volume
A	5	5A
B	8	8B
C	-1	-1C

${\Sigma \ \text{Product\ of\ Area\ or\ Volume}} = (5A+8B-1C)$

$\text{Area\ or\ Volume}=\frac{h}{12}(5A+8B-1C)$

Notes:

In the calculation of areas e.g. water planes, the ordinates are equally spaced distances. For the calculation of volumes they are equally spaced cross sectional areas.

Semi Ordinates – Since ships are as a rule symmetric about the centerline, it is common for the ordinates to be of one half only. They are then termed semi ordinates and the final result doubled to obtain the area/ volume.

Common Interval can be obtained by dividing the total length by the number of ordinates minus 1. However it should be noted that in the case of appendages, the common interval will be halved. The accuracy of the area/ volume is increased by increasing the number of ordinates.

Appendage. Ships sometimes have a change in form toward their ends e.g. bulbous bow, turn of the bilge. In these areas, the curve enclosing the area/ volume changes from the curve that is enclosing the major part of the shape be it a water plane or bulkhead. In these cases, it is usual to allot extra ordinates called intermediate ordinates at half the common interval.

When either of 2 rules may be used for the same set of ordinates, it is recommended that for reasons of accuracy the $1^{st}$ Rule to be preferred over the $2^{nd}$ Rule.

If neither of the rules is applicable, the ordinates are to be separated into two set and the area/ volume obtained separately for each section and then added up.

A less accurate way of obtaining the area of irregular shapes using geometry is the trapezoidal method. It is however not as accurate and requires many more ordinates to achieve an acceptable level of accuracy.

TPC may be calculated by using these Rules. The Full area of water plane is multiplied by SW density 1.025 and the product divided by 100.

FWA may be determined after obtaining the underwater volumetric displacement, by the use of the Rules. Subsequently the FWA is obtained from the formula $\text{FWA} = \frac{\text{Displacement}} {4 \times \text{TPC}}$

Coefficient of fineness and Block coefficient may also be determined.

Centroids

Recap of Moments

A clear understanding of the theory of moments is required as we progress beyond the area/ volume calculations and progress towards determining the centroids of water plane areas and volumes. The simplest representations of moments are weights on a single beam.

A beam AB whose center is z mtrs from xx’ weighing y mt is hinged at xx’. A weight of p mt is loaded at a point q mtrs from xx’. Find the final position of the beams center with reference to xx’

Moments about xx’ of the beam AB = yz tonnes mtrs or tm

Moments created by the additional weight = pq tm

Total moments about xx’ = (yz + pq) tm

Final centroids of the beam with respect to xx’ is the Final moment/ Final weight

$\frac{(yz + pq)}{(y + p)}$ mtrs from xx’

Similarly for calculating the centroids of Area and Volume, the product of area/ volume at a certain ordinate is multiplied by the lever to get the moment of area/ volume at that ordinate.

The sum total of all the moments of area/ volume is divided by the sum of the product of area/ volume and the position of the centroids obtained with respect to either a horizontal or vertical axis. The moments are obtained from either horizontal reference points or vertical.

In the case of a water plane the position of the centroids may be with reference to the APP, Midship or rarely the FPP. The following table indicates how the levers are to be taken for these three instances.

We can now state that:

Centroid from a reference point eg APP = $\frac{Moment\ about\ APP}{Total\ Area\ or\ Volume}$

= $\frac{(\Sigma \ Prod\ of\ Moment) \times \frac{h}{3}}{(\Sigma\ Prod\ of\ Area) \times \frac{h}{3}}$

= $\frac{\Sigma \ Prod \ of \ Moment}{\Sigma \ Prod \ of \ Area}$

$2^{nd}$ Rule numericals may be similarly treated.

Calculation of area and volume using Simpson's rule

Example:

A ships water plane area is 120m long. The half breadths measured at equal intervals from aft are: 0.1,4.6,7.5,7.6,7.6,3.7,0 mtrs. Find the water plane area.

$\text{Common\ interval(h)}=\frac{120}{(\text{Number\ of\ Ordinates-1})}=\frac{120}{(7-1)}=20\ m$

Half Breadth (Semi Ordinates)	SM	Product
0.1	1	0.1
4.6	4	18.4
7.5	2	15.0
7.6	4	30.4
7.6	2	15.2
3.7	4	14.8
0	1	0
	Sum of products = 93.9

Half Area = $\frac{h}{3}(\text{Sum\ of\ Product})=\frac{20}{3}(93.9)=626 \ sq\ mtr$

ANS: Area of water plane $= 626 \times 2 = 1250$ sq mtr

Calculation of Geometric centres of area and volume

Examples 1

Common interval $= 24 \div 6 = 4$ m.

Area of tank top $= 4 \times 3 \div 8 \times 88.3 \times 2 = 264.9\text{m}^2$

Volume of tank $= 264.9 \times 1.5 = 397.35\text{m}^3$

2. The transverse cross sectional area of the underwater portion of a barge at 12m intervals from forward are 0,13,17.5,25,19.6 sqm. The last ordinate is the AP. Calculate the position of the center of buoyancy with respect to AP.

CI (h) = 12 m

Area	SM	P O Vol	Lever about APP	P O Mmt
0.0	1	0.0	$4 \times 12$	$0.0 \times 12$
13.0	4	52.0	$3 \times 12$	$156 \times 12$
17.5	2	35.0	$2 \times 12$	$70 \times 12$
25.0	4	100.0	$1 \times 12$	$100 \times 12$
19.6	1	19.6	$0 \times 12$	$0 \times 12$
		$\Sigma\ POV\ 206.6$		$\Sigma\ POM\ 326.0 \times 12$

ANS: Distance of Centre of Buoyancy B from "Aft Perpendicular" (LCB) = $\frac{(326.0 \times 12)}{206.6}= 18.935$ m

3. The vertical ordinates of the after bulkhead of the port slop tank of a tanker, measured from the horizontal deck head downward, spaced at equal athwart ship intervals of 1m are 0, 3.25, 4.4, 5.15, 5.65, 5.9, 6.0 m.

Find

a. Centroid of bulkhead from inner boundary

b. Centroid of bulkhead from deck head

c. Thrust on the bulkhead when the tank is full of SW

This example introduces the lever and weight concept (Refer - Recap Moments). Like the GC of a lever is at its center and the moments created by it, is the product of its weight and the distance of its Centroid from the reference axis, so also if the Centroid of a section is required from an axis that is in line with the ordinates, then the Moment of Area is the Product of Area (POA) multiplied by the lever which is half the ordinate.

h = 1 m

Ord	SM	POA	Lever Inner Bndry	POM Inner Bndry	Lever Dk Hd	POM Dk Hd
0.0	1	0.0	$6 \times 1$	$0.0 \times 1$	0.0	0.0
3.25	3	9.75	$5 \times 1$	$48.75 \times 1$	1.625	15.844
4.40	3	13.20	$4 \times 1$	$52.80 \times 1$	2.2	29.04
5.15	2	10.30	$3 \times 1$	$30.9 \times 1$	2.575	26.523
5.65	3	16.95	$2 \times 1$	$33.9 \times 1$	2.825	47.884
5.9	3	17.7	$1 \times 1$	$17.7 \times 1$	2.95	52.215
6.0	1	6.0	$0 \times 1$	$0 \times 1$	3.0	18.0
		$\Sigma \ POA \ 73.9$		$\Sigma \ POM \ 184.05 \times 1$		$\Sigma \ POM \ 189.506$

ANS: Centroid from Inner boundary = $\frac{(184.05 \times 1)}{73.9}= 2.491\ m$

ANS: Centroid from Deck Head = $\frac{(189.506)}{73.9}= 2.564\ m$

ANS: Area = $1 \times \frac{3}{8} (\Sigma POA) = 73.9 \times \frac{3}{8}= 27.713\ sqm$

ANS: Thrust = GC from Dk Hd $\times$ Area $\times 1.025 = 2.564 \times 27.713 \times 1.025 = 72.833$ mt

Calculation of TPC, FWA using Simpson rules

Example 1

The half breadths of a ships water plane 100m long at equal intervals from aft are: 5.0, 5.88, 6.75, 6.63, 4.0, and 0.0.

Find the water plane area, TPC and Coefficient of fineness.

This is a case of 6 ordinates, where a single rule cannot be used, and a combination has to be selected. The options are as follows:

$1^{st} \& 2^{nd}$ Rule
$2^{nd} \& 1^{st}$ Rule

For this example we will choose the first option i.e. separate the areas in X (Area between $1^{st}$ and $3^{rd}$ semi ordinate) and Y (Area between 3rd and 6th semi ordinate)

Common interval (CI) = $\frac{100}{5}=20\ \text{m}$

Area X
Half Breadth	SM	Product
5.0	1	5.00
5.88	4	23.52
6.75	1	6.75

Area Y
Half Breadth	SM	Product
6.75	1	6.75
6.63	3	19.89
4.00	3	12.00
0.00	1	0.00

SOP = 35.27 SOP = 38.64

Area X = $\frac{20}{3}(35.27)$ Area Y = $\frac{20 \times 3 \times 38.64}{8}$

sq m

sq m

ANS: Total area of water plane $= (235.133 + 289.8) \times 2 = 1049.866$ sq m

ANS: TPC = $\frac{(1.025 \times 1049.866)}{100}= 10.761 \ \text{mt}$

ANS: Coefficient of Fineness = $\frac{1049.866}{(100 \times 6.75)2}= 0.778$

Example 2

If the immersed cross sectional area of a ship is 180 m long at equal intervals of length are 5, 118, 233, 291, 303, 304, 304, 302, 283,171 and 0 m^2

. Calculate Displacement in seawater if density of sw = 1.025 t/m^3

Ordinate No.	Ordinate Cross Sectional Area (Unit) M²	Simpson’s Multiplier	Product for Volume
1	5	1	5
2	118	4	472
3	233	2	466
4	291	4	1164
5	303	2	606
6	304	4	1216
7	304	2	608
8	302	4	1208
9	283	2	566
10	171	4	684
11	0	1	0
			$\Sigma \ V=6995$

$\text{CI} \ \text{(h)} =\frac{180}{(11-1)}= 18 \text{m}$

Volume of displacement = $\frac{h}{3}(\Sigma \ V)=\frac{18}{3} \times \Sigma \ V=\frac{18}{3}\times 6995 = 41970\ m^3$

Displacement = $\Delta = 41970 \times 1.025 = 43,019.25$ tonnes

Example 3

If the TPC for a ship at 1.2 m intervals of draught commencing from the Keel are 8.2, 16.5, 18.7, 19.4, 20, 20.5, 21.1. Calculate the displacement at 7.2 m draught.

This type of numerical can be solved by either of the following methods

Method 1

Use the given values of TPC as ordinates and apply Simpons's rule No.1

Ordinate No.	Ordinate TPC (unit) t/cm	Simpson’s Multiplier	Product for D
1	8.2	1	8.2
2	16.5	4	66.0
3	18.7	2	37.4
4	19.4	4	77.6
5	20.0	2	40.0
6	20.5	4	82.0
7	21.1	1	21.1
			$\Sigma \ TPC = 332.3$

$\text{TPC} = \text{A} (\text{W/P}) \times \text{RD} \div 100$ , Hence $\text{A} (\text{W/P}) = \text{TPC} \times 100 \div 1.025$

$\Sigma \text{Area}=332.3 \times \left(\frac{100}{1.025}\right)$

$\Sigma\ V=\frac{h}{3} \times \ \Sigma\ V=\frac{1.20}{3} \times \left(332.3 \times \frac{100}{1.025}\right) = 12967.805 \ \text{cbm}$

Displacement = 129678.805 $\times$ 1.025 = 13292 mt

Method 2

Convert each TPC to Area of W/P by the above formula, tabulate these and apply the Simpon's rule in the normal way to obtain the volume and displacement.

Effect of density: TPC, FWA, DWA calculations

Introduction:

As we are aware, a vessels weight is supported by the force of buoyancy generated by the volume of water the ship displaces. This force is directly linked to the density of the water. If the density rises the volume of displacement required to support the vessel drops and vice versa.

Learning Objectives

After completing this topic you will be able to:

Effect of the change of density of water on TPC
Calculation of draft of vessel fore and aft due to change in density

Effect of change of density of water on TPC

TPC, FWA & DWA

The following media explains about the TPC

TPC in SW (As obtained from Stability Data) = $\frac{1.025A}{100} \text{mt}$

Where A is the area of water plane multiplied by $\frac{1}{100}$ mtrs to obtain the volume of water displaced by the vessel for 1 cm rise or sinkage. This volume is multiplied by the density of SW 1.025 to obtain how much weight will be supported by 1 cm of volumetric displacement.

If the density is lower than SW, obviously a lesser weight can be supported. Therefore we can state that

$\text{TPC}\ (\text{Dock\ water\ or\ DW}) = \text{TPC}\ (\text{SW}) \times \frac{\text{Density\ of\ DW}}{\text{Density\ of\ SW}}$

Example:

A vessel floating in SW at an even keel draft of 7 mtrs has a TPC of 42 mt. Find its TPC at the same draft in DW of density 1.020

$\text{TPC}_{\text{DW}} = \text{TPC}_{\text{SW}} \times \frac{\text{Density\ of\ DW}}{\text{Density\ of\ SW}}$

$= 42 \times \left(\frac{1.020}{1.025}\right)$

ANS = 41.795 mt

FWA & DWA

The following media explains about the Fresh water allowance

Tons Per Cm. (TPC) is the number of tons of weight required to sink or raise the ship by 1 cm. It is given by the formula

Calculation of draft of vessel fore and aft due to change in density

Effect on the draft of a vessel

Consider a ship floating in dock water in even keel condition. Centre of Gravity (G), which is the centre of mass of the ship, and Centre of Buoyancy (B), which is geometric centre of the underwater volume of the ship, will lie in a vertical line, one above the other. The forces of gravity and buoyancy will be equal, opposite and in one straight line. The student should note that this straight line will not necessarily be the midship line of the ship.

If now a weight is loaded in aft part of the ship, or discharged from forward part of the ship, or shifted from forward to aft, the G will shift longitudinally aft of its original position to a new position $\text{G}_1$ . This shift will create a couple between the force of gravity through $\text{G}_1$ and the force of buoyancy through B, which will cause the ship to trim by the stern. As the ship trims, the B will also shift aft to a new position $\text{B}_1$ , which will again lie vertically below $\text{G}_1$ .

Similarly if a weight is loaded in forward part of ship, or discharged from aft part of the ship, or shifted from aft to forward, the ship will trim by the head.

If the ship trimmed by the stern now moves to sea, it will rise by an amount equal to DWA. There will be no change in the position of $\text{G}_1$ , but the underwater volume will decrease. This will cause $\text{B}_1$ to shift to a new position $\text{B}_2$ , which may be forward or aft of $\text{B}_1$ , depending on how the shape of underwater volume changes. This shift will once again create a couple between the force of gravity through $\text{G}_1$ and the force of buoyancy through $\text{B}_2$ , which will change the trim by head or stern.

Similarly if the ship moves from sea to dock water, or if it is trimmed by the head and it moves from dock water to sea, or vice-versa, the trim will change.

Hence, due to change in the underwater volume and its shape, and the resulting change in the trim, the forward and aft drafts will change.

This principle is explained more clearly with the help of example below.

Example 1

A vessel of length 200 mtrs, $\bigtriangleup$ 21000 mt MCTC 220 tm TPC 32 mt LCF 97 m LCB 104 m. Draft F 9.2 m A 10.7 m in SW. Calculate her draft on entering water of density 1005.

As shown in the sketch above, G and B were in a vertical line when the ship was in even keel condition. However due to change in distribution of masses, G shifted aft of B to a new position $\text{G}_1$ , which caused it to trim by the stern, as given in this example. This shifted the B to a mew position $\text{B}_1$ , vertically below $\text{G}_1$ .
The ship in this condition moves from SW to DW of RD 1.005, hence it will sink by an amount of DWA, i.e. both forward and aft drafts will increase by this amount.

DWA = FWA $\times$ (Change of RD) $\div$ 0.025 ; FWA = W $\div$ 40 $\div$ TPC
DWA = 21000 $\div$ 40 $\div$ 32 $\div$ (1.025 – 1.005) $\div$ 0.025 = 13.125 cm. = 0.131m. (+)
However displacement has not changed, hence the position of $\text{G}_1$ will not change.
Normally due to increase in draft, the values of MCTC, TPC and LCF should change, but in absence of hydrostatic tables it is assumed that these values given in the example do not change.
The underwater volume has increased, i.e. an extra volume “v” has been added above the original underwater volume.
$\text{v} = \text{V}_\text{DW} - \text{V}_\text{SW} = (\text{W} \div \text{RD}_\text{DW}) - (\text{W} \div \text{RD}_\text{SW}) = (21000 \div 1.005) - (21000 \div 1.025)$
$= 20895.522 - 20487.805 = 407.717 \text{m}^3$
In this example LCB > LCF, and it is assumed that the centre of v is the same as the position of COF of the ship. Hence, due to addition of volume v, $\text{B}_1$ will shift towards COF i.e. longitudinally it will shift aft of its original position, to a new position $\text{B}_2$ .
This once again creates a couple between the force of gravity through $\text{G}_1$ and the force of buoyancy through $\text{B}_2$ , which creates a change of trim by the head. This causes the stern trim of the ship to reduce i.e. forward draft will increase and aft draft will reduce.
By theory of moments --- $\text{BB}_1 \times \text{V}_\text{DW} = \text{v} \times (\text{LCB} - \text{LCF})$
Hence $\text{BB}_1= 407.717 \times (104 - 97) \div 20895.522 = 0.137\text{m}$ .

Trim change $(\text{T}_\text{C})$ = $\text{BB}_1$ $\times$ W $\div$ MCTC = 0.137 $\times$ 21000 $\div$ 220
= 13.078 cm. = 0.131m. (By head)
Trim correction aft $(\text{T}_\text{A}) = \text{T}_\text{C} \times \text{LCF} \div \text{LBP} = 0.131 \times 97 \div 200 = 0.064\text{m}. (-)$
Trim correction forward $(\text{T}_\text{F})$ = 0.131 – 0.064 = 0.067m. (+)
Final forward draft = 9.2 + Sinkage + $\text{T}_\text{F}$ = 9.2 + 0.131 + 0.067 = 9.398m.
Final Aft draft = 10.7 + Sinkage - $\text{T}_\text{A}$ = 10.7 + 0.131 – 0.064 = 10.767m.

Example 2

M.V. Hindship is floating at anchorage in SW at drafts F 7.5m. and A 7.9m. What will be the drafts on berthing in water of RD 1.002?

Unlike the previous example, in this example the student has to use the hydrostatic tables of the ship. Hence there will be no assumptions like it was in the previous example, and the final answer will be more accurate.

Mean draft $= (7.5 + 7.9) \div 2 = 7.7$ m. ; LCF for mean draft = 70.8795m.; Trim = 0.4m. Correction to aft draft $(\text{T}_\text{A}) = \text{T} \times \text{LCF} \div \text{LBP} = 0.4 \times 70.8795 \div 143.16 = 0.198\text{m}$ .
Because trim is by stern, Hydrostatic draft = Aft draft – Corr. m.
W = 15931.68 tons ; MCTC = 193.228 tm. ; LCB 72.665m.
Trim = W $\times$ BG $\div$ MCTC ; BG $= 0.4 \times 100 \times 193.228 \div 15931.68 = 0.485$ m.
As trim is by stern, LCG < LCB. Hence LCG = LCB – BG m.
Ship moves to water of RD 1.002. There will be no change in LCG.
W(virtual) $= 15931.68 \times 1.025 \div 1.002 = 16297.377$ tons
New Hyd. (DW) = 7.858m. ; LCB = 72.626m. ; LCF = 70.727m.
MCTC (SW) = 195.412 tm./cm.
MCTC (DW) $= 195.412 \times 1.002 \div 1.025 = 191.027$ tm./cm. ;
New BG m.
Because Initial LCG < New LCB, hence New trim is by stern.
New trim $= 15931.68 \times 0.446 \div 191.027 = 37.196$ cm. = 0.372m.
$\text{T}_\text{A}$ $= 0.372 \times 70.727 \div 143.16 = 0.184$ m. (+) ; $\text{T}_\text{F} = 0.372 - 0.184 = 0.188$ m. (-)
Final forward draft = 7.858 - 0.188 = 7.670m.
Final aft draft = 7.858 + 0.184 = 8.042m.

The following media explains about the Effect of Density Change on Trim.

Calculation of New forward and aft drafts when the ship moves from water of $\text{RD}_1$ to water of $\text{RD}_2$ , neither of them being SW

Given data - Forward draft (F) and Aft draft (A) while floating in water of $\text{RD}_1$ .
Mean draft = (F + A) $\div$ 2; $\text T_1$ = F $\sim$ A
For mean draft ascertain LCF from hydrostatic Tables.
Corr. = $\text T_1$ $\times$ LCF $\div$ LBP
If $\text T_1$ is by stern, Hyd. (1) = A – Corr.; If $\text T_1$ is by head, Hyd. (1) = A + Corr.
For Hyd. (1) ascertain $\text W_1$ (Virtual), $\text{MCTC}_1$ and $\text{LCB}_1$ from the Tables.
W (Real) = $\text W_1$ (Virtual) $\times \text{RD}_1\ {\div}\ 1.025$
Value of W (R) will not change when the ship moves from water of $\text{RD}_1$ to water of $\text{RD}_2$ .
$\text B_1 \text G = \text T_1( \text{cm.})\ {\times}\ \text{MCTC}_1\ {\div}\ \text W ( \text R) ; \text{LCG} = \text{LCB}_1\ {\pm}\ \text B_1 \text G$
If $\text T_1$ is by stern, sign is –ve; If $\text T_1$ is by head, sign is +ve.
Value of LCG will not change when the ship moves from water of $\text{RD}_1$ to water of $\text{RD}_2$ .
$\text W_2$ (V) = W (R) $\times$ 1.025 $\div$ $\text{RD}_2$
For $\text W_2$ (V) ascertain Hyd. (2) from Tables, and for it ascertain $\text{MCTC}_2$ , $\text{LCB}_2$ and $\text{LCF}_2$ .
$\text B_2 \text G = \text{LCB}_2\ {\sim}\ \text{LCG}$
If $\text{LCB}_2\ >\ \text{LCG}$ , final Trim $( \text T_2)$ is by stern ; If $\text{LCB}_2\ <\ \text{LCG}, \text T_2$ is by head.
$\text T_2 ( \text{cm.}) = \text W ( \text R)\ {\times}\ \text B_2 \text G\ {\div}\ \text{MCTC}_2$
$\text{Corr.} = \text T_2\ {\times}\ \text{LCF}_2\ {\div}\ \text{LBP}$
If $\text T_2$ is by stern, New Aft draft = Hyd. (2) + Corr.
If $\text T_2$ is by head, New Aft draft = Hyd. (2) - Corr.
New Forward draft = New Aft draft $\pm \text T_2$ .If $\text T_2$ is by stern, sign is –ve. If $\text T_2$ is by head, sign is +ve.

$\text{TPC} \left(\frac{\text{Tons}}{\text{cm}}\right) = \text{Area of} \ \frac{\text{W}}{\text{P}}(\text{m}^2) \times \text{RD} \div 100$

It means that TPC is directly proportional to Area of W/P, which changes with the draft of the ship. It is also directly proportional to the RD of water in which the ship is floating. Values of TPC in SW are tabulated for drafts in the Hydrostatic tables. Value of TPC in DW is calculated by the formula ---

$\text{TPC}_\text{DW} = \text{TPC}_\text{SW} \times \text{RD} \ \text{of} \ \text{DW} \div \text{RD} \ \text{of} \ \text{SW}$

Fresh Water Allowance (FWA) is the change in draft (Rise / Fall) of the ship when it moves from SW to FW or vice-versa. It is calculated by formula

$\text{FWA} (\text{cm}.) = \text{Displacement} (\text{Tons}) \div 40 \div \text{TPC}_\text{{SW}} (\text{Tons}/\text{cm}.)$

Similarly Dock Water Allowance (DWA) is the change in draft (Rise / Fall) of the ship when it moves from water of one RD to water of another RD, which may be SW, FW or DW. It is calculated by formula ---

DWA (cm.) = FWA (cm.) $\times$ (Change in RD) $\div$ 0.025

Example

A ship is loading in DW of density 1005 Kg/ cum. If the FWA is 220 mm Find the change in draft when the ship sails into SW.

$\text{DWA}=220 \times \frac{(1025-1005)}{25}$

$=220 \times \frac{20}{25}$ = 176 mm (Rise)

ANS: The draft will reduce by 0.176 mtrs

Calculation of Free surface effect

Learning Objectives

After completing this topic you will be able to:

Explain $I = \frac{LB^3}{12}$ for rectangular shapes
Explain $\text{FSC} = \frac{(I \times \text{Density of Liquid})}{\Delta}$
Explain $\text{FSC} = \frac{\text{FSM}}{\Delta}$
Calculate Moment of Inertia using Simpsons Rules

Recap

The following media explains about the Free Surface Effect

What is free surface effect and what is its impact on the ship’s stability when the ship is listed?
If a tank is completely filled with the liquid, the liquid cannot move and acts as a static weight.
So, the centre of gravity of the ship remains unchanged.
The graphic shows the listed ship when the tank is filled with liquid.
Assume that the ship floats at the same draft with the same KG and with the tank partially filled.
Now, if the ship lists, the liquid flows to the lower side causing g of the liquid to shift to g1.
This results in the centre of gravity of the ship G moving to G1.
This effect is called a free surface effect.
As a result, the vessel suffers an apparent loss of GM which is equal to GGv and hence the metacentric height is GvM.
GZ, the righting lever is also reduced by an amount which is equal to GG1.
This virtual loss of GM can be calculated. It is called a Free Surface Correction or FSC.
In order to indicate whether FSC has been applied or not the GM before subtracting FSC is called solid GM.
The GM attained after applying FSC is called a fluid GM.
In all stability calculations, you will use fluid GM.

Points to remember

If the area of the free surface is constant, the weight of liquid in the tank does not affect the FSE.
Within the ship, the position of the tank does not affect the FSE.
When a tank is either empty or full, FSE is zero.
To reduce the total FSE for the ship, the number of slack tanks must be kept to a minimum since the FSE of each slack tank contributes to the total FSE for the ship.
By fitting longitudinal divisions in the tank equally spaced, the FSE can be reduced to 1/n² times the undivided value, n = number of spaces. E.g: 3 spaces are produced by two longitudinal divisions and the FSE reduces to 1/9th value of undivided tank.
FSE will make the situation worse before the bottom weight increases to a sufficient level to bring G down, if time is taken to decide and start filling a DB tank to improve stability. The smallest tank on the lowest side should be filled first, if the ship is at an angle of loll.

Let us consider that the tank contains a solid weight. As the ship is heeled, the COB (Center of Buoyancy) shifts in the direction of the inclination and the righting lever GZ are formed.

Now, consider a tank full of liquid. When the ship is heeled, the surface of the liquid remains horizontal but results in a transfer of "a wedge of water", which is equivalent to a horizontal shift of weight, causing gravity to shift from G_0

. The effect on the position of the ship’s CG is the same as if the liquid were a solid of the same mass occupying the whole tank. The CG of the liquid inside the tank can be taken as the centroid of the tank.

When the tank is full of liquid and the ship heels, there is no movement of liquid possible inside the tank and the position of the CG of the liquid and therefore the ship remains stationery.

If the tank is partly filled, it is said to be slack. The liquid now has a surface below the tank top and this surface will slope towards the angle of heel depending upon the angle. The tank now has a ‘free surface’. The position of the CG of the liquid inside the tank will change depending upon the angle of heel and this will affect the position of ship’s CG.

The transverse movement of the CG from G_0

caused by the moving of the wedge of water reduces the righting lever G_0Z

and has the same effect as raising G_0

on the centre line. There is thus a ‘loss of GM’ by the distance GG_3

. This phenomenon leading to the virtual loss of GM is called the 'Free Surface Effect (FSE)'. The amount of virtual loss of GM is known as the Free Surface Correction (FSC).

FSC can easily be calculated as follows:

Free Surface Correction

Here,

i is moment of inertia of the slack tank (m^4)

rt is density of the liquid present in the slack tank $\left(\frac{t}{m^3}\right)$

W is the displacement of the ship (t)

i $\times$ rt is Free Surface Moment (FSM) expressed in tm.

For multiple stack tanks, we calculate the FSM of each tank and then add together to obtain the total FSM.

By subdividing the tanks free surface effect can be reduced as can be deduced from the above figure.

The following media explains about the List:

The following media explains about the Derivation of the Formula for List:

The audio presentation of the above media is written below:

By taking precautions to minimize the virtual loss of GM you will be able to maintain the stability of the ship. In the ship considered here you will fill the bottom tank so that G is lowered. Then you empty the top side tanks as shown. As you do these operations the G and M meet at a point. And this state of the ship is called the neutral equilibrium in order to bring a stable state to the ship. In other words to bring the GM value to be positive. Discharge some cargo from the deck when the ship has the positive GM she is in a stable state of equilibrium as shown on the graphic. However the general precautions to ensure the stability of a ship are planning storage in such a way that heavy cargo is stored in the lower holds. Ensuring that the number of slack tanks is kept to a minimum during passage. Keeping small tanks slack and the big tanks either full or empty, if there are different types of tanks. Sub-division of tanks reduces the free surface by four times and as a result, the GM loss can be minimized.

Case 1:

When water or ice accumulates on the deck, there is increase in the weight of the ship just above the COG of the vessel. This results in the shift of COG upwards. Ice formation increases the KG of the vessel, lowering GM.

Case 2:

When a timber cargo on the deck absorbs water, this weight is normally added above the COG of the vessel and hence the COG shifts upwards. Here, due to water absorption by the timber cargo on the deck, there is an increase in the KG of the vessel. As a result, GM is lowered.

When the ship is listed if a tank is completely filled with a liquid, the liquid cannot be moved inside the tank and acts as a static weight. So the centre of gravity of the ship remains unchanged. The graphic shows the listed ship. When the tank is filled with liquids assume that the ship floats at the same draft with the same kg and with the tank partially filled. Now if the ship lists the liquid flows to the low side causing the G of the liquid to shift to G1. This result in the centre of gravity of the ship G moving to G1. This effect is called a free surface effect. You have seen that when GM is positive, the ship is in stable equilibrium. When this GM is a large positive value, the ship is called a Stiff vessel whereas when the GM is a small positive value the ship is called a tender vessel. A tender vessel has a large angle and period of roll by stiff vessel have small angles and periods of rolls.

Derivation of the formula for lists:

In this graphic

GG1 is the transverse shift of G
The couple formed causes the ship to incline
When the COB is vertically under G1 the vessel is in static equilibrium
The angle of inclination at which this happens is the list

The forces of gravity and buoyancy is equal and opposite and separated at the righting lever and also forms a couple which tends to keep the vessel upright. The moment of this couple is measure of the tendency of the vessel to return to its upright position. It is called the righting moment or the moment of statical stability. It is given by the product of the righting lever and the righting lever for any angles of heel. The relation between them is that GM increases, decreases and becomes positive or negative with GZ.

In this graphic, GZM is a triangle. Z is right angle and theta is the angle of view. Hence, sin theta is given by the ratio of GZ and GM. Therefore, at small angles of heel at 15 degrees where GM is considered constant you can use GM as the indication of statical stability instead of GZ. For small angles of heel up to 15 degrees the value of righting moment can also be found using GM value. It is given by the final displacement of the ship multiplied by GM and sin theta.

In the right angles-triangle MGG1 tan theta is given by the ratio of GG1 and GM. Where, theta is the angle of list and GG1 is the transverse shift of G. GM is the final fluid GM before listing. Since GG1 is equal to dw divided by W, the formula becomes tan theta equals dw divided by the product of final displacement and the GM.

Here, theta is the list, dw is the final listing moment, W is the final displacement, GM is the final fluid GM in metres. In order to calculate the final list systematically the following order of work is suggested.

Find the final listing moment,
Find the final displacement (W),
Find the final fluid (GM)
Apply the list formula tan theta equals dw divided by WGM.

FSC can be calculated by the formula:

FSC equals i divided by V multiplied by di by do.

Where, i is the moment of inertia or the second of area of the slack tank surface about its centre line, in m power 4,

V is the volume of displacement of the ship, in m power 3,

di is the density of liquid in the slack tank, in tm power -3,

do is the density of the water the vessel is floating in, in tm power -3,

FSC is the free surface correction in m caused by the slack tank.

Remember, fluid GM is always less than solid GM. FSC varies directly as the density of liquid in the tank and inversely as the density of water that the ship is floating in. This means that the greater the density of the liquid in the tank, the greater the mass of the liquid that shifts, therefore greater the free surface correction.

Since the displacement of a vessel is given by the product of the volume of displacement and the density of water displaced, the formula now becomes, FSC equals i multiplied by di divided by w. This now indicates that greater the displacement of the vessel the smaller the free surface correction. Since i is in metres power 4 and di are in tonnes per metre cube idi would be in tonnes metre and is called the free surface moment or FSM.

When several tanks are slack on a ship the FSM of each tank is calculated seperately and then added together to calculate the total FSM. This total FSM divided by the final w of the ship gives the total FSC of all slack tanks. The breadth of the slack tank has an enormous effect on the FSC caused. Hence, sub-dividing the tank reduces the free surface by one-fourth, or it is four times less. If the tank is further subdivided into three compartments of equal size, the free surface will reduce by one-ninth. In other words we can say that when a tank is divided in breadth by a number of identical compartments, n, and if these compartments are slack, the free surface would be 1 by n square of the free surface that would have occurred had the tank not been subdivided. Note that, on ships, 'i' of each tank about the tank's centre line, is calculated in the stability particulars supplied by the shipyard... The audio presentation ends..

Controlling Free Surface Effect

1. Use of tank dividers

The effect of free surface of liquid will be critical in a vessel with small metacentric height where the vessel could even become unstable. In such ships, the tanks which are carrying liquid should be filled up completely and pressed up. If the ship becomes unstable with an angle of loll where the G and M are coinciding, due to the effect of free surface: and if the ship is heeling to port or starboard side and coming back up too slowly, we have a lolling or unstable condition in hand. Any attempt to fill water ballast to correct the list could reduce the stability further. Therefore before ballasting, an attempt should be made to lower the CG of the ship by pressing up the existing lower tanks at the double-bottom and lowering the CG or masses on the ship. Having tank divisions help to reduce the effect of free surface.

From calculations, you can see that the free surface effect continues to reduce by having longitudinal plate divisions forming more number of tanks of equal sizes.

The ship with the greatest free surface effect is the oil tanker, since space must be left in the tanks for expansion of oil. Originally tankers were built with center line bulkhead and the expansion tanks. Twin longitudinal bulkhead were then introduced without expansion tanks and were found to be successful, since design takes care of the loss in metacentric height due to free surface effect.

It is however, not possible to design the dry cargo vessel in the same way, since the position of the CG of the ship varies with nature and loading position of the cargo. Ships with a lot of deck cargo are thus more unstable and ore carriers are stable but uncomfortable with large righting moments where CG is well below due to the heavy cargo carried on bulk in the depth of the hold space.

2. Fitting Wash Plate in Tanks

They restrict the motion of the water in the tank, thus reducing the free surface effect. This will have a marginal effect.

3. Putting Tank Baffles

They divert the motion of the water, and this will help by creating longer time for water to reach from one end to the other. This will also have a marginal effect in reducing FSE.

I=LB³/12 for Rectangular Areas

We know $BM_T = \frac{I}{V}$

Where,

I is 2^nd Moment of inertia about the center line V = Volumetric displacement

For a rectangular water plane area, the 2^nd moment of area about the center line XX’ is

$I = \frac{LB^3}{12}$

Where L is length and B is breadth of water plane.

FSC = I x RD of liquid in the tank/W

$\text{FSC} = \frac{(I \times \ \text{Density of Liquid})}{\Delta}$

In the diagram a vessel upright at the water line WL is inclined by an angle $\theta$ to a water line W_1L_1

This causes a wedge of liquid weight w tonnes in the DB tank to shift from g on the emmersed side to g_1

on the immersed side.

Accordingly the ships G shift to G_1

and the vertical through G_1

intersects the vertical through the center at G_v

The virtual reduction of GM or the Free Surface Correction (FSC) may be considered to be GG_v

$GG_1 = W \times \frac{gg_1}{W}$

Considering the density of liquid inside the tank to be d_1

and the ship to be floating in liquid of density d_2

$GG_1 = \frac{v \times d_1 \times gg_1}{V \times d_2}$

In $\triangle GG_vG_1$

$GG_1 = GG_v \times \tan \ \theta$

$\frac{v \times d_1 \times gg_1}{V \times d_2} = GG_V \times \theta$

$GG_v = \frac{v \times d_1 \times gg_1}{V \times d_2 \times \tan \ \theta} --------(1)$

Since $I \times \tan\ \theta = v \times gg_1$

$\tan \ \theta = \frac{v \times gg_1}{I}$

Substituting the value of tan $\theta$ in ……. (1)

$GG_v \text{(FSC)} = \frac{I \times d_1}{V \times d_2} = \frac{I \times d_1}{W}$

Since $I = \frac{LB^3}{12}$ , If the breadth is divided into n compartments of equal size then,

$GG_v \text{(FSC)} = \frac{I \times d_1}{W \times n^2}$

FSC = Free Surface Moment in tonnes meters/ Displacement in tonnes (FSM/∆)

The shape and volume of the tank along with the density of the fluid within decides the quantity of free surface moment that is created by the liquid shifting within the tank.

Since the Moment of inertia (I) is about the centerline in tonnes meters and the density of the liquid is in tonnes / m^3

$\text{FSM} = \frac{I \times d_1}{W}$

FSM of box shaped tanks may be calculated by the formula $I = \frac{LB^3}{12}$ , however tanks on board ships are rarely box shaped and the FSM has to be calculated using Simpsons Rules. The FSM of individual tanks along with the liquid density for which it has been calculated is given in the ships stability particulars.

Moment of Inertia of tank using Simpson's rule

A derivation of Simpsons 3rd Rule called the 3,10,-1 rule is used to calculate the longitudinal moment of area between two consecutive ordinates of a tank.

The levers are 3, 10, -1

2nd Moment of inertia (Area between A and B) = $\frac{h^2}{24} (3A + 10B - 1C)$

Examples

1. Stability data of a ship indicates that for W = 5500 t, KM = 8.7 m, Moment of inertia (i) of No3 DBT about center line = 1428 m^4

containing diesel of RD 0.88 is slack and the ships KG is 8.5 m. Find the ships GM (Fluid).

$\text{FSC} = \frac{I \times d_1}{W} = \frac{1428 \times 0.88}{5500} = 0.228 \ m$

KM = 8.7 m

KG = 8.5 M

GM (Solid) = 0.2 m

FSC = 0.228 m

ANS: GM (Fluid) = 0.028 m (Negative)

2. A vessel of W = 16600 t, KM = 8.25 m, KG = 7.4 m has the following slack tanks.

No1 DBT (SW), $I = 400 \ m^4$ , No 3 Center (HFO), $I = 1200 \ m^4$ , No8 Port (FW), $I = 25 \ m^4$

Calculate her fluid GM, considering the RD of SW, HFO, FW to be 1.025, 0.95, 1.0

Tank		Density	FSM (tm)
No1 DBT	400	1.025	410
No3 Center	1200	0.95	1140
No8 Port	25	1.00	25
		Total FSM = 1575 tm

$\text{FSC} = \frac{1575}{16600} = 0.098 \ m$

KM = 8.25 m

KG = 7.4 M

GM (Solid) = 0.85 m

FSC = 0.098 m

ANS: GM (Fluid) = 0.752 m

3. A vessel W = 10000 t, KM = 9.3 m, KG = 7.3 m, has 2 identical rectangular deep tanks Port & Stbd, each 15 m long, 10 m wide & 8 m deep. The Stbd deep tank is full of SW and the Port tank is empty. Calculate the Fluid GM of the ship when a quarter of the SW from the Stbd tank is transferred to the Port tank.

Mass of SW in tank $= 15 \times 10 \times 8 \times 1.025 = 1230 \ \text{t}$

Mass of water transferred from Stbd to Port $= 0.25 \times 1230 = 307.5 \ \text{t}$

Since after the transfer of SW from the tank there has been a shift of weight

I.e. SW from the top of the Stbd tank has been shifted to the bottom of the Port tank

Hence

$GG_1 = \frac{w \times d}{W} = \frac{307.5 \times 6}{10000} = 0.185 \ m$

New KG

KM = 9.3 m

GM (Solid) = 9.3 - 7.115 = 2.185

Moment of inertia (i) for Port tank $= \frac{LB^3}{12} = \frac{15 \times 10^3}{12} = 1250 m^4$

FSM $= i \times \text{density} = 1250 \times 1.025 = 1281.25 \ \text{tm}$

$\text{FSC} = \frac{\text{FSM}}{W} = \frac{1281.25}{10000} = 0.128 \ m$

Total FSC (Since FSM for both tanks are same and they are now slack) $= 0.128 \times 2 = 0.256 \ m$

GM (Solid) = 2.185 m

FSC = 0.256 m

ANS: GM (Fluid) = 1.929 m

Simplified stability data

Learning Objectives

After completing this topic you will be able to:

Explain
Maximum Deadweight Moment
Minimum Permissible GM
Maximum Permissible KG Diagrams
Use Diagrams of Deadweight Moment

What is Simplified Stability Data?

While larger ships are compulsorily supposed to carry stability data both intact and damaged, it was felt that smaller ships including pleasure craft should have data which enables them to assess their crafts stability.

It was also felt that these craft would be manned by those who are not well versed with the traditional methods of calculating GM and GZ, and their compliance with intact stability criteria.

Simplified stability data originally intended for use in small ships may also be found in some larger ships

The data takes the following forms:

A diagram or table of maximum deadweight moment
A diagram or table of minimum permissible GM
A diagram or table of maximum permissible KG all related to the displacement or draught in salt water

In passenger vessels, data necessary to maintain sufficient intact stability under service conditions to enable the ship to withstand the critical damage assumptions of SOLAS would also be provided in simplified form

Merchant Shipping Notice 1122 on Simplified Stability Data gives the following information.

SIMPLIFIED STABILITY INFORMATION

Notice to ship-owners, Masters and Shipbuilders

It has become evident that the masters’ task of ensuring that his ship complies with the minimum statutory standards of stability is in many instances not being adequately carried out. A feature of this is that undue traditional reliance is being placed on the value of GM alone, while other Important criteria which govern the righting lever GZ curve are not being assessed as they should be. For this reason the Department, appreciating that the process of deriving and evaluating GZ curves is often difficult and time-consuming, strongly recommends that in future simplified stability information be incorporated into ships’ stability booklets. In this way masters can more readily assure themselves that safe standards of stability are met.
Following the loss of the Lairdsfield, referred to in Notice M.627, the Court of Inquiry recommended that simplified stability information be provided. This simplified presentation of stability information has been adopted in a large number of small ships and is considered suitable for wider application in order to overcome the difficulties referred to in Paragraph 1.
Simplified stability information eliminates the need to use cross curves of stability and develop righting lever GZ curves for varying loading conditions by enabling a ship’s stability to be quickly assessed, to show whether or not all statutory criteria are complied with, by means of a single diagram or table. Considerable experience has now been gained and three methods of presentation are in common use.
These are:
(a) The Maximum Deadweight Moment Diagram or Table.
(b) The Minimum Permissible GM Diagram or Table.
(c) The Maximum Permissible KG Diagram or Table.
In all three methods the limiting values are related to salt water displacement or draft. Free surface allowances for slack tanks are, however, applied slightly differently.
Consultation with the industry has revealed a general preference for the Maximum Permissible KG approach, and graphical presentation also appears to be preferred rather than a tabular format. The Department’s view is that any of the methods may be adopted subject to: (a) clear guidance notes for their use being provided and (b) submission for approval being made in association with all other basic data and sample loading conditions. In company fleets it is, however, recommended that a single method be utilized throughout.
It is further recommended that the use of a Simplified Stability Diagram as an adjunct to the Deadweight Scale be adopted to provide a direct means of comparing stability relative to other loading characteristics. Standard work forms for calculating loading conditions should also be Provided.
It is essential for masters to be aware that the standards of stability obtainable in a vessel are wholly dependent on exposed openings such as hatches, doorways, air pipes and ventilators being securely closed weather tight; or in the case of automatic closing appliances such as air pipe ball valves that these are properly maintained in order to function as designed.
Ship-owners bear the responsibility to ensure that adequate, accurate and up-to-date stability information for the master’s use is provided. It follows that it should be in a form which should enable it to be readily used in the trade in which the vessel is engaged.

Maximum Dead Weight moment

Deadweight Moment

The moment of the deadweight of the vessel about her keel is called deadweight moment and is expressed in tonnes metres. It is also known as the vertical moment of weight. The student will have used it in calculations to work out the vessels final KG during the 2^nd Mates Certificate of Competency. To find the deadweight moments, FSM if any is to be included.

The curve of Maximum permissible Deadweight (mt) Moments is plotted against Displacement on the y- axis and Deadweight moment (Tonnes metres) on the x-axis.

The following media explains about the Dead weight Moment

Example

Using the Deadweight Moment curve find out how much steel pipe (KG 3.5 m) may be loaded so that on completion loading the vessel complies with maximum deadweight moment curves. Load $\bigtriangleup$ is 3750 mt. Distribution of weight on board the vessel are as follows.

	Weight (mt)	Vertical KG (m)	Vertical /Deadweight Moment(tm)
Light ship	1000
H Oil	250	0.5	125
H Oil – FSM			1300
F Water	250	4.0	1000
FW – FSM			400
Cargo	2000	3.5	7000
Present $\Delta$	3500		Total Dwt moment 9825

From the curve for a $\bigtriangleup$ of 3500 mt the permissible DWT Mmt is 11800 tm.

Available DWT = 3750 - 3500 = 250

Dwt moments to be added if 250 mt is loaded with KG $3.5 = 250 \times 3.5 = 875$ tm

Final Dwt moments will then be = 9825 + 875 = 10700

Permissible Dwt Mmt = 12600 tm. (Within permissible limits)

ANS: Therefore 250 mt of pipes may be loaded

Minimum Permissible GM

The Curve of minimum GM_T

The curve of Minimum GM against moulded draught is plotted. As long as the GM falls within the minimum limits the vessel will comply with intact stability criteria. These diagrams are also available on large vessels.

Maximum Permissible KG

Permissible KG

The KG of the vessel is calculated and found to be 3.275 mtrs at a Displacement of 187 mt. We plot this on the table above and find that the KG lies below the curve i.e. the permissible area.

Use of diagrams of Dead Weight moment

Please refer the topic ‘Maximum dead weight moment’.

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Saturday, June 8, 2024

Ship Stability - I Control trim, stability and stress - Ship Stability

Approximate calculations of Area and volume

Simpson Rules 1,2 and 3

Calculation of area and volume using Simpson's rule

Calculation of Geometric centres of area and volume

Calculation of TPC, FWA using Simpson rules

Effect of density: TPC, FWA, DWA calculations

Effect of change of density of water on TPC

Calculation of draft of vessel fore and aft due to change in density

Calculation of Free surface effect

I=LB³/12 for Rectangular Areas

FSC = I x RD of liquid in the tank/W

FSC = Free Surface Moment in tonnes meters/ Displacement in tonnes (FSM/∆)

Moment of Inertia of tank using Simpson's rule

Simplified stability data

Maximum Dead Weight moment

Minimum Permissible GM

Maximum Permissible KG

Use of diagrams of Dead Weight moment

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