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ASSIGNMENT 3

Q1a.

Describe in detail (using appropriate diagrams/drawings) a common accepted

method (experiment) you would use to obtain an accurate assessment of the vessel’s Related Essays
initial GM.

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The adjustable immersion hydrometer is a tool, centered on the Archimedes Law. The tool according to
the law assists to regulate the liquid density (Dokkum, 2013). The hydrometer utilized to calculate the Research Methods in Engineering
water density where the ship floats is typically made through materials that are not easily corroded. The
materials also mainly have a subjective bulb with a thin rectangular stem that transmits a scale for Macroeconomics

gauging densities between 1000 and 1025 kilograms per cubic metre, i.e. 1.000 and 1.025t/m3.
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On the stem provided or developed according to hydrometer, the marks are positioned as follows. The
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first step involves letting the hydrometer as presented in figure 1 to float in the fresh water upright at the
part marked X. To weight the water accurately, one procures the hydrometer from the water and records InterviewCustomers name: Institution: Customers
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the measurements. To record effectively, the mass is selected as Mx kilograms (Dokkum, 2013). Once the
hydrometer has been attained from the water, it is replaced with the bulb until one can perceive it hovering Its each weeks topic discuss and then put it together as
where the position marked Y, is the higher close of the stem, in the waterline used. Then one assesses the an assignment

hydrometer once more and perceives the mass now to be measured as My kilograms. The mass is
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important for developing the initial GM of the vessel. Advantage

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Section 3

Figure: 1

The water mass of that is displaced by the stem is recorded amid X and Y, which is hence equivalent to My
— Mx kilograms. About 1000kg of the water (fresh) used inhabit a cubic of a metre, which assists measure
the stem volume. The stem (figure 1) volume is measured between X and Y as presented in figure one
above, which is equal to Related Categories

My – Mx /1000 cu. m. Engineering and Construction

In the following equation, L is the length of the stem provided. That is; the distance between X and Y is the Family & Consumer Science
stem length. On the other hand, letter A in the equation is a representative of the stems’ cross sectional
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part.

A = Volume / Length

= (My – Mx / 1000L) sq m

Now let the hydrometer float in water of density ρ kg/m3 with the waterline ‘x’ metres below Y:

Volume of water displaced = MY / 1000 – xa

= MY /1000 – x{ (MY — MX) /1000L} ……………………(i)

Mass of water displaced

Volume of water displaced =

Density of water displaced

= My / 1000 ρ …………………………….. (ii)

Equate (I) and (II).’ My / 1000 ρ = My / 1000 — x{ (MY — MX) /1000L}

Or ρ = My / MY — x {(MY — MX) /L}

In this equation, My, Mx and L are known constants whilst ρ and x are variables. As a result, to spot the
gauge it is now simply essential to select numerous standards of ρ and to calculate the corresponding
values of x.

Q1b

Using the following information provided below, calculate the following

measurements for this vessel?

Mass Displacement.

Waterplane area.

LCB (longitudinal Centre of Buoyancy).

TPC (Tons per centimeter Immersion)

Initial KG

Initial GM

The vessel is in salt water.

½(A)m2=Y ½ breadth(m)

00 0 00 00 0

0 0 0 00 0

ASMX=∑ Y*SM=∑

SM is Simpson’s multiplier

Using Simpson’s first rule,

Area of waterplane = 1/3 * CI * ∑Y * 2 where CI=h/d, 50/10

= 1/3 * 5 * 33.69 * 2

=112.3m2

Under water volume = 1/3 * CI * ∑ASM * 2

= 1/3 * 5 * 102.26 * 2

= 340.867m3

Mass displacement

= volume under water * sea water density

= 340.867 * 1.025

= 349.389 tons

Waterplane area

Area of waterplane = 1/3 * CI * ∑Y * 2 where CI=h/d, 50/10

= 1/3 * 5 * 33.69 * 2

=112.3m2

= ∑ (ASMX) / ∑ (ASM)

= 2323.6 / 102.26

=22. 722m from bow

First moment of inertia, I will be calculated

I =1/12 (Cwp) 2 * LB3where Cwp =Awp / LB

= 1/12 (0.612)2 * (50 * 6.83)

= 474.79m4

= 474.79 / 340.867

= 1.3929 m

= ( Wpa * RD ) / 100

= (112.3 * 1.025) / 100

=1.151 tonnes/m

Initial KG

= moment of inertia / weight displaced

Moment of inertia, ICL = 2/3* CI * ∑YSM

= (2/3 *5 *853.832) / (349. 389 * 9.81)

= 0.8304m

Initial GM

GM = KB + BM – KG

= 0.737 + 1.3929 – 0.8304

= 1.2995m

Q1c.

On the same vessel.

2 x weights A&B (1.8t each) are placed 2.2m each side of the LCB (v/l in equilibrium)

Weight A is moved to the position of weight B giving a pendulum movement of 75mm.

Pendulum length 3m (mounted at the LCB)

If the weight A is moved back to the original position and a new weight of 10t is loaded amidships with a
CG of 3m.

What will be the shift of G and the new KG.?

Solution

From 1b LCB is 22.72, when weight A & B are added to the vessel it remains at equilibrium, when weight A
is shifted to weight B LCB shifts by 3m to 19.72

When weight A moves back to its original position, it has a distance d1 from the new LCB

The new weight of 10t has a distance d2 from LCB

Weight B that remained at its original position has a distance d2 from the new LCB

The shift will be GG = GG1 + GG2 + GG3

Using GG = w * d / W

Where w is the weight at a distance d, W is the total weight

The total weight W = 349.39 +1.8 + 1.8 + 10

= 363tonnes

GG1 = 1.8 * 5.18 / 363 = 0.026m

GG2 = 10 * 5.28 / 363 = 0.145m

GG3 = 1.8 * 0.8 / 363 = 0.004m

Therefore, the shift of G will be 0.026 + 0.145 + 0.004 = 0.175m

Thus the new KG will be (.8304 – 0.175) = 0.6554m

Q1d

Explain how a KN (SZ) curve is used to determine the vessel’s stability data?

The processof developing a cross curve of the stability of the ship occurs through one plotting the
assumed height with the righting levers. The height is determined by the gravity center that occurs beyond
the keel (Barrass & Derrett, 2011). The constructed curves are also assumed in other cases a KG of zero.
The developed curves are identified as KN curves where KN is perceived as the righting lever that is
usually determined from the keel.

The displacement and KG values righting levers are obtained where the KN value is first attained through
the inspection of the curves displacement value. When one subtracts the KN values the appropriate
righting levers are also determined, which is an equal correction of the sin heel and the KG product.

In Figure b, let KN signify the ordinate acquired from the curves. In addition, let the ship’s centre of gravity
be at G so that KG characterizes the genuine elevation of the centre of gravity beyond the keel and GZ
embodies the distance (L) of the righting lever.

GZ =KN — G sin
θ

Thus, to determine the righting lever one must use the KN and KG. That is; the ordinate of the KN should
be subtracted from the KG sin heel correction equal.

Then using the vessel’s calculated displacement from 1b and the KN (SZ) curve provided below for this
small commercial work boat. (KS is 3.193m)

Draw the vessel’s GZ curve of statical stability and state the maximum GZ value?

KN cross curves of stability

Solution

The following data on the table below was obtained from at 349.4 tonnes on the KN curves

GZ = KN – G sinθ

Taking G as 0.8304

Angle (O)

The maximum GZ value is 1.56 m at 65 degrees

Q1e

A different vessel with the following dimensions:-

Waterline length of 100 metres

Displacement of 6300 tonnes.

Mean draught of 6 metres,

Longitudinal metacentre = 104m

A weight of 60 tonnes is moved from forward to aft over a distance of 50 metres.

Longitudinal Centre of Floatation (LCF) = Amidships

Calculate the MCT 1 cm and the new draughts forward and aft for this vessel.

Solution

MCT1cm = (W * BMl) / 100L

BML = L2 /12d

= 1002 / 12 *6

= 138. 89m

The distance BG should be recorded, which is small when correlating it to the GML and BML distances on
the stem. Therefore, for this motive, there is a high probability that BML has an appreciable error that is
occasionally replaced for GML in the method developed of calculating MCT 1 cm.

= 6300 * 138.89 / 100 * 100

= 87.5 tonnes / cm

Change of trim = w * d / MCT1cm

= 60 * 50 / 87.5

=34. 2857 cm by the stern

Change of draft aft = I/L * change of trim

Where; I is the distance of the centre of floatation from the aft in metres

L is the ships length in metres

= 50/100 * 34.2857

= 17.1429cm

Original draft 6.0000 A 6.0000 F

Change due to trim + 0.17143 — 0.17143

New drafts 6.1714 m A 5.829 m F

References

Barrass, B. & Derrett, C. D., 2011. Ship stability for masters and mates. New York: Butterworth-Heinemann.

Dokkum, K., 2013. Ship knowledge: ship design, construction and operation. New York: DOKMAR.

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