Ship Hydrostatics and Stability 2010 Part 4 ppt

Basic ship hydrostatics 65B/2 a b Figure 2.24 Monohull versus catamaran We can obtain the same displacement volume with two hulls of breadth B/2, the same length, L, and the same draught

Basic ship hydrostatics 65

B/2

(a)

(b)

Figure 2.24 Monohull versus catamaran

We can obtain the same displacement volume with two hulls of breadth B/2, the same length, L, and the same draught, T Assuming that the distance between the centrelines of the two hulls is 3B/2, the resulting metacentric radius is

19B 2

48T

The first term between parantheses represents the sum of the moments of inertia

of the waterlines about their own centrelines The second term accounts for theparallel translation of the hulls from the plane of symmetry of the catamaran

The second term is visibly the greater The ratio of the catamaran BM to that of the monohul is —- The improvement in stability is remarkable.

Catamarans offer also the advantage of larger deck areas and, under certainconditions, improved hydrodynamic performances On the other hand, the weight

of structures increases and the overall performance in waves must be carefullychecked It may be worth mentioning that also many vessels with three hulls, that

is trimarans, have been built Moreover, a company in Southampton developed

a remarkable concept of a large ship with a main, slender hull, and four side

hulls; that is a pentamaran.

Example 2.8 - Submerged bodies

Submerged bodies have no waterplane; therefore, their metacentric radii areequal to zero Then the condition of initial stability is reduced to

GM = KB - KG > 0

In simple words, the centre of gravity, G, must be situated under the centre of

buoyancy, B We invite the reader to draw a sketch showing the two mentioned

points and derive the condition of stability by simple mechanical considerations.Submerged bodies do not develop hydrostatic moments that oppose inclinations,

as they do not develop hydrostatic forces that oppose changes of depth

Example 2.9 - An offshore platform

Figure 2.25 is a sketch of an offshore platform of the semi-submersible type The buoyancy is provided by four pontoons, each of diameter b and length L The platform deck is supported by four columns The depth of the platform is H and the draught, T.

Our problem is to find a condition for the height of the centre of gravity,

KG, for given platform dimensions To do this, we calculate the limit value of

KG for which the metacentric height, GM, is zero The metacentric radius is

where we neglected the term in 62, usually small in comparison with other terms

Figure 2.25 A semisubmersible platform

Basic ship hydrostatics 67

The condition for initial stability is

The height of the centre of buoyancy above the base-line is calculated in

Table 2.6 Neglecting the term in — b 2 we obtain

Exercise 2.1 - Melting icebergs

In Example 2.1 we learnt that if an ice cube melts in a glass of water, the level ofwater does not change Then, why do people fear that the meltdown of all icebergswould cause a water-level rise and therefore the flooding of lower coasts? Showthat they are right

Hint: Icebergs are formed on the continent and they are made of fresh water,

while oceans consist of salt water The density of salt water is greater than that

of fresh water

Exercise 2.2 - The tip of the iceberg

Icebergs are formed from compressed snow; their average density is 0.89 tm~3.The density of ocean water can be assumed equal to 1.025 tm~3 Calculate whatpart of an iceberg's volume can be seen above the water and explain the meaning

of the expression The tip of the iceberg'

Hint: See Exercise 2.1.

Exercise 2.3 - Draughts of a parallelepipedic barge

Consider a parallelepipedic (or, with another term, a box-shaped) barge terized by the following data:

Exercise 2.4 - Whisky on the rocks

Instead of considering a cube of ice floating in a glass of water, as in Example 2.1,let us think of a cube of ice floating in a glass of whisky What happens whenthe cube melts?

Exercise 2.5 -A lemma about moving masses in three-dimensional

Prove the lemma in Section 2.7 for a three-dimensional system of masses and athree-dimensional displacement of one of the masses

Exercise 2.6 -A wooden parallelepiped

The floating condition of a wooden, homogeneous block of square cross-sectiondepends on its specific gravity Three possible positions are shown in Figure 2.26

(a) (b) (c)

Figure 2.26 Different floating conditions of a wooden, parallelepipedic

block

Basic ship hydrostatics 69

1 Find the ranges of specific gravity enabling each position

2 For each range find a suitable kind of wood To do this look through tables

Exercise 2.7 - B and Mcurves - variable heel

Table 2.7 contains the same data items as Table 2.4, but calculated at 5-degreeintervals With this 'resolution' it is possible to plot smooth B and M curves.First, write the data on a file lido9a similar to file lido9 Next, modify theprogramme cited in Example 2.6 to plot only the B and M curves of the vesselwhose data are called from the keyboard Run the program with the data given

at 5-degree intervals and print a hardcopy of the resulting plot

Exercise 2.8 -B and M curves - variable trim

Table 2.8 contains data of the vessel Lido 9 for constant volume of displacement

equal to 44.16 m3, upright condition, and trim varying between -0.3 and l.lm

Table 2.7 Data of vessel Lido 9 at 44.16 m3 volume of displacement and5-degree heel intervals, trim = -0.325 m

NB

(m) 1.272 1.255 1.204 1.121 1.009 0.872 0.711 0.528 0.326 0.107 -0.126 -0.372 -0.625 -0.883 -1.140 -1.393 -1.640 -1.878 -2.108

NM

(m) 4.596 4.438 4.119 3.711 3.341 3.073 2.857 2.464 2.105 1.830 1.537 1.082 0.479 -0.185 -0.543 -0.869 -1.171 -1.446 -13.314

LCB

(m) -1.735 -1.740 -1.761 -1.799 -1.841 -1.887 -1.932 -1.971 -2.002 -2.047 -2.106 -2.113 -2.072 -2.041 -2.025 -2.008 -1.994 -1.981 -1.970

NM L

(m) 23.371 23.693 24.008 23.730 23.813 23.464 23.154 22.822 22.810 23.133 21.837 19.757 17.473 16.162 15.117 14.298 13.633 13.121 12.792

Table 2.8 Data of vessel Lido 9 at 44.16 m3 volume of displacement, 0.1 m trim intervals, upright condition

NB

(m) 1.174 1.192 1.208 1.224 1.238 1.251 1.24 1.276 1.286 1.295 1.304 1.311 1.319 1.324 1.329 1.333 1.336 1.338 1.339 1.340 1.340 1.338

NM

(m) 4.536 4.550 4.564 4.577 4.585 4.589 4.592 4.598 4.604 4.610 4.614 4.615 4.614 4.612 4.610 4.606 4.603 4.599 4.599 4.597 4.594 4.590

LCB

(m) -2.777 -2.623 -2.468 -2.313 -2.157 -2.001 -1.848 -1.697 -1.548 -1.401 -1.257 -1.114 -0.971 -0.829 -0.690 -0.546 -0.407 -0.270 -0.135 0.000 0.135 0.269

NML

23.681 23.904 24.069 24.163 24.145 23.954 23.584 23.293 22.951 22.556 22.108 22.137 22.115 22.046 21.910 21.707 21.431 21.116 20.895 20.871 20.834 20.829

The LCB values in column 5 are equivalent to the KN values in Example 2.6,

Figure 2.22.

Write the data on an M-file, Iido9b m, and use the program b_curve to

plot the B- and M-curves Here the M-curve is the locus of the longitudinal

metacentre.

2.15 Appendix -Water densities

Density (tm~ 3 ) Fresh water

Eastern Baltic Sea Western Baltic Sea Black Sea

Oceans

Red Sea

Caspian Sea Dead Sea

1.0001.0031.0151.0181.0251.0441.0601.278

Numerical integration in naval architecture

car-Two methods for numerical integration are described in this chapter: the

trape-zoidal and Simpson's rules The treatment is based on Biran and Breiner (2002).

The rules are exemplified on integrands defined by explicit mathematical sions; this is done to convince the reader that the two methods of numericalintegration are efficient, and to allow an evaluation of errors The first examplesare followed in Chapter 4 by Naval-Architectural applications to real ship datapresented in tabular form

expres-Many Naval-Architectural problems require the calculation of the definiteintegral

•&

f ( x ) d x

of a function bounded in the finite interval [a, b] We approximate the definite

inte-gral by the weighted sum of a set of function values, /(zi), / ( x2) , , /(xn),

evaluated, or measured, at n points Xi G [a, 6], i = 1 , 2 , , n, i.e.

fb

I

-*a

(3.1)

In Naval Architecture, the coefficients a^ are called multipliers; in some books

on Numerical Methods they are called weights.

There are several ways of deriving formulae for numerical integration - also

called quadrature formulae - of the form shown in Eq (3.1); three of them are

mentioned below:

1 By geometrical reasoning, considering f a f ( x ) dx as the area under the curve /(x), between x = a and x = b.

2 By approximating the function f ( x ) by an interpolating polynomial, P(x),

and integrating the latter instead of the given function, so that

• The derivation is common to a group of rules which thus appears as particularcases of a more general method

• The derivation yields an expression of the error involved

In the next two sections, we shall use the geometrical approach to derive the

two most popular rules, namely the trapezoidal and Simpson's rules These two

methods are sufficient for solving most problems encountered in Naval ture The error terms will be given without derivation; however, interpretations

Architec-of the error expressions will follow their presentation

3.2 The trapezoidal rule

Let us consider the function f ( x ) represented in Figure 3.1 We assume that we know the values /(:TI), /(x2), , f(x$) and we want to calculate the definite

integral

fXS

1= f ( x ) d x (3.2) The integral in Eq (3.2) represents the area under the curve f ( x ) Let us connect the points /(xi), /(#2)> • , f(x$) by straight line segments (the dashed-dotted

Numerical integration in naval architecture 73

Given integrand

Trapezoidal approximation

X 3

Figure 3.1 The derivation of the trapezoidal rule

lines in the figure) We approximate the area under the curve by the sum ofthe areas of four trapezoids, i.e the area of the trapezoid with base #i£2 andheights /(#i), /(x2), plus the area of the trapezoid with base x2x3 and heights/(x2), /(^s)* and so on We obtain

For constant x-spacing, #2 — x\ = £3 — x2 = • • • = h, Eq (3.3) can be reduced

to a simpler form:

We call the intervals [x\, x2], [0^2, ^3], and so on, subintervals.

As an example, let us calculate

•90°

sin

The calculation presented in tabular form is as follows:

0.25880.50000.70710.86600.96591.0000

-Multiplier

1/2111111/2-

Product

0

0.25880.50000.70710.86600.96590.50003.7979The calculations were performed with MATLAB and the precision of the

display in the short format, i.e four decimal digits, was retained To obtain the

approximation of the integral, we multiply the sum in column 4 by the constant

The generalized form of Eq (3.5) is implemented in MATLAB by the trapzfunction that can be called with two arguments:

1 the column vector x,

2 the column vector y, of the same length as x, or a matrix y, with the same

number of rows as x

If the points xi, x2, , x n are equally spaced, i.e., if

the trapz function can be called with one argument, namely the column

vec-tor (or matrix) y In this case, the result must be multiplied by the common x-interval, h.

Numerical integration in naval architecture 75

3.2.1 Error of integration by the trapezoidal rule

'- [^ /(i)da

%m %l 7 o

We do not know the maximum value of the derivative in Eq (3.7); otherwise,

we would have been able to calculate the exact value of the integral We can,however, say the following:

• By substituting in Eq (3.7) the maximum value of d2/(x)/dx2 in the interval

[#i, Xm], we can calculate an upper boundary of the error.

• The error is proportional to the square of h\ if we halve the subinterval, the

error is reduced approximately in the ratio 1/4

• The method is exact if d2/(x)/dx2 = 0 This is the case for linear functions

As a matter of fact, the derivation of the trapezoidal rule was based on a linear

reduce the error by halving the subinterval h Experimenting with subintervals

equal to Tr/8, Tr/16, , Tr/128, we obtain the results shown in Table 3.1 wherethey are compared with the results yielded by Simpson's rule (see Section 3.3)

For h =' 7T/8, Figure 3.1 shows the error as the sum of the small areas contained

between the dashed-dotted line (the trapezoids) and the solid line (the given

Table 3.1 Results by trapezoidal and by Simpson's rule

Simpson's rule2.573076204287112.570930911769092.570804622318862.570796843479602.570796359059902.57079632881103

curve) This area looks really small The errors in per cent of the true valuesare shown in Table 3.2 As predicted by Eq (3.7), each time we divide the

subinterval h by 2, the error is divided approximately by 4 It is easy to see that

as h —>• 0, the trapezoidal approximation of the integral tends to the true value.

In this example, by reducing the size of the subinterval h we could make the

error negligible This was easy because we had an explicit expression for /(#),and we could evaluate as many values of /(#) as we wanted When there is noexplicit mathematical definition, as it happens when the ship lines are definedonly by drawings or tables of offsets, the number of function values that can

be measured, or evaluated, is restricted by practical limitations In such cases,

we must be satisfied if the precision of the integration is consistent with theprecision of the measurements, or of calculations involving the same constantsand variables To understand this point better, let us suppose that we want to

calculate the ship displacement mass as A = pV, where p is the density of the

surrounding water It makes no sense to be very precise in the calculation of thedisplacement volume V, if we multiply it afterwards by a conventional value of

the density p The density varies from sea to sea (see table in Appendix A of

Chapter 2), and in the same sea it varies with temperature In most calculations, itwould be impossible to take into account these variations, and the Naval Architect

or the ship Master has to use the value prescribed by the regulations relevant to

Table 3.2 Per cent error by trapezoidal and by

Simpson's rule0.088683710.005235150.000322680.000020100.000001260.00000008

Numerical integration in naval architecture 77

the ship under consideration For example, for oceans and the Mediterranean sea,various regulations specify the value 1.025 t m~3 An exception is the inclining experiment, a case in which the actual density must be measured But, even in

that case the precision of the measurement is limited and not better than that ofthe V-value calculated with the rules described in this chapter

3.3 Simpson's rule

In Figure 3.2, the solid line passing through the points B, C and D represents

the integrand /(#) We want to calculate the integral of /(#) between x = A and x — E, i.e the area ABCDEFA This time we shall approximate f ( x ) by a

parabola whose equation has the form

The parabola is represented by a dashed-dotted line in Figure 3.2 We need three

points to define this curve; therefore, in addition to the values of f ( x ) calculated

at the two extremities, i.e at the points B and D, we shall also evaluate f ( x ) at

the half-interval, obtaining the point C Let

We divide the total area under f ( x ) into two partial areas:

1 thetrapezoidABDEA,

2 the parabolic segment BCDGB

The first area equals

For the second area, we use a result from geometry that says that the area of

a parabolic segment equals two-thirds of the area of the circumscribed lelogram Correspondingly, we calculate the second area as two-thirds of thecircumscribed parallelogram BHID, i.e

paral-f

6

Adding the two partial sums yields

f ( x ) d* « [f(Xl) + 4/(x2) + /(x3)] (3.9)

which is the elementary form of Simpson's rule.

Usually we have to integrate the function f(x] over a larger interval [a, b}.

Then, we achieve a better approximation by dividing the given interval intomore subintervals From the way we derived Eq (3.9) we see that the number

of subintervals must be even, say n = 2/c, where k is a natural number Let

. a-b

Applying Eq (3.9) for each pair of subintervals, and adding all partial sums,

we get

4/(x4) + ' ' ' + 4/(zn) + /(xn+1)] (3.10)which is the extended form of Simpson's rule, for equal subintervals This form

is very helpful when calculations are carried out manually As an example, let

us calculate

sin x dx