Wave making: a low speed slender body theory

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WAVE MAKING: A LOW-SPEED

SLENDER-BODY THEORY

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Pi Jo, '

THE UNIVERSITY OF,MICHIGAN',!

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COLLEGE Of ENGINEERING,

M 2, 2028 CD Den AC 015- MOM Fax 781830

Arthur MacDonald Reed

The preparation of this paper was supported' by a grant of

the National, Science foundation (Grant GK-36848)

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cruRE

AND MARINE ENGINEERING

ag, 14

No. 169

n. laboratorlum

.-April 1975

TECHNISCHE voor V /811

NAVAL

Techn;se!,,e Hoaeseino!, Defit

DOCUis-IENTAT:E :

5 9

DATUM'

WAVE MAKING: A LOW-SPEED SLENDER-BODY THEORY

by

Arthur Macdonald Reed

A dissertation submitted in partial fulfillment of the requirements for the degree of

Doctor of Philosophy

(Naval Architecture and Marine Engineering, in The University of Michigan

1975

Doctoral Committee;

Professor T. Francis Ogilvie, Chairman Assistant Professor Robert F., Beck

Professor Chalres L. Dolph Professor Arthur F, Messiter

The preparation f this paper was supported by a grant of

the National Science Foundation (Grant GK-36848).

No. 169 April lgy5

Department of Naval Architecture and Marine Engineering College of Engineering The University of Michigan

Ann Arbor, Michigan. 48104

G

1811

ABSTRACT

A slender-body theory for the wave making of a ship in steady translational motion is developed by assuming that the length of the waves generated by the ship is of the same order of magnitude as the

beam of the ship. The exact boundary value problem is formulated and

then linearized by assuming that the wave length is short.

This linearized problem is solved to two orders of magnitude by the

method of matched asymptotic expansions. The first order solution turns

out to be a simple variation on the Tuck slender-body theory, while the second order solution turns out to be a diffraction problem accounting for the refraction of the waves away from the body.

Qualitatively, this solution shows some of the same characteristics which Adachi has observed in his experiments on models with long parallel

middle bodies. This solution is in many ways similar to the solution

which Faltinsen obtained for the diffraction of head seas by a slender

ship. Quantitatively, the wave elevations obtained using this theory

agree well with experimental measurements obtained from tests on a pointed body of revolution.

I would like to thank Professor T. Francis Ogilvie for the encourage-ment which he provided during my stay at The University of Michigan, and

for his technical guidance during the completion of my dissertation. I

would also like to express my appreciation to the National Science Foundation for supporting this work.

I would like to thank Hiroyuki Adachi, Odd Faltinsen, Ernest Tuck, and Fritz Ursell whose earlier works laid the groundwork for this

research and whose discussions and comments made the completion of this

work easier. I would like to thank Nabil Daoud for having provided an

excellent sounding board for my ideas during the analytic portion of my research and Young Shen for his interest and skeptical questions which

kept me searching for better ways to compute my results. In addition I

would like to express my appreciation to Vern Phelps, Middie Potter, and the staff of the Ship Hydrodynamics Laboratory for their assistance in the construction and testing of my model.

Finally, I would like to thank my wife, Suzanne, my parents, Maxine and Karl, and my grandfather, Macdonald Reed, for their continued

interest and support throughout my education. I would like to give

additional thanks to Suzanne for her editorial and typing assistance.

1 '6 11 17 25 iv TABLE OF CONTENTS ABSTRACT « oN o. *

.

« * ACKNOWLEDGMENTS Wfi 0 W A. * f LIST OF TABLES . . 4 4 4 A A.

.

N .0

LIST OF' ILLUSTRATIONS, ,.. . . go 1.4

.

on not

.

LIST' OF APPENDICES . , . 0 _* * .0

* * * *

OF NOTATIONS . A 4 A. '

.

4 4 4

.

INTRODUCTION on A Ol 41 A Or not ..

.

«I

.

«

.

,

A

.

PROBLEM FORMULATION AND METHOD OF SOLUTION 4. A di! .0, oi A)

Statement of the Problem Method of Solution

FIRST ORDER PROBLEM . '0 4 4

Far-Field Problem Near-Field Problem

Matching of the First Order PrOblem

SECOND ORDER. PROBLEM, 01, r. 0, ! on 4,

Far-Field Problem Near-Field Problem

Matching of the Second. Order Problem Summary of Results

NUMERICAL ANALYSIS 0

.

.4 . ft. 4 :A,

Determing the Second Order Source Strength Solving the Two-Dimensional Helmholtz Equation

COMPARISON WITH EXPERIMENTS . . 14 4

APPENDIX A APPENDIX B 1. « BIBLIOGRAPHY 11 11

35

1

.

41 P

.48

- 65 vi vii LIST viii

Table 1: Matching of First Order Solution . , .11 NW 10 Of Olif Olf 151

Figure Bla.;:

Figure Bib;

LIST OF ILLUSTRATIONS

figure 32a: Contour of Integration for k K 4 .

Figure B2b: Contour of Integration for ik < K

Figure. B2c: Contour of Integration for k < -x

vi

Comparison Between Experiments and Theory Showing Rate of

Wave Decay

as

a Function of Speed [Adachi (1973)] .

Rate of Transverse Wave Growth for Model ES-201 at a Froude

Number of 0.25 [Adachi (1973)] . , . 4

Ship with Typical Wave Pattern

Rapidly Varying Source Strength from Second Order Slender-,

Body Theory at Three Non-Dimensional Speeds . . . 28

Contours of

S0(Kr,0)

Vs. Kr. for Various Values of e

13

Contours of aSo(Kr,e)TaKr Vs. xt for Various Values

Comparison of Wave Elevations Between Second Order Slender Body Theory, Tuck Slender-Body Theory, and Experimental.

Data at a Non-Dimensional Speed of KL = 14 . 37

'Contour of Principal Value Integral, for k K

1

Contour of Principal Value Integral for

k <

-K . 4 52

1

4

-I ist 54

Figure 8: Comparison of Wave Elevations Between Second Order

Slender-Body Theory, Tuck Slender-Slender-Body Theory, and Experimental

Data at a Non-Dimensional Speed of KL = 20 . . . 3

Figure 9: Comparison of Wave Elevations Between Second Order

Slender-Body Theory, Tuck Slender-Slender-Body Theory, and Experimental

Data at a Non-Dimensional Speed of xi, 30 , . . . 39

/I Figure 1: Figure 2: Figure 3: Figure 4: Figure 5; Figure. 6: Figure 7; It 3 5 of 34 = > . 52 > 54 54

Appendix A: Inner Expansion of the Outer Expansion for a Slowly

Varying Source Distribution . . ti 41

Appendix 131 Inner Expansion of the Outer Expansion for a'Rapidly

Varying Source Distribution . .

^! N. tit til tti ti el 48

LIST OF NOTATIONS

A(x) immersed cross-sectional area of the ship

maximum beam of the ship maximum draft of the ship gravitational constant

h(x,y) function defining the surface of the ship

length of the ship

R(x) rapidly varying source strength

radial distance from the x-axis to a point in the cross

plane: r = 672+z2

ship velocity in units consistent with L , g , etc.

x,y,z _{right-handed cartesian coordinates fixed to the ship (See}

Figure 3)

d(x) Dirac delta function

slenderness parameter

(x,Y) free surface elevation

angular location of a point in the cross plane: e = tan-1(y/z)

wave number: K = g/U2

fundamental wave length: A = 27/K

fictitious viscosity (See page 51)

E(x) slowly varying component of the rapidly varying source

strength: R(x) = Re[E(x)eiKX]

a(x) slowly varying source strength

(1)(x,y,z) _{total velocity potential for the ship fixed in a uniform flow}

(x,y,z) _{normalized perturbation velocity for a body and its mirror}

Image in an infinite fluid

11)(Y,z) slowly varying component of the second order near-field

velocity potential

Van

derivative in the direction of the three-dimensional out

ward normal vector to the ship

a/aN

_{derivatiVe in the direction of the two-dimensional, outward}

normal vector in the cross-section of the ship

INTRODUCTION

For hundreds of years, one of the naval architect's goals has been to predict the resistance of a ship using purely theoretical means. William Froude (1868) took the first steps towards this goal when he

published his famous paper suggesting that ship resistance could be broken into two components, one due to viscosity and one due to gravity. Through the work of Osborn Reynolds, Froude knew how the viscous

resistance could be scaled, and it only remained for him to state how

the residuary resistance scaled. Froude developed his similarity law

for scaling residuary resistance by studying the wave patterns generated

by models and noting at what speeds the wave patterns were similar. He

also stated the well known fact that a major portion of the residuary resistance is the wave making resistance.

J. H. Michell (1898) was the first person to come forward with a theory for predicting the wave resistance of a ship which was the

fore-runner of today's thin-ship theory. Although thin-ship theory has been

rigorously studied and elaborated upon by many workers (for example, see

the many works by Wrigley, Havelock, and Weinblum), it was not until

the early 1960's that a totally new theory was developed. This was the

slender-body theory developed independently and essentially simultaneously by H. Maruo, E. O. Tuck, and G. Vosser [see, for example, Tuck (1963)].

As quickly as a slender-body theory for wave making was brought to

life, it was dealt a death blow by G. R. G. Lewison (1963). He showed

that the wave resistance given by slender-body theory was approximately twice that of thin-ship theory and the latter theory already over

estimated the wave resistance compared with experimental results. Thus,

slender-body theory as it applied to wave making became dormant for the next decade, at the end of which a number of events occurred which gave new direction to the theory.

First of all, calculations of the wave elevation alongside the ship

were completed using the Tuck Slender-Body Theory. _{These calculations}

[see Ogilvie (1974)] showed that, although the wave amplitude was too

great and the phase was in error, the theory certainly demonstrated the proper wave-like behavior alongside the ship except for near the bow.

At the same time, T. F. Ogilvie (1973) was developing a modified

slender-body theory for application to the region near the bow of the ship. This theory gave finite wave amplitudes at the bow and showed that the

waves decayed like the inverse of the distance from the bow. This agreed

with the Tuck theory which said that the wave amplitude grew like one over the distance from the bow as the bow was approached.

Another sequence of events also occurred around the same time. H. Adachi (1973) made available a number of his experimental results on

an extremely long, slender model. These results showed that the rate

of decay of the waves generated by a ship was much more rapid than the

rate predicted by conventional theory. (Conventional theories show that

the waves decay like the inverse square root of the distance from the

bow.) These results are shown in Figure 1 for two model lengths. Model

S-201 is 3.5 meters long. Model ES-201 is made by extending Model S-201

with 20 meters of parallel middle body. Adachi produced data for a

single model, ES-201, to show that the wave elevation increased linearly in the transverse direction as one proceeded away from the model, Figure

2. These results are similar to results obtained by O. M. Faltinsen

(1971) for a problem with similar geometry.

Faltinsen was solving the diffraction problem for a ship at rest in

head seas. In his solution Faltinsen found that the waves were

atten-uated in the longitudinal direction by the presence of the ship and that

the wave amplitude grew as one moved away from the ship. This led to

the conclusion [since confirmed by Ursell (1973, 1974a, 1974b)], that the waves generated by the ship were refracted by the presence of the ship.

In fact, Ursell showed that the waves should decay like 1/x3/2 as they

move away from the bow, where x is the distance from the bow.

These results have led us to formulate the following problem: let

a ship in a coordinate system as shown in Figure 3 be represented by the

sum of two source distributions. The first source distribution is a

slowly varying one intended primarily to represent flow in an infinite fluid about the ship and its double model, with the related wave motion

appearing as high order effects. The second source distribution will be

2.0 1.5 1. 0 . 5

Rate of Wave Decay Equals

0.20

0.25

Theoretical Rate of Decay

C)

Experimental Data from Model S-201

A

Experimental Data from Model ES-201

A. 4

-.6.

Froude Number

Figure 1:

Comparison Between Experiments and Theory Showing Rate of Wave Decay as a Function of Speed [Adachi (1973)].

0.30

P = 1.35

0.35

x

2.0

1.0

0.5

Linear Least-Squares Fit to Experimental Data

0 Experimental Points

0.5 1.0 1.5 2.0 2.5

y/ (L5201/2)

Figure 2: _{Rate of Transverse Wave Growth for Model ES-201 at a Froude}

5

Figure 3: Ship with Typical Wave Pattern

effects of diffraction on the waves from the first _{source distribution.}

As we observe the flow near the ship due to the rapidly _{varying source}

distribution, we will find that the waves do indeed increase like lyi

as we move transversely away from the ship. _{Although we can not show the}

rate at which the waves decay as we move longitudinally _{away from the}

bow, we do obtain excellent correlation with experimental wave

measure-ments alongside the parallel middle body of a ship model. _{These results}

are primarily due to the fact that we have now obtained _{a problem which}

allows us to have a wave-like free surface boundary condition near the

ship. _{This is a concept which Ogilvie (1973, 1974) found has allowed}

him to obtain solutions to the bow problem. _{We shall now proceed to set}

up our problem precisely and to obtain a solution by employing the _ideas

We wish to establish a theory for the wave making of a ship at low speed, where we define low speed to be a speed such that the fundamental

wave length is comparable with the beam of the ship. To formalize this,

we shall say that we have a slender ship whose beam-length ratio and draft-length ratio are of the same order of magnitude as some small

parameter, E . Now we can say that the wave length/ship length is

com-parable with c . To state this more precisely, let

= 0(c) , = ,

and then

= gL

-2TrU2

0(c) ,

where B is the ship's beam, D the ship's draft, L the ship's length,

A the wave length, U the ship's speed, and g the gravitational

constant.

The fact that we intend to develop a slender body theory implies

some constraints on the geometry of the body. We define our body by the

equation:

z = h(x,y) , 0 < x < L

where the body offsets are given by the points [x,y,h(x,y)] , and the

function h(x,y) is defined such that the ship is heading in the

nega-tive x-direction, with the origin at the bow, and with the z-axis

verti-cally upwards. Now, defining slenderness, we can say that h(x,y) is

small and we shall also demand that derivatives of h(x,y) with respect

to x should also be small. Formalizing this, we can say:

h(x,y) = ,

and

= 0(s) . Bx

An important fact to note is that in the limit as e-3-0 , the body

6

A

7

cross-sectional dimensions shrink to zero, the wave length goes to zero,

and we have an infinite number of wave lengths along the body. This

implies that the speed is now a function of the mall parameter, E

and the speed goes to zero as c-)-0 , though not linearly. In fact, from

the relationship associating A/L to ship velocity and c , we can see

that:

U =

0(E1/2)

Now stating the problem which we intend to solve as a boundary value problem, we shall assume that the fluid is inviscid and incompressible

and that the flow is irrotational. _{Given these assumptions, the flow}

can be described by a potential function. The partial differential

equation governing this problem is the continuity equation,

+ (1, + (1) = 0 , ₍₁₎

xx yy zz

in the fluid domain. _{The body boundary condition is that there shall be}

no flow through the body. This can be written as:

o on h(x,y) - z = 0 (2)

where a/an denotes the derivative in the direction of the

three-dimensional outward normal vector to the body. On the free surface, two

boundary conditions apply. The first is the dynamic free-surface

condition,

gr. cp2 4,21 =

1

u2

2 x y z 2

and the second is the kinematic free-surface condition,

(1) + - = 0 , on z = (x,y) _.

xx

yyz

Finally, we must have a radiation condition which _{states that no waves}

should be propagating upstream of the ship. Method of Solution

In order to solve the above boundary value problem, _{we must simplify}

it. To do this, we shall linearize the problem. _{By writing the}

potential function (1)(x,y,z) as follows:

(1)(x,y,z) = Ux + U(x,y,z) ,

(4)

on z = (x,y) ₍₃₎

and assuming that Ucp(x,y,z) is higher order than Ux , we may rewrite

the boundary value problem and its boundary conditions in terms of

gx,y,z) . Furthermore, we may expand the wave elevation in a Taylor

series about z = 0 and discard all high order terms to obtain a

linearized free-surface problem.* Rewriting the continuity equation (1),

we obtain:

+ + (1) = 0 outside the body for z < 0 .

xx yy zz

Rewriting the body boundary condition, equation (2), we find that hx

as = _

/1

+ h2 + h2

x y

By revising the dynamic free-surface condition, equation (3), and the kinematic free-surface condition, equation (4), we obtain:

and +.jf.- 112(2th 4. (1)2 4. 62 4)21 'x x 'y z 1.1)( + + Uct)y), - Ucpz = 0 on h(x,y) - z = 0 . on z = 0 , ₍₇₎ on z = 0 .

If we now neglect the product terms in both equations, and then

differ-entiate the dynamic free-surface condition with respect to x and

sub-tract the result from the kinematic condition, we obtain the linearized free-surface condition as follows:

cbKCPz

0 on z = 0 , (8)

where

K = g/U2 .

The radiation condition for this problem remains unchanged. Although the

problem which we must now solve is linear, it is still an extremely

dif-ficult one to solve exactly. Therefore we shall make use of the method

of matched asymptotic expansions to further facilitate finding a solution

*We should linearize the problem separetely in the near field and far field of the body, because we might expect different orders of magnitude for the

velocity components in these two regions. However, it turns out that for

this problem the linearized problems in the near and far fields are

independent of the initial linearization.

9

to the problem.

In the application of the method of matched asymptotic expansions to the linearized problem, we shall determine approximate solutions to the problem which are valid in the far field and near field respectively.

In the far field, we shall assume that y and z are 0(1) as 0

and that the derivatives of the potential function with respect to x,y,z

are all of the same order of magnitude. Because of the assumption about

the order of magnitude of y and z _{, we will not see the body in any}

detail in the far field. In fact, in the limit as c -* 0 , we will

represent the body as a line of singularities of unknown strength. In

the near field, we will assume that y and z are 0(c) and that

derivatives of the potential with respect to y and z are 0(E-1)

while derivatives of the slowly varying potential with respect to x

are 0(1) . Because of these assumptions, we will find that, in the

near field, the boundary value problem becomes two dimensional in the

cross section of the ship. This two dimensionality allows us to

for-mulate a problem involving the body geometry, which we can readily solve. In solving this problem, we will be able to determine the unknown

singularity strengths in the far field.

The final step before starting the actual solution of the problem

is to state our problem in terms of a perturbation series. In both the

near and far fields, we shall assume that the potential function,

O(x,y,z) , can be expanded in the following form:

qp(x,Y,z) =

(0)

(x,y,z;E) +(1)(x,y,z;E) +

For the far-field problem, we shall assume that each q)(n)(x,y,z;c)

can be represented by a convolution integral of a source distribution

and a Green's function over the length of the body. Let the source

dis-tributions for the first and second potentials be u(x) and E(x)e1"

respectively, where both o(x) and :(x) are slowly varying functions

of x , that is to say:

aE

= 0(c) and = 0(E)

Dx Dx

The second source distribution is chosen to be rapidly oscillating so that it can represent the diffraction effects which we know occur. Likewise, we shall assume that in the near field the second near-field

potential is of the form (1)(2)(x,y,z;c) = Re[eikxtp(2)(x,y,z;e)] , where

again

11,(2)(x,y

,z;E) is a slowly varying function of x . Here we

justify making 4,(2)(x,y,z;c) a wavelike potential because we know

that if we want 4,(2)(x,y,z) to represent the diffraction of waves, it

must be rapidly varying in the x direction. We shall now set up and

solve the first order problem.*

*In the work which follows, we shall drop the superscripts without ambiguity.

FIRST ORDER PROBLEM Far-Field Problem

In the far field we will represent the _{ship by a line of sources}

distributed on the free surface. _{First we need to determine what the}

governing equations should be for the far-field problem with a source

distribution. _{Based on the continuity equation and the assumption that}

all derivatives are the same order of magnitude in the far field, _the

partial differential equation governing the far field becomes the _Poisson

equation. We have no body boundary condition _{to satisfy.}

This leaves

only the free surface and the _{radiation conditions to be determined.}

Because we expect waves of length X = 0(c) _{in the far field, the}

deriv-ative of the potential in the far _{field must be very large due to the}

rapid changes in the potential _{necessitated by wave-like behavior.}

In

fact derivatives with respect to x,y,z must be 0(e-1) in the far

field. By _{employing this information,}

we determine that the orders of

magnitude of the terms in the _{linearized free-surface condition,}

equation

(8), remain unchanged. _{We can now write the corresponding}

boundary value problem as follows: (4)xx + (Pyy + zz) = 0(x)6(Y)6(z-c) , z < 0 and 4)

+K

=0

_{on z = 0} , xx z

where c _{is the vertical location of the source distribution}

(c < 0)

and 6(y) and 6(z - c) _{are Dirac delta functions.}

The solution to this

problem must satisfy a radiation condition so that _{waves do not propagate}

upstream of the disturbance. _{Tuck (1963) has solved this}

problem and he gives the following results:

l a*(k) _{KVk2+k2 + k2} eA2+X2

(z+c)

_421_,Q2

z_cl

(1)**(k,Z;z) = + e 2/k2+k2 _{K42+9,2 - k2}

where (1)**(k,i;z) _{is the double Fourier transform}

of _{gx,y,z) with}

respect to x and y . We can now let our line of

sources approach the

11

free surface by letting c-)-0 . If we do this, _(/)**(k,Z;z) simplifies to: /k24.12 z a*(k)e cp**(k,k;z) -(9) ik2+2.2 - k2/K

Applying the Fourier inversion theorem to _{cp**(k,R;z) , we find that the}

potential can be written as follows:*

gx,y,z) = 1

dk eikxa*(k)

dZ

eiiy'1(2+2,2 z

/k2+2

k2/K

472

CO

This is the Fourier transform form for the first order potential in the far field.

If we are later going to match this solution _{to a near-field}

solution, it will be necessary to obtain an inner expansion of the outer

expansion (IE-0E) . The IE-OE

is an asymptotic form of the far field

solution obtained by letting y and z _{become small [0(c)]}

. The

IE-OE of equation (10) is as follows:

1 1

cp(x,y,z) a(x)log r

-2T- j dC at(C)log 21x-Cisgn(x-C) (11a)

7

0(u log

0

0 0(a)

*This integral has two Cauchy _{singularities for Ikl > K}

. The proper

contour around the singularities is _{determined from the radiation}

condition (see Appendix B).

(10)

dC a'(C){1-i [K(x-E)] + [2 + sgn(x-C)}Y

_(Kix-C1)1

4

0 ₀

0

_0(aE)

(llb)

where r

= 62+z2

_{is the radial distance from the source distribution.}

Flo

and Y0 _{are the Struve function and the Bessel function of the}

second kind, respectively. _{The derivation of the IE-OE is given in}

Appendix A. _{The first two terms of the IE-OE} , line (11a), are the same

terms which we would obtain when finding the IE-OE for a line

distri-bution of sources in an infinite fluid. _{This is not surprising}

consid-ering that the usual first order approximation _{to the low speed problem}

is simply the problem in which the free-surface _{condition is (1)z = 0}

.

The third term, line (11b), is the term which contains the wave-like

13

behavior, and it is higher order than the _{terms in line (11a).}

Near-Field Problem

We shall now determine the first order near-field boundary value

problem, and establish its behavior far _{from the body. It is at this}

stage that we shall use our slenderness _{assumptions to the utmost.}

It is trivial to show that the _{zeroth order potential must be the}

same as the potential for the incident _{stream in the far-field problem,}

that is:

U0(0)(x,y,z;E) = Ux .

Therefore, we shall proceed with the _{first order near-field problem.}

The linearized boundary value _{problem was given in equations (5),}

(6)

and (8) as follows:

4)xx _YY + (1)_zz = 0 outside the body for z < 0 (12)

0(0) 0(0E-2) 0(0E-2) and (/) _{= 0 on z = 0} . (14)

OW

0(0E-2)

In our previous discussion of _{slenderness, we assumed that:}

ah _ah

h, _ax =

0(E)

and = 0(1)

oy

while in our discussion concerning the _{near field, we assumed that:}

21. _{= 0(0E-1)}

ay ' az ' ar

Applying these assumptions _{to the body boundary condition [equation}

(13)], we obtain: -hx a0 an /T-1-47 0(0E-1) 0(E)

If we apply our assumptions

concerning the orders of magnitude _{of the}

various derivatives of _{to equations (12) and (14),}

we obtain the

orders of magnitude given beneath the terms of the above _{equations. Given}

-hx

an /11-h2+h2

x y

on _{h(x,y) - z = 0}

these orders of magnitude, we can discard the high-order terms and sim-plify the near-field problem as follows:

(t) + = 0

z <0

(16) yy zz = 0 on z = 0 (17) hx -311 + h2

where

a/aN

denotes the derivative in the direction of the two-dimensional

outward normal vector to the body in the cross plane. The body boundary

condition also allows us to determine the order of magnitude of ¢ .

From

0(¢c-1) = OW

we have:

on h(x,y) - z = 0 (18)

This problem is a two-dimensional "rigid wall" problem, corresponding to the Neumann problem for a body and its mirror image, a problem which has

been studied many times [e.g., Ward (1955)]. Ward has demonstrated that

the solution far away from the body acts like:

¢ -1 A'(x)log r + b(x)

where A(x) is the sectional area of the submerged part of the ship and

b(x) is an arbitrary function of x . This is the outer expansion of

the inner expansion

(0E-IE)

.

Matching

We are now in a position to match our inner expansion of the outer

expansion with the outer expansion of the inner expansion. The process

of matching involves equating terms of similar form from the IE-OE with

those from the 0E-IE, i.e., matching f(x)log r in the IE-OE with

g(x)log r in the 0E-IE . One way to facilitate the matching is to set

up a table of the 0E-IE alongside the IE-OE with the terms becoming

higher order as you proceed down the table. Table 1 is such a table for

the first order problem.

If we compare the two sides of this table, we see that the first line

of the 0E-IE is identical to that of the IE-0E, and therefore, they

match trivially. In the second line, we see that the first terms will

match if we set c(x) = A'(x) = O(c2) . We can also match the second

,

Outer Expansion of the Inner Expansion

Ux 0(61/2)

+ - A'(x)log r + Ub(x)

7

0(c5/2log c)

terms of the second line if we set

1

b(x) = -dC o'(E)log(21x-El)sgn(x-E)

_{2 = O(2)} .

We have now matched all of the terms of the 0E-IE to terms of the

IE-OE . However, we have not matched the third line of

the IE-OE with

any other terms. _{This means that we have determined too many terms in}

the IE-OE _{and thus will not be able to utilize it until we have}

com-pleted the next order of approximation.

This matching completes the solution of _{the first order problem.}

However, before we continue on with the second _{order problem, we should}

note the physical significance of the high order _{term which we neglected}

in the IE-OE . As stated earlier, this term contains all

of the waves

generated by the body. _{These waves were obtained from the far field}

and

therefore, violate the body boundary condition. _{Thus we would expect}

that any higher order near-field solution must contain terms to match

these wave-like terms. _{It is this fact which leads us to assume that}

our second order near-field solution will be _{rapidly varying in the}

x-direction and the corresponding _{far-field problem will have a rapidly}

15

TABLE 1

Matching of First-Order Solution Inner Expansion of the Outer Expansion

L U U + -o(x)log r - _{dE al(E)1og(21x-El)sgn(x-E)} 7 27 0(GE1/2log e) ° 0(a1/2) ₍₂₎ U _{j dE cr'()4-10[K(x)]} e 0 + [2 + sgn(x-E)]Y0(Klx-E1)1 0(a3/2) ₍₃₎ Ux

0(E1/2)

₍₁₎

varying source distribution. Using these ideas, let us now proceed with the second order solution.

SECOND ORDER PROBLEM Far Field

As in the first order problem, we shall again write our far-field potential as the convolution integral of a line source distribution and

a Green's function. This time, however, we shall take our source

dis-tribution to be of the form Re[eE(x)] , where E(x) is a slowly

varying function of x . As before, our governing equations are

(Cbxx gbyy q)zz) = Re[E(x)eiKx]d(y)6(z+c) for z < 0

and

17

y/k2_14.,2 k2/K

+K

0 on z = 0 ,

along with a radiation condition which allows waves only downstream of

the disturbance. Again we find the double Fourier transform of our

solution (with c = 0) to be:

,/k2+92 z

E*(k-K) e

cp**(k,k;z) =

where E*(k-K) is the Fourier transform of Re[E(x)eiKX] with respect

to x . (E*(k-K) is not strictly the Fourier transform of Re[E(x)eiKX]

.

This matter is treated rigorously in Appendix B.) The corresponding

potential may be written as follows:

e42.4.22

O(x,y,z) = - z-TZT1 dk eikxE*(k-K) I dk ₍₁₉₎

/k24.2.2 _ k2/K

CP

If we now let y and z become 0(c) and obtain an asymptotic

expan-sion of O(x,y,z) , we will find that our IE-OE is:

x

ci5(x,y,z)

- Re

1,

_KeKzeilcx - i7/4 i

dE()

/27K(x-0 0

0(E,

-1/2)

+ KlyleKZE(x)eiKX iczE()eiKx1

0(E)

The details involved in obtaining the IE-OE are given in Appendix B.

Near Field Problem

In the near field, we shall now formulate a new boundary value

prob-lem and obtain its solution. As in the previous near-field problem, we

shall assume that derivatives with respect to y and z are 0(c-1) .

However, we shall also assume that cp(x,y,z) = eiKX14)(x,y,z) where

tp(x,y,z) is a slowly varying function of x . The linearized boundary

value problem now needs to be rewritten in terms of

Ip(x,y,z)

. First

we must determine the proper body boundary condition to be satisfied by

the rapidly varying potential. We can restate the body boundary

con-dition, equation (6), as follows:

- hx _{on h(x,y) - z = 0}

.

+ h2 + h2

x y

If we observe the right-hand side of this equation, we see that it is a

slowly varying function of x . No matter what method we use to further

linearize the right-hand side, it will still be slowly varying. From

this we may conclude that the potential which satisfies this body boundary

condition must be slowly varying. Therefore, our rapidly varying

po-tential must satisfy a zero normal velocity condition.

Using this we may rewrite the boundary value problem in terms of

I(x,y,z)

as follows:

-K211) + 2iKtpx

+

+

4)yy

+

Ipzz = 0 in the fluid domain,

0(pc-2) 0(4-1)

O(ji-

2) 0(1),-2)

_K2 _{+ +} 1 + 0 on z = 0 ,

0(p-2)

OW-1)

OM

0(c-2) an =

i

and

1

(iKlph +h +

ip h - Ipz)

₌₀

_{on h(x,y) - z = 0} .

X _{Y Y}

/1 + h2 + h2

o(

tp) ID

(E)

0(q)c-1) _0(4,6-1) y lc

Employing our previous assumptions concerning the _{orders of magnitudes of}

derivatives with respect to x,y,z _{in the near field, we find that the}

terms of the boundary value problem for _{tp(x,y,z) have the orders of}

magnitude shown beneath them. _{Discarding the higher order terms, we}

obtain the following two-dimensional _{problem for}

tp(x,y,z)

tp 4. 4, K24) = 0

yy zz

tpz - 0 = 0 on z = 0

and

This problem appears to have the trivial _{solution (y,z) = 0}

because of the fact that there is a zero normal derivative condition on

the body. _{However, Ursell (1968b) gives a multipole}

expansion for the solution of the following problem:

with --11- = 0 3N 4, + 4, K24, = 0 yy zz IPz - KIP = 0

which is valid for all r greater than some ro . Furthermore, Ursell

(1968a) also shows that this solution is nonuniform at infinity. This

nonuniformity causes no difficulty in the near field. _{In fact, it allows}

us to determine a nontrivial solution to the problem by supplying the terms which match the far-field solution.

The multipole expansion given by Ursell _{(1968b) is as follows:}

11)(y,z) = AsSo(y,z) + A(e) eKZ +

a(e)0(e)(yz)

1 M 111

The function

S0 (y' z) is the Helmholtz source, which is defined as:

co

1

So

(y,z) = -

T, t . . +

_s

19

in the fluid domain

on h(x,y) - z = 0 .

in the fluid domain

on z = 0

di cosh p KZ cosh p + iKy sinh p

,

cosh p - 1e (20)

-where 4and

are the integrals below and above the double pole at

p = 0 respectfully. The function 0(e)(y,z) is an even wave-free

m potential, which is given as follows:

0(e)(y,z) = K2m-2(Kr)cos(2m-2)8 + 2K2m-1(Kr)cos(2m-1)0 + K2m(Kr)cos 2me

where

Km(Kr) is the modified Bessel function of the second kind.

The asymptotic expansion given by Ursell (1968a) is:

11J(37,z) - A 7K

lyleKZ A(e)eKZ

as

0(Asc-1) 0(As)

The orders of magnitude given beneath the terms are the orders of

mag-nitude for the 0E-IE which are based on the fact that

As and A(e)

must be the same order of magnitude in the near field. This now

com-pletes the determination of the 0E-IE for the second order problem and

we are ready to match our second order solutions. Matching

In order to facilitate the matching process, we shall again

con-struct a table containing the 0E-IE and the IE-OE . As was done

previously, we have placed similar terms on the same line, except where

the orders of magnitude indicate that this is not valid. Lines (1), (2),

and (3) are carried over from solving the first order problem, while lines (4) through (7) are from the second order problem which we have just solved.

When we compare the terms of the 0E-IE to those of the IE-OE , we

find that the first terms which match are the terms containing Kly1

[line (5)]. This allows us to determine the value of

As(x) as follows:

Re[As(x)e ] = - 7 Re[E(x)eiKX = 0(E) .

iKX 1

If we now observe the next two lines [lines (6) and (7)1, we see that

these terms might match. However, based on the order of magnitude

relation between

As(x) and E(x) , these terms cannot match. This is

fortunate because we would not expect to be able to fix the relationship between

Outer Expansion of the Inner Expansion

Ux

A'(x)log t + Ub(x)

7

0(e5/2 log ,c) 0(E5/2)

-21

TABLE 2

Matching of Second-Order Solution

Inner Expansion of the Outer Expansion

-71TIrc(x)1og r - j dE dt()log(2fx-El)sgn(x-E)

00e1/2log

0

0 0(ac1/2) ' (2)

1

boundary condition, DOM = 0 , for arbitrary body shapes.

We have now completed the. matching, except that the _{high order term}

of the first order IE-OE [line (3)j and the. low order tett Of the

second order LE-OE [line

(4)1

have not been matched. Also we still

have not determined the second order _{source strength, E(x) At this}

- 4 'KZ 0' -UK eKZ e g t'(){Fi0[K(x-E)]

0(a01/2.)

iKx=i7/4 e f2 +.sgn(x-E)]Yo(Klx-0)}

f

dE Z(E) (3) 1127K(X-E) 0 0(E) (4.) -U7Klyle As(x)eiKX 4UKlyk"RefE(X)eiKX1

0(Asc-1/2)

O(DE1/2)

(5) iKx

fe"RefE(x)e

0(Eel/ .(6) +ueKzA(e eiKX. 0(AscliZY (7) Ux

0(E1/2)

₍₁₎

o(61/2) . +

-point we may now conclude that the sum of these two terms Must. _{tre zero,}

Which leads us to the following integral equation for _E(x)

K Re

feiKx-in/4 Id E(C)

)127K(X-C) 0 = b s( ) (21) where L, 1 bfs(x)

--f

dC ot(C)0710

[K(x-C)]

+ [2 +

sgn(x-C)]Y0(Kilx-C1)1

. (22)

This now completes the matching of our results and Also completes the.

solution of the second order problem-. We will now discuss the behavior

of our solution. to the wavemaking problem. Summary of Results

In this section, we shall study the behavior of the second order source strength near the bow and develop' equations, for the velocity

components, and wave elevation

in

the near field. If we write the second

order source strength as E(x) = _{Er(x) +} iE(x) and write only the real!

part of the integral equation, equation (21), we obtain:

x x 1

,d. Er()i dC E1(C)

x c os(Kx, ',. /4) 1

sn(Kx

n/4) - bf (X) V

Ki

Y..nic(x--e) 1/2nK(x-) _-s 0 0

The functions E (x) and E1(x) are indeterminate because we have only

one equation with two unknowns. Therefore we need to determine an

integral equation in only one unknown. In Appendix we have done this

by developing the IE-OE in terms of Re[E(x) If we return to

this form of the IE-OE , equation (B27), we can rewrite Our integral

equation. as:

iKE

f

dC Re[E(C)e lcos[K(x-C) - n/4]

41( b

_Is

(x) (23)

This is a single integral equation in one unknown, RelE(x)eicx] . We

will call this unknown R(x) and shall investigate its behavior as

-B . -/27K(X-C) 0 . =

23

KX 0 , which is the limit as we approach the bow.

In studying the limiting behavior of the source strength as we

approach the bow, we will let 6 _{be some small distance 0(c2) from the}

bow. _{We can rewrite the integral equation in terms of 6 as follows:}

6 LK J dC R(C)cos[K(6-C) - 7/4] b (6, i27K(6-E) fs' ' 0

We may now expand the cos[K(6-E) - 7/4] and rewrite the integral

equation yet again as:

6 6 I dE R(E)cos(KE) 1/27K(6-0 I dE R(E)sin(KE) bfs(6) 2K cos(K6-7/4) /271(05-0 + 2K sin(K6-7/4) 0 0

Because we have chosen 6 = 0(c2) , we know that K6,KE =

0(E)

.

There-fore we may neglect K6 _{with respect to 7/4 and set cos KE = 1 and}

sin KE = 0 over the interval of integration. This allows us to simplify

our integral equation as follows:

6

2K

COS(-7/4)

dC R(.)

bfs(6)

.

/271(0-0

0

This is the classic Abel integral equation [Tricomi (1957)] which we

can solve analytically. _{Writing the solution to this integral equation,}

we obtain: 6 hfs(0) 1 _{I dE bfs()} R(6) = + for 6 = 0(e2) . 7/71Cd 71-7-: j IT--7 0

From this we observe that R(x) has a square root singularity at the bow

unless bfs(0) = 0 , and that R(x) is not bounded near the bow unless

bfs(0) is finite.

We shall now develop the equations for the various _velocity

com-ponents and the wave elevation in the near field. _{In light of our change}

in the form of the integral equation for the _{second order source strength,}

the matching for As(x) will also change. Making the change from E(x)

to R(x) , we obtain:

iKx

In the same vein we shall redefine our multipole solution of the two-dimensional Helmholtz problem to be a series with

As = 1 so that the

entire solution of the problem can be multiplied by

-R(x)/7

.

Incorporating these changes into our formulas, we can write our total potential in the near field as:

1)(x,y,z) = Ux + Ugbrw(x,y,z) - UR(x)11)(y,z)/7 , (24)

0(c1/2)

0(c5/2)

0(e4)

where

cprw(x,y,z)

is the solution to the "rigid" wall given by equations

(16), (17), and (18). We can now differentiate this equation with

respect to x,y,z and obtain the velocity components in these directions.

We find that these velocity components are as follows:

= U + Ucprw (x,y,z) - UR'(x)11)(Y,z)/7 , ₍₂₅₎

X

o(c1/2)

₀₍₀₎

0(E5/2)

(1.37(x,y,z) = Ucprw (x,y,z) - UR(x)tpy(y,z) , ₍₂₆₎

Y

0(c312)

0(0)

and

cDz(x,y,z) = Ucprw (x,y,z) - UR(x)tpz(y,z) . (27)

O(t3/2)

0(0)

If we return to equation (7), we find that we may write the non-dimensional wave elevation as follows:

- {(1 + - R'IP/Tri2 + {(I) - R1Pyini2 + [$

RIP/71]2} +

1 .

rw z

4U2

Squaring the terms in the above equation and keeping the low order terms, we obtain:

ig = -

(4rw (x,y,0) -

2R'(x)1p(y,0)/7 + (1)2 (x,y,0)1 . (28)

rw

-2-0(E2)

0(t5/2)

0(c2)

This completes our discussion of the solution of the second order problem and summarizes our results.

=

rw

-

-

-NUMERICAL ANALYSIS

In order to make use of the above equations describing the flow near

the ship, it is necessary to resort to numerical calculations. While

most of the computations are straightforward, there are two areas which

pose some difficulty. One is the determination of the second order

source strength and the other is the solution of the two-dimensional

Helmholtz equation. We shall discuss these problems in the following

sections.

Determining the Second Order Source Strength

In the previous subsection, we determined that the integral equation, equation (23), is as follows:

2K

f

dC R(C)cos[K(x-C) -

7/4]

bfs(x)

/27K(x-E,) 0

where

bfs(x) is given by equation (22). Despite the logarithmic

singularity in Y0(Klx-C1) , the evaluation of bfs(x) is a

straight-forward problem in numerical integration. Therefore, we shall

concen-trate on the solution of the integral equation.

Let the interval from 0 to I be broken up into N subintervals.

(These intervals do not need to be uniform, and in fact, smaller

inter-vals are desirable near the bow.) Then we can rewrite our integral

equation as: xi dc R(c)cos[K(xn-C) - 7/4] 2K

f

-

b

fsn

(x ) 1=1 /27K(x-C) xi -1

for n = 1,2,...,N . We may now apply a special form of the Mean Value

Theorem for integrals, Rudin (1964), to the integrals over the

sub-intervals. Doing this we obtain the following equation:

x. 1 E _2K

R(x)

dC cos[K (xn-C) -

r/4]

= b s(xn) i=1 xi-1 /2rK(xn

wheres

xi -1 < x. . We may now make a change of variables,

1 1

u =

K(xn , and apply a trigonometric identity,

cos(u - r/4) = cos u cos r/4 + sin u sin 1T/4 . This yields:

R(x.) 1

These integrals are Fresnel integrals [see Abramowitz and Stegun (1964)], a fact which allows us to write our integral equation as follows:

E

1=1 R(x.){C21

[K(x -x.

n I-1

)] - C2[K(x -x.)] +

n 1 S2

[K(x

n-x1-1.)]

bfs(xn)

-

S2[K(xn-xi)]}

₍₃₀₎

The Fresnel integrals

C2[K(xn-xi-1)] and

C2[K(x -x.)]

n 1 come from

integrating the first integral in equation (29); the Fresnel integrals

S2[K(x

-x1.

)] and S2

[K(x

_n1

-x.)] come from the second integral in

n

1-equation (29).

Equation (30) is an exact equation, although we do not know the

points 2e! where the function R(x) is evaluated. Therefore, we shall

now make our only assumption in solving for R(x) . We shall assume

that: xi_i + xi x. = 1 2 If we set

R(x)

1 1 and K(x -x. ) K(x -x. ) n

1-1

n ril

du cos u du sin u IZT7 u-K ( x -x.)

K(x -x.)

n 1 n 1 bfs(xn)

/I

(29) C . = C

[K(x

-x )] - C

[K(x -x.)]

+ S

[K(x -x.

)]

-S[K(xn-xi)]

ni 2 n i-1 2 n 1 2 n 1-1 2

we can rewrite the integral in the following abbreviated form:

n bfs(xn)

2 R.0 . = for n = 1,....,N .

i.i. ml

27

By first separating the term RnCnn _{from the summation and then carrying}

the rest of the sum to the right-hand side, and finally by dividing both

sides by Cnn _{, we obtain the following equation:}

{

nibfs(x,) n-1 R = E R.0 .}/C , n = n . if i=1 nn

We may use this equation to determine

R(xn) in a recursive manner, and

if we have chosen the

xi close enough to each other, we may interpolate

for values of R(x) _{at additional points. Likewise, we may determine}

the derivative of R(x) _{at arbitrary points by means of numerical}

dif-ferentiation.

Three curves of R(x) _{for various wave numbers are given in Figure}

4. _{The body to which these curves correspond is a body of revolution}

with pointed ends. _{(The details of the body}

geometry are given in the

section COMPARISON WITH EXPERIMENTS.) _{These curves have been calculated}

using 114 _{points over the length of the}

ship. A spacing of 0.005

was used for 0.0 < x < 0.15 , and a spacing of

0.01 was used for

0.15 < x < 0.99 . Corresponding calculations with only

33 points over

the length of the ship resulted _{in the same degree of accuracy for}

x < 0.75 _{at which point the results started}

to wander. _{This completes}

the discussion of the numerical solution of the integral _{equation for the}

second order source strength.

Solving the Two-Dimensional Helmholtz Equation

As we found previously, the second order near-field boundary _value

problem is as follows:

For the above boundary value problem, Ursell (1968b) provides _{the follow}

following multipole solution:

eKZ _{A(e)0(e)(y,z)}

tp(y,z)

= AsS0(y,z) + A(e)

1 m m

The matching at infinity provides the value of _{A [A (x)e = - R(x)h]}

.

s s

yy zz

= 0

in the fluid domain,

tpz

-

K1I) = 0 _on z = 0 and aN = 0 on h(x,y) - z = 0 . bfs(xtd' +

-+

0.1

--..

0.0

cz,

0.1

0.2

0.3

[

I Figure 4:

Rapidly Varying Source Strength from Second Order Slender-Body Theory at Three Non-Dimensional Speeds.

i

KL

= 14

KL = 20

KL = 30

1.0

Second Order Source Strength Second Order Source Strength

0.2

Second Order Source Strength

0,4

0.6

29

With

As thus determined, it is only necessary to determine

A(e)

and

A(e) , such that the zero normal velocity condition is met on the body.

There are two methods which can be used to determine the

_A(e)

and

A(e) . One could choose a number of points at which to satisfy the body

boundary condition, and then determine an equal number of the unknown coefficients so as to satisfy the body boundary condition exactly at

these points. Alternatively, one could choose to satisfy the body

boundary conditions using fewer coefficients than points, and employ a

least-squares technique to minimize the error. The success of either of

these techniques relies on the coefficients

[A(e)(x)]

decreasing

rapidly enough for increasing in so that the solution can be

repre-sented reasonably with a small number of terms.

The method we have chosen is the former because of its relative

ease of computation and greater efficiency. The only danger in

se-lecting this approach would be if the potential is not well behaved around the body, in which case the least-squares technique might give

better results. As a check of this, a comparison was made with the

solution of the same problem as calculated by Faltinsen using the least-squares technique, and it was found that the results agreed to within

two to three per cent. Also, as a check of the rate of convergence of

A(e)0(e)

the series (r,e) , the solution was calculated using both

1 m in

seven and ten terms and the solution changed by less than one per cent. We shall now discuss the technique used to evaluate the potential due to a Helmholtz source and its corresponding normal velocity.

The Potential Due to a Helmholtz Source For the following problem,

and

72 _ 1,24, = 0 2D

- _y = 0 on y = 0

Ursell (1968b) gives the Green's function as:

S0 (KX"

K37.1( Kg) =

K0[10/(X-)L

(y-n)2]

00

dli(cosh ;1+1) eilc(x-)sinh p + K(y-g)cosh p

y < 0

cosh ;1-1

S0(Kr,e) = - 7Kr sin e e

CO

du u2cos(uKr cos 8)e-Kr sin eVi+u2

.4_ f

(l+u2)3/2

.1 du u sin(uKr cos e)e -Kr

sin

e/17.;:i-(1+u2)3/2

where r = A2+y2 and 0 = tan-1(x/y) . However, it is possible to

further simplify equation (30) by means of the following identity: 0

CO

0 where

±and

denote integration below and above the double pole

at, II = 0 respectively. (Note: We have used y-positive upwards, as

opposed to Ursell's y-positive downwards.) Faltinsen (1971) integrates

S0(Kx,Ky;0,0) by means of contour integration, and obtains the

fol-lowing form for S

0

-Kr cos 6

CO

I du cos(uKr cos

e)e-Ko(Kr) = (1+,2)1/2 0 CO

-(l+u2)3/2 Kr sin OVI+u2

du (1+u2)cos(uKr cos e)e 0

If we now both add and subtract _K0(Kr) , equation (31), from the first

integral of equation (30), we obtain:

S0(Kr,e) = K0 (Kr) - 7Kr sin e e

c°

du cos(uKr cos

e)e-Kr sin e1/1+u2

f

(1-Fu2)3/2 0 00 0 Kr sin eVi+u2 -Kr cos e -Kr sin e/i471717 .1

du u sin(uKr cos 0)e (l+u2)3/2

As a further check of our results, we can let e = 0 in equation

(32) and

(30)

-(32), which yields so(Kr,O) = K0(Kr) -31

fdu

cos(uKr)

f

du u sin(uKr) (1+u2)3/2 (ii.u2)3/2 0 0

Using equation (9.6.25) from Abramowitz and Stegun (1964), we see that

the first integral is equal to

KryKr)

. Faltinsen (1971) has shown

that the second integral is equal to KrKo(Kr) . So for 0 = 0 , we

have:

So(Kr,O) = (1 - KOK0(Kr) - KryKr) .

We can now obtain a simplified form for the radial velocity in a

similar manner. Differentiating equation (32) with respect to Kr , we

obtain the radial velocity as:

9S0- Kr cos

(Kr,e) = K (Kr) - 7 sin e e

(i -

Kr cos

e)

8Kr K1(Kr)

I du cos(uKr cos 0)e

(/1_sin

6/17;72-(6.7-17i sin

e -

u2cos 8)

(i+u2)3/2 0

4.

fdu

u sin(uKr cos 6)e-Kr sin 0/T-1717

If we now multiply K0(Kr) , equation (31), by cos e , and add and

sub-tract it from the first integral of equation (33), we obtain:

9S0Kr cos

e

(Kr,0) = - K (Kr) - K(Kr)cos 6 - 7 sin 6

e-

- Kr cos

0

dKr 1 o

J

du cos(uKr cos 8)e-Kr sin e./TIT,7

047-1--L7 sin

e + cos e)

(i+u2)3/2 CO CO 6/17-1-J sin e + cos 6) . (l+u2)3/2 0 ₍₃₂₎ f

du u sin(uKr cos 6)e-Kr sin 6474;71-1

04-..-;7

sin

e + cos 0 .

(i+u2)3/2

0 ₍₃₄₎

We can again obtain an analytic integration for

e = o

. If we set

= 0 in equation (34), we find that:

-DS()

(Kr,O) = - K1 (Kr) - K0(Kr) +

dKr

j du cos(uKr) j du u sin(uKr)

Evaluating these integrals in the same manner as for So(Kr,O) , we

obtain: 9S0

(Kr, 0) = (Kr - 1)[K (Kr) + K (Kr)]

dKr 0 1

The integral for S0(Kr,6) , equation (32), and the integral for

DS0(Kr,0)/DKr , equation (34), can both be evaluated by means of

nu-merical integration. Figures 5 and 6 show the variation of So(Kr,O)

and _{3S0 (Kr,6)/DKr} as functions of Kr _{for several values of} 6 . The

asymptotic curves shown on these figures are the functions

-Kr COS

6Kr cos

-TrKr sin e e and sin e e - Kr cos

0

respectively. These asymptotic limits are valid for Kr sin 6 .

These calculations complete the section _{on numerical analysis.}

2.0 1.0 0.0 -1.0 -2.0 -3.0 33 Asymptote for

e =

7/2 Kr -0. 0.2 0.4 0.6 0.8 1.0 0 = 0.0 e = 0.357/2

Figure 6: Contours of

no r,'9/Kr

Vs.

Kt

for Various Values of 6 0.0 2.0 4.0 -6.0 12.0 14.0 -16.0 18.0 -20.0 Kr 0.6 0.4 0.0 0.2 0.8 1.0

COMPARISON WITH EXPERIMENTS

In order to obtain a measure of the validity of our results, it was

decided that a comparison should be made with _{experimental data. As we}

saw earlier, the second order source strength behaves _like:

x

bf5(0)1

bfs()

and as we stated earlier, _{our solution depends critically on the value}

of

bfs(0) . If bfs(0) is infinite, then the source strength is

infinite, at least near the bow. _{Experience with bfs(x) shows that for}

an arbitrary body, bfs(x) has a term which behaves like _a(0)Y0(Kx)

which comes from the discontinuity of _{a(x) at x = 0}

; this term tends

to infinity logarithmically as x 0 . However, if we demand that the

body be pointed, a(x) has no discontinuity at x = 0 , and the

corre-sponding bfs(0) is finite. _{With this information as a guide, it was}

decided to compare our results with _{experiments on a body of revolution.}

After a brief investigation, it was determined that there was no existing data for bodies of revolution with pointed ends and a parallel

middle body. _{Therefore, a decision was made to conduct our own}

experi-ments to obtain the necessary data. _{For a model we chose a parabola of}

degree 2n _{to represent the radius of the body}

as a function of

longi-tudinal position. _{This function is as follows:}

r(x) = r0(1-x2n) for -1 < x < 1 .

This representation has the advantage of allowing us to introduce an

apparent parallel middle body (by choosing n _{large enough) and still}

maintain a continuous function _{throughout. We found that n = 5}

would

give about 60% of the body _{as parallel middle body. We chose}

_{ro = 0.10}

to give us a beam-length ratio of _{one tenth (B/L = 0.10).}

The radius distribution given above was then scaled to give _{a model}

15 feet in length with _{a beam of 1.5 feet. This model was built of}

35 R(x) = 711

I

for 7)/ 71 TX 0 = 0(c2)

fiberglass stretched over a plywood frame. The model, which was fixed

to the carriage so that it could not sink or trim, was towed in the

towing tank and wave measurements were taken 2 inches from the side of

the model using a resistance wire wave probe. These wave records were

recorded on a strip-chart recorder, digitized manually, and then

non-12

dimensionalized by

0-f Ti .

These non-dimensionalized points are plotted in Figures 7, 8, and 9 along with the wave elevations predicted by our second order

slender-body theory and the Tuck Slender-Body Theory. The second order wave

elevations are obtained by evaluating equation (28), while the Tuck

calculations are obtained by evaluating the following equation:

gc = - [2c1) (x,y,0) + 2b

(x)+

e (x,y,0)]

rw rw

4u2

U2 where bfs(x) is the derivative of bfs(x) given by equation (22). The curves of wave elevation from the second order theory have been truncated in the stern region because they start to oscillate

rapidly with increasing amplitude. This oscillation, which is caused

by R(x) becoming irregular, probably has two causes. First, the

values of R(x) will tend to drift as we move toward the stern due to

an accumulation of errors in the calculation of previous values. Second,

as we approach the stern, b,s(x) starts to grow due to the waves

generated by the stern section. These waves may be out of phase with

the waves from the bow, which in turn results in R(x) having to change

phase. This is further complicated by the fact that the formula for

R(x) contains a memory term which includes information on the earlier

phase. The combination of these two effects could well cause the

irregularities found in the second order wave elevation.

If we disregard the stern region and compare our three sets of

curves, we see that the phase of the waves from the second order theory

lag behind the waves from the Tuck theory by about 7/2 . At the same

time, we see that the second order phase agrees very well with the

experimental one over the middle region of our curves. The amplitude of

the second order waves over the midbody is much less than that due to the Tuck theory and agrees fairly well with the experimental curves. Both the second order theory and the Tuck theory grossly overestimate

0.20--0.2

-0.4

-0.6

0.6

0

Experimental Wave Elevation

Second Order Slender-Body Theory Wave Elevation Tuck Slender-Body Theory Wave Elevation

0.9

1.0

0

0

Figure 7:

Comparison of Wave Elevations Between Second Order Slender-Body

Theory, Tuck Slender-Body

Theory, and Experimental Data at

a Non-Dimensional Speed of KL = 14 .

0

0.4

0

0.2

0.0

-0.2 0.4 0.6

Figure

Second Order Slender-Body Theory Wave Elevation Tuck Slender-Body

Theory Wave

Elevation

C)

Experimental Wave Elevation

I0.3

0.6

/0

0

/

Comparison of Wave Elevations Between Second Order Slender-Body Theory, Tuck Slender-Body Theory, and Experimental Data at a Non-Dimensional Speed of

KL = 20 . 0.6 0.4

0

0.2

I/

--\

\

/

\

/

\

/

I

1'

-0.4 0.0 -0.4 -0.8 _-1.2 -1.6 Figure 9: 10.2

0

\c)

II

\

11

/

\j/

\

\

I

\

\

Second Order Slender-Body Theory Wave Elevation Tuck Slender-Body Theory Wave Elevation

0

Experimental Wave Elevation

1.0

\-/

Comparison of Wave Elevations Between Second Order

Slender-Body Theory, Tuck Slender-Body Theory,

and Experimental Data at a Non-Dimensional

Speed of

KL = 30

wave amplitude in the bow region. This is not unexpected of the second

order theory since it has a square root singularity at the bow. These

findings point out the necessity of developing more fully the theory for

flow about the bow of a ship. This completes our discussion of the

APPENDIX A: INNER EXPANSION OF THE OUTER EXPANSION FOR A SLOWLY VARYING SOURCE DISTRIBUTION

In this appendix we shall show how the inner expansion of the outer

expansion equation (11a 45, b) for the potential due to the slowly varying

source distribution, equation (10), is obtained. In the body of the

paper, we gave this potential as:

ity

117.4iTz

cl)(x,y,z) = - z-r-Tr

j

dk ekxc*(k) (Al)

1 di e e

422

k2/K

We shall do, most of our work with the Green*s function*--really Its Fourier transform--which we can write as:

/k2+2,2 z

G**(k,i;z) = - e

ATFITT-

k2/K

If we invert G**(k,i;z) with respect to

y 4

we have

ikv

A2412

z

co

G*(k;y,z) - 1 di e le

A2412 k2/K

and if we. note, that the. integrand of G*(k;,y,z) Is even With respect to

we can rewrite the e as cos(2y) By doing this and by

making

a change of variables,

,m ikl

di.= Ikl we can rewrite equation (A2) as:

sinh, cosh p dp

*By Green's function, we mean the fundamental solution to a partial

dif-ferential equation. _{This solution may satisfy some. of the boundary con=}

ditions of the problem. In our particular case, the Green's function

satisfies both the linearized, free-surface boundary condition and the radiation condition,

41

(A2)

CO

dp cosh p

1)

cos(1kly sinh

p)e11(.1z cosh p G*(k;y,z) =

-7

COSh p -

lkl/K

where this integral must be interpreted as a contour integral, and the

proper path chosen depending on whether k is positive or negative.

For small values of 1k1r , Ursell (1962) gives the following asymptotic

expansion of this integral:

sgn k-a)ctnh

a}

G*(k;y,z) = -(7-e)ctn 1 - -

K0(11(10

7 { cosh ma Io(lkir) + 2 (-1)mIm(lkir)cos me 1 cos me

where the upper or lower expressions are valid as follows:

{cosh

a > 1}

cos < 1

Using the identity below [Ursell (1973)],

(cosh

ma}

I0(11(10+ 2 _{(-1)mIm(1k1r)cos me} = eKZ cos(yVK2-k2)

1 cos me

where cos(y/K2-k2) becomes cosh(yvk2_,2N) for 1k1>K , we can

rewrite equation (A3) as:

(-iii sgn k-a)ctnh a

G*(k;y,z) = - -1 eKZ cos(y/K2-k2)

(71.-e)ctn IT

-

-1 Ko(lkir) Tr (A4)

LI

2 + - (-1)m 7 1 a

-Tv(11(10cos ve

(A3)

/

1:

sinh ma ctnh a v=m sin me ctn e 2

+ -

₇ 7 1 (-1)m _{-57)-v(11(10cos ve}

{sin

ma ctnh

Isin

me ctn e V=M 1

-We shall now place this expression for G*(k;y,z) back in equation

(Al). _{However, first let us discuss the expected behavior of}

a*(k) .

According to our assumptions, a*(k) is a slowly and smoothly varying

function of x . Correspondingly, we can use the theory of Fourier

transforms [eg., Lighthill (1964)] to show that if a(x) is

suffi-cently smooth, 0*(k) must decay like an inverse power of k for large

k . Using this information, we can conclude that _{a*(k) is peaked near}

k _{equals zero and that it falls off rapidly}

away from the origin. Furthermore, we shall take advantage of the fact that we are interested in evaluating the potential in the near field, which implies that

r = 0(e) .

Returning to the evaluation of the potential, _{we can now write:}

(x,y,z) = (71

a()G(x-y,z)

cc = I dk eikxu*(k)G*(k;y,z) 27 cc = I dk [!

2j

_ik )]

Now substituting the value of G*(k;y,z) , equation (A4), into this

equation, we find ¢(x,y,z) = 1 I dk eikx[-ika*(k)] 272 e"cos(y/K2-k2) ik cc 1,2 I ikxa*(k)K0(1kIr) + 1 (-1)m cc dk ekx[-ika*(k)]

-j

Iv(lkir)cos ve 7 1 v=m sinh ma ctnh 1 ]Tc. sin ctn 43 sgn k-a)ctnh (7-)ctn f3, (A5a) (A5b) (A5c) dk . a (-in

_a

Setting (1)1(x,y,z) equal to line (A5a),

cp2(x,y,z)

equal to line (A5b),

and ¢3(x,y,z) equal to line (A5c), we can work with one term at a time.

Evaluating cpi(x,y,z) first, we can write

CO ikx 1 (-i7 sgn k-a)ctnh a eKzcos(3,4-217i(2) ik cpi(x,y,z) = -2-7rT I dk e [ika*(k)] (7-f3)ctn 6 1 = - cO(E)f(x-C,y,z) 7 where f(x,y,z) = eKZ at

dk eikx (-i7 sgn k-a)ctnh a

cos(y42_k2)

27 -ik

(7-6)ctn 6

By expanding f(x,y,z) into three integrals,

KZ e

dkikx

f(x,y,z) ( 271 I -ik i7 sgn k-a)ctnh a cos(y/T-Tic2) KZ dk eikx e

+

---ik

(7-6)ctn 6 cos(y.2)

271 J -K -K eKZ dk eikx

+ (-i7 sgn k-a)ctnh a cos(y42_k2)

27 -ik

and by

combining

terms and reducing the interval of integration to

0 < 6 < m , we find that f(x,y,z) reduces to:

at

f(x,y,z) = -eKz dk cos kx ctnh a cosh(y/k2-K2) (A6a)

K

at

e dk sin kx

- a ctnh a cosh(y/k2_,(2) (A6b)

7 KZ

I

dk sin kx

+ e ctn 6 cos(y42-k2) 0 eKZ I dk sin kx 6 ctn 6 cos(y42-k2) . 7 0 (A6c) (A6d)