First- and second order forces on a cylinder submerged under a free surface

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First- and second-order forces on a

cylinder

submerged under a free surface

By T. FRANCIS OGIL VIE

U.S. Office of Naval Research, London Branch Office Received 2 October 1962 and in revised form 1.3 January1963

Introduction

Several years ago Dean (1948) showed how to solve the linearized potential problem of water waves passing over a submerged circular cylinder. He

dis-covered the remarkable fact that there is no reflexion from the cylinder; the transmitted waves have the same amplitude as the incident waves, but they suffer a phase shift in passing the cylinder. Soon after the publication of Dean's

paper, Ursell (1950) investigated the problem anew. He placed the solution on

a rigorous basis, supplied a uniqueness proof, and developed a form of the solution with which it was reasonable to perform calculations.

Ursell's procedure is here applied and extended toseveral specific problems. In particular, the first-order oscillatory force and the second-order steady force are calculated for the following situations: (a) the cylinder is restrained from moving under the effect of incident sinusoidal waves; (b) the cylinder is forced

to oscillate sinusoidally in otherwise calm water; (c) the cylinder, which is neutrally buoyant, is allowed to respond to the first-order oscillatory forces. In all cases the problem is treated by two-dimensional methods. The water is

considered to be infinitely deep.

It is first proved that knowledge of the first-order potential supplies informa-tion sufficient to solve these problems. The solutions are obtained and then

numerical results are presented, the relevant quantities being plotted as functions

of depth, with wave-number as a parameter. Althoughthe main purpose here is to calculate these forces, it is a simple matter also to extend the results of Dean and Ursell on the transmission of waves. It is found that in situation (c), as in (a), there are no reflected waves. For both (a) and (c), curves are presented for

the phase shift of the transmitted wave. In case (b) it is shown that outgoing waves are generated in one direction only, if the cylinder centre follows a circular

orbit.

The

second-order problem

Assume that a circular cylinder is located under afree surface, with its centre at z = (t), y = - h ± 77(0. The instantaneous surface of the body is then specified

by

S(x, y,t) = (t)]2 [y+ h --77(t)]2 a2 = 0, (1) where a is the radius of the cylinder. We also define a surface So:

So(x. Y) x2+ (Y + h)2 a2 = 0 (1') 29-2 4. 17, 451 1.

452 _{T. Francis Ogilvie}

(see figure 1). The undisturbed free surface is taken _{as the x-axis, and the}

in-stantaneous free surface will be specified by

y Y(x, t) = 0.

(2) The y-axis is positive upwards.

FIGURE 1. Geometry of the problem.

We seek a velocity potential, j (I)(x, y, t), which satisfies (see Stoker 1957. or

Wehausen Laitone 1960)

01),,+ = 0, ₍₃₎

OrYz - °Dv+ = 0 on y (x,t), (4)

g Y + icbt + ?,-(c1)1 + =- 0

on y= Y(x, t),

₍₅₎

(DzSz.4-0,S7., = 0

on S = 0.

(6)

Equation (3) must hold for all time, t, and for all (x,y) in the fluid domain, i.e.

for y < 17(x, t) and S(x, y,t) > 0. Equation (5) is, of course, Bernoulli's equation.

and (4) and (6) express the usual kinematic condition at _{a boundary of a perfect}

fluid. There will also be conditions at infinity, which will be formulated later. We assume that all of the dependent variables _{can be expressed in power} series in terms of some small parameter, e:

y, t) = eclo,a)(x, y, t)+ e20(2)(x, y, t) + (x, t) = er(i)(x,t)+ 6.21,72)(x, t)

(7)

= 601)(0+ e2(t)+

(t) = 677(1)(t) 6277(2)(t)

1- We choose the potential so that its positive gradient is the velocity. x, y)

(The superscrir

potential funct

t and that the

daries can be e: of these bound Let us define a Every point (x,. is on S = 0, the where clearly (, equivalent to so that, as the v The potential Va certain circlev the potentials a of the tangent

then hi < a.

:454-CDT ± gin) *- etc.; r.[vo(2)_ (2)( etc.; we have w

Iwith i and j uni

If one finds and (10a), ther

Similarly, furth in each pair of and the right-ii

To find the fc

the boundary in

= g(0

Farces cm, a submerged cylinder ₄₅₃

(The superscripts in parentheses are strictly indices.) It is then assumed that the t. potential function can be continued analytically into the region y < 0, So >

=:* and that the value of (I)(x,_{y, t) (and values of its derivatives) on the two}

boun-,

daries can be expressed in terms of Taylor series about the undisturbed positions of these boundaries. The following conditions are found.:

(1)(rd +1:1421,? = 0 for y < 0, So > 0 (p 1., 213 .) ₍₈₎

(14P+

=0 on

y =10, _(9a)

+ g - 2 (1)(P (1)gt) - (IV ) clnit) + g -Lc* + 01412,'

on y' = 0,

(9b)

etc.;

r .1V(1)(1.)- 6-)(t)] =0 on _{So = 0,. (10a)}

r.[V0(2)_*2)(,)] ga)(01. v4)(1)] (i)(t). [vow' _{) j on S0 = 0,}

(10b)

etc.; we have written

r = xi±(y+h)j,

_l(Ha)

73)kt) = (P)(t)i-F77(9)(0,i (11b)

with i and j unit vectors along the_{x- and y-axes, respectiVely.}

If one finds a potential function, (1(1)(x, y, t), which satisfies equations (9a)

and (10a), then the right-hand sides of equations (9b) _{and (10 b) are known.}

Similarly, further conditions could be foundon clo.(P)(x, y, t) and (73)(t), for p > 2,,

in each pair of which the left-hand sides would be of the form of (9a) and (10a) and the right-hand sides would depend onlyon the lower-order solutions.

To find the force on the cylinder, we integrate the pressure, p(x,y,t), around - the boundary in the clockwise direction:

X(t)i-i Y(t) = if

p(x, y, t)(dx - idy). s= 0

Let us define a system of polar co-ordinates,(r,

x r sin 67,. y + r cos 8:

Every point (x y) on E = 0 can be referred to a point on So = 0; that is, if (x, y),

is on S = 0, then

x = E+ a sin 0 , _{It+ 77 +a cos, 0,}

where dlearly casino, -1.1-1-a_{'cos 6)1 is a point on So =0. These relations are} equivalent to

'

so that, as the: variable of integration follows the circle S_{= 0, we have}

,dx - idy at° dt 9.

_ _{The potentials to be fOund will be. harmonic in the lower half-plane, outside} 73.

a 'certain circle with centre at (0, - h In fact, it can be shown that, if g 0,

the potentials are harmonic outside' X2 _{±h)2 = (h -i)2,02, where 1 is the length}

,of the tangent from (0,0) _{to the circle x2+ (y +1)2 = a2. Clearly,, if h}

_{- a > 0,}

then

-1 < a. If the motions of the cylinder

_{are sufficiently small, a similar}

". (12 et, ,b) the hi', (2) 057. OT 04) uation. perfect Iter. power 0, = ... ; ((;) Lin. i.e. (7)

=

r

8): = = y =

x -

= - i97)- ia ei° , = (y h

454 _{T. Francis Ogilvie}

result can again be proved, and the potentials can be expanded in Taylor series

in a finite neighbourhood about every point of S or S. The same must then be true of the pressure functions, so that

P(x, t)!s--0 = p(. ± a sin 0, - h, + + a cos ,t)

= p(a sin 0, - h, + a cos , t) + .(t) p x(a sin 0 , -11 + a cos 0, t) y(t) p v(a sin 0, - 11+ a cos 0, t) + .

= + + n(t)Py +

We now have for the forceon the cylinder

X (t) (t) = ia

f

el° [p + -(t)px+ 77(0 + 0_0 d0 (13)

To the expansions in (7), we add another:

p(x, y, t) = ep(')(x, y, t) ,2 -(2)_{p (x, y, t) +}

where p(x , y ,t) is the hydrodynamic pressure (the difference between the actual

pressure and the hydrostatic pressure measured from y = 0). From Bernoulli's equation and from (7),

P(x, t) = PcDt P[CD1 + In]

= e[

_{+62[PO2)--iP(0))2}

_{.i.P(0,1))2] ±}

0(0)-Also,

p x(x, y,t) = e[- p] +

(e2), _{pv(x, y,t) = e[- p(1)(,)t)] + 0(69.}

When these relations, as well as (7), are used in (13), we obtain

X(t) _iY(t) e[x(n(t) _i ya)(t)] 62[x(2)(t) ip2)(t)] + 0 (63)

where _{X(')(t) - iYa)(t) = - iap j} eie) d0 ; (15)

X(2)(t) iY(2)(t) = - iap e20 [V,2)+ i(c1)(2))2 E(IV))2+ (2-)(1:g +770.)(1)(y1,... de. (16) In the problem to be considered presently, 01)(1)(x. y, t) will vary sinusoidally in time. From (9b) and (lob), it is clear then that (1)(2)(x,y, t) will have two com-ponents: (1) a time-independent (` d.c.') part and (2) a part that oscillates

sinusoidally- at twice the frequency of the first-order solution. Ifwe calculate the time-average of X(2)(t) - il7(2)(t), using (16), we see that (1)(2)(x,y, t) does not

contribute at all. Thust

firx(2)(ot

I-(2)()1 = _ iap _{eM(1)1))2t ±;(c1),--(PP`±P)(1)(1/ +77('t1 ir_a (10}

.

(16')

If we let (1)4)(x, y ,t) be the real part of a function of a complex variable, say f(z,t), where z = x + iy, then

1T

...Y(2)(t) 1 (2)(t) = - iap _{e1° [if ' (z,}t) (z,t) +Re f( ' (')+ Sf ' (z, 6t}11.-ad 0 .

IT

(16-)

The prime denotes differentiation with respect toz.

t An over bar with a t following it will be used to indicate time averages. A bar without index will indicate complex conjugate.

(14)

General solut

The first-order p

contains only a res-Except for the 1

infinity, it is assum cylinder must appE in Appendix A, eqt also represent outg in terms of which t

is generated inside t

in such a way that

Let the only ext

potential for which fo(z, t) = A e:

where v = a-2/g, and

incident wave is the

Thus the amplitude

Iand

71(1)(t) vary sinu result of some force

60.)(t) = Then the point on

(to first order) ocos 8(

Let the entire first-c

m± (1)2m n=i in! ( Inz (va)2'n n=1. m! ( Sm+ (va)27, f(z,t) = f o(z, tI The functions f,, fp!

-equals the e pressio:

am + (va)"

y

,

le 11 i 4 actual C(M -nilatv Ill)i de. ( 1 (it (1(r) ithout

Forces on a submerged cylinder 3.

General solution of the first-order problem

The first-order problem has been entirely solved by T:rsell and this section contains only a restatement of some of his results which will be needed later.

Except for the possible presence of incident sinusoidal waves coming from infinity, it is assumed that the only disturbance of the free surface far from the cylinder must appear as outgoing waves. The pulsating-singularity potentials in Appendix A, equations (56), all satisfy the free-surface condition, (9a), and also represent outgoing waves as x On r = a they form a complete set

in terms of which the normal fluid velocity can be expanded (provided no fluid

is generated inside the cylinder). So it is necessary only to combine them linearly

in such a way that (10a) is satisfied.

Let the only external hydrodynamic disturbance be an incident wave, the

potential for which is the real part of

f 0(z,t) = A exp [ i(vz + o-t)] = A exp [ exp [vr e-'°] exp [ jot], (17)

where v = o-2/g, and A is a real constant. [The elevation of the free surface in the

incident wave is then

1E A o- .

e--R {f0(z, t)}1y=0 = sin (vx + crt). g ct

Thus the amplitude of the incident free surface wave is H, = Ao-Ig _] Let E(1)(t)

and ya)(t) vary sinusoidally also, either in response to the wave action or as a result of some force applied directly to the cylinder. Specifically, let

= sin crt + E2 COS o-t, 77(1)(t) = 771 sin o-t + 712 COS crt. (15) Then the point on the cylinder. at the angle 0 has an outward radial velocity

(to first order)

o- cos 0(?71 cos o-t 71, sin at) + cr sin 0(E, cos crt E2 sin o-t). (18')

Let the entire first-order potential be the real part of

CO

= fo(z,

{an f,(z,t)+,6fn2(z,t)+7g,(z,t)+6g2(z,t)}.

(19)

n=1

The functions f f ,n15 and gn2 are given in Appendix A. By (10a), a Re {f(z, t)}/Er

-.1- equals the expressions (18'). Thus the following sets of equations are obtained: --*

:m, _{am ± ( va)""int1 (m! (n+11 )1!) !(Am+. an Bm+n fin) = 2( va)2 8m1,}

v (20a) CI 7.n (m n)! o- A e-Ph (va)2m (m)2m V 455

nt1m! (n 1)! (A"±nfin +13m±n = 771(Pa)2 6mi m!

7

(m n)! o A e-'h (va)27.

7,,,+ (va)27. _{-, On) = E2(va)2 8,].}

m!

(n-1)!(A

m v m!

(m + n)!

o-+ (va)2m + B yn) = E,(va)2 ona, (20d)

.

-

- vh]

-t)

t)+

-(200

+ ) ₊ 2

456 T. Francis Ogilvie where Si; is the Kronecker delta:

11 if i = j,

6ij = 10 if i

j,

and the A; and B; are known constants, given in Appendix A by equations (55). Thus we have four infinite sets of algebraic equations for the four infinite sets of

coefficients, am, ym, sm. The conditions under which solutions exist were

discussed by Crsell and will not be considered here. We shall assume throughout that unique solutions exist. The method for solving these equations, and, in

particular, for uncoupling them, was also discussed by Ursell. Further mention of this problem will be found in Appendix

If the formulae (56) of Appendix A are substituted into (19) and the order of

summations is reversed,t (20) can be used to eliminate the double sums, and the complex potential becomes:

fire±iS ) einth

(/?-ik)(vr)ne-im°

f(z,t) = cos crt5:, le (1r)r" 1 ± (va)27n ± (77,+

n[An(fin-ia)+.13(x-iyn.)]}

(c.

+sin crt v + iy,) einia V(71. 2'.7t) (vr)me-ilne

PT)m -r " (a)lm

- gr

(772 + '42) n[An(a. -

-

Bn(13-i6.)]}

The last sum in each bracket is a constant which does not affect any of the sub-sequent force calculations. The constant A does not appear explicitly now, but the values of am, fi ym and all depend on it, as well as on E2, 7/1, 772.

At this point, the problem is solved, in principle.

If the cylinder is restrained from moving, then E, = =- = = 0.

Equations (20) yield the coefficients for f(z, t), as defined in (19). The desired

forces are then given by (15) and (16").

If there are no incident waves and the cylinder is forced to oscillatewith

given amplitude, direction, and frequency, then 7, and 7/2 are known.

and A = 0. The coefficients are obtained from solution of (20), and the force

components are obtained from (15) and (16").

If the cylinder is assumed neutrally buoyant, equations (20) give the

coefficients in terms of the unknown motion parameters, E.1, E, 772, and then

the motion is determined by solution of the equations of motion of a rigid body,

with the force again given by (15). Then the coefficients are completely known

and (16") gives the second-order steady force.

It is of interest to carry these solutions further. The next three sections treat the above cases in more detail.

t The singularity potentials in Appendix A are expressed by Laurent series. It is

legitimate to add sums of such series term-by-term if the resulting series converge

uni-formly. But the potential is itself analytic and single-valued in a <z+ ih <2h a. So it must have a Laurent expansion, which must then be (19').

(19')

4. Cylinder restra

We assume here ti

That is, Ei = = 77, -- that

The first-order force

Thus, from (19'), .f(z, t) ft(i.

= it

Xil)(t) Yil)(t) = -The second-order st, X7,27;- i 1TP (t) = - 2iniov In Appendix B it is and if

These results can be and also that HF, =

Then, from (23), where From (24), ---xi2)(0` = Numerical results ... curves in figures 2,

where 2vh = 2va. kaT

meaning, because ti

r* the part of each Cu: siderable. care. The r

clearance between t

the linearized theor:\ a

then

4. Cylinder restrained.

We assume here that the cylinder is rigidly held in placewhile waves pass by. That is, = E2 = = '2. 0, and fo(z, t) is given by (17). From (20) it is clear

that

Ym = An) m = (21)

Thus, from (19'),

f(z,t) f,(z,

(am + ifirtt) eirn°

"id (17)m

CO

m(A 1.) (a, - i An). ,(22) The first-order force is found from (15):.

XP(t)- i IT)(t) - (27rpo7v) [(alsin at + ft1 coso-t)- ice1 sin crt -al cos at)]. (23)

The second-order steady force is found from (16"):

XT-7(1)`

- i

Y(t)'

-

22npv

m(m + 1) [(am am-a. + fimfim,-1) + i(fim anr-i am fim+i)). ₍₂₄₎ 2)2m+2.

1

In Appendix B it is shown that if em satisfies

(m, + n)I (pa)2m 1(25) em +(va)2m 1 (n - 1)!I n= m! ' L.?-4,-__

Forces on a, submerged cylinder .457

S 27e-2j' n (25')

(n -1)!!'

= A e-vh S +SD', 71m A e-Ph em/(1 +S!). (26a, b.)

These results can be usedto simplify (23) and (24). We note again thatHo = A cr g

and also that Ha =. v A2 g ,-where Ho, is the amplitude of the incident surface wave. Then, from (23),

X(t)

Yil)(t), -(27rpgHolv) evh 1(1. + S2c)i} exp {- Vri_).}., (27)

.where

. tan-1 S

?...,; From (24),

.x12)(0' 0; 3-7-(1,2-)-(61, 277 pgHg e-2Ph nz(m + 1) erne,' _{(28 a, b)}

1+82 .Q2

7- 'le (va)2m1-2

Numerical -results from equations (2'i) and (2'8) are presented as the solid curves in figures 2, 3 and 4. The curve for each value of vaterminates at the left where 2vh = 2va. An extension beyond such an abscissa would have no physical

meaning, because the cylinder would not be completely submerged. In fact,

the part of each curve near the left-hand end must be interpreted with

con-siderable care. The amplitude of the incident waves Must be much less thanthe

clearance between the top of the cylinder and the undisturbed freesurface if

the linearized theory is to have meaning.

e;

/ Pm) '1 V7 ri (va)2m 111 II and if then E, = t) (am -= (1 =

- i

= i(a-t (27')

tiD 150 125 100 75 50 '25 ,01 3 4 6 -2vh

FIGtRE' 3. Phase lag of ,oscillatory force on restrained cylinder,

It is of interest to estimate the Tesults for va < 1. Let

(va)2m

= 7 en, (va)2#.

m!

8 9

This expansion is justified by Ursell. Also, let

ymn {(m +701/(n 1)1} Am+7,. (30)

I Note that (and tilt

I approximately

XI90

ITV) -G

Irmo

FIGURE 4.1

where I1(2va) is the modi:

all coefficients of (va)24 az

Thus the first term in bra

conditions is satisfied: w

5. Cylinder forced tc

In this case we assum'

there are' no incident way The unknown Icoefficieri, first-order force comport

T213(t) Y(211(t) = (277.11

(29)

The second-order stead\

XV)(t)ip /722T r 1 I \ \ va = 1-0F I %

\

\

, 1 I 1111 1

.

_.

--1, ill y ',.., ,va= 0.5 va = 2.

.

-,.. 0 ...,t -... L.,.= 4.0 ,

/

I1 i I I , , ', ! va =1-0 va = 2-0 1 va = 4.0 1 0.5 10-3 \ 1o./ 1 --, , N.,..._ I, , 458 T. Francis Ogilvie

If we take the undisturbed incident potential wave (see equation 1(17)) as a reference, we see from figure a that the phase of the oscillatory force lags more and more as the cylinder is considered to be closer and closer to the surface. An interpretation of this situation will be presented later when transmission of the

wave is considered.

3 4 5 6 7 11 13, D4

2vh

FIGURE 2. Amplitude of oscillatory force on restrained cylinder. 0.15 ,CO &05 '0'0 0-05 = ein = 10 11 12 13 1-0 10 12 0.20 0-10

(33) \ 1 I \ ,

IIIMINIMINIEL

, SMOMMIS=111=1111M111

MNWIIIMMINI

MINEMMIRWMNIMININNOMMININM

suiummemmsmo

11111111111111111 1111111K111=MIMINNIIMMIIIIMINIMMIIMIIMINMIIIIIIM

111=1EINIIMMNIMININIIMINIIM

,

INK

_,

,

IMIIMINI

%

.

INUIIIMML',

W%

cp , IMINIIIINL%

,

,2 _{1111/111111.11.11M1}

,

% % i

,'

0

\ ,

:, 6

11111LVIIIIMI11111111

1111.11.111L

Forces on a submerged cylinder 459

Note that A,_,. (and thus yn) is a function only of 21..h. For the forces we have

approximately

X1(t) -

(t) = - (27pg Hol (va)2 e--'h

X ft[1 yii(va)" + (Y11-4,-Y12-4,2 eIvh (va ± ...]

± 2,-i e--2,h (va)2 [1+ _{- 2yii) (va)" +...]}} (31 a) Y Ty; = 27TpgHge-2" {(va) I1(2va) - (va)4 (711+Y21)± (31 b)

4 6 8 10 12 14

2 vh

Plot-RE 4. Steady vertical force on restrained cylinder.

where I1(2va) is the modified Bessel function of the first kind. In the lastequation, all coefficients of (v0)2" after the Bessel-function term approach zero as 2viz co. Thus the first term in brackets provides an approximation if either of the following

conditions is satisfied: va 4 1 or 2vh > 1.

5. Cylinder forced to oscillate

In this case we assume that E 7/, and 7/2 are known (i.e. given) and that

there are no incident waves. The complex potential is given by (19)withf0(2, t) 0.

The unknown coefficients in (19) are found from (20), where now A = 0. The

first-order force components are given by

r21)(t) - i I-(21)(t) = (2irpo- v) {(al sin - 71 cos al) - i(1 cos at -131 sinTo}

6

_Th90.2a2fi sin at/E E2 cos crt) - I sin a-t 7/2 cos crt)} . (32)

The second-order steady force components are given by

°' 772(m + 1)

XT;27;

-

i 7276 = -irrpv

ar2

[(a. ± (am-i (v

+ Gem + jam) (A.+1 10 001 0-001

-o-t + i7m) I 0-1

n -

,

, UP-A°

2-1-6

g

1Z 0-4 0-2 +0-2 +0-4 ±06 t 9aw" 460 _{T. Francis Ogilvie}

It is shown in Appendix B that, if _m and S are defined by

( (va)274

'In±n

A

r

7.-71 m! (n- 1)! 6;n1,' cc Cn S 97T e-2Ph 1)!' Figures

-

t

In partial

(.34)1 4:, , -7-..-7-1 in phase vl ll '-1": this force. (341 ,,,...,,- --As. =di ri - -1 1 .1 va=0-05 , [ - i 11 va=10-1 1 1 va=o--2 I ! samsva-o3.

INS.

va.=

I 1-0 'a=2-0' I 0-5 va=

AIMIPPI

I 1

Ill

va= 0-4

1111r

11111111E1111111.1111

; 1 , , , i I ir 3 4 5 a '9 lo 11 12 13 14

FIG7RE 5. Oscillatory forde on cylinder moving

sinusoidally:-component in phase with cylinder acceleration.

then am, flm, -ym and d'm can be expressed in terms of em, 77, and of Ei, E 77 11

(see equations (62)). One obtains for the force components

V,1)(t)-

-npa2([2S

_-1 [,.5(1)(t)- 7(1);(t)1 1+ST [2ci-SCcil [E(1)(t) 2:77(t11),_1+ST 1.77.1 _{27rpgS2711.- :12).c.,`"' m(m + 1) (e,,./}

-

₎ (//a)2m-2 _77-pg(.1+ 41+ 91+7* 77i(m+ 1) 2 2 U,4

1+S

(va)2771-2 {. .777...m-1-1+ (S cm S em)1 (Se

It is interesting to note that the horizontal steady force vanishes if the cylinder oscillates along a straight line, no matter what the orientation of this line.

1-8

14

1-2

1-0

0-6

2vh = 1] (35) (36a) 1 FIGURE

( :14 )

.4.

Forces on a submerged cylinder 461

Figures 5, 6 and 7 show numerical results obtained from equations (35) and (36 b).

In particular, figure 5 presents the component of hydrodynamic force which is in phase with the acceleration. If we define added mass as the negative ratio of

this force component to the acceleration, we observe that negative added masses

1.0

0-1

00001

0-00001

FIGURE 6. Oscillatory force on cylinder moving sinusoidally:

component in phase with cylinder velocity.

2 4

2vh

8 10 1? 14

FIGURE 7. Steady vertical force on cylinder moving sinusoidally vertically.

c-IMI&

111E1111M

=`...61

Ivim wumwrow

VIM 1111K

'

-

'

\ 1

--.

,

0 0 0 */

..mm.

wm.

Milia6.

MIINIMImm - WEI 1111=NPrh,

.Millk...

...

.

,

_z 0 csi 1 II II :...1. I I "v I a =0-!1 I v a = 1.0 Pa=1.0

Ugra

v (35) (36,7) (36 b) !incicr .114 N 05 0-4 0.3 02 01 0 1 02

03

0-4

0.5 10 11 12 13 14 6 2vh 0 3 4 5 0'01 0.001 7 8 9 ii I 6

c.

Cif

'

462. T.. Francis Ogilvie

exist under a few conditions where the cylinder is very close to the free surface, Figure 6 presents the force component which is in phase with the velocity. The ratio of this force component to the velocity is the negative of the conventional damping coefficient. Of courses, this coefficient is always positive, although its value is very small for certain small values of the depth. At large values of the

depth, the damping force naturally approaches zero, since the wave-making

capability of any oscillating body vanishes as the depth becomes very large. Figure 7 presents the steady vertical force on the oscillating cylinder. For any value of va, the force is upwards for small submergence, becomes negative for

larger submergence, and approaches zero from below as the submergence.

becomes infinite. For very small values of va, it can be shown analytically than

the curves all cross zero at approximately 2vh -.-,-.- 2.8. This is confirmed by the

calculations, although it is not apparent from the. figure because of the scale of

the ordinates.,

6. Cylinder free to respond to waves ;

1

With the incident waves given by (17), we can write the potential as the sum, of the potentials found in §§ 4 and 5, but with "i, E V ?i2 now unknown. The

first-order force is the sum of the two forces found previously, but of course this

is not true for the non-linear second-order force. Because of this last fact, it is just as convenient to return to the formulation of § 3 for solving thisproblem.

The complex potential is still given by (19'), with the unknown coefficients

to be be found by solving (20). Such solutions for the coefficients are expressed in

terms of E E,,, 7/, 772 -also still unknown,, and the first-order force is then found

in terms of these motion parameters:

XS11).(t) YS1)(t) 77 pcalsin T42(8, + Iva cra(6.1 4772)]

cos crt[2(y1+ ial)/va + o-a(E.2 i712)]}. (371).

(See also equations (62) in Appendix B.) This complex force is set equal to the corresponding inertial reaction giving the equation of motion: .

npaz[Em(t) i77-001 = xsi(t) _i IT) (t).

It is easily seen that this equation requires that

= =

=0.

From Appendix B then,

-A e-vh 61(' S 1)1

i.'= 712 = cra(va)

_(Se

se c.1)2]',

A eh

'"C' 2 771 ,o-a(va) + (Se

These results can then be used in the expressiOns for fim, ym. and 8,,, equations

(62), for m > 1, yielding

A evh (Se S61) ('16m 61, m)

± (S Se .ci)2 A e-I'" i(Ci ern

= (-S-63. Se.C)2 -%-!.. -IThe second-order st oot o.o3ooi 10' IGLI1E 8. "Steady

We nate again that

10-ave..1 For ,small va

aun

3-77

0-01 N = I 'Z','" i, , II.? q21-001 0-0001 (3S) (39) .(.40,a) (401Y) (41a) (41 b)

- i

--

_±

ei- S

-27rp 1-0 this is a goo(

Forces on sa submerged cylinder

The second-order steady force is found to have components

XV)(t)( = 0; m(m+ 1) ('lem.-1-1 el ct+1) 1-0 0-1 0-01 NC Fe, 0-001 0-0001 0-000010 20 4-0 60 2vh 8-0 10-0 12-0 14-0

FIGURE S. Steady vertical force on a free neutrally buoyant cylinder under waves.

7:;We note again that t42 gif, where H, is the amplitude of the incident surface

have. For small values of va, this result can be approximated

, t , 1

\

wommin6uNvoomm

_{41111111111MIL}

1

11111t

\`,

\%, \ % \ ,

1111110111i

MOW

lumammaisomm

_{1111111111111111111111111111} .. MIIIMMEWMVIMI MMINIIIIMINIMIll

WIIMIENIMINILIIIIIIIIIIM

NIL

'

,,

IIKIIIIIILMIIIMINIIINIMIL

Im_

IIIIMINL'...101=1

IMMO

LIIIIIIIIM1111111111111111

Tim

amommossslammommoNom

IWINIMIIIIMIIIIIIIIIN

1k

lailmos

IL

(410 (41/0

"77pgHge (va[Ii(2va) va]+ 0 Again, this is a good approximation if either va is small

(l'a)8 (2vh,)41}' (43) or 2vh, is large. 463 (42 a) (420 cc. nal its thy

rt.

any for 1-1C.t :1141 T111. sum Thy his. it i in. ed in ound (37 (39) (40,0 (40b) itions ['he (va)2.-2

4 04 T. Francis Ogilvie

Figure 8 shows the results of calculations from equation (42b). The steady force in this case is always upward, as in the case of the restrained cylinder under waves If va is very small, we note that the steady force is much smaller than was the case for the restrained cylinder. However, for larger values of va, there is not

much difference between the two cases. The approximate expressions for the

steady forces, equations (31 b) and (43), show that this is reasonable if 21.,h is large.

It is of some interest to consider the motion of the cylinder further. Clearly,

from (40),

and so the cylinder follows a circular path. In fact,

Hoeie-Ph

. :(11(t)- iqa)(t) =

_{exP -}

_crt

(va)2 + (Se 'S` e,)214

= tan-1{ Ci where 150 125 100 75 50 25 = (t _77) 3 va= 0-3 wa -5 pa = 2-0 va =4-0

FIGURE 9. Phase lag of free cylinder motion.

13

For small va, the amplitude can be approximated

Hoele-'h

= Ho e-'h {1 + 712(va)4+ 0[(va)6il. ( va )2 + S;e,)2}i

The corresponding water particle (in the absence of the cylinder) would have an orbit given by

(')(t)- i77(1)(t) = 110

Thus the amplitude of the cylinder orbit differs from that of the water particle

by a quantity of fourth order in pa. The phase of the cylinder motion lags behind

that of the corresponding water particle by the angle

= tan-' [77 e-2vh (va)4+ ...],

again a quantity of fourth order in Pa. The quantity is plotted in figure 9.

Two points may

3- (1) Since there is '1! vertically. In such

at all, and so we

restrained from res - a steady drift of w

4-4.- expected to produ

zontally. However after it operates fo:

velocity of the cylii include a steady se

the other quantitie

If there were no frE could be used to f

cylinder into the e: cylinder, the comp

c:c tinder present is

7. Approximate

according to the condition, but since h is large, one mig range of a and h. In the forces and a col As before, we ha here now

Then

Similarly, - X'

where Ii(2va) is a ir of the incident way

X1"2

' These are to be cot

order forces are cor

'horizontal steady f The second-order vc

results are shown as

phase lag, ;!/-, prese' 30 10 11 12 8 9 2 3 4 5 6 7 2vh = 1 = 1-0 14 I -:' -Then

Forces on a submerged cylinder 465 - forc( Two points may be mentioned with respect to this description of the motion.

uivitr (1) Since there is generally a steady upward force, the cylinder should accelerate vertically. In such a case it is impossible to formulate a steady-motion problem

w;is

at all, and so we have assumed that by some artificial means the cylinder is is hot

restrained from responding to this force. (2) In a second-order theory, there is

o thi.

a steady drift of water in the direction of wave propagation, and this could be expected to produce a steady force which would accelerate the cylinder hori-zontally. However, such a force must be of higher order than second, so that after it operates for an infinite time it will have produced a steady translational

velocity of the cylinder. Thus, strictly, a description of the cylinder motion should

include a steady second-order horizontalvelocity, but its absence does not affect the other quantities calculated.

7 7. Approximate solution

(a) Restrained cylinder

- If there were no free surface present, Milne-Thomson's (1960) 'circle theorem'

could be used to find the change caused by introducing a restrained circular

4 cylinder into the externally-produced potential flow. If, in the absence of the , cylinder, the complex potential is A e-i("+°"0, then the flow with the restrained

.

cylinder present is given by the complex potential

fi(z, t) = A e-i("+°1) + A evh exp {i[crt + va2/(z ih)]}, (44)

s according to the circle theorem. This potential does not satisfy the free-surface condition, but since the second term becomes unimportant if either a is small or ii is large, one might expect it to yield reasonable results over an appreciable

range of a and h. In this section, such approximate solutions are used to calculate

the forces and a comparison is made with the theorypreviously developed. As before, we have

where now

XT(t) -

(t) = - iap

eie (IV) I,._a de ,

cV)1,..= = Re {Aix/ e--'h [- exp (va e--1°) e-ic4+ exp (vaeie)

Then Xil)(t)- i Y(t) = - 27rpo-a(va) A e- e-1°'. (45)

Similarly, X.).2(t)'-ir,71-; = - 27ripvA2 (va) 11(2v1), (46)

*whereI1(2va) is a modified Bessel function of the first kind. So finally, in terms of the incident wave amplitude I-10, we have

2rrpg1-10 e---'h

ril) (t) i y(11) (t) _(va)2e-i'', (47)

v

; behind

ire 9.

Xi2)(t)t = 0,

/TV;

27TiogH`02 e-2'h (va) I1(2va). (48 a, b) particle *These are to be compared with (31 a), (28a) and (31 b), respectively. The first-_{l_order forces are correct to the lowest order in powers of va. The second-order} thorizontal steady force is given correctly (zero) by the approximate solution.

4The second-order vertical-force expression is only approximately correct. These

results are shown as broken lines on figures 2 and 4. In this approximation, the

Phase lag, ;!ci, presented in figure 3 is identically zero.

30 Fluid Mech. 16 13 Id lin ve -±

-,., =

,466, T.. Francis Ogilvie (b) Cylinder forced to ,oscillate

In § 7 (a), the mathematical expression for the effect of the cylinder is equivalent to a set of multipole potentials, the singularities being located within the cylinder.

That part of the potential of § 4 which represents singularities in the upper half,

plane is neglected. In other words, the effect of the free surface on the disturbance

due to the cylinder is not considered. In the present case, where the only

dis-turbance is due to the oscillating cylinder, the corresponding complex potential is

f2(z,t) a2

+i

rto.)(0_,ko.)(0.1.

zh`

(49)

This is just the classical potential for a cylinder oscillating in an infinite fluid. The only first-order hydrodynamic force on the cylinder is, the familiar added.

mass force,

..X.(21-)(t)

irj)(t) = - rrpaq(1)(t)-

(50)

The second-order force is easily found to be zero. These results are shown by the

broken lines in figure 5. In figures 6 and 7, this approximation yields answers identically equal to zero.,

(c) Cylinder free to respond to waves

The potential is now the .sum of the potentials in the last two subsections, Viz.

f, (z, t) ..7-= A ,e-i(P:÷°0+ A e-Ph exp Mat +,va2/(z + ih.)}1- [a2/(z a)] [01(0 - iijal(t)] with 60)(t) and 7)(1)(t) now unknown functions.. To find the motion, we solve the

equation

7.rpa2(v1) i;ja.))= [xy.)(0, +,./51)..(t)] i[ )11

- 2npAcra(va)e-vh va 2 ( 4:(1) 3.5 (1))

.= -.7p4o-a(va)-e-.-vh e-iat.

We obtain (1),(t)i- i'>(t) -= (4v e=l'h Id) e-icrt H, e-vh (52)

This is identical with the motion of the equivalent water particle in the absence, of the cylinder. We find the second-order steady forces directly from (16"):

XV)(t)''= 1:30, = 2-5pgHS.e.-2vh (va)[I,(2va) - (va)].. (.53..a,,b)'

This result is to be compared with (43). Again the first terms of the exact solution, are given correctly. This approximate value is indicated on figure 8 by the

broken lines.

8. Reftexion and transmission of waves

Dean and trsell showed that in the problem of § 4 (cylinder restrained) there

is no reflected wave, but that the transmitted wave has a phase shift afterpassing the cylinder. This phase shift is readily calculated from the quantities already

'40

-involved in th

, singularity pot

_ These are

subs-777"

(59) then gives

Thus the way defined in &ILL

The fact thE SUggests an in va = 4-0, wher shallow the st: cylinder, none is similar to th and so when it the undisturbE phase lag. If

"of the lac, OCCi

reasonable thE the total trail, phase of the p; When the v: small iubmer&

ould not ex: 'argument rela

Next, for th find the ampli cylinder, we s The procedurE V. into ( 19 ). and 1, Re {fo(z. t)1 -[ -3 ,T H(x7t) 4

--

-yii)(t) = (51)

'

=

I1 0,..77 .I. Re {f,(z, t)} -->- -T

(n-1)!

ev-h) sin (vx -7 crt), .--,--. Hal i, tt): .5 I) 2 c 11, hcrc -Ady

Forces on a submerged cylinder 467

involved in the force calculations. To show this we first note that the basic singularity potentials of Appendix A have the following asymptotic behaviour :

Re {f.,(z,t)) Re {gn,(z, t)} -7T -7T (n 1)! cos (vx T-(n 1) !ev-h) sin (vx Re {g..,.(z, t)} -->-

7-(n -1)!ePoi-h) cos (vx -T- o-t).

These are substituted into equation (19), together with (63). Use of the definition (39) then gives for the wave shape

Ho sin (px o-t) (x + gD),

H(x,t) Ho

_.,[(1,S1)sin (vx+ o-t) 23, cos (vx + at)]

(x>gD).

Thus the wave shows a phase lag of tan-' {2Se/(1 SI)} = ./,, where Vr, was

defined in equation (27') and plotted in figure 3.

The fact that VI!, seems to increase monotonely as 2vh, decreases (for fixed va)

suggests an interpretation of what is happening physically. Consider the case pa = 40, where the diameter-to-wavelength ratio is about 1-3. No matter how shallow the submergence, essentially all of the wave energy passes above the cylinder, none below. Thus when a wave approaches the cylinder, the situation is similar to that of a wave entering shallow water. Its phase speed is reduced, and so when it emerges on the opposite side it exhibits a phase lag compared to the undisturbed wave. The smaller the value of (h - a), the greater will be the phase lag. If we suppose further that, because of the geometrical symmetry, half of the lag occurs before the wave reaches x 0 and half afterwards, then it is reasonable that the oscillatory forces should show a phase lag just one-half of the total transmission phase lag. The force in a sense depends on the average

phase of the passing wave.

When the value of va is quite small, the picture is not so clear-cut, for then at small submergence some of the wave motion occurs under the cylinder. We

should not expect the phase lag to increase indefinitely as h, a 0, but the

argument relating force phase and transmitted-wave phase is still valid.

Next, for the problem of § 5 (cylinder oscillating in otherwise calm water), we

I find the amplitude of outgoing waves. In particular, for circular orbits of the

cylinder, we show that progressive waves are produced in only one direction. The procedure is exactly as before. Substitute the asymptotic expansions above into (19) and use equations (62), with A = 0. For the velocity potential we find

Re {f2(z, t)} o-aS(va)eP(11-10/(1+3){[ 712 +7/18, -T T- E,Se] cos (Lx T- o-t) [ 71j ± 712S t52 sin (VX T crt)}. 30-2 -4 it ,o-e), +

-= +

-

E`iSe]

408 T. Francis [Ogilvie

If the orbit of the cylinder centre is counter-clockwise, that is

97(1),(t) (1)(t - 7Ti2g),,

then

Re ;rif2(z, t)}

e S .(va) eli-h)1 (1 + S"..,=[)[

x

-

772 S,),cos 0.,x[ +[[crt)-(77i 0S }sin (vx +

and waves are generated only to the ,left. If the orbit is clockwise, that is,

na)(t) = 5.9t +1712471i

then

Re {f,(z, t)} 2caS ;(va) ell -101 (1 + [S)

f(

--

COS

(a"

Grt) ₀₇₁ 772SE sin ,px - o-t3

co

and waves are generated only to the right,

This unilateral production of waves can be made to appear plausible in the

following way. If the cylinder oscillates only vertically, then the generated waves

are symmetrical in x. If the motion is only horizontal, the generated waves are I anti-symmetrical in x. If now equal vertical and horizontal cylinder motions. I

are combined, the relative phase of the two components of motion can be

adjusted so that the outgoing waves on one side just cancel each other. But 11 because of the different symmetry characteristics of the two waves, they will certainly not cancel on the other side. It is seen above that the circular cylinder paths provide just the appropriate phase differences for this condition.

Finally, in the case of the free cylinder, we find that again there are no reflected .waves. The velocity potential in this case is a linear superposition of the potentials

of §§ 4 and 5. At both infinities the potential of § 4 represents waves to

the left, as already shown. The induced cylinder motion orbit is circular, and,

from equations (40), the sense is seen to be counter-clockwise. According to the

results above, this motion produces outgoing waves to the left only. Thus there

are no outgoing waves to the right for the combinedpotential, which is equivalent

to saying there is no reflected wave. The actual phase lag of the transmitted wave is again easily calculated. It is found to be

(S - S e1

1 c 1

tan-1

-

(Se - Sc602 y-2,

where7fr2was already defined as the phase lag of the motion of the cylinderand

was shown numericallyin figure 9. Comparison of curves of 0.2 and 1,k1 shows that

for large 1,,a there is practically no difference, but for small va the free cylinder causes practically no phase shift compared with that caused by the restrained cylinder. This is quite reasonable. A free cylinder which is small compared to wavelength will respond to the waves very much as if it were simply made of

water particles, and so it will not greatly disturb the wave motion. A large free

cylinder on the other hand undergoes much less motion that the equivalent

water p ance,ai cylinde: Most. Model I Newm a prograt comput

Appen

Let F (7. The imy Re {I' (z , located equal tc similarl: potentiz Since sin.gulai about Pcz, t) Where If eit: part of the free -, = {( + 7h S) + 0, moving 2C ) -G(z, t) = z

I tin. (,,, s are thins n 'Int will in der :cted ltia to and. o lie there alent itted and that where us. nder ined d to le of free lent

Forces on a submerged cylinder 469

water particles, since it responds to more or less of an average pressure

disturb-ance, and so it affects the waves in much the same manner as the restrained cylinder. A similar situation was noted in the comparison of figures 8 and 4.

Most of this work was done while the author was employed by the David Taylor

Model Basin, Washington, D.C. Personally the author thanks Dr J. Nicholas Newman for valuable discussions of the subject, Mrs Patricia M. Monacella for

programming and carrying out the lengthy calculations on the IBM 7090 computer, and Mr James M. Newman for checking many of the analytical results.

F(z,t) = irloov(z+ih)

k

- h)

dk sin al + {2.7r cos al, (54a)

2,(Z iv

r .-ih)

G(z, t) =

log v2(z+ih) (zih)+2ig

dk' sin o-t {27ri e--19 cos at.

J.

k

-(54b)

The improper integrals are to be interpreted in the Cauchy principal value sense.

-Re {F(z, t)} is the potential function (see Wehausen _{Laitone 1960) for a source,}

located at z = ih, which is pulsating in time with an instantaneous outflow

equal to 27 sin ct. The undisturbed free surface lies in the x-axis. Re {G(z, t)}is similarly the potential function of an oscillating vortex at

_{z = ih. Both}

potentials represent outgoing waves at right- and left-hand infinities.

Since both functions are analytic in the half-plane y < h except for logarithmic

g. singularities at z = _{ih, the analytic portions can be expanded in Taylor series}

about z = ih. The following expansions are obtained, valid in jz+ ihj < 212,: F(z, t) = flog 1,(2; _{A,[iv(z + ih)]id sin o-t +}

_{B[iv(z+ih)]mli}

_cos

(54a')

G(z,t) =Ci_{log v(z +ih) l',L47,[iv(z+ih)]m} sin o-t}

-

iB,[ iv(z + ih.)]7"1 cos o-t, _(54b') A

m(2vh)"z+m! [e-1Ei(2vh) \-1 (2vh1) (772 1), (55a)).7

A, = log ( 2ivh) + 2 e-2'1Ei(2vh); _(55a')

--77" e 9 -vh

B,

m ' (55b) 2vh et' du Ei(2vh)

f

If either F(z, t) or G(z , t) is differentiated n times with respect to z, the real part of the resulting expression still satisfies the_{same boundary conditions, viz.} the free-surface condition and the condition that only outgoing waves exist at

Appendix A

Let Pulsating singularity potentials

&

=

470 T. Francis Ogilvie

infinity. After relabelling these derivatives, multiplying by some real constants.

and adding complex potentials in time quadrature to these, we obtain the following basic sets of singularity potentials:

( eine _(m+n)!

(pry"e-17"°1,. sin at

It is of interest to note the form of Am(2vh) for large values of 2ph. As 2ph

(j - 1)!(j- 1)!

1

\

e-2vh Ei(2vh)

(2ph)i (2vh)5 (2vh)--v '

where N is any integer larger than (m 1) (see Jahnke & Emde 1945). Thus.

asymptotically,

1 1(m-1)! 9 (j 1)! 1

\

Am = m! (2vh)m (2vh)j °(2ph)2v '

as 2vh oo.

Appendix B Solution of infinite sets of equation's

As shown by Ursell, the four sets of equations, (20), can be effectively

un-coupled. Let

_{yrn = ((m+n)!1(n- 1)1)A,,;}

(58)

.11

if m = n,

rnn

₀

_{if m + n;}

= 2e'

(59)

(n-1)!'

where {x} is any sequence suchthat this sum exists. From Appendix A,

0?,h

Bm+n

_(m+n)!

-:=`: Thus the unknov (pal-m (va). m (va)'

Y.+

Let c,, and be

Then, if the set o zero solutions (w

sets of equations

,We then form ti

equations in S.

stituted into the

yin 6, In the special ca see equation (2 (.7, fn,(z,t) =

-'A

(1,7-)" 7n= 0-1 ( 1718 CO t) 4 " m! (n - 1)! (m +n)! (56a)

(56)

(56c) (56d) (57) 7-"m! (n - 1)!(vr)ineime)1, cos crt;

{ m= o (m+n)! "m! (n - 1)!(17)me-im')cos ct i2(z, = 77`+ (vr)n i ei"° . (m +n)! ,11 fl { 7n= 0 (n - 1)! (vrr e-im8) sin ct; (m +n)!

(n- 1)1(1.7)"ae-ini.9}sin crt t) 9711(2,

= (vr) +

mf20Am,,m1 { {2:eine gn2(z, t) = (vr)n ,= 0

-iii) has been expressed in polar

z+ ih (m+n)! iB (vr)"' e-im6)COS 01;

m+n! (1)!

m= 0

mn

-(m + n)! n m! (n - 1)!

()m

0_1m0)cos crt (m+n)! ( E (vr)"'e-im°) sinct. m'n m! (n - 1)! 7n= 0 co-ordinates as

= r ei°) = ire.

m+

-m 1

-+ i

-m

). Thus, 5(-1) (see equation (21)). (va)2m Ym '1. Jr:

co (yarn' e-vh (va)2ra

E -ymn'yn z'8 o-a(va) 8,7,1+

n=i 772.1 m

(va)2mco oco2rn

67,i 1- 1

n1ymnSi, = ' mit 37+ f..1 o-a(va)18m1., Let em and (;;; be defined by

ii _{e +(va)2m ..:-..., Yrnne n =} (varin

m f 9n1 m n= 1 Az-(pa )2771 . .171. .-n = ml 171! 1L-1

Then, if the set of homogeneous equations corresponding to any of these has only zero solutions (which isgenerally the case), linear combinations of the above six (,513,di sets of equations can be formed to give the results:

-S fie,n+ 772 o-a(va) cis; 01 a)

1(57 /3, = ( - S A e-Ph)e,77,-- Ilicra(va)

.01 b)

vh (S 8-+ A e-Ph)e,n- t.52 o-a

(61c)

-

+ 51 a-a (va) 61d)

We then form the sums, S, etc., by (59), which provides four linear algebraic, equations in S, Sfi, Sy and S, which we can solve. The results are then sub=

& stituted into the above expressions for am, ym and am, yielding

traSe(va) (77 + 72 S) + A e'h

972cra(va)

am _{1 +}

cragc(va),( - 772 +77, Sc)+ A e-vh

flrn,=

_1+S

o-aS c(va) (El+ 28e)+ A e-vh

ern, .E2ca(va) ,r4;

1+

-am o-aS (va)11'(,z52 A e-vh _{e6 + 63. o-a(va).,}

1±S

-OS) r.

the ,special case of no cylinder motion,these results simplify greatly to A e-Ph S am = - _{1+ IS!} Cm,; A e-P.h 13rn nt ± 4.3.626M =!' (62a); (62 b)

(620

(624) (6,3 a) C6,3 Lill I

Forces on a submerged cylinder

Thus the unknown coefficients satisfy

II ; 471 ( Va )27n °3' ymn 71= 1 (va)2m o-a(va) r: tr.

am+ nzt

= mt

Sfi+ (va)2m fim + (va)2m S .4 e-vh (yarn' Ymn fin n -1 !? At-t.

nil nica(va) (3mi. + 7n,!

(60a) (60 bo, ,(543e4-5 (504 = -ely Ym 1 Ymn =

= -

m. stains. (56

471 T. Francis Ogilvie

The sets of equations (60a) and (60b) were truncated for solution. First they were each cut off with only ten unknowns and ten equations and the desired

forces were all calculated. Then the procedure was repeated with twenty equations

and twenty unknowns and the forces recalculated. In cases where hla was only slightly greater than one, there was generally some discrepancy, and then the

procedure was repeated again with forty equations and forty unknowns. For

the results reported in the figures of this paper, the last two calculations agreed to at least three significant figures.

REFERENCES

DEAN, W. R. 1948 On the reflection of surface waves by a submerged cylinder. Proc. Camb. Phil. Soc. 44, 483-91.

JAHNKE. E. & EMDE. F. 1945 Tables of Functions. New York: Dover.

Mn...NE-THomsoN, L. M. 1960 Theoretical Hydrodynamics, 4th ed. New York: MacMillan. SroKER, J. J. 1957 Water Waves. New York: Interscience.

I:ESELL, F. 1950 Surface waves on deep water in the presence of a submerged circular cylinder. I. Proc. Camb. Phil. Soc. 46, 141-52

WEHArsEN, J. V. 8.7 LAITONE, E. V. 1960 Surface waves. Handbuch der Physik-, Vol. L.N. Berlin: Springer-Verlag.

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