UNCLASSIFIED AD 295. 094 · Timman [3-], has made an attempt at solving this problem. In his work...

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1295 094,

I TECHNICAL RESEARCH GROUP2 AERIAL WAY 0 SYOSSET, NEW YORK

REPORT NO.

TRG0453-SR- 1

Submitted by:

TRG, Incorporated2 Aerial Way

Syosset, New York

LINEARIZED THEORY OF THE UNSTEADY NOTION

OF A PARTIALLY CAVITATED HYDROFOIL

Prepared under Contract Nonr 3434(00),pponsored under the Bureau of Ships

Fundamental Hydromechanics Research Program,Project S-4009 01 01,

technically administered bythe David Taylor Model Basin

Author: Approved: K-

Submitted to: AST"A

Commanding Officer and Direc nDavid Taylor Model Basin 3

Washington 7, D.C. 2.I

TtSIA A

Reproduction in whole or in part is permitted

for any purpose of the United States Goveruent.

March 1962

TICHNICAL RESEARCH GROUP

TABLE OF CONTENTS

Page

List of Figures --------------------------------------------- ii

Acknowledgement --------------------------------------------- iii

Abstract ---eeeeeeeeeeeeeeeeeeeeeeeeeeeeeiv

INTRODUCT IONeeeeeeeee---- eee-------- ee------- eee------ e1

GENERAL STATEMENT OF PROBLEM -m ---- m -------- meeeee-eeeeeee4

STEADY STATE SOLUTION ---- m--------- Cc -------- e e e e e e e e e e9

UNSTEADY CASE ----------- m ------ M --- ft -- -- eeee eee eeee eee13

THE FORCE AND MOMEN ON THE HYDROFOIL-(UNSTEADY CASE) ------- 25

NUMERICAL ANALYSIS FOR UNSTEADY SOLUTION --- c---e--c ------ 30

APPENDIX ' * 30 03 * 40 04--m e -eeeeeeeeeeee- e eeeeee36

APPENDIX I I In 1 (a) , Jn (ax) c---e- e ee- en n ---- 64

APPENDIX III -Steady State Integrals --------------------- 67

APPENDIX IV -Condition (4) Integrals ----- e--- m ------ m---- 68

APPENDIX V -Force and Moment Terms -- ee-----------~ ec 75

~'k 'okAPPENDIX VI - *ko Ok' Ta 1 z- Tabulated --------- m ------ 94

B IBLIOGRAPHY e----- m ---eeeeeeeeeeeeeeeienc m - -100

TECHNICAL RESEARCH GROUP

LIST OF FIGURES

Page No.

FiLgure 1---------------------------------------------------------- 5Figure 2 --------------------- 26


ACKNOWLEDGEMENT

Acknowledgements are due to Professor Samuel Karp

for his assistance in clarifying the procedure for the

development of these results, to Messrs. Edward Grenning,

Edward Friedman, and Jacques Bresse for checking the calculations,

and to Mrs. Thelma Bourne for typing the report.

iii

TECHNICAL RESEARCH G.ROUP

ABS2&AC

A linearized theory for the unsteady motion of a

partially cavitated flat hydrofoil in two dimensions is carried

out. A second linearization procedure is ~sed, based on ideas

of Tiuunan and Guerat, to obtain the unsteady pressure distribu-

tion around the hydrofoil and the resulting force and moment,

as functions of the cavitation number and Strouhal number

for given pitch and/or heave.


NOMENCLATURE

normalized pressure (or acceleration potential) -

-a- , where p is the actual pressure and d is the density

a normalized cavity pressure. (cavitation number - al U2)

G Green's function for regular term for pressure

I one-half the length of the hydrofoil

c position of the rear end of the cavity (assuming hydrofoil

between -A and 1)

a velocity potential

U free strean velocity

v oscillation frequency

a ZA (reduced frequency)

M(t) slope of hydrofoil

B(t) y intercept of hydrofoil

f force perpendicular to hydrofoil

m moment on hydrofoil

v

TECHNICAL RESF.ARCH GROUP

10

HYDROELASTIC EFFECTS OF UNSTEADY MOTION"OF A HYDROFOIL

1. INTRODUCTION

In order to analyze the hydroelastic effects of

unsteady motion of a hydrofoil it is necessary to determine

the pressure distribution on the hydrofoil. The problem of

particular concern here is that of a partially cavitated hydro-

foil, experiencing simple harmonic oscillations in vertical

position (heaving) or angle of attack (pitching).

In order to solve this problem, it is first

necessary to determine the flow pattern around the oscillating

hydrofoil. Furthermore, in order to produce a solution rel'atively

quickly, it is desirable to make certain simplifying assumptionso

First a two-dimensional flow pattern will be assumed, i.e., the

hydrofoil will be assumed to have infinite Qpan. Secondly, the

fluid will be assumed infinite in all directions. Third, thickness

effects and initially, cauber effects will be neglected. Moreover,

the angle of attack will be assumed sufficiently small that

linearized theory may be used. Finally, the unsteady motion

will be assumed to consist of small oscillations around a steady

motion so that a 'second linearizatono may be made. Therefore,


2.

the flow pattern is assumed to fluctuate at the same frequency as

the hydrofoil oscillation, i.e., higher harmonics are neglected.

Timman [3-], has made an attempt at solving this problem.

In his work the assumptions made to complete the definition of the

problem have consequences which are physically unacceptable.

Guerst [l] has formulated the problem in such a way that this contra-

diction is resolved. However, he has not carried through the problem.

to any extent. In. our treatment, Timian's analytical machineii is

used throughout, but the physical assumptions used to determine

the unique solution are those of Guerst. Although Guerst's set

of conditions is not beyond question we felt it is the best

possible in the light of present knowledge.

To solve this problem, it was first necessary to obtain

a steady state solution by Timman's method. This was done, and

the result compared to that of Guerst (2]. Preliminary calculations

indicate complete agreement, except.. for discrepancies which can

be attributed to numerical inaccuracies (of less than 17) in

Gueratts calculations. Tiiuan's approach, as developed here,

gives much simpler closed-form expressions for the pressure and

velocity potentials than Guerst was able to obtain.

The next step in solving this problem was to develop

an explicit mathematical nepre'entation for the physical assumptions


3.

made. This was done, and resulted in six simultaneous linear

equations, involving six unknowns. The coefficients are functions

of reduced frequency and cavity length. Some of them are defined

by integrals which must be evaluated numerically. To do this a

program is being written for the IBM 7090, which will compute

these integrals as functions of the two parameters. The expressions

for pressure and velocity are simple combinations of these

coefficients and elementary functions.(see Section 4)

Finally the lift and moment of the hydrofoil were

calculated as linear functions of the above mentioned coefficients.


20 GENERAL STATEMENT OF PROBLEM

Timman has derived the general expression for the acceleration

potential t 1o be:

= - -i-f f(x',t)G(x',o,x,y)dx' + cl0I+c

2v wetted

surface

(see Figure 1).

In this expression c1 and c2 are functions of time which are

determined, together with the position of the rear end of the cavity,

by additional conditions given by the physical model used.

If h(x~t) profile of hydrofoil

then w(xt) ht +Uht x

and f(xt) wt+Uwx

For a flat plate, the above integration may be done explicitly. (See

Appendix I),

Let h - M(t)x+B(t) 0

Then w = M'x+B' +MU

and f = M"x+B"' +2M'U

Therefore

It -0 4- A 2 MF 2 - 1.(B 4 2M'U)F 1 + c 1 + c2 2

M". (B"'+2M'U), c 11 c2 are functions of t only. Fk' #k are harmonic


PlYSIC.L PLANE 5.

yflow

U =velocity

cavity c

hydrofoilz X'-Liy

IMAGE PLAN

image of end of cavity leading edge

cavity image

trailing edge imageimage

w /\2 2-a

r A-qt x axis image

4-i

ainJn

Figure 1

6.

functions of x and y in the plane slit along the hydrofoil.

Let ~-/f

where z - .x + iy

and - + .in

21 - length of hydrofoil

c - x-coordinate of rear end of cavity

Fk' *k (see Figure 1) are harmonic and rational functions of t and n

only, and are time dependent only through o.

The solution of the general problem requires specifying

three conditions to determine the three parameters Cl, c 2 , and a.

there are four conditions which may be used:

Boundary conditions (general)

(1) *(-) - 0

(2) 1y - w(x,t) on wetted surface, where I - the velocity potential.

(3) h(c,t) - h(-At) - f c 0I(X,O+,t - c dx' (closure)

(4) 1 y(X,0+t) + V y(X,0-,t), x > I (continuity of verticalcomponent of velocity)


7.

Conditions (1) and (2) seem quite satisfactory on physical

and mathematical grounds, so that the choice lies between (3) and (4).

For the steady state problem (4) is automatically satisfied so that

(3) is the only condition left. However in the unsteady problem,

the physics of the problem seem to us more reasonable if (4) is used

rather than (3). This is the view of Guerst, Wu and others, and this

approach is used here.

Using condition (1), and following Timman, we write

I=1 x 4u f + x Ck~ -x $~k k( )dx'

where

c4 = jt'2M", c3 -(B"+2M'U), $3 -F 1, *4 = F2

x 1 x 41 M,... dx, k f E dx'-k = kkky - k-l

But

4'k d*k 1 21k d ( t - - )

T4NICA R AC kc~~~ 4xk.l 'k)


Therefore

f= c( k cst( -Y()dxj

y Ck(t)k(x,y,(t)k()k(X'

The harmonic functions are listed in Appendix VI, Table I

where those with subscripts 1 and 2 were derived by Timman and those

with subscripts 3 and 4 are calculated in Appendix I.


9.

3. STEADY STATE SOLUTION

Since c3 M C4 - 0, we have

- + c1 1 + c2#2

y-U (c,* + c2*2)

The boundary conditions become (1, 2, 3)

(1) ' CLA' - c2A +'a -0

(2) MU2 + + c2B - 0 ( - 0 on wetted surface)

(3) u2M(cj) c - (I - ,)da c2 + (-B)dz

where A and B are defined in Appendix VI.

If we substitute (2) in (3) we find

CC(3') c1 f - dx + c 2 ,f dx 0

Since x - £ 24)2

1+(t2 ) 2

and 1 - 0 on the cavity,


10.

(31) is replaced by

01 f dtd +c 2 o0 d -

ora* t2 + Q dt 2 U2 4242C, f 2 dt 2 22 dt 0

0 [l+(Q2 -) 21] 0 (1+ 240) 2

The integrals are evaluated in Appendix III and we obtain:

(3) (ja/+1+ 2) - + ( V/o'+l) C2 O

The solution of (1), (2), and (3*) is as follows:

Let p 2U2Ma

Consider the-.equation y3 .Y + p - 0

1T'hen a = (-y

C - y+

-p+i /4- P2 2

Let WT 2 0p2


11.

Then Y W AI/3 +

The only allowed roots are those which give positive y. In general

there are two solutions y1, Y2 where 1 ? yj > 0 and1

1 Y2 k r The only solution which is physically meanl gful

(see (2]) is given by yl.

Special Cases:

1) p2 4 y (apparently the maximu angle of attack for which

linearized theory will hold.)

y .. (2 positive roots identical)

I

c

.75 (cf. Geurst value of .7485)

2) p = 0 (no angle of attack)

S c

0 -s (no cavity, physically real solution)

1 0 A (cavity full length; anomalous)


12.

Since y is only an intermediate variable, it is of interest

to express everything in terms of a.

By solving a - - y), we obtain

Since p y y 3 we have

c 1 - -oB( +2) (. a+l - a)

C 2 M vB(q/0a21+1 + 1 - a2)

C 2 2

S-- -+ + )2

-u2 (v'r 5 + 2a +0

Therefore

C . U (a2 +1)(a+ 4o+1)

EcN C= E2CMB GU2+ P


13.

4. UNSTEADY CASE

The general expression for the velocity 1y is given by

= z cfk(t)*k(X'Y$Q(t))kf

1 d x xyxG t x )dxfU dt f :.Ck(t U ,)k ,yY(t-U"

Assume the time dependence is given by a small simple harmonicoscillation

around the steady state

= o + a 1 cos(vt-e0 )

ck = Cko + Ckl cos(vt'ek)

where a0, c10 , c20 are the steady state values.

Let M(t) = M° + M1 cos vt,

B(t) = B1 cos(1 t-OB)

where Mo Ml , Bl , 0B are all given quantities.

Then M'(t) = - vM1 sin vt

At M V- V 2M1 cosvt

-2B"(t) - B1 cos(Vt-0B)

c 3 -Vi(vB 1 cos (vt-8B) + 2MlU sin vt)

- vj(vBlcoseB cosvt + ( 2 M1U+vBl sin OB) sin vt)

1(vB)2.+ 4(M U)2+ 4vBM sin 0. coo(vt-tan'1 (tan + M1V 1VENU+ A E coSeRU))


14.

C3 0 -0

31 vsV(V3J ) +4(M1U) 2 +4VB U &in 9B

63 ta 1 (tan 0 + U

iB'Ecos %C4 -- 2 J2 M1 Cos vt

C40 0

a41 2 . 2Ml

94 .0

The problem is to find aI, C119 021, ell *20 ' *

We now carry out the second linearisationt

(Aover a symbol denotes the perturbation in that quantity Siven by the

symbol without the A)

Ck(t)#k(xzym(t)) - CkOk(yo) + CklCOs(vt-tk)#k(zLy .g.)

b k

ECHI RS(xyAC )ocoU(vt-9)


- *I 'P

15.

+ U Z k os (Vt-ek)*k(x'yt ) + kO yi Cost - )kni -0

V

x

U -jck sin - - ., ) k ('Y o)

+co# (xI s)a sin(v(t- dx']

The above integrations are carried out explicitly for the cases

where y - 0+ or y = 0", thus giving the vertical component of the pertur-

bation velocity along the hydrofoil. The steady state term has been

derived previously. Similarly the following equation gives the pertur-

bation term for the pressure distribution along the hydrofoil when y - 0+

or y =0":

A 4 k, (c ( kl co p (Vt -9 k)'k(x~'' o) + k0 -- al cos (Vt-ee)ak-l 0

6of lF@IP2 and are given in Appendix,7II, Table 1.0

of 03,l3p,4 and ?p are not needed since c3 0 -c 4 0 -0.

The boundary conditions (1), (2), (4) of Section 2 are explicitly given

as follows:


16.

Condition (1)

4z c k1COS{,vt-9k )4kc) + kO a k acos(vt-e a)) -O

k=1. 0

where the values at co of and, are given in Appendix V1, Table I..

Condition (2)

w M'x .+ BI' U(M-1M )

-xvMl 5 j0.- Bsair fVT-.) lJMc - .IyP B 1.

-xvM~ s -'r Vt + Um ±vB1 sin -,B) cosv-t -vB cose sin vt

~(X sL i-t -+/ t 4 o L -t I*

Therefore Condition (2) become~s

Uc,, sin + K o (vt-e3

4 x-XZ.3Cl1~ ~ Cl3~tk c~ Oa)sin(v(t- 'j)-V~dx 1

2 c)*k

+U -'k1 " U)ky-k, 0.

c -?k X',0,a )a Sin (.1"t- X.x)e)d0


17.

2 Xlie)x Ckl*k (OOOo) rcos(vt-e k-Cas(v(t---=)-8)

+ C kO ; (O,90o0)a, [Cos (It-%a)-cos'(t -24-4).9)

k c~ are constants along the wetted surface for kml,2].

This may be-simplified to

k 3 k kk + I sixiov t) UFU k)*k(x 0 )d)

2 Uk

Ick C- (x'O Czsin n.v(t dx

+ 7k-kO, U)O(~ kl±k 0) k

Uc1 -Cc

+= - aj- (j x in t + c~v~t) +.-x , sin (vtx1'k1'i

+Let (ks 0- PC ) Ckx o sinV (x'+9) - k t'

k1~ kU 0 U ko8 %)cos( 0(0 a Ocdz*'C *

-kc (As dein t usin Co t + ,sn v

TEHNCA AESEARC 3ROU

These integrals must be evaluated numerically.

The tanid -needed are listed in Appendix V1, Table V (in the form

6;k. 0 *

* ~ and the non-constant parts of anrd - defined in Appendix I)..0

Let

p - V(t - ),

x

W-f sin(p + - e3) ,(xo~l)dx

Ao~ + )(xe I+ Vdxtx

sV + 3) x'+ +71Uf cos(P + - 83

"+*-) -2 0U j~)csv- 3)+2 2 1

Then

x X

f sin(p +- X

= W *~3 (-2O_1 o)(C05(P + - 03)-O( - -3s

U -)cos (Vt-03)+ U ~.sin(t-9 3 -invt LU~~3j

+ UX+,SV*3 (-AIO-90 0)cos(v('t - j- 3)


xLet T =f sin(p + vx' )dx

2f sin(p + x'+A 2 __2

ucos(p + -w [(")2 L Xl~

v 229

+ 2 - COS(P + vlXJ

=1 U 21r x 2 2-A1

2 Cos (p + EX-j) -r

-+ 22Xlx

v I~ + ) -j sin Tp + T

T2xsn]+f i~ x dxl)]XI -2 -1

f x ws in(p + -)+ Asin (p -~ i + (csp-M) o~A v Co Up

Note that T -Cos (p + E-) *

Therefore

x

T -T )((AO Chos (p + 4x-csp!)


20.

U -_U 2 x U= - 4 (x,o ,%) cos vt + -- (_ sin vt + C7 Cos vt)

u 2 ) I+ -- (sin(p - Cos - cos-( + K (-Y,o- )%-c>v2A

Condition (2) then becomes

where s = t -U

0 Z Ckl [(Jks+*k(,o ao))COS(vSe k) + Jk, csin(vs-ek)]

+ aick0[(Js + k (-Io ,ao))cos(vs-e )+ Jksin(vs-9

-c -i(s9 +c Usnv) ( U) cos(vs))31 V 3 41 - n

Condition (4)

0= f cklin(v(s + U ) ek) [ k(x I,o+,%)-Pk( ' ,ao',1]k=l -1

x'in(v(s + .U+) ea 67P (x1 k +",oa)I dx'

Integration along the wetted surface in the z plane is equiva-

lent to integrating along the q axis ih the C plane. Similarly integrating

along the cavity surTace in the z plane is equivalent to integrating

along the g axis in the plane. Therefore (4) can be stated as follows:


21.

4 0 x'+ ( x+1. )*. Ixo

SCklsin(v(s + k ,)+ckoli7Jn(v(s + -'- d

1 0 X1 a~ '

+ c QklS.l(V(6+ +'),.a xto U0"

, Let k - non-constant part of *k' Since the above equation is

satisfied by the constant part, it is sufficient to consider only the

non-constant parts, which are given in Appendix VI, Tables III and IV, for the

wetted cavity surfaces respectively.

On inspection of these tables we see that along the wetted

surface # k and all equal 0 for k-l and 2. Furthermore for k-3 and 4,

*k on the wetted surface are both polynomials in x, while on the cavity

surface are equal to a quantity defined in Appendix I, plus the

same polynomials.

Therefore condition (4) can be simplified to the following:

0 = X f cc sin(vs )k lsin(vs 0 + (2 ) dj1 - kkPin-;Tck&ps. ] 2 "

where 1 = .(reduced. frequepcy)

P = + Q2 + a)2

and is replaced by * for k-3 and 4.


22.

LetIk,c ' f 0kCS) 2)( 2

Iks =00 ksin(2P) (Q2+a) d t

where is used instead of for k-3 and 4.

Let Ik s be the corresponding integrals involving 5--.

These integrals are reduced to infinite series in Appendix IV.

Then condition (4) can be stated

X ckl(Ikcsin(vs-9k) + Ik,scOS(vs-Ok))

* *+ lco(I* sin(vs-e) + I* cos(Vs-9 )). 01 °ikO( kc a k's. a

Since the conditions must hold for all time, each condition

is expressed as two linear equations, given by the coefficients of the

cosine and sine of vt or vs, respectively.

Let dk,c w cklcs Ok k - 1,4

dk s " CklSinA Qk

0 = 01co s ecs 1 sn Q


23.

Condition (1) becomes

4 C 1I d+Cko --1 Lk~c k~ o 0

4 (

x ssO~o + ckO (;? aJ -o0


x dk c +d ( + ?k(Io a )) + ck J* a~ ~ic dk~(k,s k ''0 k k,c c

+Ck c J + 'r (-I~o,a ))a]

4,

Jks k,c±+ dk,c(Jk,s + *k-~ ,ao))-CkO ik,c as

+ cko(Jk s + 0~-( A 0 % o)

+ 77d3 ,s - 77 c4,s + (V7 d4,c 0


4**

1 fdkcIk,c + dk,slk,s + CkOIkca + C~ks

x dk$6ik~ +d kclk,s -c kOikc 8 + C kOIksal


24.

After solving the above set of equations for dl,c, d1,89 d2 ,cs

d2 ,s' ac and a , we then can explicitly write the unsteady pressure

term as:

A 4kSZ dk'c'k + cCfko -a. ) j CosvtLkl 0C'

+ 4 C) 0 k

+ k: dk,sk + cas(ckO -) sinvtk-l 0

From this the unsteady lift and moment can be obtained.

These are explicitly developed in Section 5.

Since the closure condition was dropped from the calculation

it is of interest to obtain the cavity shape. To obtain the horizontal

component of velocity along the hydrofoil, needed for such a calculation,

we would have to evaluate certain integrals which are not otherwise

needed. Therefore the solution of this problem is postponed.


25.

5. THE FORCE AND MOMENT ON THE HYDROFOIL (UNSTEADY CASE)

Let g(xt) - *(x,Ot)- *(xO+,t). Since * - p)/d,

where d is the density, the pressure difference at each point x of

the hydrofoil at time t is given by -dg(xt). Thus She force f normal

to the foil and the moment m (around the leading edge) are given by

(see Figure 2):

f - -d f g(xt)dx

m - -d . (x +A)g(x,t)dx.

The lift - f cosP, where 1 is the angle of attack (given by cosp I ).

The moment m' around any other point x0 is given by:

m'(xo) - m (x0 +A)f.

ASince 4 = 4=0 + 0,

Aand since 0 can be expressed by

A* = *ccosA + i8sinvt

where 4 kc E dk ,c'k +O(CkO k

k-i '0

s '1ksk + as(ckO S-)

it follows that we can similarly resolve g(x,t) and the resultant f and m:


26.FORCE AND MOMENT

lif t

pivot aboutwhich themoment is taken

Force ()anid moment (in) are positive in direction of arrow.

Vigure 2

27.

g(xt) = go(x) + gC(x)cosvt + gs (x)sinvt

f - f0 + f cosvt + f sinvt

m - m° + mccosvt + msSinvt

For problems of heave alone or pitch alone, f is the magnitudeC

of the unsteady force in phase with the displacement (assume for heavealone, aB - 0), while -fs is the magnitude of the component of force in

phase with the velocity. Similarly mc and -m are the components of the

moment in phase with the displacement and velocity. We have

2go - k~l ckO(*k O)

k-l ck c -

9+ + ac - oo

a~~ dkS( 0+) +- s ckOk=l s'o 0

z+Let * = 2 k - *jk)dx

-2 f2 (x+1)(,0- - *j+)dx

and let R , S * be defined similarly using k instead of The integralsk k inta fk Teiterl

above can be considered as litte integrals (in the '(x,y) plane) ovek the

surface of the hydrofoil in a counter cl6ckwise direction. This is

equivalent to integrating (in the (4,n) plane - see Figure 1) from -aoto 0


283.

along the axis and then integrating from 0 to - along the 4 axis. Since

all k and ;a- are zero along the g axis, Rk. Sk, R and Sk can be

expressed by

R Rk 2 0f d k(O ld

0 3' (k0 ,T) i lk l

Ski (X+ t )*k( 0 ,9,n - diak d d-m

Rk and Sk are found by replacing *k by B0 in the above. Note that allI k dx-k' and d are odd, while x+ I is even as functions of T1. Forpurose o£ 0 neIpurposes Of integration:

1 '( -c 0) + 1

d cx. -8 1j(r1 -C ,1),01'2 " .. 00; . + 1) 2

The integratilons are carriLed out in Appendix V.2

+i. ' k-l

2

k-l

TECHNICAL RESEARCH GROUP'

29.

Then -d2

fo = 2 F ckORkk=l

-d4

k=l

and m, mc, and ms defined similarly using Sk and Sa instead of

Rk and R . fI mo R , and S are calculated in Appendix V. The steadya 0a a

state terms are:

fo ="rdU2M (1 + F )

m = - 'd'A2UM (4+ 2 a( + a)2 )o ( 2+i3/2

J


30.

6. NUMERICAL ANALYSIS FOR UNSTEADY SOLUTION

The "I" integrals may be evaluated using the results of

Appendix IV, while the "J" integrals require direct numerical

integration, using Euler's methocd for increasing the rate of con-

vergence of alternating series. Given these integrals in terms of

a and 0, wne problem is reduced to solving the system:

Qv = q

where

Q Q1 Q2

and Q and Q2 are each 3 x 3 matrices:

40 (00) 0 2(W

Q= 1 *J1 (W) J 2 s- 2 () 0 *a(=)

lIs 12,s Ias

0 0 0

Q2 l ,c J2,c

I ,c 12,c IcaJ


where 31.

() - 1 2()a ckO adUMD0 1

() 1 2 Ok(00)=pa -= Z ckO ad

1 *ac = z c kOJk) c

Similarly J ' , and I are defined in terms of J I* The * * * k~ s kc'

and Ik,s* The vector v is (d15, d2 , as dlc, d2 ,.0 , aa) and the

vector q is given by:.

di, s~k(-)dkl k -

kc(Jk, c + a + k,s(Jk,s + -*k())

4 dkclkc + dk slksq- 2k-3 *,=()

3-k kc(ks + - k(.))-k: ,Jk c + I

THc ks Eks kRc


32.

where

UNM

d d ksk, sUNo

Since any solution to a problem of pitching and heaving can be

expressed as a combination (taking into account phase) of pitch

alone and heave alone, it is necessary to find only the solutions

for unit amplitude pitch alone and unit amplitude heave alone.

Let qp and qh be the unit pitch and heave vectors respectively,

given by M, - M° for pitch alone and BI = m° for heave alone.

For pitch alone:

c3 1 - - 2viM U - - 2n(Miit 2)

93 = r/2

c4 = - v 2I2 M1 . a 2(M1U2)


33.

Therefore (for MI - Mo)

+3 +- 0 3 ~c)

d * 023,cs

d 3) 2LB - - a2nU

d4,J c =_

d*3 - 0

d 0TS

n 2(j4, c + 1i) + 21l (J 3, s .3 3(0)

n 2 14, c + 2M 13, s

qP 02 04(')

n 2 (1 (c) - )~J4, s 77 *- 4( ) "2J3, c

a 214,s -2al3, c

For heave alorie:

=.V2 ..~ a 2( 1U2)c31 TB

9 3 -0

'41 -0


34.

Therefore (for BI= IM )

d* 23,c

d* = d* =d* =03 s 4,C 4,s

0

~20-. 2(J3,c

2- .l213, c

qh = 2

(J3, sa23,s

Let

vs = (dl s d2, a)

c = (d2, d2,ce)

and let vSIP V be the unit solutions for pitch alone andc,p

Vs,h, Vc,h the unit solutions .for heave alone. Let vs,h, vh

be the unit solutions for heave alone and e3 = r/2. Then

/ch = Vs,h and V, h = v c,h . The unit solution for heave alone

at an arbitrary phase 0B is given by


35.

Vs= VshCOSB + VchSin

vc , VC,hCOS B Vsh B

Therefore for unsteady pitch - MIcosvt and unsteady heave - Bicos(Vt-qB)

M1 BI

v rnv cos6 ins M 0 'sp + - (,h B + VhsOB

H1 BIO- + CO)


36.

APPENDIX I - 'P' 3) 40 04

Calculation of 4/31 031 44 04

A. General

0 f k--lk-2~dk 2rI - wettedsurface

Where G = Re(In(L. LUZ )

2-zcand - +Z - a

Along wetted surface C' is imaginary

and x = 21 2l+ (u 2-a)2

Therefore G =,I(u+C . ui-) since argument of In is apositive realu-ia c i

number. Ui

As a result, G = F(Q) + Q

-where F(Q) = In( )~tu-ic"

Since as a function of C~, the 0kare the imaginary parts of an analytic

function of C, where *kare the real parts, each *kcan be expressed as:


37.

k 21 (k " k)

Let Ht) 2iF(Q)

1Then G(Q) (HQ)

Therefore f - f xk 3 H(Q,iu) dxkri 2k wetted

surface

i x k-3 1n(U+iC)rlk-2 fu-ic"d

where f *k i~okbeing determined only up to an additive constant,

so that k -w 0

The wetted surface of the hydrofoil corresponds to the interval

f0,) of u,

-i 0 k-3 dx .uiTherefore f' f x k - x u i du

k ri 0

- (xk'2- ( )k'2) u+ 1(k-'2) ,rek'2 '--- j

i 00 k-2.+ i f°°xk'2 (_I)k- (u- 1 1-- du+(k-2)TrJk-7 0oTC U1

At u , x =-, in ("+- ) f 0u-ic

at u . x , a- 2 .n( u+i) ri

Sin ce is only known up to an additive constant, the first term for f

nvay be dropped.


38.

Let *k be the non-constant part of *k"1. 1 2ui+_

Since - - - , the integrand is even andUTiC u-ic

fk f f " above constant) can be represented by:

f __ r 2 k-2 _lk-2) 1 1 dk 2(k-2)r f +- 1) -(1 _ du

In particular:

3 f 1+(u 2 u2-_0) du

ff= f ( 2 2 2 2Ju.iu )du4 -L l+(U2 -2a)2 1+(u -a).2

Let t = A+Bi, A and B positive.

2where t = a + i

Then 1 + (u 2-) 2 = (u 2-t 2)(u 2-t )

Since C = +in, iC = -q+ie, where t > 0.

Therefore for purposes of determining residues to evaluate f3 and f4 by

contour integration, the upper half plane poles are t, -t, and iC.

Since 1 1 1 1(u2.t 2 ) (u2-t 2 t2.E2 (u2_t 2 u _

1- (_1 - 12 2 2_2 -T-- 2 )

T C Ou-t


39.

f4 3 2 f ( 2 2 +-2 duooQU -t u -t

f f - (. l ,1 -12 1d14 - ~ ~+ 2 2 2- I ici ~ d-. u'-t2 u2 (u -t )(u 2-t2 -

-i11 2 1 1.u u2.u 2 i

In addition to the general expressions, the following special

cases are desired

1) at x =-*(or f=B and T=i-A), * and ,

since the desired ?p is 0 at this point and the function must be adjusted

accordingly.

2) at = 0, (the wetted surface) * and

3) at i = 0, (the cavity surface), , and separated into odd and

even functions of e,

4) 12 2 +,(for Jk c and J integration)

B f3 Calculation

f11 l1 1(1 1 1 du3 2 2t ut u+t) 2E u-E u+t ui u-it

1 2 1 R -2 1 ]t~i~ i~t J~r

t F (t(-iC) ( t+r)


40.

t+ic 1 2 I A+Bi-Tn+it 2= (A-11) 2+ (B+t)2

t+i~t 12 .-iTli - (A+Tl) 2+ (B+t)2

t(t+i7C) -(A+Bi)(A+rn+i(B+4)) - AA,)BBt+(A+tBI

- al + A1I-Bt+i(l+At+BTI)

-(A-Bi) (A-Tj-i(B+t)) - 0 - A I-Bt-i(l+At-B1)

Therefore f 1= At- -i ! -B S+~(A-n)2 + (B+t)z (A+rT) + (B+t)

+ q-&T-BE _g______-___

(A-I) + B~t 2 (A+TI)2 + (B+t)2

Therefore ~- 1 1 +*3 (ATI)2 + B + 7t (A+YI) 2 + Bt2

22 2 221"- '1) + (B+t) (A+ij) + (B+t)

3 a-BE2 1 2- 'AT 21 2~ (A-TI + (B+t) 2 ( +Tj2 (B+t) 2 "

cl (+)1 (B+T)2 + (B (+t) 2


41.

Special cases

1) 3 + 1 +2) *3+ \4A+B2) 4B} 2 ~ 4(A 2+B 2 4B2

1 1

2(o2+1) 2 +( J'I -2 )2 c, + -1

= a 2

2(a2+1)( _ -a)

2 ''27

a, + + V,+

- +

21.7+1 4 f2+1(-

a 12 +

2(a +1)

V,;2

2(o 2+1)

1) (_____B1 q) 2 + -B2 22) +3 B2 + (A-r) 2 B (A+*q) 2i * B + (A-q) 2 B 2 (A*-n)

2(B 2+A 2+T2) - 4ABTI2.

lk,7 ((A2+B2+12) 2 - 4A2 2)


42.

273 2 2 1,4 _.2 2

2 x+j2 -a) +1

B +(A-n)+ 2 -2 A+ 1 7+- T-- B)+''

r 2- (2) 21) (4A, + 2A (P IOY + 11))

2AT-1(2a - . 21-T12)

1 2 2+1)

3) 2 (1-40 2

((B+) 2+A2)

2(1+At) ( 1 + e2 -2BQ)

= 2 " + t 2 +2B) ( + 2 -2B)

2 2A( 2 +1+ - -U+)

2 2 E

S(4 4-a 2 + I +2c 2 ) (( 2) 2 +1

C) 3


43.

where *• Q 2 +2 -h 3 [o a27+(1 + (e2 +c0),2

4 )Let D = ((A-TI) 2+ (B+t) 2) ((A+n) 2+(B+t) 2) (A2 -TI2 ) 2+(B+t ) 4+2 (B+ 2 (A2 + 2)

Substitute t 2 + 0 for j 2

D - (B+t) 4 + 2(B+ ) 2 (A 2+2 4a) + (A - - )

B4 +4B3 e +6B2 e2 +4B 3 +t 4 + 2(B 2 +2Bt+ 2 )( 2 + A2 )

+ A4 +2+ t4+ 2 jJ2.2A 2 A 2t2

c+l - +4Bt(o+ + (6B2+ 2o+ 2A2+2B2+2a-2A 2

+ 8BJ 2 + 4t4

=1 + 4BWcs Y+ 2 ~(4 clyJ) + Ut3+ 4t 4

2 ((I+At)( a2 +1 + 2t2 +a+2BE) - (e 2+a))

'(1+4(o + )2+i) + 4t2(10 +j) + 8B 3 +444)

2(2e2+ +1+ 2A,r1-ca)E +At(a +1+o +2 2 ))

T (1 + 4Bt (a + P2T++ + 422+ + 8B3 + 44

2(2t 2 + I7Y + At(3 4T1 - a) + 2A3)

TECHNICAL2 ESARt 3 t4O


44.

2 (k12 2 t-2-. 2 2f -

(1 + 4t2 T + 44)2 - 16B2 t2 (o + Z= + 2j2)2

2 = 4o2 4t4x+ 2

(1+4+2 0 + 4t (4+4 + 42. 1+4( - )+2 + 2)2

, (4t4 + 4t2 - 1)2 + 4(44+ 1 ( 1 )t2 + 1

+ 4(2( - )02 + 1)2

= (4+ 4c2 1)2 +4(2( 1 +1 _ cz02+ 1) (4t4 + .) 2)

= + 4i2 21)2+8 j2( 2-+1_-Q)( 2 a2+ +0) 2

2p I T-,-+ 2j ) 2 1+ 4e"...+46 --*3 2Ai + v

+ A((.3 a.-c+22 , )(1+4 l e2+44,- 4(FU -c)(o ct 2+t 2(2+22>

= 2 (x)2a2 +t2(2+4(o2+1) -2(3(o 2+1)-a 2 +2o/ ')

TECHNCAL +RSEA+R -OUP


45.

Co f Calculation2__ i "1 22

"- 1a2 -2 +- (u t2 t t u+

4 2 (u-) (u+t)2 6u

+ ( 1 _+ 1( + O + (

Le -1ir\ut) (u-t u+ \u-tY (uE

-7 +-)2 (-)u

FRC

TECH4 CA REEAC GROU

Let h* f 4ff* 46.

fi *)(~~(u-) 13 (4r-do 1uiv (4t. + it t-

-go (uE)45 2t t -t t+7t/

+(u-iC)(u+t) (4t t (t-

+ + . + L du

(4 (-t1\ (1i- 1+

= -- ++ ~ ~ +

=~ , 4 y Z3)i-i r4~ i itiy

1 4z t 1)

-M + n

+ E2 1_ ( -:.1 + 1 1z 1 d

+ 2(u+i) Ec i +C E+iC u+i "t d

Therefore - 1 ~1 + 1


47.

F t (ti) 2 t (tii

E-2 -i 2 t2 -ti)2

4(tt7 (t~ic(E~W 2 -im~iD 2

1 f[a-Arj-Ba-i(l+A-B~] ta+A~ -Be+i(l+A+Bi)1

4(a,21) L 9[ (A 2 + (B+) 2 + (A+TI)2 + (B+ .)21 2

1 _ rCaATI )2 -j1Ae-B)2 +_(____-_ E _- (1+AE+Bi) 2

4(a2 +) (A- q) 2+ (B+ )2)2 (A+T) 2+ (B+ )2)2

2(a2 +1) '.((A-ri)) + (B+ ) 2) 2 ((A+Tr') + (B+ ) 2)2

f 1 (a 2 _ +a(n 2 2 n, rE -2 E(A-iaB) +tifl (B -aA)

4(a 21) ((A-TI) 2 +(t 2 )2

(tT)2t(B4 ) 2) 2)

2_2i_(2a.+;2 (.A-B) E-2(aB*A)n-2a'E+'q a

4(az 2 + 1 ((A-TI) 2 4 (B+0 ) 2

-2a+2 (ciA-B) k+ a.A q2giE12

((~n 2+ (B+ t) 2 ) 2 2


48.

p t 7 +4(- 3 (tC

4(tE+) (ic(-i) + 17+~

-1 (ct21 /2 (A-i-n) + (B+e AT)2+(~ )2

4(a 2 +1) 3/2 (A-TI) 2 + (B+t) 2 + (A-i-r) 2 + (B+) 2

-i Afl+Be-2a-cS(AE-BTl) +Ari-Bk+2o-so(AM-,Bri

4(a 2+1) / 2 (A-ri) 2 + (B+t) 2 (A+r1) 2+ (B+V) 2

ti~e ((t~i) T-;t ) - JC (t_____

f 3 = 1 I ~ l-T1 + 1+AE+BI2 29+1 (A-l )2+ (Bt)2 (A+i)' + (B+t) 2

_____ a-ArI-BE ag+ATI-BE

2 +12i (Aq) 2 + (B+J) 2 (A+ri) 2+ (B+j)2

A =B( , 1+ a)

B =A( ]c21- a


49.

2 223/ 2 2 2

4 (a2 +1) 32(A-TI)2 + (B+ )2

2 2 - 12+1a

I(A+rl) 2+ (B+t)2

j-i (a+Ali(l1a 4ava +) + BEa j

4(a21) 3/2 N(A-TI) 2+ B 2

I-2c8-ATI(1-a a + Be (,_-a F j

(A+T )2 + (B+e)2

a 2 f a ( BTI+AE A -BT4 (a2+1) 34 (a 2+1) (A-Ti) 2+ (B+e) 2 (A+i) 2 + )

a + /a2 ~+1' B E+ An + fa24(a 2+1) \\(A-i) 2 + (B+t) 2 (A-Tlr) 2 + (B+ )2 /

Let r =q a 2 - f3((21jj 4( (a1))

* * a 2 -1 fIs + 2 34a +1)

Therefore f 4 = f + r + s


50.

Special Cases

1) Ref* o -1- (2 -1+22 + o -1+ 0 2 o 14 (o 2 +-)16 2+1 - .- 4( 2

. oT1 a ~2 ci

4(o2 +1) \4(o2i) 2

J 2 F2

1-0 * 0- a +1 +,+-16 (02+1) 8(o +1)

i. ______+___)

Re(s) or2 ~48(c2+1)

8(42 +1)

Re.,,/"- +) 7+1 21 r +2 )k4 (o +1) 2(c 2+1) 2 I T - ) /

2-t [- a 31 0)

2

3o (+1) / + " a

,22+1 028TE +1 )C HR

-22 2 2c( a ++ 0). 7o +1 -3o +1 -~ +

*4 4c 2 +1 (,2+1) 8(ci +1)3-4o+) 16(o +1)2 2

2 2_1

8(o 2 +1) 3284(o 2+1) 16(c *1)2

1(02 +____3/ + (21) 2+1)


51.

2~ + - 2 2 - t+a+aa4(a2 +1) 16,(ci 2+1)

2at + i - + a a a

! ~~4 8 (2+) a4.1 a 1~

8(o2+1) '7

41

IM(s)= - 204(a 2+1)/2

+21)4Fa2/r,+

82+1)2 r2

4 2 2 a- /8(a +1) 8(t+)4(0 +1)

* 1 (a2 1-i+.12 + 2q(B-oA) a 2 -1n22(B-)4(2)+1) \ ((A-) 2 +B) + 2nB,2

Re_,f) .2 2 2 2 + 2 2

M(-1) 2 + B2)(A.+T) 2 + B 2 ) B 4 + (A 2_-q2. 2 + 2B 2(A 2+1 2)

(a 2 + 1 + 4 . 22


2 2Re( *) a21+2n2 OTav+-l + q 22 +2 (L2+, +0)112 + 2"q(B'2A)4An( a'+1 * ,q2)

2 2 (+1) (602 +l+n1.0 2 2) 2.

2- 1+ 2) ( + 1 2(29+1+ +n+4n + A2)

o2~14012 2-___ 4___ 2__2-

2(0 +1)(M) 4 2_20n

2 222

* 1401

22

(o ) ( + 1) B (M+ ) +-

262l~2+l 42a+1) (+ ) + .20,m +1 )zc

2 2 2+ 2-(a+ .12)2 + " 2(~ a2+1 1) ( ++Ta 2 y

=4(o2+1)( (a+1+TI -2ot12 )

2

2- 2 2 2 4=c 22+10) a+l+ 2(o+2 ) o+ 2o

Re (a' ') , - 32+! 2 22 24(+1) ( (A-n) 2ta+ (+l

* ~ 2

2 4 214(a +1) 71 _240 +1

iTECHNICAL RESEARCH G;ROUP

53.

2 2*4 (a +1+,q 4_2ayi )2 a 2+1+rj4 -2cij2

1 x+2 2 x+j

Im (f) 1 2o - 2(A+Bo) i+Ti 2 2a+2 (A+Ba) TI+'l2 1

4(oc2+1) ((A-ri) 2+ B 2) 2 ((A+rh ) 2+ B22) 2

2 2 42 22(o +1) (T) -2al' +a 2+ 1)2

-(A+Bca)TI2 4 22

(ar t +1 ( 2Ia +1)r2?/i- (iBaiiYa)24 2 4+ 2

+ C(a +1) (TI -2ari + a02+1)2

= A( I02+1 + 2a)

lxn(f) = 2 22TI + 2ATI (ri 4 a c t47(2 +1) (TI 4_2ail 2 +0 +1) (a 2+1) (T 4 -2ol 2 -ta2 12

= 2 4 22 * +1 2 2*(o +1) (TI -124, (n-tal-a +) ~Yr2~i. 1)2


54.

Im(s) = l (0 +1 -A ;4(a2 1) \\ (A-n) 2+B 2 (AMn)2 + B 2

-- a (An y- 2)2T 0+

2 4 22

2 22

im~ (3~llAn( 2 Q1Ti n

~2(a +1) \ +(,_0T2 0 1

+4 2 ,1A, -2T(1 2o+ 2to +1

2_0 -o -- 7

2(oZ1+1/ (fI -20fl0 2+41)+1

2 22/24 22

2( +1)13/ (iio -2r 40o +1)1

a +,(T-4_20'12402+1


55.

3 1 a 2 -2 (AaB) )3 , \ (A2+(B+t)

r2 2 f2F2

= 2+1+4+ 2 2( 2 J"-) - 4B( l/' +2)

Rf*) (2-1.-2)422(2TYY+ -) , 02+1 ) +4 21 + 2)(-a vl- 2) .

2 4 22 2a(c +1 _ 2 ) 4 +1

+ 2(A-aB) (ci2+1+t +2t (2V - a1 + 1 2)

+ ~2a1 +a 2+ 2+1 2 -2

I= f 1 + O

2 + 1-o 2 2e2((22)2)( +- ( 2J)(.2-Ta+)fe2(o2+1) +2c,2.402+1) ( a+1) 4+2 a22+1)2

4 2 222( (2+) (4+2o~ 42.2+1) (c2+i)( +2cot 40+1)

2 2 2= 2 f4

( 2+ ( 2_2 2). L2+i (_4+) +,. o2+i)2

2(a +1)(t +2ot4 +1) +() Q2o +20+1)o+

a ( 2(+1 2 A2+(B+t)2 ,)at _ - 22

a( +1) t

T(-2 2 2

+2o 20 +21 Q +2t 0+)


56.

S* 28E 222

a(c +1) Q 4 +2a +0c +1)

S 3a 2 +1 2

) +2ot +a +1

2 -4a 2-4(pE +2ae 2 a 2+71) 2(02+j)(j 4+2aj2 402 +1)

(a +1)(Q +2a +0~ +1) 21

-1_a) (0o +(17+1 + 1- (-

_ ~(42-I) 2j~ _(1a 2+ ~) F2 2 +2 ( 2 1X2+1U)

(2+I) +21) e2+02+1)2+)

* A2s-2 4 22

s0 2 (a2+1)(e4 +2oe 2 +0 +1)

r* -(3o 2 1)o 4(a +1) *

2v F2 2 2 T 2 2 2fA7 (e4+2- a j) (oE -aia(1+2(~-a +1)) 3a +1)


57.

4) -4(o 2+1)f ((0 2_1-fa(TI 2~ 2 2t(A4oE))(D21q~)+D 2(-TI))

D1(Ti)D 1(-1)

2 2+ 2ij(Q+B-oA) (D 1(TI) -D1 (-TO))

where

D2() A 2 j2 (BO2 2)2

122 2 2 2 2 2+2-(A: + Y +(B+4) ) + 4A 11 +4ATI (A +1+(B+t))

-2(a +) f * 1 ((o_1 Ao)( ~ +

+ 2( 0 +1+ cl)(Q 4a)) + 8Q4)QB-A +-02+Bt

Di (71)D I(-TO) . (J2+ +o+2t 2 +2BW+AI) ( cl+1 ++2t+2Bt-24Tl)

- o (J~+ +22 +2Bt) 2 -2(I +1+)(t2+0)

- 8B +4m+ t2 +4(To'+G +)Bt+1

222o -1-2(B) 42

D 1 (TI)D 1 (-Yl) DI T)1(n

+ 2(t 0( +% 2tlo( + o)


58.

2 2 2_-1 -2t(A4clB) 4(tam 2 (12( GcB)-2(a +1)f - +2 2 o12(Pi+)tD T) () D 1 (tI)D 1 (- 1)

2o 2 1-2t(MB) + (2_o 2

D (nD (T)) D2 (r)D 1(2

4o2 #(CM 0 ) 43j+B~,Q-.-- ~ +--~ 24 2 a ( 4 o B)t+

22 1-2t(A4*B)+4+4( o Y1o)BE

D 1(TI)D 1 (-Yl)

432 22+ 2 (-6Bt +Bo-4+1 t +(G +L +AG 2o)Bt-o) l

D 1 (rI)D1 (-)

2o ~ TCHIA R2_AC 1-2GROUP44( + 0

59.

-2(a 2+1) f 2o12(-a)4 ++)E2~ioD 1 (71)D 1 (-11)

242(24 3_ cg2 20B+ 2D- (T) - &~-2B4

2 Iik21 (4B (vi.r a~ - +2I (1_) e2

+ (1+2o(c-fo2Y))EC-S'

-- 1-2o~o+ T+ 2 BE,

+ I.4 MtB +(2Y Y.)2+1-* a4 +-30)Bt- GQ~ja ))

Let D1 (TI)D 1(-q) - D(

2( jf = 2P+1(Be-ca)-l 2 (4Bg3+22~.LG~

D2 ") D 2()D 2(-g) -

+ 2(([1_0c) -2)Bt- Di 7 jza)lD 2(-_0

2Ji72T090)1 44D2 (-t) 2- 2 (2B4 -at-2cz)

2")2 2)4

D 2 (-4) -D 2 (4)- 16B 3 _8 ();+c,+) B


60.

-2(a2+1) f* 2,,- 2iiB.K) -1 i (B 2-- c

32B 2(2E2+~iI) 2+ -(2 9++)-(2Bt -cst-2oB)

D 4 2+B-e (9)DThu(2+~*) 42~

22 2"

= L

+ ~ (ovlo (4 (4C,-B(~i~c)

2 2

N 32t(2t + 0 1 TECH(NICAL_0 tE2AC GR~tOUPF1-

61.

N 1 + N 2 16( 0 10(2D2 )4t2

-~~ +6((_0-) t -Bt)D 2 ("))

-2(o +1)f __ ___

+ 4tK (2£-a£-2sB) -2(2i+-a) (4£2+2B-)+16 (;2Tia) £2_-a)

2- 2

Let Q .14t 4 +±o 2

-2( +1) f*Q 2 -Q(292TiB-2j 1i-t)+2 2Y(-4(foi-*) t4

-2 2 t34


62.

+ 4;+(m+ )t4(21

4o) *1 1 1 1

A( + D(T)+ B(1(T D

22 +1) At 0-102 2E+ 40

2AA +2g AT710 4

2(02+1)D*20

(U (A3 +2t 2+A(TM2+u) t~)18684

+ (8 (4*Y-u) -8((4GTY-. 0)) t4+(4A.-8A+..-SOB)t3

(44( ri-c)-2(a i40) 22t

+ (24faTY40l)-4Qx)Ag+2o

-(2M3+22 +A(lT+,i4) t4)Q-8t 6-8Be-16mt4 _8m~t3

+ (2-8a 2)t ~2+2(4s~l-)At+2o


63.

2 3 4 2 22(+)*2-(2A +(A4o14)-21-R44a)Q-8at _80 t +(29o +-a)A-2B)t+2oa

-(2.4 3+3-cTlA-

2(o 2+1) (f + s AQ 2 Mt3-(c-G+~A2121)t0 12(21

;Y-t 8 2 04~iI2a14)At+4(2+1)

(2Aot +(( +

4 4/1 g~)At+4(021)

2 (302 r O 1) 1(1 ---- (+~.)A)

2( ~ 2 O 2(21o3 2 0(3m 2 +1))A

+)3( +1) r ~

2( -)4 (2 (cs73I21 -± P_ At3+ H(72+) _,At

+ (0+) Q~~(243(~iG +)4(02+1)

TECHNICAL- RESEl(ARC GROUP)A

64.

APPENDIX II -I n(a), Jn(a)

Let duLet 1 ( 0) f m® -

u2 dLet () f du

where P(uo) - 1 + (u 2+a) 2

P= P1 = 4u(u2 +a)

Integrate by parts

I -nf -s-P' duIn = n p d

ca 2 2 + ) d4n f u2 (u +a) du- P pn+l

u2 (u 2 + a) =P- u2 _ 1 2

ThereforeIn I ,,(1 2)4'n In' an+I - (la n+l

or

(1- n~) =In , Jn+l+ (1+ )ln+l


65.

FinallyS 3

Jn 3 f -n'_ ' du

Therefore

W J + 2 - 2 2 u2 (u2 0) du4n " n n (++l

Since

U 2(U2"40)du = n_ n+l 4n

and

(140 )I (1 - -

(ltt)n+1 = ( f -) n - tn+1

therefore3 CAI

(1 - fn)Jn ,, Jn+1 + 4

For recursion purposes

Jn+l = - 4 + (1 -n)Jn

11n+1 THC2 REn-A)n - Jn+G)


66.

n - 1 Determination

P - (u-z) (u-) (u+z) (u)

Where z and - are in the upper half plane, and z2 -G+i

7 (u-z) u+ Q dz - 0f dzw

Therefore J " z zI = 0

i.e. J = 1 -I eO

f du

(u 2z 2)(u2-Z)

I 21. 1

Since z2 22 222

Ii 1

T(Z+Z)2 zi2za -z

Ii (,.a) - ira1


67.

APPENDIX III - Steady State Integrals

Let L.k(0) - f* x2(k-1) x 2 ) dx

where P - P(xa) is defined in Appendix II.

We wish to evaluate L1 and L2 .

Using Appendix II and suppressing the argument a in Lk, Ik, Jk"

L1 = J 2 +a1 2

+ 3112 1J + -f (-- " aJ2 )

2 21fa 4(l+ 2 )

-aI J 3cI1

4(1+a 2 ) 4(1+a ) 4(1+t 2)

2a + V0+1

4(l+ ) 1

rB 2a

4- + 2r (1 24(lic )

IiL2 4

VB


68.

APPENDIX IV - Condition (4) Intezrals

Let P( ,) be defined as in appendix II.

Let M Im "f -cos ( d

.o pmP1 (22) 2nf d

(_ 1)n(2.) 2n

0 (2n) ! i2n+m0

Let

N Co dNC m = Pm

0(_l)n(2a) 2n

= (2n)! J 2 n+m0

Let

M, =f - sin (2-) d

00(_. n (2,) 2n + l

= (2n+l) 2n+l+m0

Let~~22

N 8 , _ Lm sin ( dt

.1 n (2,.2n+10T (2n+1) : 2 n+l+m


69.

1,c f Ef pd-v p

n22

(2n (J J o-0, (2 2n(2n+2- o2n+2)

n 2) n+- (+r (J 2n+3 "4GtI 2

n+3 )

= z ' (2 )OD 2 2.t

I2, t Q 40 C os 2n d

09(-Qln(20 2n 2 Q2 dt

0

Sii.lar ly E . 1n( ) 2 n+1 In+21 2 8 - Z (2nl)! 2n+2

0

1 00 2 2

Ii'c - ot 2"

M + Fi-rCos2ndt2 c22f2 Pt

TEP

m~Mc+ofdP~ d0d1 2fl1 dP dt


70.

Since P - ( 2) + 2a(Q 2 ) + +1

dP 24 2 + 2o

d 1 2 2 +2f

P- PI- P- p -w -

Therefore

* Q 40)2" in) Cos 2n-c T 2+ 3 (a. o-) dt

PI

" Mc.2 + 40 [n(Ns,4 " M 9,4) ' Nc, 3 +°c,3)]

22S(-l)n(2) 2

2 0 (2n)! I2m+2

1 n(2n)2n+2 2n+2

+ 4a(P (2+l)! (J 2n+54I 2n+5)' 2n+20

0 42n)l ! (2n+3 + *I23- 2nI

S (2n) 2n+2 - 4o (n+l) (J2 n+3 +01

an+1 +m M2+l 0(1 ' n)In-I+1 n -2t,

I* . ( j')( 2m) 2f1 1,c o~ (2) ! (( 4n+4 )I 2 +3 - (4n+3)i2n+ 2)

TICHNICAL RIESARCH GROUP

71.

40fs sin iT

=ll MS4 + . p

2ms,2 +0 f ( sin T)( *2a)dA

1 co (a2 4)( n 2d 21 m s,2 _ -- p3 cos - + sin

1 • . Ms,2 _ c,4 +cMc,4 ) +Ms, 3 + c 3

1 (-n+n(2, 2 n+ l

(2n+l)! I 2n+3

._ - 4a[ Z (23)' (J 2n 4. + 2n+4)" 2n+L0

n (2n+1

+ £ (2n+1) ! (J2n+4 + 012n-4)]0

0 1 2n+ 2n1 2n+3)I- ~ -4o(n + f(1~2n+4 + 4

. l)! 2n )4n+n n) 2n+l1 2n+3 3___1

(2n-) - 4(n + 1)(- 12 4(2n+3)212n+3 2+)0

(-) 21 2n+1 (4+)(2n+1)! 2n+4" (4n+5)12n+3)

112, c 2 - l,c

2, s = - l1,s


72.

Let E - 2oc, +1

22(Q2"Q(2+E) cos 2 dI3,c -= " 2 22

2 A vp (.1) n(2n) 2n A.2 2 -1)(Q 2+E)E (2) ! 2+3 dt

2A a -1 2n W2+ El2~

z 2n! 4(2n+2 + 4(2n+2)

Q (2n) 2 n J2n+2 2 l 2n+2)

E - £ (2n) ! 2n+2

Similarly

A 1 (-1)(2n)2n+ l W3 2n+3 + (2a a-2+l)I 2n+ 3

3,s (2n+l) 2n+37 0

22Let F =a 2p+1- 2( 2+1)

= 2A 2 2 40 ) (12+E) 2n' (

I4,c 4 2A o d

A e Q22 to ) 2+

2 - Cs d 3M 2+1-3c2(a +1) _ P - 4(p +1)

- U1 + U2 + U3


73.

2n 2 2 2U1 2A (-ln)", 2 n Q 2 40) (a2+E) dt0 o (2n) . 00 p 2n+4

2A (-1(2f) 2 (2n+3 EI 2 n+3

S(2n) 4(2n+3) + 4(2+3)- -

A 00 (_l)n(2,) 2 n o [2(

2 2(02+1) o 2n! _ 2n+3

A c(- 1 )2n(2fl) 2 n 3o2n+2 F12n+2S2(ct2+1) 2n! "4(2n+2) - 4(2n+2))

U3 2 "(! 2n+2 + 2n+2"

8(a +1) 3/a2U2 -U3 m +A 2 _, .)n(2 2 n .

A ~, 2nU2 +U) (2+2 (3(c 2s+1" (30 2+1))J 2 n+22 3 8( (2+1) 3 / 2 0o2)(n+)2+

- (Fi[i+ + (3m 2+1)E)12n+ 2 )

F 2C+i+ (3o 2 +1)E =a(12+1) - 2(o2+1)4a2+1 + (352+1) (2a-a2+)

2 2

=l (7d2+3) - a +1 (5c,2+ 3)

4,c 8 2 I 3 / 2 o (2n)! "2n+2 " 2 +

(5%2+,3g+l _ cl(7,m2+3)1

+ 2n+2 2n+2

4(02+1). 2+),23)

+ 2n+3 (W 2n+3 + (2c&


74.

Similarly

A (-in(2,)2n+l 3(o (302+1))s 3/(2n+). .2n+3 2n+314s 8 (c2!+.) 3 / ..

+ 502 +3) 12 +1- 0(7m2+3)+ 2n+3 I2n+3

4(o2 +1) a 2n++)T N2n+4 R C2n+4 GROUP


75.

APPENDIX V - Force and Mmnent Terms

Part 1.

LdxLet Rk a f 0 k 2- dT

Rk -kSdx,a Tdil

S dx d

sk - (x+1 ,k d d-0

* sk dxSk - - + ) 77-

Let P . P(Ti,-o) - 1 + (2a)2 (See Appendix II)

-81 f nL 20 d~j1 00 P 2

1 81 (J2(-a)-aI2(-))

Henceforth we will suppress -o in in and In

311

2 l +ci

14EO

R- 21 2o 2-


76.

* r TI2"-41 1 d OT

2m T -P, 11 I P

fTIP TIP.2 -2 .T

--8 2 !I0 dTj

- -8(J 3-1 3 )

R- -41(12 + 8o(J 3 - GI3))

013 - -i (aj 3 + 8 12)

3 ] 52 +012

R - -41 (12 + c' (5J 2 - 6a12))

6o2, .1-5a 2 31T 1- -(1- 6+-Q0 - 2 (1 o2 2 2+4

1-a2 %J60 01(5 + + - (1 +.2 1.2 "2 (.2)2 + -)

i4CL 140o (i14o )

R 2*3a 1-3m2

R1 2-J ((.23/2 2 (.22 1

(14Go) (140O)


77.

R 2 =-88 f ,L2(, 2. dj- .e P2

= -21I

. 1R - R,

= 2a 40l0 R2 2

.. ~~~~ 3 . _A ,f 2(.q2-G) (,q2-E)

~09

where E -29 _-O+

16AJ 3

E - G + I8(10: 2)

(3 CE 3(' 2 "GE C&I i1~-)J (T.. + -i-)

8(14a2) 8(1.40

2) 4

AR 2 2 2R.3 -7 ((3(1a)"- E)J1 + (3c(1-*a2)-(2+3)E)I1 )

2(1lto2)

(01J1+ (c 2+ 3)11)E - Toh+ / + a+2+ 3)(2- PM+l)Ii

- (2( 2+3),-(a2 ) (23))I1

- ((2 +5)+ (2-3))I1

2 2)[213(14a (J1 1 I1) - 3(1i t a2 ) l


78.

All 1 ( 2 f7R 3 (2m -2) + /C,+1 (202+6))

UIL~I~2) 2 W 1) +/i (i+3))2(14o )" r 2.a22 2 2 2 T

2(1403)2 (a 2(21) + (a 2+')(a2+3) + a 1 a+ 1 ( 20 2 +2))

A , 2, 2m+ 3m2 2o(a2+1)3/2)2(142)

' 16AJ 20) -n 2- d

4A1

+ R+

(c,2+i) -t p3 4(c +1)

where F -a 7,+- 2(14a2)

16 J 3 El 3 4AI 3oJ2 Fl 2 o12+I2it+++ '+ " -"+" ( -"T/ + ("i-- +-5:--<-,T-+.,W) + 1 4 8 (c,2+1)

- 1 + T2 + T3

El 3 E 7

- 12(1-2 ) 3 8 ( '2)

12 (14 12(1 )


79.

2__3/_ 2 2 2 )T1 6(1+ct2)3/2 (5(3(1+cB2)- cE)J 2 + (3cs(1+a 2 ) - E~co2+7))12)

A6 ((1 -i

T +T A - ((9(O+I+3 5(3(1+a2 -E))a12 6(14o ) /

+03(A -. F+ o(1+o2)) - E(ok2+7))12 )

-22

9 + +5(3(1+ -2OE) 14oq + 5(o 2 +3)

3(J7l I .(11)) - +(2 +7)=3(2,(1+ 2 ) - 2(1+a2)3/2)+( 2 )(2+7)

2o (22-4) + I+ (1-5o 2)

2_2 cf 2 311VT3 1

(2o(2a 4)+ cl +(1_50a2)),I2 - 2+ vj(1i ( 2 +o +1

2,2( 2_ (152))

(14afa+1 + 5(a 2+3) + 4 22 +1 2

9 + 12a 2 + 15 + T T+(9 0 2+ 15) (T +1

- a2+i

+T (04 21 2((235/ + 4a~t + 5 + a +1 (302+5))J1- 8(140 )5/2

+ (a(3a4 + 8c 2 -3) + f (3a 4 + 1)),1)AJI 1 4 2f2

- 1 (60 + 16a2 + 2) + C2+1 (64 +402 +6))8(.Z)5/2

8(1+0)


80,

-(3a2 +1)AI 1 2 2) + (22+-" 3 ( 5/ 2" +

.8 .t25/2 (a(6 4 - 2) "AmI .12 (2 +) 2 + )

-All 1 4 2 (64+2+)(o7~ ((6cs _4a - 2) + T+ 6 0 )

Alil1 2 2R4 2 5/2 (o(20a + 4) - 16oa2 +1)

8(14+0)

. S c, +,) (16,2(,. , ii+) + r,(2+1))

16(14o2 )3

n-i 222 3 (00 ct+l)(c +1 + ot) -' 4 2 )

4(1-ta2)

2 .R - Z kRk

*3iaBI 1 2 2- ---- (2(o 0+1) + 1 3 )'F+1 a)

a 1

c 3RoBI1 ((5-2+1) + [o+ (1 - 2))

20 23

aBl l 2 2c2R = (2a - 3o ( + + 1 -30 2 )

N +1


81.

R aBlI 1 2 2 2R= 2 (a(12a +4) + a +1 (2- 12a))

2 3/2 (a (6a2+2) + a (1- 62))(a +1)

2- -U27r (J 2 +) 2 (1- 2)( 2-i) + 3a )

-u Mir 2(-6aaa (a2+1) 3/2 (a +1+a)

2-U M2r (fa2+ + aii) a2+3 F 1a(a 2+1) 3/2

a (a2 +) 3/2a j

S1 = 1612 f 2 3a d I

= 162 (J 3 - a13 )

_e3 = -a( 73 2 (8 2 + aJ3 )

222 2

a5 J21--8-2(i- 2+ )3 -° 2 8 + -

SI -+0 (2 2 - 6a12 )

32 212J22 R 1- 2

2(1ra2 ) l+a

3= 32 1 2a \ I 2 (2+1 + c) 1i322 1 2a 22 2 4_j )

= 2 1 a 3 6a

2 + - 2(+ 2 ) (+2)3/2 + 22


82.

SI = -8 f d2 3d

=8 82(1 -a f~ .4l3)-3 23 P

f ]-3 3 -i-0 1 P -l 0- TIP

=-12 f (12-_)di

P 4

* 2SI = -812 (13 + l2ct(J4 - 044))

-a4 _ -a 11

4 +a2 (2 13 + aJ4)

:2 1 a 3

* + 2 a Z2(-

S= -82 2 (1 + -l 2 (9J - lomi))2 4 3 3

-98 2 s . .. 1 32(1-a2 140 3

1 1 713 i 2 (t712 + aj 3)

1 (7+021-2 2 + 8 J2)

S+a2 3(7-ia)12 _ 2 (t 1 + aJ 2)

17IA


83.

(5+7-Fa2 a2 a J

a J(5 + 7=2aJ 6(1 2)+(4 +"--

1. 2 2l2 2

3 1 2( 2 (6(2+a2 )aJ 1 + (3(7+a) + 6a 2(2+a2))l)32(1--c )

-3 2 7- 2 4

32(1-+ac (2(22+a2 ) a\a+1 + 7+5 + 2a

S* 121 [ 92 + 1 3 6a)- 2(1-a 2 2 1i2-2 2(1i+a2) (1+a2) 3 / 2 (1+a 2)

32 (2(2+a 2 )a Va+1 + 7 + 5a2 +4(1i c)

3i (3 (1 - 2 ) 3 /2+3 2(1i- 22) '18 a l 2 2" 2 3/2" Fi - ,2

4 1a2 43 1a+6 _2 1a2

-7 - 5a 2 - 2a 4)

- 2)1 (a (lgaa 2 )3 / 2 - 20 TI 2 - 7 + 34a2 + a 4 )

4(1iac2 )3

S2 -16 2 f OD 23 dq2 - p

- 21 1

-2 31I-2 + aJ2)

S= 2

2(i1+ac2) ((3+a 2 )1 + aJ, )

= _ 2 22i (3 + + a + )Ii

2 C( I+ H


84.

* i 's 2 -- " s

2 1 3 6a2 ( + 3 / 2 +22 2 /i 2(-) ('-*) "("1+oi) 2

2 +a- 2rj

= 32A.2 2-a) (T2_E) dS3 fa 2V+1 . p 4

32AA2 J 3 El 332A 1 (. -4)

= 2T 1 (See R4 )

A12 3/2 (5(3(1+2 )-aE)J 2 + (3a(l-a )-E(a 2+7))12)

(3a(l-a 2 ) -E (a +7))1 2 (3a E(a 2 + 7))(._1._ €j2E = 2 42+2

(15(15+a 2)-5aE+3a 2 Ee+)J 2 = (15+ 18a 2 + 4i) 1

A +1 2 1 1+1 -

s= -, (3a+(5+6a2) (a+ +j) E (c&2+7+2a(c 2+2) (cc+ [a2+)))4(ia) 3 / 2 a 2+1


85.

E (a2+7+2a (a2+2) (o+[+l))) =c (c,2+7)2 a2+2) (a+c X+)+(02+7) (a-Fa )

-2o(o 2+2)

- 2a(2+7)+2o(o 2+2)(a 2-1)+ /2+(2a2(22) 2

C, +2 __a-7)

2o(o4+2a2 +5) +fa (2a 4+302 -7)

2

Al .1 1 2 +)a2 +)a4 _22_5S3 (2ci((3a2+4)(ci2+)-a= 2a2 -5)4(1. 2)5/2 ".

+ a +f((62 +5)(l 2) - 204 - 3&2 + 7))

2r ( /~io 2 4a(2(2 4 + 5a -) + 1 (4o2+84 + 12))

22- (2(2a4 + 5a 2 -1) + (a2 +1) (2a 4 + 4 2 +6)4(1"t2)

+o Y2c4 52_ 1+ 2a 4 42+6))

• j: = .23 (4c 6+lla4+9a2+6+a 2 1 (4+ 4 9c2+ 5))4(1-t2)

4 2B co 2 3+1 p (a +1) p. P 4(a +1)

32- A 3"M4 EI4 8At2 l3 FI3 32+('T-2 ) 2 (__ + ( 2+1) $2+1+ 4(a2+)

=X 1 + X2 + X3


86.

-EI = -E 114 = 13 + OJ 4 )

2)J4 E 7) +3( +- - c 1+0j -- "-

x At 2 (93- oE 7J "+ (3o - (1+0 2) )- 3 2 31-g -fa 4

At 2

x+X = A 2 ((12a + 9 o2+i,(3 - aE))J126 (i-2+) o2 3

+ (4F + a (3a - (1I+c.2 )E)) 1 )232

l 2a + 9[2+1 (-E 2=1+ +9a

9+2 +227= i21a+92+2 7V~ &

2

= 2Lx + c +27

-_____2 (11-ct2- 8l )(10 2 224F +ia +(303 (140)E = aVZi81 - a 2 2 +11+-

2a 21

= 0 - (5 2 - 15)- 7a2+ 3

25a (2.3) 3-7a 2 712

( ( 23)- 7a +3)13 =( 2 +.aj3

TECH ICA+1) a + 1

TECH.NICAL RESEARCH GROUP

87.

(21a+ 9 (+3 .+ 5a 2(a23) +(3"7a2))j( 2 +1 ) 3 / 2 a 2+ 1 3

4 l 2 5J 2

a (14 +24) + 14a 4+21a 2+27('1 2 _.J 2a2 +1 ( 2+ i)3/2 8 +8

+ X At 2 ((5a(14a 2 +24 + 5(14a 4 +21a2+27)48 (2 1)2 +)

4 2a 4 2

+ (14a 4 - 25a 2+ 21+ a (14a 4+ 56a 2 - 78))12)

a +1

121= ', 2 1 + aJ)1+a

a+ a +2 4 11

2 12 (3 + a 2 + aV2+)I I4(1a)

1+AI 1 '(Fa + a )

2 F+

x x r,92(+1+a) (52(14a2+24)+ 5 (14a 4 + 21a 2 + 27)1 2 384(a2+1) 5/2

+ (5(a +1)(14a 2+24) + 5 (14a 4+21a 2+27))

+ 2-t (.14a4.25a2 +. 2 (!4a 4 +56a 2 -78)

+ 2 3/2 ((a 2+1)(14a 4-25a 2+21)+ (a 2+3)(14a 4+56a 2-78)))


88.

S2 ( a2+1+0) 2(168256a2+78+ l4o4-1062+120384(02+) / 2 a 2+1

+ 0 (168a 4 +326a 2 +198 + 2 84+112oJ -1 56

aa +1

71 a~hii 4 2422402 512 (168a +270m -4 240384(o +1) a +1

+ t (16804+354a 2+282 -. 240F2- a 2 +1

r( 2 4 +624a 2 + 240)

2 5 24 2

384(o2+1) 5 /

+ 1 ((o 2 +1)(168o4+ 270 2 42)

+ 240+a (168a + 354o + 282 240)))a +1

r'i2 (48o(7a 4 + 13m 2 + 5)384(2 +1)5/2

+ 1 (336I6+ 792d4+ 510o2_ 42 2.40))S2/+1 a +1

3x 2 (3c 2 +I) (36k -l 2 4 6 +ay+1 (404+902+ 5))

16 (c2+1)4 .4


89.

38(c2 .& (-24(3a 2+1)(4a 6 +Il4 +9a 2+6) + 240$4 = 384(a 2 +11/

+ ( a2+1) (336a 6+792a, 4+510a 2-42)

+aVa +1 (-24(3 2+1)(4ct+9a2+5). + 48(a2+1)(7q4+13a2+5)))

ff 2 48 a8+.40a 6 + 65,4-30. 2+ 964(a2 +1)

4

+ 4a +1 (2 6+9934+12a2 +5))

S = S + c2S'

3aB2 1 Va+ a(a 4 +341 2 - 7 + - 2 (422 -19))c10 S1 4(22+)2

-3al 2 2 2 4 2 TT4 2 2 2-25/2 (a((a +1)(a -19)-a -34a2+7)+c t+1(a4+34a -7-a (a -19)))

8(a2 +1)

_____2_7 2_ 72 2-21 5/ (a(-52a -12) + a +1 (53a -7))

.BA 211(ay a2+1+ i- 2 2)c20S 2 = 4(2+1) (a( 2 +13) + V§'I (a 2 --5))

at2 2 22 2= 8(a2. 5/ (a+213)(1-a2)+(a 2+1)(a2_5))+ 2+((a2_5. a 2)+a 2(as2+13)))

atw 2 2 2228(c2+i)5/2 )c*(+2))

-it (ca(-16a 2 +8) + a2+1 (19a 2 -5))I_8(a 2 +1)5/2 +

II TECHNICAL RESEARCH GROUP

900

a 2 +": 5/2 (a(140cs2+44) + 4 (-140a 2 16))

_ -uM 2 (i 2 ) r 2 2 (2-)+1 +2a( 2+1)5/2 y+a (35a (a - lla 4 +1)

-U2M12 r (2-i+c) 2+ 4+ 15ca) 2+i)2a (a2 +) 5 / 2 +4+15

22r ((-5a + 19) + a (-5a2 +4))2a (a2+1) 5/2+

IIII

91.

Part 2: (Steady Force and Moment)

-d 2

I c10 R1 = -2acBI 1 ( a2+i.-0)( /a2+1-2af= 2- 2 0R

i -2X2aI(32 + 1 - 3aV +l)

20R2 = -2 laBII( a+1+ i-a 2)

Ifo = d 2a2

f 0 = B1 (2ca +2 -2~

I = dair(f/i+1- a)

I -U 2Mdir(1 + )I m -m -d (

0 2 (C10 S1 + c 2 0 S 2)

C0S1 a 2 (2)

I 2(c*o 2 +1) ((

+1) 2 2(a +1 - 6 (a +1) - 12a + 18a Ul)

- r37 (-17a 2 - 5+ l8a \2+1)f 4(a 2+1) 3 / 2


92.

2 V2c 20S -2-aBl 11 (a 2+1-) (3a aa l

-C T 2 2J 2 2 -21

4(a ) (+2 (a2+1)+ (1- a2)(3+2) + 4a a +i)

42 3/2 (-a 2 + 3 + 4a

4(a21)3/ 34 1

m 8da23 (16 2 8- 14a \2/+1)

- 8e2+i)3/2 2 2 2

-2Md 2 ( +1 +) (7Va2+1 ( 2+1 -a) +a 3)4e-U 2 +) 2 " + -a+4 2++aa

4a (a +1)-3)

-2 d2 2 2 22 2 .2= -(a2M 2 (F-2+e)((a +4) a ++a( - 3))

-u 4 2 2 2 2 2= -O2d3/2 ((a 2+4) (a2+) a (a2 3 ) (2a2+1) a~a +1

U 2Mdl 2 r (2a4+ 2a2+ 4 22+i -)= 4a (a2 +i) 3/2 +4+( a +) a j

2 2- -U2Mdl2r + 1 ( 2+1+24c(a 2+1) 3/2 (4 + a )


93.

Table I

Index R S

-2+2 2 ( 2 +l) 2

.22(3+ 2+ 2+)(.cs)2) -2ei(-) 22a+1 2 (a +1)

a +3+2+ (a12 +l) 3 /2)2( 2+1)2 4( 2+1)( +

+ (2+ (4a 4 + 9a 2 +5))

4) 72 ((e2+j)(fe2+l+€,)4 2) 72 4 ( 8c 8 + 40 6 + 65e-30Z+4 (2 +1) 64(a2 +1)

+ 4a2t+1( 246+ 94+12a2 +5 ))2i 22 -2-- 2 i 4 2d

a) "U27 (4ct+Vc2a+l) UM 2 7r 2(CS,2, - CSe+

2 3/2

(4a ++1) f 2 (c2+)5/


4k Ok 94.APPENDIX VI - 7k' 7k' - Tabulated

Notes for Tables

B -

*k -k *

;)*k airk ak

+ N+

+i

M UP ~ N N

w +N .44

m+ + ~C14I C1 +1.0 N T N +

+ +3tis Nc* + "f+

gop +1 N4 + Nl

N1 0 6 N +Kw 04 NN +

04- N, 1 L46+ 5r

1 N4

IIti~l+4-04

+- N4I.+ N

u~ N

cq +1 + -w1 A +

N= Ni Nq +*n W W + N+Nq % +N c

+ Nr+

NNN N 4I + AN+

N* N W. N ALP NI N+ + N

N N .CI + 6 -

+ up

r-4N

+ ++C+

4 %.04W4.CA i

TECNICL RSEACH ROU

96.

ci c

+ ".4

+

r-4 i-4 M I-

+

04 N.N

N

r-4

-:-

+w+ N C

0~~ +N-,-,4 + 04HH N1C4 1

H- 0 d

0)

r-4

0 r

I..H

P4 +C.N

:104

-~ - 0

H N C -4 -"


97.

0 0

~00

C'wc

r-4 U

"4

r-4 rZ44

.5.4 04

s -4 e4

N 0 +

I ~ ~ -C' HtC-4

+%+1 + N

10 i-INM I

TC NA RESARC GUP

98.

r4

+ +'-V4

0 c

U. ,.J+

4-4

N 04

-- rJ ,-*

,.-4 0 I.I€

-,

-4 + 0

cc 04 qq-0

-'-4J

j r

004

+ + N

+ N4


99.

+cne4

ci +,+ .l

0 04!

N N

I'

0 cn

'-44

r14

0 C4

-e-4

+I +

04 N

0 ci

rCoE NCLREERH RU

u Co

.1-4

r. ++0 0*

C)*

04 Ic.1P4 1

-l +oC~r~ r- -

/111


1000

Bibliography

1. Guerst, J. A. A Lineaxi~zed Theory for the Unsteady Motion

of a Supercavitating Hydrofoil, Instituut voor

Toegepaste Wiskunde, Technische Hogeschool te Delft,

Nederland, Rapport Nr. 22

2. Guerst, J. A. Linearized Theory for Partially Cavitated

Hydrofoils, International Shipbuilding Progress,

Vol. 6, No. 60, Aug. 1959

3. Timman, R. A General Linearized Theory for Cavitating

Hydrofoils in Nonsteady Flow, Second Symposium on

Naval Hydrodynamics, Washington, D.C., 1958


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