lab.
y. Scheepshouwkunde
u PJ I
Techmsc!e i ocoo
iFSTEdllrflK
ML
FORSCHUNGSIJEFTE FtR SCHIFFBAU UND SCHIFFSMASCIIINENBAU
Afl
CHIEF
High- and Low-Aspect Ratio Approximation
of Planing Surfaces
Prof. H. Manso National University of Yokohama
introduction
The ship wave problem is a kind of boundary value problem of the classical hydrodynamics. Its essential feature is the linearization of the boundary condition at the free surface. There is a requirement to the ship form in order that the above linearization becomes possible. Since the disturbance given to the fluid by the movement of the ship is to be small, the ship should be slender in the direction of the movement. This re quirement is fulfilled by several types of ships. The thin ship is most familiar in the ship hydrodynamics since Midiell's classical solution [1]. The slender ship is another type which receives attention quite recently [2]. There is another type that enables the linearization of the free surface condition. That is the so to speak a flat ship. Its definition is that the draft of it is much smaller than the length and the breadth. It represents a ship of very shallow draft, and the high speed motor boat is a typical one. When the motor boat is running in full speed, most of its weight is supported by the dynamic lift acting on its bottom. This condition is usually called the planing and we name the bottom surface which is in contact with water the planing surface. Although the thin ship and the slender ship have been discussed quite a lot, the flat ship has not received a due consideration in the field of ship hydrodynamics. The main reason is that the boundary value problem is reduced to an integral equation of two variables whichlooks intractable. The fluid motion generated by the flat ship is equivalent to the disturbance given by a pressure distribution applied on the surface of a stream. The solution for the surface disturbance of a stream is given in Lamb's text [3] in the two-dimensional case, and Hogner [41 developed the therry for the
three-dimen-ìm I1ut:mL rF -LC V"C _ta
¶-quadrature. A notable contribution was zuade by W JAwv 1]. He calculated the wave resistance for various types of prescribed pressure distribution and put emphasis upon the application of the fiat ship theory to the problem of ship wave resistance. However the requested direct problem is the determination of the pressure distribution and associated wave resistance for a ship or a planing surface of a given shape. in this case, the
Heft 72, Mai 1967 (14. Band)
pressure distribution is expressed as a solution of an integral equation. In the two dimensional case, the integral equation was solved fIrst by Sretenski [6] and later by Maruo [7]. The latter showed a good agreement between the theoretical result and the measured pressure distribution on a gliding fiat plate [8]. On the other hand, the progress in the three-dimensional case was poor chiefly because of the difficulty of the integral equation. Though the solution of the integral equation of two variables looks formidable, some sort of approximate method may be possible. The integral equation of the planing surface has a resemblance with the integral equation for the circulation distribution over a lifting surface of a thin wing. It is well known that there are two approximations for the lifting
sur-face. One of them is the high aspect ratio approximation which is equivalent to Prandtl's lifting line theory [9]. The other is the low aspect ratio approximation which was originated by Jones [10]. These approximate methods can be applied to the planing surface. The intention of this paper is to show the
integral equation is simplified in both extreme cases and how the solution is found approximately.
z
Fig. t
Die in diesem He/i abgedruckten.Beiträge wurden von den Verfassern ebenso wie die Au/sätze in Heft 70 ± 71
Prof. Dr.-Ïng. Dr-ing. E. h. Georg TVeinbluni zum 70. Geburtstag gewidmet.
Formulation o/ the boundary value problem
An irrotational motion of an inviscid incompressible fluid is assumed. Instead of a flat ship moving on a calm water, we consider a uniform stream upon which the ship is floating at a fixed place. Then a steady motion is obtained. Take
car-tesian coordinates with the origin on the still water surface and choose the x-axis in the direction of the uniformstream. The z-axis is taken vertically upwards and the y-axis athwartships. Because of the irrotational motion, the velocity potential exists
and is expressed by Ux + (J, where U is the velocity of the uniform stream. The components of the fluid velocity are
13 + 3(JI3, v,
= a/ay,
= 3(J)/3z. (1)The velocity potential satisfies the Laplace equation in the space occupied by the fluid.
a2(J
aJ
aI
ax2 3y2
+--=0.
3z- (2)Because of the basic assumption that the ship isvery flat and the fluid disturbance generated by the ship is very small, the deformation of the free surface is very small so that the boun-dary condition at the free surface is consideredon the horizon-tal plane z 0. By the deletion of the quadratic and higher order terms, the free surface condition is expressed by the linearized form as
ai
g+
U2
at z = 0. The fluid motion should satisfy the condition of the rigid boundary on the wetted surface of the ship or the planing surface. Because of the flatness of the planing surface, tile boundary condition there can be considered on the projection of the wetted surface on the plane z = 0. If the ship's sur-face is expressed by the equation
z==f(x,y) (4)
the boundary condition on it can be written in the form as a
ar
U
atzü
(5)ax 3z
because the quadratic terms are omitted.
We assume that the water has an infinite stretch and an infinite depth. Then the boundary conditions both on the free surface and on tile rigid surface are defined on the horizontal plane z 0, but they take different forms, because the
boun-dary condition inside the region S that is the projection of the wetted surface is the equation (5), while the equation (3) holds outside S. There is a condition at the infinity that the dis-turbance motion vanishes. In order to solve this mixed boundary value problem we begin with a general expression for the solution which satisfies the boundary condition outside S. Let us define a function (x ,y, z) by the equation
g
4 (x, y, z) =
+
-- . (6)3x2 U2 3z
This is harmonic function in the lower half space, so that
+
a+
=0.
3x2 3y2 3z2
It vanishes at the infinity and also on the plane z = O outside S because of the equation (3). It may be finite on S and let us designate the value of i4, on S by i4, (x, y). These conditions can be satisfied by a double sheet potential suchas
Schlffstechntk Bd. 14 - 1997 - Heft 72 (7) 58
-=
1 Ç' Ç z Wi (x', y") dx' dy'. 2 tJ J {(x - x')2
+ (y - v')2 ± z2}1/Now let us consider the Fourier transform as follows:
* (w,y, z) =$(J (x, y, z) e" dx
(w, y, z) =j (x, y, z) eb0( dx
We oeprate the Fourier transformation on both sides of eq. (6) and integrate by parts tile first term of the right hand side, making use of the fact that (J' and its derivatives vanish at the infinity. Then we obtain
4,* (w, y, z)
= - w2
(w, y, z) + x - -- (10)where we put x = g/U2 A particular solution of the above differential equation is i ,»'
r
* (w, y, z) = - e Z j e IQ* (w, y, u) du X (11)whidi must be added by the complementary function
y* (w,y, z) that is a solution of the homogeneous equation
_w2X*+x '.__=O.
(12)Then the inverse transform of eq. (ii) added byx' (w, y, z) is
1I (x, y, z) -j ett»x dw
2tx j
10
e - (, y, u) du w! -+ x (x, y, z) (13) where x (x, y, z) is the inverse Fourier transform of v. (w,v,z) and is a harmonic function in the lower half space,a.
a ay2
32
satisfying the boundary condition
a!x
+ x O
ay3x2 3z
on the whole plane z 0.
In order to express the velocity potential in terms of
'i (x, y), we introduce the integral representation as follows
in eq. (8). z
{(x x')2 + (y-y')2
+ z2}'/ i(exp
[Vu2 + 2 z + i a (x' - x)2t3
+ i (3 (y'y)} du d(3After a straight forward reduction making use of the Fou rier integral theorem, we obtain
The function (x, y, z) is determined by the condition that there is no wave motion in the upstream side. Then it becomes
x (x, ,z)
=
(x',y') dx' dy's
xz (1u') sin {x i/i + u2 (x'x)} cos {xu
i + u2 (y'y)} du.
Next, putting
i)' (x', y')
-and integrating by parts, we have i y (x', y') dx' dy' y (x', y') 3x' ., 00
=
Ç (x',y') dx' dy' 4,tJ J
s -00 -00where the exponential factor which secures the convergence is omitted for simplicity. This is the integral equation to deter-mine the unknown function y (x, y) when the function w (x, y)
is given.
Let us next consider the pressure distribution over the sur-face S. We omit the second order terms from the pressure equation. Then
pp0
a)
gz.
ax
(21)
There is a relation (5) on the surface S. Since the free surface and the planing surface consist of common streamlines, we can integrate the eq. (5) along the surface streamline. Then we
have f (x, y)
J!!_(x'
y, o) dx'. exptVw2+ (30 z + iw (x'x) + i (3 (y'y)] dw d(3 + X (x, y, z). xVw2 + (32 w2 (22) orpP0
= -.
ou
axthe pressure is expressed by
p -
= -
9U j14 (x', y) dx'. (24)Making use of the definition (18), the following relation is obtained.
p - p0
= QU y (x, y) (25)This result indicates that the function y (x, y) is equivalent to the circulation density on the thin wing. Consequently, the eq. (20) can be regarded as the integral equation for the deter-mination of pressure distribution or the circulation. The verti-cal component of the resultant pressure is the lift of the plan-ing body, and is given by
L =
Ufly (x,y) dxdy.
(26)s
On the other hand, the resistance is given by
R
= -
(p - p0) R=
U $j' y (x, y) w (x, y) dx dy - (27) s (x, y, o) -(16) (17) g -- (x , y,o) dxU j
X (x', y. o) + x (x', y, o) =ou
a}dx'.
(23) Ia azBecause of the relation
14 (x. y, z) = --- +
X-h--ax2
- 59 - Sch!ffstechnik Bd. 14 - 1957 - ISeft 72
00
Jex2
(1+u')cos {x
I + u2 (x'x)} cos {xu i/i + u2 (y'y)}Vl + u2 du.
(19)o
There is a boundary condition on the rigid surface given
by the eq. (5). af/ax in the left hand side of the equation gives the slope of the planing surface. Designating it by - w (x, y) and substituting eq. (19) in the right hand side, one obtains
w (x, y) 00
2 u Jy (x', y') dx' dy'
J J
(VJ
sin {w (xx')) cos {(3 (yy'} dco d(3.+ :2
1'
(x',y') dx' dy'J
cos {x Vi + u2 (xx')) cosu Vi + u2 (yy')) (1 + u2)'/ du
(20) (18) Hence the pressure on the surface S is given by
w sin {w (x'x)} cos {f3 (y'----y)}eV0'+1'
dw d(3
w2 - x 1/w2 + (32
circulation distribution of a lifting surface of a thin wing.
rnwt,
n(b
in the first integral of eq. (20). Then the integral equation
becomes
i
2 u .chlffstechntk Bd. 14 - 1567 - Heft 72 0000sin {m (')} cos {n (n_n')} drn dn
w (, ) =
_L
y (',
') d' d'Sf
m m2 ± n2 (Lib)2m2xt
Vm2 + n2 (Lib)2 00 x2Lb"1'
+
y (',
') d' dn'S cos { xi Vi + u2 (')} cos {xbu /i + u2 ('}
(1 ± u2)3/ du. (29)ItU )
S 0
As the aspect is assumed large, the above equation is
ex-panded with respect to a small factor Lib. If the speed
para-meter b is assumed to be of the order of unity, L is also a small quantity. Taking the first term of the expansion, one
obtains
ï (' i') d'
d1i'jJ
sin {m (')} cos {n (i')} dm dn
00
0000
+
y (', ii') d'
dii'$J
sin {m (')}cos {n
dm dnx2U
J
s00
00+
Ç « y (t',') d' d'
cos {xh ui + u2 (')} (1 ± u2)'f du
(30)xU 3J
s oThe first two integrals are evaluated at once and the last one
s expressed by the modified Bessel functions. To show tile method of solution, we take a special case where the surface S has no camber, for example. Then to is a
=
-L_
$
y
d' +
y (,
) d function of i alone. Instead of the eq. (34), we consider the U-'
2U J
simultaneous equations as follows:ti
t1 t2i C2g(',y) d'
t2wi(TflJ
xi
,,-,
t1 (35)-
2Uj
w (i') w1 () + x0
g (2,
') K (ni') di1'.
(36)+
$$
(" ii') [3K0 { xb
')}The former one can be solved by well.known transformation.
s We have the following theorem:
K0 { xb (1i1')}J d' dr'
Ifwhere and are the coordinates of the extremities of ìe section i = const. of the surface S. The third term involves diverging integral but the principal value is taken in the
nse of
- 60
-It has been shown that the solution of the integral equation
Now we write
(20) i essential to find out the lift and tile resistance of the -flat ship. Though the kernel of tile integral equation is corn- i
(a',
) d' = g (,
) (32)plicated enough to frustrate any attempt of solving it, we can
u
find an analogy between it and the integral equation for the xb
We know there are approximate methods to find out the lift [3 K0 {
, b ( - i") }
K0 { x b (i - i") } J =
and the drag of the lifting surface. As pointed out before,we 16x
can consider a similar way in the case of e planing surface.
K (-1') .
(33)
In the first place, let us consider the çao of large aspect
Then we obtain an Integro-differetitial equation ds follow! ratto, Thare tha integral equation is expanded with respect to
the redtwt1 of the aspect ratio. We adopt dimensionless
tnrrrThìutes dainting the half length by i and half breadth to
i) =
g + x0 g (, i)'
-'
by b, uoh g
yib (28)
1
and change the integration variables as
g (, i) +
Çg (, i) K (iî1') drj'
(34)1
where x0 = xi.
t! High aspect ratio approximation
then
Multiplying u (x) =
There is also a result i
r
y
on each term of eq. (35) and integrating with respect to
we find
- x g (E2, i)
(37)To solve the eq. (36), one needs just the value of g (E TI) at
=
. Then integrate the eq. (37) with respect to X from to E2 and we obtaint
t
(E-E1) toi (TI) g(E,TI)
-
;:g (E, TI) dt
-
(E2-E1) g (E2TI)4
g (E' TI) 0i (TI)
V
-E1 f(E)dE
-E xE
E-E1 g(E,n)
dEç X-E
Substituting in the second and third terms of eq. (38), we ob-tain
or, putting
E2-Ei=2c(TI)
(40)g(E2,)=c(TI)wl(){7t_4x0c()}.
(41)Then eliminating w1 (TI) from eq. (36) and the above and omitting the second order term, we obtain another integral equation sudi as
g (E2 TI) e () (I) (ji)
{it 4 c (TI))
x0c ()$g
(E2,TI)K (TITI') dTI'This integral equation may be solved numerically. Examples of numerical results for the spanwise distribution of lift are shown in Fig. 2. One of them is a rectangular planing surface and the other is the triangular one, when the aspect ratio is
four. O
-+"
RECTANGULAR 5UFFACE Fig. i o'l
TRIAN6-ULAR SURFACE Fig. 2The total lift of the planing surface is
(38)
LpU2b$g(2,)d.
1
I
dX. (39) R = p U2 b £g2 (E2 TI) dTI { 1
} dTI
+
xo1
Itc(TI) IC2I t
+ p U2 b £ x g ( TI
f
g i') K (i - TI') dTI'. (45)1
1
As we have the relation
=
xb (
=
tJ
cos {xb u 1/1 + u2 (TITI')) (1 ± u2)'/ du the second term of eq. (45) can be writtenR1 pU2 x2 (b £)2 $(M2 + N2) (i ± u2)'i du (46) (43)
- 61 - Schiffstechnlk Sd. 14 - 1967 Heft 12
We need only an approximate solution which is linear with The resistance is given by respect to x0. So we take the first approximation for g (E TI)
which is given by omitting the term with multiplier x in eq. R = p U2 b £ $ g (E2 TI) w (TI) dTI . (44)
(37). That is
1
Because of the relation (36) and (41), it becomes i :Tt
(9X) (-)
i
(XE1) (E2-E)
X-(42) whereM}
j'
sin
N g (E0, ) cos(x b u j/i ± u2 'i) dl. (47)
This portion of the resistance gives the wave resistance. There is another term in eq. (45) that is
L
U2 b £ g2 (E' ){ 1+
4x0 d1.
1 (48)1
c(q) t JThere is no energy consumption by the wave corresponding to
this portion of the resistance and this suggests the existence of another component of resistance different from the wave
resistance. This term is a consequence of the singular behavior
of tite pressure distribution at the leading edge, but the
linearized theory does not prepare its physical interpretation. However, twodimensiona1 ow theory for a gliding plate [11] - obtained without regard to tile effect of gravity - indicates a drag originated by the momentum change due to the forn-tian of spray. It can be proved that the resistance component expressed by eq. (48) is a linearized form of this sort of drag. Therefore, it may be called the spray drag of a planing body.
Low aspect ratio approximation
The planing of an elongated body is of much importance in the problem of high speed craft. As the width of the wetted surface is much smaller than the length, the beam length ratio or the aspect ratio is much less than unity and the first appro-ximation may be obtained by the expansion of the integral equation with respect to the aspect ratio. The slender ship theory is equivalent to this idea, but the free surface
condi-tion near the ship becomes different in the case of tile planing hull because of the small value of the speed parameterx L.
Though the direct expansion of the integral equation (20) will give the first approximation, an easierway to get the lead-ing term of the expansion is to begin with the function i4
de-fined by the eq. (6). Employing dimensionless coordinates
= x/I,
'1 = y/b, = z/b (49)the Laplace equation for14' becomes
2p
!_ / b
\2 3214, 312+
32
L ) 3E2Then, if the function ip is expanded with respect to b/I, the leading term should satisfy the equation
(51)
312
32
and the next term is of the order of (b/I)2. Then omitting
higher order terms, one obtains a plane harmonic function in the E.plane, or in the y z.plane. Since 14 is a double sheet potential in a two-dimensional plane, it is given by
Making use of tile relation
z
- (yy')2+z2 -
cos (X (y - y')) dA (50)cJ) (x, y, z)
=
2,fj'? (x', y') dx'
dy'00
(' (
wexp{Xz + iw(xx')} cos {X(yy')}
dw
.j
dXJw2_x?
There is a relationexp {iw (xx'}
do) 5000) (02 - xX -00 when x> x'= xicos (1/Ax (x.x')}
when x < x'-Then the function x (x, y, z) is determined by the condition at the infinite upstream where there is no wave.
00
ecos (1/Ax (xx')) cos (X (yv')} dA.
(54)Hence tile velocity potential becomes
00 X
(J) (x, s', z) =
-
i
Jdy' J
(x',y') dx'00
fcos {1/
(xx')) cos (A (yy')) e
d).(55)
Differentiating by z, the slope of the wetted surface becomes
00 X
w (x, y)
'
J dy' J j (x',y') dx'
.fcos
{1/(x - x')} cos (X (y y')} A d..
(56)Integrating by parts and employing the function g (x, y) de. fined by eq. (32), or Lg(x,y) (x', y) dx' U one obtains L 14' (x, y, z) = -_±_ Ç zi4i (x, 3" dv'. (52) w (x, y) = J
(yy')2+z2
It 00- J sin (VAx (x - x') } cas (A (y - y')) X'1' cIA
s
00 X
L1/xC , C
j
dyj
g(x,y)dx
g (x, y') dy'jcos (A (y - y')) A dA
-00 0
and employing the function y (x, y) defined by the eq. (18), we find the velocity potential in the form
(57) Schiftstechnik Bd. 14 - 1967 - Heft 72 62
Making use of the dimensionless representation, one can write To show he solution explicitly, we take the simplest case that the surface has no twist. Then
(,
) is independent of î,i £ a / and we write w () instead of (s) (, ri). It is convenient to w 1) = g (, ii') ( ) dr' change the variable, such as
t b J
s
then
w here
s (a') Then one can write
+
V(L)'/2
J d' $ g (a',
') F ( - ', 1 ') d'
1
s (')
(58)where the contour of the wetted surface S is expressed by the equation
The equation (58) is the integral equation which determines the function g (, ) for a given function w (, Ti). In order to solve it, we make use of the following theorem. If
11
u(Y)= w(Ti)!n
2!.i
= - Ln
11 = ± s ().
iai
i\
u(Ti)---(---Jdll
JI JEfl \fl J
2 ()- Y +
/{s2 () - Y2} {s2 () 112}s() (i1Y)
on both sides of eq. (58) and integrating with respect to , we obtain
s2Yii + l/(s2__Y2) (s2ì2)
s (i - Y)
---(sinsin*'+sin2Osin2'+
(a9)+
sin 3 * sin 3 *' +. . . .).
d1
As (s) () is independent of , eq. (61) becomes
f (, Y) = w () s () sin *'
Jsin {Ax Lib ( - ')} J1 {As ()} J1 {As (')} A' dA.
(65)
Since we take tise term of the lowest order with respect to
xL/b, wemayput
S
sin {l/Ax0 Lib (')} J {As ()} J1 {Xs (')} A" dA
= Vx0Lib()
J1 {As ()) J1 {Xs (')} dAI)
+ O { (x0 Lib)"i}
(60) The above integral can be expressed by tise complete elliptic
integral of the first and the second kind, as follows: s () Y) =Sw(,Ti)W(Y,1i)dTi (61) 2 r JJ1 (As ()} J1 {x (')} dA K K (, Y; ', = F (
- ', - ')
W (Y, ) d1. ° E{s }] JI (62)S()
Ti = S () cos* ,
Y = s () cosO'
s () G ()=
s
J g (, Y) dY when s (g') <s (E)1f we assume that the maximum width is at the trailing edge, we can write the eq. (65) as follows:
It b
G(E)
2£
t
+ 2xoJ w (E') (EE')
[K{'
}_E
{s dE'a(E) s(E)
(66)
From the definition of the function G (), the lift coefficient
(63) becomes - 63 SchiffstechnIk Bd. 14 - 1987 - Heft 72 Multi ply in g
=
s () w () +
L (a') W (') d' W (Y, 1) =Because of the complicated and singular behavior of the ker-nel, to solve the above integral equation is a formidable task. However, an approximate solution can be obtained when the Froude number is very large. It has been assumed that the Froude number i so large that the speed parameter x = xL is at most of the order of bi!. Here we consider the case of still higher Froude number and assume x Lib muds smaller than unity. Then the first approximation is
g(,Y)
f (, Y) -i$
d'$
I (a', i') K (, Y; ', ') ds'' s (t')
Ln 2 sin (* + *')
2 sin (O - O')
= w (E) i/sa (E) Y2. (64
To find out the total lift, eq. (63) is integrated with respect toY.
g (, Y) =
bf (,Y)
Ls()
Çd' $ g (t', Ti')K (, Y;
',') d'
t s (e')
2L CL
-QU2S S- Xw (1)
42btG (1)
(67)where 2 is the aspect ratio 4 b2 / S and u. is the area ratio
S/4b
of the planing surface. Since the complete elliptic integral of the first kind K (k) has a logarithmic singularity at k = 1, the formula (67) does not hold when the planing surface has parallel side edges. Another type of expansion is needed but is not yet obtained. The equation (67) shows that the smaller the aspect ratio is, the more important the effect of gravity becomes. Therefore, the gravity-free approximation does not hold in the case of low aspect ratio planing bodies. The resistance is given by the integralR= QUff?(xY)w(xY) dx dv
=
QU2bEId()d.
(68)Substituting eq. (66) and after some reduction, we find
R = U2 b 1w2(1) + 2 4 1 J d
O)2()d}
I d s ()1
Et
d'
(69)s ()
since we assume s (1) = 1, s (-1) = O. It is of some inter-est to observe that the resistance does not vanish at infinite
Froudo number. It can be proved that the fluid motion at
Schiffstechn!k Bd. 14 - 1967 Heft 72 64
-infinite Froude number is identical with the fluid motion around a thin wing. So the wave resistance is equivalent to the induced drag. The induced drag of a wing is calculated from the energy in tile trailing vortices. In the case of the planing surface, the induced drag due to the vortices should be halved, because the fluid motion exists only in the lower half space. On calculating it, we find it to be ('/4) t g U2 b2 co (1) . Then the residue of the resistance gives the spray drag, and is given
by
i
R9 = g U2 b2
Jds()
w- () d.
(70)2 d
As an example, let us consider the lift of a triangular or delta planing surface, for which we have the relation
It is assumed further that there is no camber, so that the slope of the planing surface is constant.
Then the lift coefficient given by the eq. (67) becomes
1 1 CL =
ha
+
8g1«
{J
K(t)dt__$ E(t)dt_}
2t
U2 ¿I =cs1+0664
\2t
U2/ (73)This result indicates that the effect of gravity increases the lift, while it can be shown by tile eq. (42) that the lift is
re-duced by the effect of gravity when the aspect ratio is large. These reverse tendencies have been proved by experiments [12].
References: J. H. M i e h e il Phil. Mag., (1898).
V o s s e r s , Schiffstechnik, (1962).
L a m b, Hydrodynamics, 6th edition.
E. H o g n e r, ist Internat. Congr. App. Mech. Delft, (1924). W e i n b i u m, Z.A.M.M., (1930).
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