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Motion o f a Stream o f Finite Depth past a Body. 107

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The Motion o f a Stream of Finite Depth past a Body.By Robert J ones, M.A. (1851 Exhibition Scholar of the University College . of North Wales, Bangor).

(Communicated by Dr. R. T. Glazebrook, C.B., F.R.S. Received June 8, 1915.)

When a circular cylinder moves uniformly in an ideal fluid (i.e. frictionless and incompressible) at rest at infinity, the resultant force acting on it is zero, if no external forces act. This is, however, only true when the motion is the usual potential motion. Supposing that in addition to the potential stream produced by the motion of the cylinder a circulation around it be considered, the velocity of the fluid is increased on the one side, and decreased on the other, and this produces a force acting on the cylinder perpendicular to the direction of motion.*

Kutta has applied this method of considering the motion of an infinite fluid to determine the thrust on a lamina, and systems of laminae, plane and circular.2- The cyclic constant of the circulation he leaves arbitrary, and

Chaplin"' Hydrodynamics’’ PP' al*fl 'a (1906). Lanchester, ‘Aerodynamics,’

+ Kutta, -‘ fiber eine mit den Grundlagen des Flugproblems in Bezieliung stehende Z mduu.M onale Strbmung,” ‘ Sitzungsberichte d. k. Bayerischen Akademie d. Wissen- ■schaften, Math. Phys. Klasse,’ Jahrgang 1910, 2 Abhandlung.

VOL. XCII.— A, K

on July 15, 2018http://rspa.royalsocietypublishing.org/Downloaded from

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108 Mr. R. Jones.

determines it ultimately, by introducing a condition that the velocity of the fluid at one edge of the lamina be finite. He obtains, for the total thrust on the circular lamina, the expression

AirpVWs in -s in (J + 0 ) ,

where Y is the relative velocity of the fluid and body, rthe radius of the circular arc,

2a the angle at the centre,/3 „between the direction of motion and chord,

b the length of the lamina, . p the density of the fluid.

He claims that his results agree fairly well with experimental resultsobtained by Lilienthal. _

The same formula has been obtained independently by Dr. Tschapligin for a more general case. He considers an aerofoil with a section of knowncontour. ^

Dr. Joukovsky has also investigated the problem,■]“ and has obtainedinteresting contours for the sections of an aerofoil and a strut by representing a circle conformally. He obtains the force acting perpendicular to the motion of the immersed body in the form pJV, J being the circulation around the known contour, Y the velocity, and p the density of the fluid.

Blasius applies the methods of the Theory of Functions to hydrodynamical problems related to the above.* The section of the body to be examined is conformally represented as a curve, the motion around which can be determined.

All these problems treat, however, of motion in an infinite fluid. Prof. L. Prandtl (of Gottingen) pointed out that it would be advisable to consider the motion of a stream of finite breadth flowing past a body, and the author’s thanks are due to him for suggesting the lines on which the investigation should be carried out. The problem resolves itself into two cases, which will in turn be considered. First the boundaries of the stream are supposed to be free, and secondly the fluid is taken to be flowing in a channel with rigid boundaries. The method follows somewhat the lines taken by Blasius, the motion is two-dimensional, and is supposed to be symmetrical with respect bo the plane of symmetry of the immersed body. We start wuth the hodograph of a hypothetical motion and proceed to expressions giving the contour of the section of the body consistent with this motion.

* 4 Mbscauer Mathematische Sammlung,’ vol. 28 (1910).t 4 Zeitschrift fur Flugtechnik u. Motorluftschiffahrt,’ November 26, 1910.J 4 Zeitschrift fiir Math. u. Physik,’ 1910 and 1911.



Motion of a Stream of Finite Depth past a Body. 109

(i) Stream with Free Boundaries.Let the stream and the immersed body be represented by fig. 1, which will

be called the z (= x+iy) plane.The chord AB of the body is parallel to the x axis and the figure is sym

metrical with respect to the y axis. Let V be the velocity of the fluid at infinity.

F ig. 1.

The image of the hodograph (inverted with respect to the x axis) is showrn in fig. 2, w (== u —iv) plane.

F ig. 2.

The curve ACB represents the upper surface of the body, and AC'B the lower surface. EDF represents the upper free stream and ED'F the lower. EDF and ED'F are coincident circular arcs with O as centre and V as radius, for the velocity is constant and equal to V along the free stream. The interior of the slit FDED'F represents the space exterior to the free boundaries, whereas the exterior of ACBC' represents the interior of the body.

Let fig. 2 be now transformed conformally on to the t plane, fig. 3, so that the circular arc EF lies along the imaginary axis of t.



n o Mr. R. Jones.

The required relation between w and t is easily seen to bew = d(w — Y)/(w + V), (1)

where d is a constant ( = to the distance of the point in fig. 3 corresponding to 0 (fig. 2) from the origin in t). CO7 lies as before on the real axis, the positions of A and B are not known.

Now transform fig. 3 so that the slit EF becomes a circle, radius 11 in the

F ig. 3. F ig. 4.

f (= f +iv) plane (fig. 4). If the length of the slit be 2a, the required transformation formula is easily shown to be

2 tja = £/B — R/£. (2)From the symmetry of the problem, the arcs D'F and D'E are equal, E

and F represent — and + infinity in the z plane, i.e. the regions from which the fluid comes and into which it goes respectively.

Hence we assume a source and a sink to exist at E and F (fig. 4) respectively, and find a fluid motion in the f plane containing the circle DED'F and Ctj'C't) as closed stream-lines, and to be made up in part by the source and sink.

We first of all suppose a circulation about the origin introduced; this will not disturb the stream-line EDFD' of the soqrce and sink. To obtain the closed curve Crj'C'rj, we assume a uniform stream to flow from — i oo round the circle EDFD'. These three motions, superposed, satisfy the required conditions, and the complex potential of the motion in the f plane is

' + = A log ^ E ^ r e + Bi log £+ Gi(?— (3)

where (K, 0) are the co-ordinates of F, A the strength of the source and sink,



Motion o f a Stream of Finite Depth past a Body. I l l

B the cyclic constant of the circulation, and C the velocity of the uniform stream. The exact meanings of the constants A, B, C, with regard to the s plane we proceed to investigate.

Now w = dx/dz,

therefore z — . -yCr, (4)) w J

v) being obtained as a function of £ from (1) and (2) and dx/d£ from (3).If then the form of the contour Cij'C't can be found, the section of the

body can be obtained by integrating (4).The equation i Jr = constant for the stream-lines gives us the equation for

Grj'C't] for a value k of the constant corresponding to that curve.

A tan-1 — 2B sin B cos 9)(£—it cos 6 f + (j]2—B2 sin2 6)

+ 4B log(P + 7?2) + C k. (5)

The curve Crj'Cr] being a closed stream-line, no fluid crosses it, and the velocity vanishes at two corresponding points on it. We make use of this fact to determine k.

wdx/d£ = 0 gives us these points.

dx/dl £ = A /(£ -E e* ) - A / ( ? - R e- ’*) + Bi/£-f Ci-f- (XR2/£ 2, (6)

' = i*+ ( n - ac 003 + ¥ (c+ u sin 0 “ 1 003

. + t § : ( i r 2Ccos9) + 1 = 0’ <7>a reciprocal biquadratic.

Let £ i±ivibe the required roots. The upper sign gives the point B, the lower one the point A ; the other two roots are merely the images of these points in the circle £ = R.

We obtain immediately

2£i + 2£i/(£i2+ j?i2) = - 1 /C . (B/R —2C cos

and ^24. 2 + __ 1__ , _ M i 2 _ 2 /n A sin 6 B cos 0\ {Q,* Zi2 + Vi* £x2 + V i * ~ C \ K / ( 8)

and having found £1 and rj1} k is obtained on substituting in (5). e now proceed to find a criterion for determining B.

In Kutta s problem B is determined by the condition that the velocity at one edge of the lamina is finite. This is assumed to be the case in the present problem, but another condition presents itself. Terms involving tan 1 are present in the expressions for x and y after integrating (4), which



112 Mr. R. Jones.

change by 2ir when the integration extends round the contour, and thus z is not left single-valued. We rid ourselves of this difficulty by so chopsing B that these terms vanish.

From (1) and (2)— = 1 (£+<*)(?+ff) /Q\w v ( r - a) ( r - /3 ) ’ w

where « and /3 are the roots of f2 + 2<fB^/a—B2 = 0; hence, on integrating (4) we have terms involving

f dZ f dK r <*r [dK W a n d f] ( ? - « ) ’ K - B e * ’ J f - B * - " ’ J ? ’ JC* J

d£ ( 10)

Either £ = a or f3corresponds to the point 0 (fig. 2) outside the contour, and, since a/3 = — B2, the other must lie within the circle £ = It. Let that point be f = a.

The four points f = 0, a, Re10, and Be-1'9, lie within the contour around which we integrate, and consequently each of the first four integrals (10) is equal to 2m, whereas the other two vanish. Collecting together , the coefficients of these terms, and remembering that Jcfe has to vanish when taken round the contour, the following equation for B is derived:—

4adAsm 6 _2a<fAsin0 _ -na2 sin2 0 + d? /(/ —asin#) — (11)

where l2 = a2 + d2.Let us write E Then

—Vz

1 (this only fixes our scale) and integrate (4).

A (a2 sin2#—d2) a2 sin2 6 + d2~ adiA sin 6

a2 sin2 6 + d2

log:

log 1

-eie , 4adiA sin 6■e~ie a2 sin2

. 1 2 cos 9\r2 r r

log?,

+ ilog^(B + 4C^/a) + fC (^ - l /f ) . (12)The terms in log r vanish in virtue of (11).When f = cos 0 + i sin 9,t = ia sin 0 by (2). This corresponds to F, fig. 3.If COl (fig. 2) then 1 (fig. 2) is the point (Y cosX, V sinX), hence

by (1)sin 9 . a /d = tan | X. (12a)



We shall limit ourselves to cases where is nearly equal to 7t/ 2 and X is small, hence afd is small. This gives us a small and /3 large.

For a finite A, B will be small by (11), and for a finite C, fi will be small

by (8).

Further, from (12a),


a2 sin2 0—d2 a2 sin2 9 + d2

—cos X.

If then squares and products of A ad sin 6 l2 + al cos O’

and a be neglected a goodL

approximation for Z is given by

—Yz + constant = — AcosXlog + Aisin Xlog^l + ^ —

- iC ( £ - l / £ ) + ;C «(t2+ l / £ 2), (13)the constant being the term involving log/3 in (12).

[The origin of z will henceforth be supposed shifted, so as to eliminate this constant.]

Equating the real and imaginary parts of (12), and making use of the equation ^ = k in the expression for y, we obtain

—Yx = - ^ cosX log i t - c o s ^ + - s i p Of 2 ( f —cos 0)2 + (?7 + sin 6)2

and

+ A sin X j^tan 1

+ c (’,+ F T v

‘f j j y —y cos #)P - ^ + l -

— 2Cx (^7]-

- 2 f cos- 2 tan 1 -

f-

(ID

- Y y = /ecosX + JA sinXlog {(£2—??2+ l —2£cos 0)2 + 47?2( f -c o s -(A s in X -iB co sX ) log(£2 C(cosX—l) [ f - £ / ( £ 2+c « [ p +( p - ^ ) / ( f +vy i (15)

For the values of x and y at the free boundaries we have to put = 1,and substitute for k, A (7r —0) for the one boundary and —A0 for the other.

Then

~ Y yi = A(7t — 0)cos X + A sinXlog 2(£'—cos 0) + 2C a(f:- Y y 2 = A# cos X + A sin X log 2(f" — cos 0) + 2Ca(g"2—

- V* = - 1 A cos X log _ A sin x ton- , S + 2C1 — gCOS 0 + 77 s m 0 | '



114 Mr. R. Jones.

Consider yi—3/2 and xx—x2at infinity, where £' = f " = cos#, andr)' = 7)" = sin #.

Then V f a —y2) = -A rr cosY(xi—X2) = — A7t sin X,

and (y\—3/2)/(X] —£%) = eotX,

F ig. 5.

and if the thickness of the stream he T, thenT = Att/Y . (16)

Hence if Y be given, the constant A depends on the thickness of the stream.

= Y ? + 2 d y a - lt,2—2d^/a — l

M /a ,

Again consider •IV = w

(Z2 + rj2 — l)2 ■+4y2—M 2 / a2 . (17)

On substituting £ = cos #, and rj = sin#, v ju = tanX (fig. 2), and this equation gives, as before, ajd , sin 6 = tan the direction of the stream at infinity.

Now, if instead of substituting £ = cos #, rj = sin #, we substitute the values of £ and rn obtained from (7) or (8), we can find the angle between the tangent at the edge of the body and the line joining the extremities of the section. To simplify the formulae, let us put 9 = r.

We have seen that B is small and of the same order as a (A and C being finite), hence fi is small by the first of the equations (8); we therefore neglect £2 in (17).

If fi be the angle between the tangent at the edge of the body and the chord

tan fi = —v/u ___ 4dja . (77 + 1 / 77)

(v + l / v)2 - 4cd2/ a 2 ‘

But from the second of the equations (8), .v2+ l / v2 = 2(1 + A/C) hence 77 + 1/77 = ^ [2 (2 + A/C)]r

ajd = tan |X.also



Motion o f a Stream, o f Finite Depth past a Body. 115'

Therefore

tan yu. =

but

thereforeor

4cot^Xv/[2(2 + A/C)] = 2y/[2(2 + A/C)]/2cot 2(2 + A /C )—4cot2 £X, 1—[2(2 + A /C)/4cot2|X] ’

tan fi — 2 tan \)x 1—tan2 \ f

2tan£yu = tan AX, . y/[2(2 + A/C)] A/C = 2 tan2 cot2 £X — 2, (18)

so that, given X, fi and A, C is determined.For constant A and X, //, decreases as C increases and the body becomes

flatter.When X is small compared to /a the term tan2A-//, cos2 |X becomes large,

and either A becomes large or C small. The thickness of the stream increases with A for a constant V, or V increases with A for a constant thickness, hence when p, is small we have the case of a body in a large stream or in a swift one.

From (14) it will be seen that the term that contributes most to V# is C[v + vl(v2 + ¥ ) l hence it appears that the effect of increasing C is to make the body larger and flatter (C does not contribute so much to 77, both terms- in C being small since a and (cosX—1) are factors).

When # = i7r the equation (11) reduces to

AsinX = B(2cos JX — 1).

When X vanishes altogether, A becomes infinite by (18), but if we assume B to be finite A sin X remains finite, and we have the problem of a solid moving in an infinite fluid.

Finally, we have to examine the effect of varying 6.£ is small when it corresponds to the edge of the body, and (£—cos 6) is

small by (8); hence it appears that varying 6 does not appreciably alter the position of the body, but from (15a) it is evident that for points on the free stream lines, y varies appreciably with 0. For points in the plane of symmetry rj = 0 and f = 1. Hence as 0 varies, the position of the body relative to the stream varies.

The above observations have been confirmed by numerical calculations, but it has not been considered worth while to introduce the figures here.

It now remains to determine the thrust on the body.The more straightforward and simple way is to consider the vector

difference of the momentum of the fluid moving towards the body on the one side, and moving away at infinity on the other. The quantity flowing past a fixed section of the stream at infinity in unit time is pYT, and its



116 Mr. R. Jones.

momentum is PV2T, hence by the parallelogram law the thrust perpendicularto the x axis is ' ,inN

P = 2pTV2 smX. (19)As a test on the analysis, the alternative analytical method has also been

worked out.P = £ I w2dz around the contour.

Therefore

cI x _ (lXdz d £ dz *

and w are known in terms of hencedK

■ pi =1 1 2 (? + « )(£ + /3 )< f r5J (? + « x r+ /8 ) r f r

Splitting up by partial fractions, and remembering that

- If-whereas the others vanish, P reduces to —27ri pX^ 0n making usea2snre/ + a2 2of equation (11).

Therefore

P 27rApY sinX (19a), since 2ad sin 6 sin X,a2 sin2 6 + d2

but A = thereforeIT

P = 2Tp V2 sin X.We have seen that when X = 0, A is infinite and, assuming the circula

tion B to be finite, that A sin X is finite and equal to B. Substituting in (19a) we obtain P = 27rpYB* for an infinite fluid (19&), or for finite values of A and X

P = 27r/>YB(2 cos ^X—1).

(ii) Stream with Straight Rigid Boundaries.We proceed to treat in a similar way the motion of the fluid in a channel

with fixed walls.Pig. 6 represents the z plane, denoting the section of the body and the

channel. Tlie hodograph (w diagram) is given in fig. 7.

* Compare Joukovsky, loc. cit.



Motion of a Stream o f Finite Depth past a Body. 117

________________ D__________________

CE A" C1 b F

D1F ig. 6.

In fig. 7 OE = OF = Y represents the velocity at infinity.

F ig. 7.

The two lines DED and DFD along which the direction of the velocity is constant represent the fixed boundaries of the channel.

I t will be seen that the present problem differs from the other in that the direction of the velocity is now constant along the external boundaries, whereas in the previous case the magnitude was constant.

The transformation corresponding to (1) § 1 is not needed, since the two coincident lines DED and DFD are already straight, consequently we use directly the transformation equivalent to (2) § 1 to obtain the f diagram from fig. 7.

In fig. 7 let M be the middle point of DD', DD' = 2a, FD = b, FM = (a — b)/2 = e/2 (say).

Then OM = Y —<?=/; let f / a = d.Then the required transformation is

2 ( i* - /) /a = l / f + £ (20)The f diagram is the same as fig. 4, the radius K of the circle being taken

to be unity. % = yfr = k, and d^jdw = 0 are the same as in § 1{equations (3), (5), and (6; ).

When £ = cos 6 + i sin 6, 2( w - f ) / a = 2 cos 0,or when w = Y (i.e. at E and F) this reduces to (a — b)/a = cos 6. (21)

1 2 £■ ow~ ~ where a and /3 are the roots of £2 + 1 = 0.

“ J s f f a8“ (4)-z



118 Mr. II. Jones.

This has to vanish round the contour as before, and the condition- corresponding to (11) is •

A sin d + cos

2Ca B. (22)

For purposes of approximation, we will assume that the change of velocity in the neighbourhood of the body is small compared to the velocity at infinity, i.e.a and b small compared to V, and neglect terms of the second order in a ; a is as before taken to be the smaller root of = 0,.hence a. is small.

The approximate expression for z reduces toaz2

A2 (d + cos

log %-eief - e - " ’

(23)-

the origin being changed as in (15), and use being made of (22). This equation after equating real and imaginary parts gives

axT c « ,+v A . log a/ [ (£ ~ cos fl)2 + (?7 —sin fl)2] (24,

2(cf + cos0) ° \ / [ ( ^ —cos 0)2 + (i7 + sin v

[x = 0 when r] = 0, and changes its sign with 77],

and - f = C { * ) { “ + 2(rf + cos« )}

log ({8+V )+ 2 {T f Z Z e y <25>k having the same meaning as in (5).

Multiply both sides of equation (24) by 2 + cos 6), and consider thevalue of x when 8 — %tt.

ad is then equal to V, for a = when = §7r and 0, also 2<xd = — a.(a + /S) = — a2— a/3 = — 1 when a2 is neglected. We want to-find x at the edge of the body, i.e.at the point where £2 is supposed to be negligible compared to rf, i.e. the point f i —

This gives us* Va: = — C ^ + l / ^ + A l o g ^ l ^ l ,

= -C < ,1 + l / „ ) + A l o g 2 l ± i ) a = |

which reduces to

. -C a /[2 (2 + A /C )] + A Iog^ -| + ^ ; v ; 2 . (26)

in virtue of (8).



Similarly consider (25), and y is given byY y = - |B lo g ( £ 2+i?2) + *. (27)

We are justified in putting k — IB log (tjf + r/f) (to the same degree of approximation), the point being the point at whichA tan"1 [2£/(P + »?2- l ) ] equals Cf(£2 + r,2- l ) / ( f 2 + ??2) when f 2 is neglected in comparison with if.

Hence (27) gives y = 0 at the edge.Let £' and f " be the values of f in ^ = when = 0, f ' being positive

and £" negative. These correspond to the maximum ordinates of the upper and lower surfaces respectively, then

Yy' = 4 B { - lo g P } = B ( - lo g f ) ,Y y " = I B j - l o g f '2} = B ( - l o g r ,)>

and the thickness of the body y '—y"

= B /Y . log(r/n- (28)The exact value of y has been found to be given by

\ay+ Clog/3 = £C[log(f2 + •f?2) + log { (f—/3)2+ 1?2} — log { (£ -« )2 + ??2}]A x —2 sin #(£—cos fl)

2{d +cos 0) | 2 + i?2—2£cos 0 + cos2 sin2Now to obtain y for the sides of the channel put = 1 and this

reduces to\ay + C log £ = log (1 — 2/9 £ + /32)- lo g (1-- 2 « £ -« 2)

+ o/h ,A— m tan_1(± tan 0),2(6? + cos 6)or since a/3 = 1,

^ = ■ /T- —— ^ tan_1( + tan 0).2 2(6? + cos0) '

But a (d + cos 6) = V, by (21),therefore Yy = A# for one side of the channel and = A( — nr+6) for the other side.

The thickness T of the channel is given byT = A tt/Y , as in § 1. (29)

Since the edge of the body is always on the straight line y = 0, it is seen that the distance of the body from one side of the channel varies directly as 6.

The equation (22) reduces, when 6 = J77-, toA -2C W = Bd,

when a2 is neglected — 2a.d = 1,

d = Yfa, therefore (A + C)/V = B/a.


(30)



120 Mr. R. Jones.

Hence if A, C and Y be fixed, and the body be placed in the middle of the channel, the increase of velocity on the upper boundary, or the decrease on the lower boundary, varies directly with the circulation round the body.

Again, since cos 6 = (a—b)/a,it follows that 2a sin2 10, which shows that b vanishes when 6 — 0 and that it increases with 6. Hence, the nearer the body is to the top of the channel, the faster the fluid flows above it, as was to be expected.

Let us now consider tan /x, where /x has the same meaning as before,

_ the imaginary part of w ^[1 — 1 /(P + ??2)] an^x— the real part of w |[1 + 1 /(£2+*?a)] + 2<2

This vanishes when £2 + y2 = 1, i.e. along the fixed boundaries.We shall consider the case of 6 = t. Equations (8) give us, on

neglecting £2,V/(2A/C) _ 2X/(2AC) _ 2av /(2AC).

-B /2 C + 2d 4C V -aB ’tan fjl

aB we neglect, being of order «2.

Then tan /x = ^ \ / (31)

and as before decreases as C increases.We now see that, given Y, T, %, and tan fi, we have sufficient data to

determine a A, B, and C, if we make use of our condition (22).Equation (29) determines A, and after substituting in (26) for A, C is found ;

a is then obtained from (31) and finally B from (30).As already stated 8 depends on the position of the body in the stream, and

knowing a and 0, b is immediately obtainable. *We have still to consider the thrust on the body.This is done in the same. way as before, viz., by considering the integral

P2 j

w*dz taken round the contour.

P is, in this case, equal to j" —— ~ d£, which gives on being

reduced P = Trap^C+ Bd +A sin$), ' (32)when 6 = \ tt,A + C = Bdand ad — Y,hence P = 2irpVB. (33)*

When 6 ^ \nr, A sin 6 = 2C a(d + cos 6)+ B(<f + cos 6),* = — d + /(d?— 1), a(ef + cos0) = Y and cos

therefore * d = ( y + b)ja—l,* Cf. (196) and Joukovsky.

i



121Motion o f a Stream o f Finite Depth past a Body.

and x/(d2- l ) = (V + h)/a • v/ [ l - 2 a / ( Y + J)]= (V + 5)/a — 1 — |a / (V + + higher powers,

- ^ a/ (Y + b) approximately.therefore « Hence

A sin 6 = — Ca(d+p™d)+ CV BV 'V + 6 + a '

Substituting in (32) we obtainp = 7rp{B(arf + Y) + &&C/(V + &)},

or = 27r1oBY — 7rpacosdB + 7rpabG/(Y + b),

which shows that for small displacements from the central position in the channel, the thrust on a fairly flat body (/jl small and of the same order as a by (31)) differs from the thrust at the central position only by quantities of the second order in a.

Equation (32) gives the exact thrust on the body without making any assumptions as to the value of a.

A note may be added concerning the motion when the chord AB of the body is not placed parallel to the axis of x in the z plane.

In the free stream case; if Xi and X2 be the direction angles of the stream at + and — oo respectively, and /xi and /z2 the inclinations of the tangents at A and C respectively to the axis of x, the magnitude and direction of the resultant thrust are easily found, if /xi—Xi = fi2—X2. The case discussed is extended to include this case by merely turning the z plane about the origin, and moving the origin in the t plane (fig. 3). The resultant thrust has then two components parallel to the x and y axes, of magnitudes,

2TpY2 sin |(X2 —Xi) sin |(X2 + Xi) and 2TpY2 cos -|(X2—Xi) sin |(X2 + Xi) respectively.

When, however, 9 ya2—X2, the problem becomes more complicated,and fig. 4 no longer remains symmetrical with respect to a line parallel to the axis of x. Nevertheless it may be possible to introduce further terms into the expression (3) for cfr + iyjr that will give a velocity potential consistent with the new conditions, for example a term corresponding to a doublet on the 7] axis between E and y would make yai —Xi and fi2—X2 unequal. Various systems of sources and sinks would apparently give us a variety of contours similar to Such a procedure has already been used with advantagefor comparing the efficiency of balloon and airship models.*

The latter remarks apply also to the rigid boundary problem.

* Fuhrmann, ‘ Jalirbuch der Motorluftschiff-Studiengesellschaft,5 vol. 5 (1911-12).



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