Uniplanar motion alone is so far amenable to analysis; the velocity function φ and stream function ψ are given as conjugate functions of the coordinates x, y byw = ƒ(z) where z = x + yi, w = φ + ψi,(1)and thendw=dφ+ idψ= −u + vi;dzdxdx(2)so that, with u = q cos θ, v = q sin θ, the functionζ = −Qdz=Q=Q(u + vi) =Q(cos θ + i sin θ),dw(u − vi)q2q(3)gives ζ as a vector representing the reciprocal of the velocity q in direction and magnitude, in terms of some standard velocity Q.To determine the motion of a jet which issues from a vessel with plane walls, the vector ζ must be Constructed so as to have a constantdirection θ along a plane boundary, and to give a constant skin velocity over the surface of a jet, where the pressure is constant.It is convenient to introduce the functionΩ = log ζ = log (Q/q) + θi(4)Fig. 4.so that the polygon representing Ω conformally has a boundary given by straight lines parallel to the coordinate axes; and then to determine Ω and w as functions of a variable u (not to be confused with the velocity component of q), such that in the conformal representation the boundary of the Ω and w polygon is made to coincide with the real axis of u.It will be sufficient to give a few illustrations.Consider the motion where the liquid is coming from an infinite distance between two parallel walls at a distance xx′ (fig. 4), and issues in a jet between two edges A and A′; the wall xA being bent at a corner B, with the external angle β = ½π/n.The theory of conformal representation shows that the motion is given byζ =[√ (b − a′·u − a) + √(b − a·u − a′)]1/n, u = ae−πw/m;√ (a − a′·u − b)(5)where u = a, a′ at the edge A, A′; u = b at a corner B; u = 0 across xx′ where φ = ∞; and u = ∞, φ = ∞ across the end JJ′ of the jet, bounded by the curved lines APJ, A′P′J′, over which the skin velocity is Q. The stream lines xBAJ, xA′J′ are given by ψ = 0, m; so that if c denotes the ultimate breadth JJ′ of the jet, where the velocity may be supposed uniform and equal to the skin velocity Q,m = Qc, c = m/Q.If there are more B corners than one, either on xA or x′A′, the expression for ζ is the product of corresponding factors, such as in (5).Restricting the attention to a single corner B,ζn=(Q)n(cos nθ + i sin nθ) =√ (b − a′·u − a) + √ (b − a·u − a′),q√ (a − a′·u − b)(6)ch nω = ch log(Q)ncos nθ + i sh log(Q)nsin nθqq= ½(ζn+ ζ−n) =√b − a′√u − a,a − a′u − b(7)sh nΩ = sh log(Q)cos nθ + i ch log(Q)nsin nθqq= ½(ζn+ ζ−n) =√b − a√u − a′,a − a′u − b(8)∞ > a > b > 0 > a′ > −∞(9)and thendΩ= −1√ (b − a′·b − a′),dw= −m,du2n(u − b) √ (a − a·u − a′)duπu(10)the formulas by which the conformal representation is obtained.For the Ω polygon has a right angle at u = a, a′, and a zero angle at u = b, where θ changes from 0 to ½π/n and Ω increases by ½iπ/n; so thatdΩ=A, where A =√ (b − a·b − a′).du(u − b) √ (u − a·u − a′)2n(11)And the w polygon has a zero angle at u = 0, ∞, where ψ changes from 0 to m and back again, so that w changes by im, anddw=B, where B = −m.duuπ(12)Along the stream line xBAPJ,ψ = 0, u = ae−πφ/m;(13)and over the jet surface JPA, where the skin velocity is Q,dφ= −q = −Q, u = aeπsQ/m= aeπs/c,ds(14)denoting the arc AP by s, starting at u = a;ch nΩ = cos nθ =√b − a′√u − a,a − a′u − b(15)sh nΩ = i sin nθ = i√a − b√u − a′,a − a′u − b(16)∞ > u = aeπs/c> a,(17)and this gives the intrinsic equation of the jet, and then the radius of curvatureρ = −ds=1dφ=idw=idw/dΩdθQdθQdΩQdudu=c·u − b√ (u − a·u − a′),πu√ (a − b·b − a′)(18)not requiring the integration of (11) and (12)If θ = α across the end JJ′ of the jet, where u = ∞, q = Q,ch nΩ = cos nα =√b − a′, sh nΩ = i sin nα= i√a − b,a − a′a − a′(19)Thencos 2nα − cos 2nθ = 2a − b·b − a′= ½sin22nαa − a′a − a′·u − bu − bsin 2nθ = 2√ (a − b.b − a′) √ (u − a·u − b′)a − a′·u − b(20)= sin 2nα√ (a − a·b − a′);u − b2nc(1 +b)√ (a − b·b − a′)φρu − b√ (u − a·u − a′)(21)=a − a′ + (a + a′) cos 2nα − [ a + a′ + (a − a′) cos 2nα ] cos 2nθ×cos 2nα − cos 2nθ.(a − a′) sin22nαsin 2nθAlong the wall AB, cos nθ = 0, sin nθ = 1,a > u > b,(22)ch nΩ = i sh log(Q)n= i√b − a′√a − u,qa − a′u − b(23)sh nΩ = i ch log(Q)n= i√a − b√u − a′,qa − a′u − b(24)ds=dsdφ=m=cQdudφdtπquπqu(25)πAB=∫abQdu∫ [√ (a − b) √ (u − a′) + √ (b − a′) √ (a − u)]1/ndu.cqu√ (a − a′) √ (u − b′)u(26)Along the wall Bx, cos nθ = 1, sin nθ = 0,b > u > 0(27)ch nΩ = ch log(Q)n=√b − a′√a − u,qa − a′b − u(28)sh nΩ = sh log(Q)n=√a − b√u − a′.qa − a′b − u(29)At x where φ = ∞, u = 0, and q = q0,(Q)n=√b − a′√a+√a − b√−a′.q0a − a′ba − a′q(30)In crossing to the line of flow x′A′P′J′, ψ changes from 0 to m, so that with q = Q across JJ′, while across xx′ the velocity is q0, so thatm = q0·xx′ = Q·JJ′(31)JJ′=q0[ √b − a′√a−√a − b√−a′]1/n,xx′Qa − a′ba − a′b(32)giving the contraction of the jet compared with the initial breadth of the stream.Along the line of flow x′A′P′J′, ψ = m, u = a′e−πφ/m, and from x′ to A′, cos nθ = 1, sin nθ = 0,ch nΩ = ch log(Q)n=√b − a′√a − u,qa − a′b − u(33)sh nΩ = sh log(Q)n=√a − b√u − a′.qa − a′b − u(34)0 > u > a′.(35)Along the jet surface A′J′, q = Q,ch nΩ = cos nθ =√b − a′√a − u,a − a′b − u(36)sh nΩ = i sin nθ = i√a − b√u − a′.a − a′b − u(37)a′ > u = a′eπ/sc> −∞,(38)giving the intrinsic equation.41. The first problem of this kind, worked out by H. v. Helmholtz, of the efflux of a jet between two edges A and A1in an infinite wall, is obtained by the symmetrical duplication of the above, with n = 1, b = 0, a′ = −∞, as in fig. 5,ch Ω =√u − a, sh Ω =√− a;uu(1)and along the jet APJ, ∞ > u = aeπs/c> a,sh Ω = i sin θ − i√a= ie−1/2 πs/c,u(2)PM =∫∞ssinθ ds =∫e−½πs/cds =ce−1/2 πs/c=csin θ,½π½π(3)so that PT = c/½π, and the curve AP is the tractrix; and the coefficient of contraction, orbreadth of the jet=π.breadth of the orificeπ + 2(4)A change of Ω and θ into nΩ and nθ will give the solution for two walls converging symmetrically to the orifice AA1at an angle π/n. With n = ½, the reentrant walls are given of Borda’s mouthpiece, and the coefficient of contraction becomes ½. Generally, by making a′ = −∞, the line x′A′ may be taken as a straight stream line of infinite length, forming an axis of symmetry; and then by duplication the result can be obtained, with assigned n, a, and b, of the efflux from a symmetrical converging mouthpiece, or of the flow of water through the arches of a bridge, with wedge-shaped piers to divide the stream.Fig. 5.Fig. 6.42. Other arrangements of the constants n, a, b, a′ will give the results of special problems considered by J. M. Michell,Phil. Trans.1890.Thus with a′ = 0, a stream is split symmetrically by a wedge of angle π/n as in Bobyleff’s problem; and, by making a = ∞, the wedge extends to infinity; thench nΩ =√b, sh nΩ =√n.b − ub − u(1)Over the jet surface ψ = m, q = Q,u = − eπφ/m= − beπ2/c,ch Ω = cos nθ =√1, sh Ω = i sin nθ = i√eπ2/c,eπ2/c+ 1eπ2/c+ 1(2)e½π2/c= tan nθ,½πds=2n.cdθsin 2nθ(3)For a jet impinging normally on an infinite plane, as in fig. 6, n = 1,e½π2/c= tan θ, ch (½πs/c) sin 2θ = 1,(4)sh ½πx/c = cot θ, sh ½πy/c = tan θ,sh ½πx/c sh ½πy/c = 1, e½π(x + y)/c= e1/2 πx/c+ e1/2 πy/c+ 1.(5)With n = ½, the jet is reversed in direction, and the profile is the catenary of equal strength.In Bobyleff’s problem of the wedge of finite breadth,ch nΩ =√b√u − a, sh nΩ =√b − a√u,au − bau − b(6)cos nα =√b, sin nα =√a − b,aa(7)and along the free surface APJ, q = Q, ψ = 0, u = e−πφ/m= aeπs/c,cos nθ = cos nα√eπ2/c− 1,eπ2/c− cos2nαeπ2/c=cos2nα sin2nθ,sin2nθ − sin2nα(8)the intrinsic equation, the other free surface A′P′J′ being given byeπ2/c=cos2nα sin2nθ,sin2nα − sin2nθ(9)Putting n = 1 gives the case of a stream of finite breadth disturbed by a transverse plane, a particular case of Fig. 7.When a = b, α = 0, and the stream is very broad compared with the wedge or lamina; so, putting w = w′(a − b)/a in the penultimate case, andu = ae−w≈ a − (a − b)w′,(10)ch nΩ =√w′ + 1, sh nΩ =√1,w′√ w′(11)in which we may writew′ = φ + ψi.(12)Along the stream line xABPJ, ψ = 0; and along the jet surface APJ, −1 > φ > −∞; and putting φ = −πs/c − 1, the intrinsic equation isπs/c = cot2nθ,(13)which for n = 1 is the evolute of a catenary.Fig. 7.43. When the barrier AA′ is held oblique to the current, the stream line xB is curved to the branch point B on AA′ (fig. 7), and so must be excluded from the boundary of u; the conformal representation is made now withdΩ= −√ (b − a·b − a′)du(u − b) √ (u − a·u − a′)(1)dw= −m1−m′1,duπu − jπu − j= −m + m′·u − b,πu − j·u − j′b =mj′ + m′j,m + m′(2)taking u = ∞ at the source where φ = ∞, u = b at the branch point B, u = j, j′ at the end of the two diverging streams where φ = −∞; while ψ = 0 along the stream line which divides at B and passes through A, A′; and ψ = m, −m′ along the outside boundaries, so that m/Q, m′/Q is the final breadth of the jets, and (m + m′)/Q is the initial breadth, c1of the impinging stream. Thench ½Ω =√b − a′√u − b, sh ½Ω =√b − a√u − a′,a − a′u − ba − a′u − b(3)ch Ω =2b − a − a′−N,a − a′u − bsh Ω = √ N√ (2·a − u·u − a′),u − bN = 2a − b·b − a′.a − a′(4)Along a jet surface, q = Q, andchΩ = cos θ = cos α − ½sin2α(a − a′) / (u − b),(5)if θ = α at the source x of the jet xB, where u = ∞; and supposing θ = β, β′ at the end of the streams where u = j, j′,u − b=½sin2α,u − j½sin2αcosθ − cosβ,a − a′cos α − cos θa − a′(cos α − cos β) (cos α − cos θ)u − j′= ½sin2αcos θ − cos β′;a − a′(cosα − cos β′) (cos α − cosθ)(6)and ψ being constant along a stream linedφ=dw, Qds=dφ=dwdu,dududθdθdudθπQds=πds=(cos α − cos β) (cos α − cos β′) sin θ,m + m′dθcdθ(cos α − cos θ) (cos θ − cos β) (cos θ − cos α′)=sin θ+cos α − cos β′·sin θcos α − cos θcos β − cos β′cos θ − cos βcos α − cos β·sin θ,cos β − cos β′cos θ − cos β′(7)giving the intrinsic, equation of the surface of a jet, with proper attention to the sign.From A to B, a > u > b, θ = 0,ch Ω = ch logQ= cos α − ½ sin2αa − a′qa − bsh Ω = sh logQ=√ (a − u·u − a′)sinαqu − bQ=(u − b) cos α − ½ (a − a′) sin2α + √ (a − u·u − a′) sin αqu − b(8)Qds= Qdsdφ= −Qdwdudφduqdu=m + m′·(u − b) cos α − ½ (a − a′) sin2α + √ (a − u·u − a′) sin απj − u·u − j′πAB=∫ab(2b − a − a′) (u − b) − 2(a − b) (b − a′) + 2√ (a − b·b − a′·a − u·u − a′)du,ca − a′·j − u·u − j′(10)with a similar expression for BA′.The motion of a jet impinging on an infinite barrier is obtained by putting j = a, j′ = a′; duplicated on the other side of the barrier, the motion reversed will represent the direct collision of two jets of unequal breadth and equal velocity. When the barrier is small compared with the jet, α = β = β′, and G. Kirchhoff’s solution is obtained of a barrier placed obliquely in an infinite stream.Two corners B1and B2in the wall xA, with a′ = −∞, and n = 1, will give the solution, by duplication, of a jet issuing by a reentrant mouthpiece placed symmetrically in the end wall of the channel; or else of the channel blocked partially by a diaphragm across the middle, with edges turned back symmetrically, problems discussed by J. H. Michell, A. E. H. Love and M. Réthy.When the polygon is closed by the walls joining, instead of reaching back to infinity at xx′, the liquid motion must be due to a source, and this modification has been worked out by B. Hopkinson in theProc. Lond. Math. Soc., 1898.Michell has discussed also the hollow vortex stationary inside a polygon (Phil. Trans., 1890); the solution is given bych nΩ = sn w, sh nΩ = i cn w(11)so that, round the boundary of the polygon, ψ = K′, sin nθ = 0; and on the surface of the vortex ψ = 0, q = Q, andcos nθ = sn φ, nθ = ½π − am s/c,(12)the intrinsic equation of the curve.This is a closed Sumner line for n = 1, when the boundary consists of two parallel walls; and n = ½ gives an Elastica.44.The Motion of a Solid through a Liquid.—An important problem in the motion of a liquid is the determination of the state of velocity set up by the passage of a solid through it; and thence of the pressure and reaction of the liquid on the surface of the solid, by which its motion is influenced when it is free.Beginning with a single body in liquid extending to infinity, and denoting by U, V, W, P, Q, R the components of linear and angular velocity with respect to axes fixed in the body, the velocity function takes the formφ = Uφ1+ Vφ2+ Wφ3+ Pχ1+ Qχ2+ Rχ3,(1)where the φ’s and χ’s are functions of x, y, z, depending on the shape of the body; interpreted dynamically, C − ρφ represents the impulsive pressure required to stop the motion, or C + ρφ to start it again from rest.The terms of φ may be determined one at a time, and this problem is purely kinematical; thus to determine φ1, the component U alone is taken to exist, and then l, m, n, denoting the direction cosines of the normal of the surface drawn into the exterior liquid, the function φ1must be determined to satisfy the conditions(i.) ∇2φ1= 0. throughout the liquid;(ii.) dφ1/dυ = −l, the gradient of φ down the normal at the surface of the moving solid;(iii.) dφ1/dυ = 0, over a fixed boundary, or at infinity;similarly for φ2and φ3.To determine χ1the angular velocity P alone is introduced, and the conditions to be satisfied are(i.) ∇2χ1= 0, throughout the liquid;(ii.) dχ1/dυ = mz − ny, at the surface of the moving body, but zero over a fixed surface, and at infinity; the same for χ2and χ3.For a cavity filled with liquid in the interior of the body, since the liquid inside moves bodily for a motion of translation only,φ1= −x, φ2= −y, φ3= −z;(2)but a rotation will stir up the liquid in the cavity, so that the χ’s depend on the shape of the surface.The ellipsoid was the shape first worked out, by George Green, in hisResearch on the Vibration of a Pendulum in a Fluid Medium(1833); the extension to any other surface will form an important step in this subject.A system of confocal ellipsoids is takenx2+y2+z2= 1a2+ λb2+ λc2+ λ
Uniplanar motion alone is so far amenable to analysis; the velocity function φ and stream function ψ are given as conjugate functions of the coordinates x, y by
w = ƒ(z) where z = x + yi, w = φ + ψi,
(1)
and then
(2)
so that, with u = q cos θ, v = q sin θ, the function
(3)
gives ζ as a vector representing the reciprocal of the velocity q in direction and magnitude, in terms of some standard velocity Q.
To determine the motion of a jet which issues from a vessel with plane walls, the vector ζ must be Constructed so as to have a constantdirection θ along a plane boundary, and to give a constant skin velocity over the surface of a jet, where the pressure is constant.
It is convenient to introduce the function
Ω = log ζ = log (Q/q) + θi
(4)
so that the polygon representing Ω conformally has a boundary given by straight lines parallel to the coordinate axes; and then to determine Ω and w as functions of a variable u (not to be confused with the velocity component of q), such that in the conformal representation the boundary of the Ω and w polygon is made to coincide with the real axis of u.
It will be sufficient to give a few illustrations.
Consider the motion where the liquid is coming from an infinite distance between two parallel walls at a distance xx′ (fig. 4), and issues in a jet between two edges A and A′; the wall xA being bent at a corner B, with the external angle β = ½π/n.
The theory of conformal representation shows that the motion is given by
(5)
where u = a, a′ at the edge A, A′; u = b at a corner B; u = 0 across xx′ where φ = ∞; and u = ∞, φ = ∞ across the end JJ′ of the jet, bounded by the curved lines APJ, A′P′J′, over which the skin velocity is Q. The stream lines xBAJ, xA′J′ are given by ψ = 0, m; so that if c denotes the ultimate breadth JJ′ of the jet, where the velocity may be supposed uniform and equal to the skin velocity Q,
m = Qc, c = m/Q.
If there are more B corners than one, either on xA or x′A′, the expression for ζ is the product of corresponding factors, such as in (5).
Restricting the attention to a single corner B,
(6)
(7)
(8)
∞ > a > b > 0 > a′ > −∞
(9)
and then
(10)
the formulas by which the conformal representation is obtained.
For the Ω polygon has a right angle at u = a, a′, and a zero angle at u = b, where θ changes from 0 to ½π/n and Ω increases by ½iπ/n; so that
(11)
And the w polygon has a zero angle at u = 0, ∞, where ψ changes from 0 to m and back again, so that w changes by im, and
(12)
Along the stream line xBAPJ,
ψ = 0, u = ae−πφ/m;
(13)
and over the jet surface JPA, where the skin velocity is Q,
(14)
denoting the arc AP by s, starting at u = a;
(15)
(16)
∞ > u = aeπs/c> a,
(17)
and this gives the intrinsic equation of the jet, and then the radius of curvature
(18)
not requiring the integration of (11) and (12)
If θ = α across the end JJ′ of the jet, where u = ∞, q = Q,
(19)
Then
(20)
(21)
Along the wall AB, cos nθ = 0, sin nθ = 1,
a > u > b,
(22)
(23)
(24)
(25)
(26)
Along the wall Bx, cos nθ = 1, sin nθ = 0,
b > u > 0
(27)
(28)
(29)
At x where φ = ∞, u = 0, and q = q0,
(30)
In crossing to the line of flow x′A′P′J′, ψ changes from 0 to m, so that with q = Q across JJ′, while across xx′ the velocity is q0, so that
m = q0·xx′ = Q·JJ′
(31)
(32)
giving the contraction of the jet compared with the initial breadth of the stream.
Along the line of flow x′A′P′J′, ψ = m, u = a′e−πφ/m, and from x′ to A′, cos nθ = 1, sin nθ = 0,
(33)
(34)
0 > u > a′.
(35)
Along the jet surface A′J′, q = Q,
(36)
(37)
a′ > u = a′eπ/sc> −∞,
(38)
giving the intrinsic equation.
41. The first problem of this kind, worked out by H. v. Helmholtz, of the efflux of a jet between two edges A and A1in an infinite wall, is obtained by the symmetrical duplication of the above, with n = 1, b = 0, a′ = −∞, as in fig. 5,
(1)
and along the jet APJ, ∞ > u = aeπs/c> a,
(2)
(3)
so that PT = c/½π, and the curve AP is the tractrix; and the coefficient of contraction, or
(4)
A change of Ω and θ into nΩ and nθ will give the solution for two walls converging symmetrically to the orifice AA1at an angle π/n. With n = ½, the reentrant walls are given of Borda’s mouthpiece, and the coefficient of contraction becomes ½. Generally, by making a′ = −∞, the line x′A′ may be taken as a straight stream line of infinite length, forming an axis of symmetry; and then by duplication the result can be obtained, with assigned n, a, and b, of the efflux from a symmetrical converging mouthpiece, or of the flow of water through the arches of a bridge, with wedge-shaped piers to divide the stream.
42. Other arrangements of the constants n, a, b, a′ will give the results of special problems considered by J. M. Michell,Phil. Trans.1890.
Thus with a′ = 0, a stream is split symmetrically by a wedge of angle π/n as in Bobyleff’s problem; and, by making a = ∞, the wedge extends to infinity; then
(1)
Over the jet surface ψ = m, q = Q,
u = − eπφ/m= − beπ2/c,
(2)
(3)
For a jet impinging normally on an infinite plane, as in fig. 6, n = 1,
e½π2/c= tan θ, ch (½πs/c) sin 2θ = 1,
(4)
sh ½πx/c = cot θ, sh ½πy/c = tan θ,
sh ½πx/c sh ½πy/c = 1, e½π(x + y)/c= e1/2 πx/c+ e1/2 πy/c+ 1.
(5)
With n = ½, the jet is reversed in direction, and the profile is the catenary of equal strength.
In Bobyleff’s problem of the wedge of finite breadth,
(6)
(7)
and along the free surface APJ, q = Q, ψ = 0, u = e−πφ/m= aeπs/c,
(8)
the intrinsic equation, the other free surface A′P′J′ being given by
(9)
Putting n = 1 gives the case of a stream of finite breadth disturbed by a transverse plane, a particular case of Fig. 7.
When a = b, α = 0, and the stream is very broad compared with the wedge or lamina; so, putting w = w′(a − b)/a in the penultimate case, and
u = ae−w≈ a − (a − b)w′,
(10)
(11)
in which we may write
w′ = φ + ψi.
(12)
Along the stream line xABPJ, ψ = 0; and along the jet surface APJ, −1 > φ > −∞; and putting φ = −πs/c − 1, the intrinsic equation is
πs/c = cot2nθ,
(13)
which for n = 1 is the evolute of a catenary.
43. When the barrier AA′ is held oblique to the current, the stream line xB is curved to the branch point B on AA′ (fig. 7), and so must be excluded from the boundary of u; the conformal representation is made now with
(1)
(2)
taking u = ∞ at the source where φ = ∞, u = b at the branch point B, u = j, j′ at the end of the two diverging streams where φ = −∞; while ψ = 0 along the stream line which divides at B and passes through A, A′; and ψ = m, −m′ along the outside boundaries, so that m/Q, m′/Q is the final breadth of the jets, and (m + m′)/Q is the initial breadth, c1of the impinging stream. Then
(3)
(4)
Along a jet surface, q = Q, and
chΩ = cos θ = cos α − ½sin2α(a − a′) / (u − b),
(5)
if θ = α at the source x of the jet xB, where u = ∞; and supposing θ = β, β′ at the end of the streams where u = j, j′,
(6)
and ψ being constant along a stream line
(7)
giving the intrinsic, equation of the surface of a jet, with proper attention to the sign.
From A to B, a > u > b, θ = 0,
(8)
(10)
with a similar expression for BA′.
The motion of a jet impinging on an infinite barrier is obtained by putting j = a, j′ = a′; duplicated on the other side of the barrier, the motion reversed will represent the direct collision of two jets of unequal breadth and equal velocity. When the barrier is small compared with the jet, α = β = β′, and G. Kirchhoff’s solution is obtained of a barrier placed obliquely in an infinite stream.
Two corners B1and B2in the wall xA, with a′ = −∞, and n = 1, will give the solution, by duplication, of a jet issuing by a reentrant mouthpiece placed symmetrically in the end wall of the channel; or else of the channel blocked partially by a diaphragm across the middle, with edges turned back symmetrically, problems discussed by J. H. Michell, A. E. H. Love and M. Réthy.
When the polygon is closed by the walls joining, instead of reaching back to infinity at xx′, the liquid motion must be due to a source, and this modification has been worked out by B. Hopkinson in theProc. Lond. Math. Soc., 1898.
Michell has discussed also the hollow vortex stationary inside a polygon (Phil. Trans., 1890); the solution is given by
ch nΩ = sn w, sh nΩ = i cn w
(11)
so that, round the boundary of the polygon, ψ = K′, sin nθ = 0; and on the surface of the vortex ψ = 0, q = Q, and
cos nθ = sn φ, nθ = ½π − am s/c,
(12)
the intrinsic equation of the curve.
This is a closed Sumner line for n = 1, when the boundary consists of two parallel walls; and n = ½ gives an Elastica.
44.The Motion of a Solid through a Liquid.—An important problem in the motion of a liquid is the determination of the state of velocity set up by the passage of a solid through it; and thence of the pressure and reaction of the liquid on the surface of the solid, by which its motion is influenced when it is free.
Beginning with a single body in liquid extending to infinity, and denoting by U, V, W, P, Q, R the components of linear and angular velocity with respect to axes fixed in the body, the velocity function takes the form
φ = Uφ1+ Vφ2+ Wφ3+ Pχ1+ Qχ2+ Rχ3,
(1)
where the φ’s and χ’s are functions of x, y, z, depending on the shape of the body; interpreted dynamically, C − ρφ represents the impulsive pressure required to stop the motion, or C + ρφ to start it again from rest.
The terms of φ may be determined one at a time, and this problem is purely kinematical; thus to determine φ1, the component U alone is taken to exist, and then l, m, n, denoting the direction cosines of the normal of the surface drawn into the exterior liquid, the function φ1must be determined to satisfy the conditions
(i.) ∇2φ1= 0. throughout the liquid;(ii.) dφ1/dυ = −l, the gradient of φ down the normal at the surface of the moving solid;(iii.) dφ1/dυ = 0, over a fixed boundary, or at infinity;similarly for φ2and φ3.
(i.) ∇2φ1= 0. throughout the liquid;
(ii.) dφ1/dυ = −l, the gradient of φ down the normal at the surface of the moving solid;
(iii.) dφ1/dυ = 0, over a fixed boundary, or at infinity;
similarly for φ2and φ3.
To determine χ1the angular velocity P alone is introduced, and the conditions to be satisfied are
(i.) ∇2χ1= 0, throughout the liquid;(ii.) dχ1/dυ = mz − ny, at the surface of the moving body, but zero over a fixed surface, and at infinity; the same for χ2and χ3.
(i.) ∇2χ1= 0, throughout the liquid;
(ii.) dχ1/dυ = mz − ny, at the surface of the moving body, but zero over a fixed surface, and at infinity; the same for χ2and χ3.
For a cavity filled with liquid in the interior of the body, since the liquid inside moves bodily for a motion of translation only,
φ1= −x, φ2= −y, φ3= −z;
(2)
but a rotation will stir up the liquid in the cavity, so that the χ’s depend on the shape of the surface.
The ellipsoid was the shape first worked out, by George Green, in hisResearch on the Vibration of a Pendulum in a Fluid Medium(1833); the extension to any other surface will form an important step in this subject.
A system of confocal ellipsoids is taken