direction 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)
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 = 1/2π/n.
The theory of conformal representation shows that the motion is
given by
ζ = [
√ (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
)
n
cos nθ + i sh log (
Q
)
n
sin nθ
q
q
= 1/2(ζn + ζ−n) = √
b − a′
√
u − a
,
a − a′
u − b
(7)
sh nΩ = sh log (
Q
) cos nθ + i ch log (
Q
)
n
sin nθ
q
q
= 1/2(ζn + ζ−n) = √
b − a
√
u − a′
,
a − a′
u − b
(8)
∞ > a > b > 0 > a′ > −∞
(9)
and then
dΩ
= −
1
√ (b − a′·b − a′)
,
dw
= −
m
,
du
2n
(u − b) √ (a − a·u − a′)
du
πu
(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 1/2π/n and Ω increases by 1/2iπ/n; so
that
dΩ
=
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, and
dw
=
B
, where B = −
m
.
du
u
π
(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 curvature
ρ = −
ds
=
1
dφ
=
i
dw
=
i
dw
/
dΩ
dθ
Q
dθ
Q
dΩ
Q
du
du
=
c
·
u − b
√ (u − a·u − a′)
,
π
u
√ (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)
Then
cos 2nα − cos 2nθ = 2
a − b·b − a′
= 1/2sin2 2nα
a − a′
a − a′·u − b
u − b
sin 2nθ = 2
√ (a − b.b − a′) √ (u − a·u − b′)
a − a′·u − b
(20)
= sin 2nα
√ (a − a·b − a′)
;
u − b
2n
c
( 1 +
b
)
√ (a − b·b − a′)
φ
ρ
u − 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′) sin2 2nα
sin 2nθ
Along 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
,
q
a − a′
u − b
(23)
sh nΩ = i ch log (
Q
)
n
= i√
a − b
√
u − a′
,
q
a − a′
u − b
(24)
ds
=
ds
dφ
=
m
=
c
Q
du
dφ
dt
πqu
π
qu
(25)
π
AB
= ∫ab
Q
du
∫ [
√ (a − b) √ (u − a′) + √ (b − a′) √ (a − u)
]
1/n
du
.
c
q
u
√ (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
,
q
a − a′
b − u
(28)
sh nΩ = sh log (
Q
)
n
= √
a − b
√
u − a′
.
q
a − a′
b − u
(29)
At x where φ = ∞, u = 0, and q = q0,
(
Q
)
n
= √
b − a′
√
a
+ √
a − b
√
−a′
.
q0
a − a′
b
a − 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 that
m = q0·xx′ = Q·JJ′
(31)
JJ′
=
q0
[ √
b − a′
√
a
− √
a − b
√
−a′
]
1/n
,
xx′
Q
a − a′
b
a − 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
,
q
a − a′
b − u
(33)
sh nΩ = sh log (
Q
)
n
= √
a − b
√
u − a′
.
q
a − 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 A1 in an infinite
wall, is obtained by the symmetrical duplication of the above, with
n = 1, b = 0, a′ = −∞, as in fig. 5,
