(8)

Q ds= Q ds = − Q dw
du du qdu
= m + m′· (u − b) cos α − ½ (a − a′) sin2 α + √ (a − u·u − a′) sin α
π j − u·u − j′
π AB= ∫ab (2b − a − a′) (u − b) − 2(a − b) (b − a′) + 2√ (a − b·b − a′·a − u·u − a′)du,
c a − 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 B1 and B2 in 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 the Proc. Lond. Math. Soc., 1898.