A simple construction gives the form of the channel which fulfils this condition, for it can be shown that when m = 1⁄2d the sides of the channel are tangential to a semicircle drawn on the water line.
Since
Ω / χ = 1⁄2 d,
therefore
Ω = 1⁄2 χd.
(1)
Let ABCD be the channel (fig. 115); from E the centre of AD drop perpendiculars EF, EG, EH on the sides.
Let
AB = CD = a; BC = b; EF = EH = c; and EG = d.
Ω = area AEB + BEC + CED,
= ac + 1⁄2 bd.