Ω/d2 = 2 cosec. β − cot β.

(4)

or

d = √ {Ω sin β / (2 − cos β)}.

From (3) and (4),

b/d = 2 (1 − cos β) / sin β = 2 tan 1⁄2 β.

Proportions of Channels of Maximum Discharge for given Area and Side Slopes. Depth of channel = d; Hydraulic mean depth = 1⁄2d; Area of section = Ω.

Inclination
of Sides to
Horizon.		 Ratio of
Side
Slopes.		 Area of
Section Ω.		 Bottom
Width.		 Top width =
twice length
of each Side
Slope.
Semicircle .. .. 1.571d2 0 2d
Semi-hexagon 60°   0′ 3  : 5 1.732d2 1.155d 2.310d
Semi-square 90°   0′ 0  : 1 2d2 2d 2d
75°  58′ 1  : 4 1.812d2 1.562d 2.062d
63°  26′ 1  : 2 1.736d2 1.236d 2.236d
53°   8′ 3  : 4 1.750d2 d 2.500d
45°   0′ 1  : 1 1.828d2 0.828d 2.828d
38°  40′ 11⁄4 : 1 1.952d2 0.702d 3.202d
33°  42′ 11⁄2 : 1 2.106d2 0.606d 3.606d
29°  44′ 13⁄4 : 1 2.282d2 0.532d 4.032d
26°  34′ 2  : 1 2.472d2 0.472d 4.472d
23°  58′ 21⁄4 : 1 2.674d2 0.424d 4.924d
21°  48′ 21⁄2 : 1 2.885d2 0.385d 5.385d
19°  58′ 23⁄4 : 1 3.104d2 0.354d 5.854d
18°  26′ 3  : 1 3.325d2 0.325d 6.325d
Half the top width is the length of each side slope. The wetted
perimeter is the sum of the top and bottom widths.

§ 114. Form of Cross Section of Channel in which the Mean Velocity is Constant with Varying Discharge.—In designing waste channels from canals, and in some other cases, it is desirable that the mean velocity should be restricted within narrow limits with very different volumes of discharge. In channels of trapezoidal form the velocity increases and diminishes with the discharge. Hence when the discharge is large there is danger of erosion, and when it is small of silting or obstruction by weeds. A theoretical form of section for which the mean velocity would be constant can be found, and, although this is not very suitable for practical purposes, it can be more or less approximated to in actual channels.