_Problem I._--Given the transverse section of stream and discharge, to
find the slope. From the dimensions of the section find [Omega] and m;
from (1) find [zeta], from (3) find v, and lastly from (2) find i.
_Problem II._--Given the transverse section and slope, to find the
discharge. Find v from (2), then Q from (3).
_Problem III._--Given the discharge and slope, and either the breadth,
depth, or general form of the section of the channel, to determine its
remaining dimensions. This must generally be solved by approximations.
A breadth or depth or both are chosen, and the discharge calculated.
If this is greater than the given discharge, the dimensions are
reduced and the discharge recalculated.
[Illustration: FIG. 114.]
Since m lies generally between the limits m = d and m = ½d, where d is
the depth of the stream, and since, moreover, the velocity varies as
[root](m) so that an error in the value of m leads only to a much less
error in the value of the velocity calculated from it, we may proceed
thus. Assume a value for m, and calculate v from it. Let v1 be this
first approximation to v. Then Q/v1 is a first approximation to
[Omega], say [Omega]1. With this value of [Omega] design the section
of the channel; calculate a second value for m; calculate from it a
second value of v, and from that a second value for [Omega]. Repeat
the process till the successive values of m approximately coincide.
§ 113. _Problem IV. Most Economical Form of Channel for given Side
Slopes._--Suppose the channel is to be trapezoidal in section (fig.
114), and that the sides are to have a given slope. Let the
longitudinal slope of the stream be given, and also the mean velocity.
An infinite number of channels could be found satisfying the
foregoing conditions. To render the problem determinate, let it be
remembered that, since for a given discharge [Omega][oo] [cube
root][chi], other things being the same, the amount of excavation will
be least for that channel which has the least wetted perimeter. Let d
be the depth and b the bottom width of the channel, and let the sides
slope n horizontal to 1 vertical (fig. 114), then
[Omega] = (b + nd)d;
[chi] = b + 2d [root](n² + 1).
Both [Omega] and [chi] are to be minima. Differentiating, and equating
to zero.
(db/dd + n)d + b + nd = 0,
db/dd + 2[root](n² + 1) = 0;
eliminating db/dd,
{n - 2[root](n² + 1)}d + b + nd = 0;
b = 2 {[root](n² + 1) - n}d.