Best trapezoidal section and half-hexagon proof

Condition for maximum discharge

From Manning or Chezy, $Q=\dfrac{1}{n}AR^{2/3}S^{1/2}$. For a given area $A$, slope and roughness, the discharge is maximum when $R=A/P$ is maximum, i.e. when the wetted perimeter $P$ is minimum.

Part 1: Proportions for a given area $A$ and side slope $z$ (H:V)

Trapezoid with bed width $b$ and depth $y$:

$$A=(b+zy)y\;\Rightarrow\;b=\frac{A}{y}-zy$$ $$P=b+2y\sqrt{1+z^2}=\frac{A}{y}-zy+2y\sqrt{1+z^2}$$

For minimum $P$ at constant $A$ and $z$:

$$\frac{dP}{dy}=-\frac{A}{y^2}-z+2\sqrt{1+z^2}=0\;\Rightarrow\;\frac{A}{y^2}=2\sqrt{1+z^2}-z$$

Since $A/y^2=b/y+z$:

$$\boxed{b=2y\left(\sqrt{1+z^2}-z\right)}$$

(The second derivative $d^2P/dy^2=2A/y^3>0$, so this is a minimum.)

Hydraulic radius:

$$A=y^2\left(2\sqrt{1+z^2}-z\right),\quad P=\frac{A}{y}-zy+2y\sqrt{1+z^2}=2y\left(2\sqrt{1+z^2}-z\right)$$ $$R=\frac{A}{P}=\frac{y}{2}\qquad\text{(for every most efficient trapezoid)}$$

The condition can also be written as $b+2zy=2y\sqrt{1+z^2}$, i.e. the top width equals the sum of the two sloping sides, so a semicircle of radius $y$ can be inscribed in the section.

Part 2: Side slope varied (best of all trapezoids)

Now keep $A$ and $y$ constant and vary $z$:

$$P=\frac{A}{y}-zy+2y\sqrt{1+z^2}$$ $$\frac{dP}{dz}=-y+\frac{2yz}{\sqrt{1+z^2}}=0\;\Rightarrow\;\sqrt{1+z^2}=2z\;\Rightarrow\;z=\frac{1}{\sqrt3}=0.577$$

So the sides make an angle $\theta$ with the horizontal where $\tan\theta=1/z=\sqrt3$, i.e. $\theta=60^\circ$. Then

$$b=2y\left(\sqrt{1+\tfrac13}-\tfrac{1}{\sqrt3}\right)=2y\left(\frac{2}{\sqrt3}-\frac{1}{\sqrt3}\right)=\frac{2y}{\sqrt3}$$

The sloping length is $y\sqrt{1+z^2}=\dfrac{2y}{\sqrt3}=b$. Hence the bed and both sloping sides are equal, and the sides are at $60^\circ$ to the horizontal: the section is half of a regular hexagon.

 _______________________   water surface, top width = 2b
 \                     /
  \  side = b         /    sides at 60 deg to the horizontal
   \                 /     each side = b = 2y/sqrt(3)
    \_______________/
            b

The hydraulic radius from Part 1 is $R=y/2$. Since this section also has the least perimeter over all side slopes, it gives the largest $R$ of any trapezoidal section of the given area, so the half-hexagon is the most efficient of all trapezoidal sections.

Result: $b=2y(\sqrt{1+z^2}-z)$, $R=y/2$; for the best section, $z=1/\sqrt3$ (half hexagon).