---
title: "Best trapezoidal section and half-hexagon proof"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:trapezoidal-proportions-half-hexagon-proof"
source: https://www.bctengineer.com/q/semester_4/hydraulics/trapezoidal-proportions-half-hexagon-proof
site: BCT Engineering
---

# Best trapezoidal section and half-hexagon proof

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2070 Magh, 6 marks; 2069 Poush, 10 marks (asked 2 times)
- Video: [Most economical trapezoid channel section //open channel flow //](https://www.youtube.com/watch?v=9PJhN517foA) — Civil Engineer FBH
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/trapezoidal-proportions-half-hexagon-proof
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Find the proportions of a trapezoidal channel which will make the discharge a maximum for a given cross sectional area of flow and given side slopes. Show also that if the side slopes can be varied, the most efficient of all trapezoidal sections is half-hexagon (i.e. for the section of greatest hydraulic efficiency, hydraulic radius R = y/2).

## Short answer

For a given area and side slope z, the most efficient trapezoid has b=2y(√(1+z²)−z) and hydraulic radius R=y/2, with top width equal to sum of sloping sides. If side slopes can vary, the absolute best has z=1/√3, giving a half-hexagon with all three sides equal and sides at 60° to horizontal.

## Answer

#### 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**.

```text
 _______________________   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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a given area, slope and roughness, discharge is maximum when:
   A. Wetted perimeter is minimum (R is maximum)
   B. Wetted perimeter is maximum
   C. Top width is minimum
   D. Depth is minimum
   Answer: A. Q = (1/n) A R^(2/3) S^(1/2) is largest when R = A/P is largest, i.e. P is smallest. (Key point: Condition for maximum discharge)

2. What bed width gives the most efficient trapezoid for a side slope z (H:V)?
   A. b = 2y(sqrt(1+z^2) + z)
   B. b = y(sqrt(1+z^2) - z)
   C. b = 2y(sqrt(1+z^2) - z)
   D. b = 2zy
   Answer: C. Setting dP/dy = 0 gives A/y^2 = 2 sqrt(1+z^2) - z and hence b = 2y(sqrt(1+z^2) - z). (Key point: Best proportions)

3. What is the hydraulic radius of every most efficient trapezoid?
   A. R = y
   B. R = y/4
   C. R = 2y
   D. R = y/2
   Answer: D. With A = y^2(2 sqrt(1+z^2) - z) and P = 2y(2 sqrt(1+z^2) - z), R = A/P = y/2. (Key point: Hydraulic radius)

4. When the side slope is also varied, dP/dz = 0 gives which z?
   A. z = 1/sqrt(3), i.e. 60 degree side slopes
   B. z = 1
   C. z = sqrt(3)
   D. z = 0
   Answer: A. sqrt(1+z^2) = 2z gives z = 1/sqrt3, so the sides are at 60 degrees. (Key point: Best side slope)

## Related in Hydraulics

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