What This Solves
Calculates the most hydraulically efficient channel cross-section dimensions that maximize flow capacity for a given cross-sectional area.
Best Used When
- You want to minimize excavation while maximizing channel capacity
- You are designing a new channel and want the most efficient geometry as a starting point
- You need to compare the efficiency of rectangular, trapezoidal, and triangular sections
Do NOT Use When
- You already know the channel geometry and need to calculate flow capacity — Use Manning's Channel Calculator
- You need to evaluate channel lining adequacy for erosion protection — Use Channel Lining Calculator
Key Assumptions
- The optimal section is derived from minimizing wetted perimeter for a given area
- Side slopes are stable and can be constructed at the optimal angle
- Roughness is uniform across the entire wetted perimeter
- The channel is prismatic (constant cross-section) along its length
Input Quality Notes
The theoretical best section may not be practical due to soil stability, construction constraints, or right-of-way limits. Use it as a design starting point, then adjust for site conditions.
Design the most hydraulically efficient open channel for your flow. This tool returns the optimal flow depth, bottom width and side slope for rectangular, trapezoidal and triangular sections — the geometry that carries the most water for the least excavated area by minimizing the wetted perimeter.
Optimal Section Proportions
| Shape | Optimal Proportions | R/y Ratio | Efficiency* |
|---|---|---|---|
| Semicircle | Theoretical optimum | 0.500 | 100% |
| Rectangular | b = 2y (width = 2 x depth) | 0.5 | ~95% |
| Trapezoidal | z = 0.577, b = 1.15y (half-hexagon) | 0.5 | ~97% |
| Triangular | z = 1.0 (45-degree sides) | 0.3536 | ~83% |
* Efficiency compared to semicircular section for same flow area
Ready to Calculate
Select a channel shape and design approach to find optimal dimensions.
For educational purposes only. Not a substitute for professional engineering judgment.
How it works
Open-channel flow is governed by Manning’s equation, which expresses discharge as a function of the cross-sectional area, the hydraulic radius and the bed slope:
Q = (k / n) · A · R2/3 · S1/2
The hydraulic radius is defined as R = A / P, the ratio of flow area to wetted perimeter. For a fixed flow area A, slope S and roughness n, the discharge Q is therefore maximized when the wetted perimeter P is as small as possible. The cross section that achieves this is called the best (or most efficient) hydraulic section. Minimizing P for each family of shapes leads to fixed optimal proportions:
- Rectangular: b = 2y, so A = 2y2 and R = y/2. The optimal depth is y = √(A/2).
- Trapezoidal (half-hexagon): z = 1/√3 ≈ 0.577 and b = 2y/√3, giving A = y2√3 and R = y/2. The optimal depth is y = √(A/√3).
- Triangular: z = 1.0 (45° sides, 90° apex), so A = y2 and R = y/(2√2). The optimal depth is y = √A.
When you design for a required discharge, the calculator first solves Manning’s equation iteratively to find the flow area that delivers your target Q at the given slope and roughness, then applies the optimal proportions above. An efficiency ratio compares the resulting wetted perimeter to that of an equal-area semicircle (the absolute optimum).
Method after Chow, V.T. (1959) Open-Channel Hydraulics, §7-6, and USACE EM 1110-2-1601 (1994).
Variables & symbols
| Symbol | Quantity | US units | SI units |
|---|---|---|---|
| Q | Discharge (flow rate) | cfs | m³/s (cms) |
| A | Flow (cross-sectional) area | ft² | m² |
| P | Wetted perimeter | ft | m |
| R | Hydraulic radius (A/P) | ft | m |
| y | Flow depth | ft | m |
| b | Bottom width | ft | m |
| z | Side slope (horizontal : vertical) | H:V | H:V |
| S | Longitudinal channel slope | ft/ft | m/m |
| n | Manning’s roughness coefficient | — | — |
| k | Manning conversion factor | 1.486 | 1.0 |
Optimal proportions by shape
Geometric criteria that minimize the wetted perimeter for each channel family, with the resulting area and hydraulic-radius relationships used by the calculator.
| Shape | Optimal side slope z (H:V) | Bottom width b | Area A | Hydraulic radius R |
|---|---|---|---|---|
| Semicircle | — (theoretical optimum) | — | πr²/2 | r/2 |
| Rectangular | 0 (vertical walls) | 2y | 2y² | y/2 |
| Trapezoidal (half-hexagon) | 0.577 (1/√3) | 1.155 y (2y/√3) | √3 · y² | y/2 |
| Triangular | 1.0 (45° sides) | — | y² | y/(2√2) ≈ 0.354 y |
Values from Chow (1959), §7-6 and Table 7-4. The best rectangular and best trapezoidal sections both achieve R = y/2; the half-hexagon trapezoid is the most efficient of the straight-sided shapes.
Worked example
Best trapezoidal section for a required flow area A = 100 ft²:
- Optimal depth y = √(A/√3) = √(100 / 1.732) = 7.60 ft
- Bottom width b = 2y/√3 = 1.155 × 7.60 = 8.78 ft
- Side slope z = 0.577 H:V (sides at 60° from horizontal)
- Wetted perimeter P = 3b = 26.3 ft (half-hexagon: three equal wetted sides)
- Hydraulic radius R = A/P = 100 / 26.3 = 3.80 ft (= y/2, as expected)
The equal-area best rectangular section gives y = √(100/2) = 7.07 ft, b = 14.14 ft and P = 28.3 ft — a larger wetted perimeter, confirming the trapezoid is the more efficient choice.
Frequently asked questions
What is the best hydraulic section?
The best (or most efficient) hydraulic section is the channel cross section that carries the maximum discharge for a given flow area, slope and roughness. Because Manning’s equation gives Q in proportion to A·R^(2/3) and the hydraulic radius R = A/P, maximizing R for a fixed area A means minimizing the wetted perimeter P. The theoretical optimum is a semicircle; among practical shapes the half-hexagon trapezoid is the most efficient.
Why is the optimal trapezoidal channel a half-hexagon?
For a trapezoid the minimum wetted perimeter occurs at a side slope of z = 1/√3 ≈ 0.577 (sides at 60° from horizontal) with a bottom width b = 2y/√3. At those proportions the three wetted sides are equal and the section is exactly half of a regular hexagon. This shape gives a hydraulic radius of R = y/2, the same ratio as the best rectangular section.
Should I always build the best hydraulic section?
Not necessarily. The best section minimizes excavation and lining area, but the optimal trapezoidal side slope of 0.577 H:V (about 60°) is too steep for most unlined earth channels and ignores sediment transport, freeboard, side-slope stability and maintenance access. It is most useful for lined channels and for preliminary sizing; final geometry should be checked against geotechnical and constructability constraints.
How is discharge capacity calculated for the optimal section?
Discharge is found with Manning’s equation, Q = (k/n)·A·R^(2/3)·S^(1/2), where k = 1.486 for US customary units (cfs, ft) and k = 1.0 for SI units (cms, m). The calculator computes the optimal A, R and the section factor AR^(2/3) for your chosen shape, then applies your slope S and Manning’s n to return the capacity.
Related tools
Was this calculator helpful?
Last verified: February 2026