---
title: "A3 curve meets critical depth line normally"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:a3-curve-meets-critical-line-normally"
source: https://www.bctengineer.com/q/semester_4/hydraulics/a3-curve-meets-critical-line-normally
site: BCT Engineering
---

# A3 curve meets critical depth line normally

Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF)

- Asked in: 2072 Magh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/a3-curve-meets-critical-line-normally
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

Justify analytically that A3 curve meets the $y_c$ line and channel bottom normally.

## Short answer

On an adverse slope, as y→y_c, Fr→1 so dy/dx→∞, meeting the CDL vertically. As y→0, dy/dx ≈ gn²y^{-1/3}→∞, so the A3 curve also meets the channel bottom vertically.

## Answer

The A3 curve occurs on an **adverse slope** ($S_0 < 0$) in zone 3, where $y < y_c$ (supercritical). Write the bed slope as $-|S_0|$ in the dynamic equation:

$$
\frac{dy}{dx} = \frac{S_0 - S_f}{1 - Fr^2} = \frac{-(|S_0| + S_f)}{1 - Fr^2}
$$

In zone 3, $Fr > 1$, so the denominator is negative and $dy/dx > 0$: the depth increases downstream.

**(i) Near the critical depth line ($y \to y_c$)**

$Fr \to 1$, so the denominator $1 - Fr^2 \to 0$ while the numerator $-(|S_0| + S_f)$ stays finite and non-zero. Hence

$$
\frac{dy}{dx} \to \infty
$$

The tangent to the curve is vertical, so the curve meets the $y_c$ line **normally** (at a right angle). (In practice a hydraulic jump forms before this point is reached.)

**(ii) Near the channel bottom ($y \to 0$)**

For a wide channel, with $S_f = n^2q^2/y^{10/3}$ and $Fr^2 = q^2/(gy^3)$, as $y \to 0$ both $S_f$ and $Fr^2$ become very large and the $|S_0|$ and $1$ terms are negligible:

$$
\frac{dy}{dx} \approx \frac{-S_f}{-Fr^2} = \frac{n^2 q^2 / y^{10/3}}{q^2/(g y^3)} = g\,n^2\,y^{-1/3}
$$

$$
\text{As } y \to 0:\quad \frac{dy}{dx} \to \infty
$$

So the tangent is again vertical, and the curve meets the **channel bottom normally**.

Hence the A3 profile starts vertically at the bed (or at a small depth just after a gate) and ends vertically at the CDL.

## Check yourself

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

1. The A3 curve occurs on an adverse slope in the zone where:
   A. y > yc
   B. y < yc (supercritical)
   C. y = yn
   D. y > yn
   Answer: B. A3 is zone 3 of an adverse slope, with depth below critical depth. (Key point: Where A3 occurs)

2. Why is dy/dx positive in A3?
   A. Numerator -(|S0| + Sf) is negative and denominator 1 - Fr^2 is negative (Fr > 1)
   B. Numerator is positive
   C. Fr < 1
   D. Sf is negative
   Answer: A. Negative divided by negative gives a positive slope; the depth increases downstream. (Key point: Sign of dy/dx)

3. As y tends to yc, why does the curve meet the yc line normally?
   A. Numerator tends to zero
   B. Denominator tends to zero while the numerator stays finite, so dy/dx tends to infinity
   C. Sf tends to zero
   D. Fr tends to infinity
   Answer: B. A finite numerator over a vanishing 1 - Fr^2 makes the tangent vertical. (Key point: Meeting the yc line)

4. As y tends to 0, dy/dx is approximately g n^2 y^(-1/3). What does this imply?
   A. The curve is horizontal at the bed
   B. The curve never reaches the bed
   C. The slope equals S0
   D. The slope tends to infinity, so the curve meets the bed normally
   Answer: D. y^(-1/3) grows without bound as y tends to 0, giving a vertical tangent at the bed. (Key point: Meeting the bed)

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