---
title: "2-h and 4-h UH from an S-hydrograph"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:s-hydrograph-2h-4h-uh"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/s-hydrograph-2h-4h-uh
site: BCT Engineering
---

# 2-h and 4-h UH from an S-hydrograph

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2071 Shrawan, 8+6 marks
- Video: [S-Hydrograph](https://www.youtube.com/watch?v=wSvFrX48JOo) — Seema Shah-Fairbank
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/s-hydrograph-2h-4h-uh
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

An S-hydrograph is given such that at time t = 0, its ordinate is 1 cm/h and it remains so for an indefinite period of time. Determine a 2-hour unit hydrograph. Using this unit hydrograph, determine a 4-hour unit hydrograph.

## Short answer

Lag the S-curve by2 h and divide the difference by2: the 2-h UH is a rectangle of height0.5 h⁻¹ from0 to2 h. Adding two 2-h UHs lagged by2 h and dividing by2 gives a 4-h UH rectangle of height0.25 h⁻¹ over0–4 h.

## Answer

The S-hydrograph is the response to a **continuous net rain of 1 cm/h** that continues indefinitely. Here its ordinate is 1 (unit: cm/h of runoff) from t = 0 and stays constant, i.e. $S(t)=1$ for $t \ge 0$.

#### (a) 2-hour unit hydrograph
Lag the S-curve by $D = 2$ h and subtract. This gives the response to 1 cm/h for 2 h = 2 cm. Divide by $D\times 1$ = 2 to get a UH for 1 cm:

$$
U_2(t) = \frac{S(t)-S(t-2)}{2}
$$

| t (h) | S(t) | S(t−2) | S(t)−S(t−2) | 2-h UH = difference / 2 |
|---|---|---|---|---|
| 0 | 1 | 0 | 1 | 0.5 |
| 1 | 1 | 0 | 1 | 0.5 |
| 2⁻ (just before) | 1 | 0 | 1 | 0.5 |
| ≥ 2 | 1 | 1 | 0 | 0 |

So the 2-h UH is a **rectangle of height 0.5 per hour (cm/h per cm of rain) from t = 0 to t = 2 h**, and zero afterwards. Check: area = 0.5 × 2 = 1 cm. ✓.

```text
 U (1/h)
 0.5 |#########
     |#########
   0 +---------+--------- t (h)
     0         2
```

#### (b) 4-hour UH from the 2-hour UH
Add two 2-h UHs, the second lagged by 2 h, and divide by 2 (4 h/2 h = 2 blocks, and the sum is for 2 cm):

$$
U_4(t) = \frac{U_2(t) + U_2(t-2)}{2}
$$

| t (h) | U₂(t) | U₂(t−2) | Sum | U₄ = Sum/2 |
|---|---|---|---|---|
| 0 to 2 | 0.5 | 0 | 0.5 | 0.25 |
| 2 to 4 | 0 | 0.5 | 0.5 | 0.25 |
| > 4 | 0 | 0 | 0 | 0 |

**Answer: the 4-h UH is a rectangle of height 0.25 per hour from t = 0 to 4 h** (area = 0.25 × 4 = 1 cm ✓). The same result comes directly from $[S(t)-S(t-4)]/4 = 1/4$.

## Check yourself

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

1. What does the S-hydrograph with constant ordinate 1 cm/h represent?
   A. Response to 1 cm of rain in 1 hr
   B. The 2-hr unit hydrograph itself
   C. Base flow of the river
   D. Response to a continuous net rain of 1 cm/h applied indefinitely
   Answer: D. The S-curve is the hydrograph due to continuous rain at unit intensity. (Key point: S-curve definition)

2. How is the 2-hr UH obtained from S(t)?
   A. [S(t) - S(t-2)] / 2
   B. S(t) + S(t-2)
   C. [S(t) - S(t-2)] x 2
   D. S(t) / 2 only
   Answer: A. Lagging and subtracting gives response to 2 cm; dividing by 2 gives 1 cm. (Key point: UH from S-curve)

3. What is the resulting 4-hr UH?
   A. Rectangle of height 0.5 from 0 to 4 h
   B. Rectangle of height 0.25 per hour from 0 to 4 h (area 1 cm)
   C. Rectangle of height 0.5 from 0 to 2 h
   D. Triangle with peak 0.25 at 2 h
   Answer: B. U4 = [U2(t)+U2(t-2)]/2 = 0.25 for 0-4 h; area 0.25 x 4 = 1 cm. (Key point: 4-hr UH)

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