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2-h and 4-h UH from an S-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.

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.

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)=1S(t)=1 for t≥0t \ge 0.

(a) 2-hour unit hydrograph

Lag the S-curve by D=2D = 2 h and subtract. This gives the response to 1 cm/h for 2 h = 2 cm. Divide by D×1D\times 1 = 2 to get a UH for 1 cm:

U2(t)=S(t)−S(t−2)2U_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
01010.5
11010.5
2⁻ (just before)1010.5
≥ 21100

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. ✓.

 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):

U4(t)=U2(t)+U2(t−2)2U_4(t) = \frac{U_2(t) + U_2(t-2)}{2}
t (h)U₂(t)U₂(t−2)SumU₄ = Sum/2
0 to 20.500.50.25
2 to 400.50.50.25
> 40000

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[S(t)-S(t-4)]/4 = 1/4.

✓ Answer checked by Claude Sonnet 5.5

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