Storm hydrograph from 6-h UH, 3-hr storm

The storm lasts 3 h but the UH is for 6 h, so first derive the 3-h UH from the 6-h UH by the S-curve method, then apply the rainfall excess.

Step 1: Rainfall excess

$$ER = P - \phi t = 4 - 0.5\times 3 = 2.5 \text{ cm}$$

Step 2: S-curve and 3-h UH

The S-curve is the sum of the 6-h UH repeated every 6 h. Its values at 6-h points are 0, 8, 22, 50, 66, 76, 82, 86, 86 (running sum of the UH ordinates). A smooth S-curve is drawn through them and read at 3-h intervals, so that the curve can be lagged by 3 h:

$$S(t) = U_6(t) + S(t-6)$$

The 3-h UH is the S-curve lagged by 3 h, subtracted, and multiplied by 6/3:

$$U_3(t) = \frac{6}{3}\,[S(t)-S(t-3)]$$

Step 3: Storm hydrograph

$$Q(t) = 2.5\,U_3(t) + 2$$

The S-curve reaches a constant value of 86 m³/s (sum of the 6-h UH ordinates), as it should.

Answer: peak of the storm hydrograph ≈ 73.1 m³/s at t = 18 h; it returns to the 2 m³/s base flow at t = 45 h.