SCS Curve-Number Runoff

S = \frac{1000}{CN} - 10 Q = \frac{( P - 0.2 S ) ^{2}}{P + 0.8 S} (P > 0.2 S)

NRCS (SCS) rainfall-runoff: storage S (inches) from curve number CN. Direct runoff depth Q uses initial abstraction Ia = 0.2S. P is rainfall depth. Q = 0 if P ≤ Ia. CN from soils/cover tables (higher CN → more runoff).

$C N$: Curve number
$S$ (in): Maximum retention
$P$ (in): Rainfall depth
$Q$ (in): Runoff depth

Step-by-step solved example

CN = 80. Storm P = 3.0 in. Find S and runoff Q.

Rainfall excess from CN: think of a catchment section storing S.

1. S
S = 1000/80 − 10 = 12.5 − 10 = 2.5 in.
$S = 2.5 in$
2. Ia and Q
Ia = 0.5 in. P > Ia. Q = (3−0.5)²/(3+2.0) = 6.25/5.0 = 1.25 in.
$Q = 1.25 in$

Answer: S = 2.5 in, Q = 1.25 in

10 practice questions

0/10 correct

1.CN = 100 means

2.CN = 50. S =

3.If P < 0.2S, Q =

4.Higher CN produces

5.Ia in the handbook method is

6.Q and P in the US handbook formula are

7.A hydrologic soil group D (clays) tends to have

8.CN = 80, P = 0.4 in. Since 0.2S = 0.5 in, Q is

9.Antecedent moisture (wet) effectively

10.Runoff volume is Q_depth × area. 1.25 in on 1 ac is nearest