Horton Infiltration

f = f_{c} + (f_{0} - f_{c}) e^{- k t}

Infiltration rate decays from f0 toward fc. Cumulative F=∫f dt. Horton is capacity; actual f cannot exceed rainfall intensity (excess = i−f when ponded).

$f 0$ (in/h): Initial capacity
$f c$: Ultimate capacity
$k$ (1/h): Decay constant

Step-by-step solved example

f0=3 in/h, fc=0.5 in/h, k=2 /h. f at 1 h?

Infiltration capacity decaying toward fc vs time.

1. Exponential
e^{−2}=0.135.
2. f
0.5+2.5×0.135=0.84 in/h.

Answer: f ≈ 0.84 in/h

10 practice questions

0/10 correct

1.As t→∞, f →

2.At t=0, f =

3.Green–Ampt uses

4.If rainfall i < f_capacity

5.Cumulative Horton F(t) =

6.φ-index is

7.k has units

8.Urbanization / paving reduces

9.f0=4, fc=1, k=1/h, t=0.693 h: (f0−fc)e^{−kt}=

10.SCS CN method estimates excess by