---
title: "Derive total infiltration from Horton's equation"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:horton-total-infiltration"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-total-infiltration
site: BCT Engineering
---

# Derive total infiltration from Horton's equation

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2074 Asoj, 4+2 marks
- Video: [Horton's infiltration method | With sample problem | Hydrology Lesson 10](https://www.youtube.com/watch?v=GMTWEqz-Osg) — Tutor Jack Ph
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-total-infiltration
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Starting from Horton's equation, derive an expression for total infiltration in time "t". Also draw a graph showing infiltration and total infiltration vs time.

## Short answer

Total infiltration from Horton’s equation: F_p(t) = f_c t + (f_0 – f_c)/K (1 – e^{-Kt}). Graphically, infiltration capacity decays from f_0 to f_c; cumulative curve rises with initial slope f_0, then constant slope f_c.

## Answer

#### Horton's equation
The infiltration capacity (rate) at time $t$ is

$$
f_p = f_c + (f_0 - f_c)\,e^{-Kt}
$$

where $f_0$ is the initial capacity, $f_c$ the final (constant) capacity and $K$ the decay constant (h⁻¹).

#### Total (cumulative) infiltration
Total infiltration in time $t$ is the integral of the rate (assuming rain is always above capacity):

$$
F_p(t) = \int_0^t f_p\,dt = \int_0^t \left[f_c + (f_0 - f_c)e^{-Kt}\right]dt
$$

$$
F_p(t) = f_c\,t + (f_0 - f_c)\left[\frac{e^{-Kt}}{-K}\right]_0^t
$$

$$
\boxed{F_p(t) = f_c\,t + \frac{f_0 - f_c}{K}\left(1 - e^{-Kt}\right)}
$$

As $t \to \infty$, $F_p \to f_c t + (f_0 - f_c)/K$, so the cumulative curve becomes a straight line of slope $f_c$.

#### Graph
Infiltration capacity $f_p$ falls from $f_0$ to $f_c$ (sketch drawn for $f_0 = 4$, $f_c = 1$, $K = 1.5$ h⁻¹):

```text
 f (cm/h)
  |*
  |
  | *
  |  *
  |   *
  |    *
  |     *
  |      **
  |        *****
  |             ****************************  fc
  +----------------------------------------- t
```

Total infiltration $F_p$ rises quickly at first (slope $f_0$) and then increases at the constant slope $f_c$:

```text
 F (cm)
  |                                      ***
  |                                  ****
  |                             *****         final slope = fc
  |                         ****
  |                     ****
  |                *****
  |             ***
  |         ****
  |      ***
  |    **
  |  **           initial slope = f0
  | *
  |*
  +----------------------------------------- t
```

The area under the $f_p$-$t$ curve up to time $t$ equals $F_p(t)$.

## Check yourself

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

1. Total infiltration is obtained by:
   A. Differentiating fp
   B. Integrating fp over time
   C. Multiplying f0 by t
   D. Averaging
   Answer: B. F = integral of fp dt. (Key point: Total infiltration)

2. F(t) equals:
   A. fc t + (f0 - fc)/K (1 - e^-Kt)
   B. f0 t
   C. fc + (f0-fc)e^-Kt
   D. (f0 - fc) K t
   Answer: A. Integrating gives that expression. (Key point: Result)

3. For large t the F curve:
   A. Flattens
   B. Becomes straight with slope fc
   C. Falls
   D. Is zero
   Answer: B. e^-Kt tends to zero. (Key point: Late time)

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