---
title: "Horton constants from double ring infiltrometer data"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:horton-constants-double-ring-data"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-constants-double-ring-data
site: BCT Engineering
---

# Horton constants from double ring infiltrometer data

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2076 Chaitra, 6 marks
- Video: [Horton's infiltration equation parameters | Hydrology](https://www.youtube.com/watch?v=CXOPaeeWrmw) — The Study Sphere
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-constants-double-ring-data
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

The infiltration rates observed during a test on a double ring infiltrometer are as given below:

| Time (hrs) | 0.0417 | 0.125 | 0.333 | 0.75 | 1.5 | 2.5 | 3.5 | 4.5 | 5.5 | 6.5 |
|---|---|---|---|---|---|---|---|---|---|---|
| f (cm/hr) | 0.781 | 0.747 | 0.662 | 0.535 | 0.370 | 0.255 | 0.224 | 0.218 | 0.207 | 0.207 |

Determine the constants $f_0$, $f_c$ and k of Horton's equation which fits the above data.

## Short answer

f_c = 0.207 cm/h from the final steady readings. Linearizing ln(f - f_c) = ln(f_0 - f_c) - k t and fitting a straight line yields k ≈ 0.95 per hour and f_0 ≈ 0.82 cm/h.

## Answer

Horton's equation: $f = f_c + (f_0 - f_c)e^{-kt}$

#### Step 1: Final capacity
The rate becomes steady at the last readings, so $f_c = 0.207$ cm/h.

#### Step 2: Linearise
Take logs: $\ln(f - f_c) = \ln(f_0 - f_c) - kt$

| t (h) | f (cm/h) | $f - f_c$ | $\ln(f-f_c)$ |
|---|---|---|---|
| 0.0417 | 0.781 | 0.574 | -0.555 |
| 0.125 | 0.747 | 0.540 | -0.616 |
| 0.333 | 0.662 | 0.455 | -0.787 |
| 0.75 | 0.535 | 0.328 | -1.115 |
| 1.5 | 0.370 | 0.163 | -1.814 |
| 2.5 | 0.255 | 0.048 | -3.037 |
| 3.5 | 0.224 | 0.017 | -4.075 |
| 4.5 | 0.218 | 0.011 | -4.510 |

The readings at 5.5 h and 6.5 h equal $f_c$ and are not used.

#### Step 3: Straight-line fit (least squares of $\ln(f-f_c)$ on $t$)
- Slope $= -0.948\ \text{h}^{-1}$, so $k = 0.95\ \text{h}^{-1}$.
- Intercept $= -0.493$, so $f_0 - f_c = e^{-0.493} = 0.611$ and $f_0 = 0.611 + 0.207 = 0.818$ cm/h.

**Answer: $f_0 \approx 0.82$ cm/h, $f_c = 0.207$ cm/h, $k \approx 0.95\ \text{h}^{-1}$.**

Equation: $f = 0.207 + 0.611e^{-0.95t}$ cm/h.

## Check yourself

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

1. How is f_c taken from the data?
   A. The maximum
   B. The steady final readings, 0.207 cm/h
   C. The first reading
   D. The average of all readings
   Answer: B. The curve flattens to the constant final capacity at the last readings. (Key point: Finding f_c)

2. What is plotted to get a straight line?
   A. f_c against t
   B. ln f against ln t
   C. f against t
   D. ln(f - f_c) against t
   Answer: D. ln(f-f_c) = ln(f_0-f_c) - kt is linear in t. (Key point: Linearisation)

3. Why are the 5.5 h and 6.5 h readings not used?
   A. f - f_c = 0, so the logarithm is undefined
   B. They are too large
   C. They are initial
   D. They are errors
   Answer: A. They equal f_c, so f - f_c = 0 and ln cannot be taken. (Key point: Unusable readings)

4. What are the fitted constants?
   A. f_0 about 0.82 cm/h, k about 0.95 per h
   B. f_0 about 0.61, k about 0.207
   C. f_0 about 0.207, k about 0.61
   D. f_0 about 1.5, k about 0.5
   Answer: A. Intercept gives f_0 - f_c = 0.611 so f_0 = 0.818; slope -0.948 gives k = 0.95. (Key point: Results)

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