Infiltrometers and deriving Horton's constants

Infiltrometers are devices used to measure the infiltration rate of soil in the field. There are two types.

1. Flooding-type infiltrometers

Water is ponded on the soil, and the rate of water addition required to keep a constant depth is measured.

• Simple (single) ring infiltrometer: a metal cylinder, 30 cm diameter and 60 cm long, driven about 50 cm into the ground. Water is kept at a constant depth (about 5 cm) and the volume added per unit time is the infiltration rate. Drawback: water spreads sideways below the ring, so the rate is over-estimated.
• Double-ring infiltrometer: two concentric rings, inner 30 cm and outer 60 cm in diameter. Both are kept filled to the same depth. The outer ring forces the water in the inner ring to move vertically; only the inner ring is measured.

  |~~~~~|~~~~~~~~~~~~~~~~~~~~~|~~~~~|   <- water kept at the same level
  |~~~~~|~~~~~~~~~~~~~~~~~~~~~|~~~~~|      in both rings (about 5 cm)
==|=====|=====================|=====|== ground level
  |     |                     |     |
  v     v    v    v    v      v     v   water enters the soil
 outer  |<--- inner ring ---->|  outer
 ring        (measured)         ring

2. Rainfall-simulator type

Artificial rain is applied by sprinklers on a small plot (about 2 m by 4 m). Runoff is collected and measured; rainfall minus runoff gives infiltration. It represents the impact of real rain on the soil better than the flooding types.

Deriving Horton's constants

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

Procedure:

1. Plot $f$ against $t$; the curve flattens to the constant $f_c$ (read it from the final stable readings).
2. Compute $(f - f_c)$ for each reading and plot $\ln(f-f_c)$ against $t$ (or $f - f_c$ on semi-log paper). This should be a straight line:

$$\ln(f - f_c) = \ln(f_0 - f_c) - kt$$

3. The intercept at $t = 0$ gives $f_0 - f_c$, so $f_0$. The slope of the line is $-k$.

Assumed data

Let $f_c = 0.5$ cm/h and the readings be:

The points lie nearly on a straight line. Taking the two end points: slope $= (-1.715 - 1.386)/3 = -1.03$, so $k \approx 1.0\ \text{h}^{-1}$. The intercept is $\ln(4.0)$, so $f_0 = 4.0 + 0.5 = 4.5$ cm/h.

Thus the fitted equation is $f = 0.5 + 4.0e^{-1.0t}$ cm/h.