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Horton constants from double ring infiltrometer 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.

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

Time (hrs)0.04170.1250.3330.751.52.53.54.55.56.5
f (cm/hr)0.7810.7470.6620.5350.3700.2550.2240.2180.2070.207

Determine the constants f0f_0, fcf_c and k of Horton's equation which fits the above data.

Horton's equation: f=fc+(f0−fc)e−ktf = f_c + (f_0 - f_c)e^{-kt}

Step 1: Final capacity

The rate becomes steady at the last readings, so fc=0.207f_c = 0.207 cm/h.

Step 2: Linearise

Take logs: ln⁡(f−fc)=ln⁡(f0−fc)−kt\ln(f - f_c) = \ln(f_0 - f_c) - kt

t (h)f (cm/h)f−fcf - f_cln⁡(f−fc)\ln(f-f_c)
0.04170.7810.574-0.555
0.1250.7470.540-0.616
0.3330.6620.455-0.787
0.750.5350.328-1.115
1.50.3700.163-1.814
2.50.2550.048-3.037
3.50.2240.017-4.075
4.50.2180.011-4.510

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

Step 3: Straight-line fit (least squares of ln⁡(f−fc)\ln(f-f_c) on tt)

  • Slope =−0.948 h−1= -0.948\ \text{h}^{-1}, so k=0.95 h−1k = 0.95\ \text{h}^{-1}.
  • Intercept =−0.493= -0.493, so f0−fc=e−0.493=0.611f_0 - f_c = e^{-0.493} = 0.611 and f0=0.611+0.207=0.818f_0 = 0.611 + 0.207 = 0.818 cm/h.

Answer: f0≈0.82f_0 \approx 0.82 cm/h, fc=0.207f_c = 0.207 cm/h, k≈0.95 h−1k \approx 0.95\ \text{h}^{-1}.

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

✓ Answer checked by Claude Sonnet 5.5

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