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Formulas / Soil Resistivity Calculations — Wenner and Barnes Methods (Imperial)

Soil Resistivity Calculations — Wenner and Barnes Methods (Imperial)

Barnes Method CP 3 Field Tests Imperial Units Soil Resistivity Wenner Method

What this formula explains

This page covers Imperial Wenner and Barnes soil resistivity calculations. The Wenner four-pin method is used to determine average soil resistivity at a given pin spacing, and the Barnes method is used to determine the resistivity of individual soil layers from successive resistance readings.

This Imperial version uses the constant 191.5, with spacing and layer thickness in feet and resistivity in Ω-cm.

Imperial Barnes analysis worksheet

Layer
(n)		 Pin Spacing
(a)		 Resistance
(R)		 Avg. Resistivity
(ρ)		 Layer Thickness
(t)		 Layer Resistance
(RL)		 Layer Resistivity
(ρL)
1 a1 R1
ρ1
= 191.5 × a1 × R1
 —
RL1
= R1
ρL1
= 191.5 × a1 × R1
2 a2 R2
ρ2
= 191.5 × a2 × R2
t2
= a2 − a1
RL2
= (R1 × R2)
/ (R1 − R2)
ρL2
= 191.5 × t2
× RL2
3 a3 R3
ρ3
= 191.5 × a3 × R3
t3
= a3 − a2
RL3
= (R2 × R3)
/ (R2 − R3)
ρL3
= 191.5 × t3
× RL3
4 a4 R4
ρ4
= 191.5 × a4 × R4
t4
= a4 − a3
RL4
= (R3 × R4)
/ (R3 − R4)
ρL4
= 191.5 × t4
× RL4
5 a5 R5
ρ5
= 191.5 × a5 × R5
t5
= a5 − a4
RL5
= (R4 × R5)
/ (R4 − R5)
ρL5
= 191.5 × t5
× RL5

Reference formulas

1) Average Soil Resistivity (Wenner)

\[ \rho = 191.5 \,\times\, a \,\times\, R \]
\(\rho\) = soil resistivity (Ω-cm)
\(a\) = pin spacing (ft)
\(R\) = meter resistance reading (Ω)
\(191.5\) = Imperial Wenner constant

2) Barnes Layer Resistance

\[ R_{Ln} = \frac{R_{n-1} \,\times\, R_n}{R_{n-1} - R_n} \]
\(R_{Ln}\) = layer resistance between spacings
\(R_{n-1}\) = resistance at previous spacing
\(R_n\) = resistance at current spacing

3) Barnes Layer Resistivity

\[ \rho_L = 191.5 \,\times\, t \,\times\, R_L \]
\(\rho_L\) = resistivity of the soil layer (Ω-cm)
\(t\) = layer thickness (ft)
\(R_L\) = layer resistance (Ω)
\(191.5\) = Imperial Barnes constant

4) Finding Missing Resistance Readings

\[ R_n = \frac{R_{n-1} \,\times\, R_L}{R_{n-1} + R_L} \]
\(R_n\) = resistance at current spacing
\(R_{n-1}\) = resistance at previous spacing
\(R_L\) = layer resistance

Practice problems

Problem 1 — Average Soil Resistivity

A soil resistivity survey was conducted. Calculate the average soil resistivity at a pin spacing of 10 ft.

Layer Spacing Resistance
15 ft12.0 Ω
210 ft5.0 Ω
315 ft3.0 Ω
420 ft1.5 Ω
Reveal solution
\[ \begin{array}{c} \rho = 191.5 \,\times\, a \,\times\, R \\[10pt] \rho = 191.5 \,\times\, 10 \,\times\, 5.0 \\[10pt] \rho = 9575\ \Omega\cdot cm \end{array} \]
Answer: 9,575 Ω-cm
Problem 2 — Layer Resistance

Calculate the layer resistance for the soil layer between 5 ft and 10 ft.

Layer Spacing Resistance
15 ft10.0 Ω
210 ft4.0 Ω
315 ft2.0 Ω
420 ft1.0 Ω
Reveal solution
\[ \begin{array}{c} R_L = \frac{R_1 \,\times\, R_2}{R_1 - R_2} \\[10pt] R_L = \frac{10 \,\times\, 4}{10 - 4} \\[10pt] R_L = \frac{40}{6} = 6.67\ \Omega \end{array} \]
Answer: 6.67 Ω
Problem 3 — Deep Layer Resistance

Calculate the layer resistance between 15 ft and 20 ft.

Spacing Resistance
5 ft20.0 Ω
10 ft15.0 Ω
15 ft12.0 Ω
20 ft10.0 Ω
Reveal solution
\[ \begin{array}{c} R_L = \frac{12 \,\times\, 10}{12 - 10} \\[10pt] R_L = \frac{120}{2} = 60\ \Omega \end{array} \]
Answer: 60 Ω
Problem 4 — Barnes Layer Resistivity

Calculate the layer resistivity between 5 ft and 10 ft.

Layer Spacing Resistance
15 ft10.0 Ω
210 ft4.0 Ω
315 ft2.0 Ω
420 ft1.5 Ω
Reveal solution
\[ \begin{array}{c} R_{L2} = \frac{10 \,\times\, 4}{10 - 4} = 6.67\ \Omega \\[10pt] t = 10 - 5 = 5\ ft \\[10pt] \rho_{L2} = 191.5 \,\times\, 5 \,\times\, 6.67 \\[10pt] \rho_{L2} = 6386\ \Omega\cdot cm \end{array} \]
Answer: 6,386 Ω-cm
Problem 5 — High Resistivity Layer

Calculate the layer resistivity for the layer between 10 ft and 15 ft.

Layer Spacing Resistance
15 ft20.0 Ω
210 ft8.0 Ω
315 ft7.0 Ω
420 ft6.0 Ω
Reveal solution
\[ \begin{array}{c} R_{L3} = \frac{8 \,\times\, 7}{8 - 7} = 56\ \Omega \\[10pt] t = 15 - 10 = 5\ ft \\[10pt] \rho_{L3} = 191.5 \,\times\, 5 \,\times\, 56 \\[10pt] \rho_{L3} = 53620\ \Omega\cdot cm \end{array} \]
Answer: 53,620 Ω-cm
Problem 6 — Imperial Deep Layer

Calculate the layer resistivity for the layer between 7.5 ft and 10 ft.

Layer Spacing Resistance
12.5 ft15.0 Ω
25.0 ft7.0 Ω
37.5 ft4.0 Ω
410.0 ft2.0 Ω
Reveal solution
\[ \begin{array}{c} R_{L4} = \frac{4.0 \,\times\, 2.0}{4.0 - 2.0} = 4.0\ \Omega \\[10pt] t = 10.0 - 7.5 = 2.5\ ft \\[10pt] \rho_{L4} = 191.5 \,\times\, 2.5 \,\times\, 4.0 \\[10pt] \rho_{L4} = 1915\ \Omega\cdot cm \end{array} \]
Answer: 1,915 Ω-cm
Problem 7 — Barnes Layer Analysis Table

Complete the table below by calculating the missing values [A] through [F].

Pin Spacing (a) Measured Res (R) Avg. Resistivity (ρ) Layer Res (RL) Layer Resistivity (ρL)
5 ft [A] 3,830 Ω-cm (Same as R) 3,830 Ω-cm
10 ft 1.50 Ω [B] [D] [E]
20 ft [C] 2,298 Ω-cm 1.00 Ω [F]
Reveal solution
\[ \begin{array}{c} [A] = \frac{3830}{191.5 \,\times\, 5} = 4.00\ \Omega \\[10pt] [B] = 191.5 \,\times\, 10 \,\times\, 1.50 = 2873\ \Omega\cdot cm \\[10pt] [C] = \frac{2298}{191.5 \,\times\, 20} = 0.60\ \Omega \\[10pt] [D] = \frac{4.0 \,\times\, 1.5}{4.0 - 1.5} = 2.40\ \Omega \\[10pt] [E] = 191.5 \,\times\, 5 \,\times\, 2.40 = 2298\ \Omega\cdot cm \\[10pt] [F] = 191.5 \,\times\, 10 \,\times\, 1.00 = 1915\ \Omega\cdot cm \end{array} \]
Final Answers: A = 4.0 Ω; B = 2,873 Ω-cm; C = 0.6 Ω; D = 2.4 Ω; E = 2,298 Ω-cm; F = 1,915 Ω-cm

Topics

Barnes Method CP 3 Field Tests Imperial Units Soil Resistivity Wenner Method