What this formula explains
Soil resistivity is one of the most important parameters in cathodic protection design because it directly
affects current distribution, anode bed performance, and overall circuit resistance.
The Wenner four-pin method is used to determine the average soil resistivity to a specific
investigation depth. The Barnes method is used to determine the resistivity of individual soil
layers by analyzing the parallel resistance paths that exist between successive electrode spacings.
Reference formulas
1) Average Soil Resistivity (Wenner)
\[
\rho
=
K \,\times\, a \,\times\, R
\]
\(\rho\) = soil resistivity (Ω·cm or Ω·m)
\(K\) = 191.5 (Imperial) or \(2\pi\) (Metric)
\(a\) = pin spacing (feet or meters)
\(R\) = meter resistance reading
\(191.5\) = 2 \(\times \) \(\pi\) \(\times \) 2.54 cm \(\times \) 12 in
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
\[
\rho_L
=
K \,\times\, t \,\times\, R_L
\]
\(\rho_L\) = resistivity of the soil layer
\(K\) = 191.5 (Imperial) or \(2\pi\) (Metric)
\(t\) = layer thickness (feet or meters)
\(R_L\) = layer resistance
\(191.5\) = 2 \(\times \) \(\pi\) \(\times \) 2.54 cm \(\times \) 12 in
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
Problem 1 — Average Soil Resistivity (Imperial)
A soil resistivity survey was conducted. Calculate the average soil resistivity
at a pin spacing of 10 ft.
| Layer |
Spacing |
Resistance |
| 1 | 5 ft | 12.0 Ω |
| 2 | 10 ft | 5.0 Ω |
| 3 | 15 ft | 3.0 Ω |
| 4 | 20 ft | 1.5 Ω |
Reveal solution
\[
\begin{array}{c}
\rho
=
K \,\times\, a \,\times\, R \\[10pt]
\rho
=
191.5 \,\times\, a \,\times\, R \\[10pt]
\rho
=
191.5 \,\times\, 10 \,\times\, 5.0 \\[10pt]
\rho
=
9575\ \Omega\cdot cm
\end{array}
\]
Answer: 9,575 Ω-cm
Problem 2 — Average Soil Resistivity (Metric)
Calculate the average soil resistivity at a spacing of
3 meters.
| Layer |
Spacing |
Resistance |
| 1 | 1 m | 25.0 Ω |
| 2 | 2 m | 12.0 Ω |
| 3 | 3 m | 7.5 Ω |
| 4 | 4 m | 5.0 Ω |
Reveal solution
\[
\begin{array}{c}
\rho
=
K \,\times\, a \,\times\, R \\[10pt]
\rho
=
(2 \,\times\, \pi )\,\times\, a \,\times\, R \\[10pt]
\rho
=
6.28 \,\times\, 3 \,\times\, 7.5 \\[10pt]
\rho
=
141.3\ \Omega\cdot m
\end{array}
\]
Answer: 141.3 Ω-m
Problem 3 — Layer Resistance
Calculate the layer resistance for the soil layer
between 5 ft and 10 ft.
| Layer |
Spacing |
Resistance |
| 1 | 5 ft | 10.0 Ω |
| 2 | 10 ft | 4.0 Ω |
| 3 | 15 ft | 2.0 Ω |
| 4 | 20 ft | 1.0 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 4 — Deep Layer Resistance
Calculate the layer resistance between
15 ft and 20 ft.
| 5 ft | 20.0 Ω |
| 10 ft | 15.0 Ω |
| 15 ft | 12.0 Ω |
| 20 ft | 10.0 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 5 — Layer Resistance (Metric)
Calculate the layer resistance between
4 m and 6 m.
| 2 m | 5.0 Ω |
| 4 m | 2.0 Ω |
| 6 m | 1.2 Ω |
| 8 m | 0.8 Ω |
Reveal solution
\[
\begin{array}{c}
R_L
=
\frac{2.0 \,\times\, 1.2}{2.0 - 1.2} \\[10pt]
R_L
=
\frac{2.4}{0.8}
=
3.0\ \Omega
\end{array}
\]
Answer: 3.0 Ω
Problem 6 — Barnes Layer Resistivity
Calculate the layer resistivity between
5 ft and 10 ft.
Reveal solution
\[
\begin{array}{c}
R_L
=
6.67\ \Omega \\[10pt]
t
=
5\ ft \\[10pt]
\rho_L
=
191.5 \,\times\, 5 \,\times\, 6.67 \\[10pt]
\rho_L
=
6386\ \Omega\cdot cm
\end{array}
\]
Answer: 6,386 Ω-cm
Problem 7 — High Resistivity Layer
Reveal solution
\[
\begin{array}{c}
R_L
=
\frac{8 \,\times\, 7}{8 - 7}
=
56\ \Omega \\[10pt]
t
=
5\ ft \\[10pt]
\rho_L
=
191.5 \,\times\, 5 \,\times\, 56 \\[10pt]
\rho_L
=
53620\ \Omega\cdot cm
\end{array}
\]
Answer: 53,620 Ω-cm
Problem 8 — Metric Deep Layer Resistivity
Reveal solution
\[
\begin{array}{c}
R_L
=
6.0\ \Omega \\[10pt]
t
=
2\ m \\[10pt]
\rho_L
=
6.28 \,\times\, 2 \,\times\, 6.0 \\[10pt]
\rho_L
=
75.4\ \Omega\cdot m
\end{array}
\]
Answer: 75.4 Ω-m
Problem 9 — Imperial Deep Layer
Reveal solution
\[
\begin{array}{c}
R_L
=
4.0\ \Omega \\[10pt]
t
=
2.5\ ft \\[10pt]
\rho_L
=
191.5 \,\times\, 2.5 \,\times\, 4.0 \\[10pt]
\rho_L
=
1915\ \Omega\cdot cm
\end{array}
\]
Answer: 1,915 Ω-cm
Problem 10 — Barnes Layer Analysis
Complete the Barnes analysis table by calculating the missing values.
Reveal solution
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
Problem 11 — Metric Spacing Analysis
Reveal solution
A = 10.0 Ω
B = 5,027 Ω-cm
C = 2.0 Ω
D = 6.67 Ω
E = 4,191 Ω-cm
F = 5,027 Ω-cm
