Soil Resistivity for Cathodic Protection

Soil resistivity describes how strongly soil resists electrical current flow.

In cathodic protection work, soil resistivity is useful because CP current must pass through soil or another electrolyte. Lower-resistivity soil generally allows current to flow more easily than higher-resistivity soil. But one value does not fully describe a site. Soil conditions can change with depth, moisture, temperature, location, and backfill.

This page explains a basic Wenner-style calculation. It does not teach a complete soil survey procedure and it does not define CP protection criteria.

Quick Definition

A common learner formula for soil resistivity is:

ρ = 2 × π × a × R

Where:

  • ρ is soil resistivity.
  • π is pi, approximately 3.1416.
  • a is probe spacing.
  • R is measured resistance.

If a is in centimeters and R is in ohms, the result is in ohm-cm.

If a is in feet and R is in ohms, the result is in ohm-ft.

When Soil Resistivity Is Used in CP Work

Soil resistivity may be used when reviewing:

  • expected current flow through soil;
  • relative corrosivity discussions;
  • groundbed or anode-bed planning context;
  • possible variation between areas;
  • basic CP design or review assumptions.

This page does not say what resistivity value is acceptable.

This page does not say whether CP is adequate.

A single soil-resistivity value should not be treated as a complete site characterization.

Formula Reference

Solve forFormulaUse when you know
Soil resistivityρ = 2 × π × a × Rprobe spacing and measured resistance
Measured resistanceR = ρ ÷ (2 × π × a)resistivity and probe spacing
Probe spacinga = ρ ÷ (2 × π × R)resistivity and measured resistance

Variable and Unit Table

SymbolMeaningCommon unitUnit symbolCP field meaning
ρSoil resistivityohm-centimeter or ohm-footohm-cm or ohm-ftHow strongly soil resists current flow
πPiunitless—Approximately 3.1416
aProbe spacingcentimeter or footcm or ftSpacing used in the resistivity calculation
RMeasured resistanceohmΩResistance measured by the instrument

Use consistent units:

  • cm × Ω gives ohm-cm;
  • ft × Ω gives ohm-ft;
  • do not mix centimeters and feet in the same calculation unless a conversion is made first.

Common conversion:

  • 1 ft = 30.48 cm

How to Use the Formula

Use the probe spacing and measured resistance from the same reading.

Check the spacing unit before calculating.

For ohm-cm:

  1. use probe spacing in centimeters;
  2. multiply 2 × π × a × R;
  3. report the answer in ohm-cm.

For ohm-ft:

  1. use probe spacing in feet;
  2. multiply 2 × π × a × R;
  3. report the answer in ohm-ft.

Do not use one soil-resistivity value to make broad conclusions about an entire site without additional review.

Worked Examples

Worked Example 1 — Soil Resistivity in ohm-cm

A Wenner-style reading uses probe spacing of 100 cm. The measured resistance is 25 Ω. What is the soil resistivity?

Known values:

  • a = 100 cm
  • R = 25 Ω
  • ρ = ?

Use: ρ = 2 × π × a × R

Substitute: ρ = 2 × 3.1416 × 100 cm × 25 Ω

Calculate: ρ = 15,708 ohm-cm

Answer: The soil resistivity is approximately 15,708 ohm-cm.

Field meaning: This is the calculated resistivity for that spacing and reading. It does not fully characterize the entire site.

Worked Example 2 — Soil Resistivity in ohm-ft

A reading uses probe spacing of 5 ft. The measured resistance is 12 Ω. What is the soil resistivity?

Known values:

  • a = 5 ft
  • R = 12 Ω
  • ρ = ?

Use: ρ = 2 × π × a × R

Substitute: ρ = 2 × 3.1416 × 5 ft × 12 Ω

Calculate: ρ = 376.99 ohm-ft

Answer: The soil resistivity is approximately 377 ohm-ft.

Field meaning: Because spacing is in feet, the result is in ohm-ft.

Worked Example 3 — Solve for Measured Resistance

A soil resistivity value is 6,280 ohm-cm using 100 cm probe spacing. What measured resistance does that represent?

Known values:

  • ρ = 6,280 ohm-cm
  • a = 100 cm
  • R = ?

Use: R = ρ ÷ (2 × π × a)

Substitute: R = 6,280 ohm-cm ÷ (2 × 3.1416 × 100 cm)

Calculate: R = 9.99 Ω

Answer: The measured resistance is approximately 10 Ω.

Field meaning: This rearrangement checks the resistance value that corresponds to a resistivity and spacing.

Worked Example 4 — Convert Probe Spacing

Probe spacing is 4 ft. What is the spacing in centimeters?

Known values: a = 4 ft

Use: 1 ft = 30.48 cm

Substitute: a = 4 ft × 30.48 cm/ft

Calculate: a = 121.92 cm

Answer: The spacing is 121.92 cm.

Field meaning: Convert spacing before calculating if the desired resistivity unit requires a different length unit.

Common Mistakes

  1. Mixing spacing units. Use centimeters for ohm-cm or feet for ohm-ft.
  2. Forgetting that resistance and resistivity are different. R is measured resistance. ρ is calculated resistivity.
  3. Treating one reading as the whole site. Soil can vary by location, depth, moisture, and temperature.
  4. Forgetting the 2 × π factor. The formula is not just spacing times resistance.
  5. Using probe spacing from a different reading. Use spacing and measured resistance from the same reading.
  6. Confusing ohm-cm with Ω. Resistivity and resistance use different units.
  7. Using too many decimals. Resistivity results are often approximate.
  8. Treating soil resistivity as proof of CP protection. Soil resistivity supports review and planning. It does not prove protection.

Practice Problems

  1. A reading uses probe spacing of 100 cm and measured resistance of 10 Ω. What is the soil resistivity in ohm-cm?
  2. A reading uses probe spacing of 50 cm and measured resistance of 20 Ω. What is the soil resistivity in ohm-cm?
  3. A reading uses probe spacing of 5 ft and measured resistance of 8 Ω. What is the soil resistivity in ohm-ft?
  4. A soil resistivity is 12,566 ohm-cm using 100 cm spacing. What measured resistance does that represent?
  5. Convert 3 ft to centimeters.
  6. If measured resistance doubles and probe spacing stays the same, what happens to calculated soil resistivity?
  7. Does one soil-resistivity value fully characterize an entire site?
  8. Does soil resistivity by itself prove that a structure satisfies CP protection criteria?

Practice Problem Answer Key

ProblemAnswerCalculation / explanation
16,283 ohm-cmρ = 2 × 3.1416 × 100 cm × 10 Ω = 6,283.2 ohm-cm
26,283 ohm-cmρ = 2 × 3.1416 × 50 cm × 20 Ω = 6,283.2 ohm-cm
3251 ohm-ftρ = 2 × 3.1416 × 5 ft × 8 Ω = 251.33 ohm-ft
420 ΩR = 12,566 ohm-cm ÷ (2 × 3.1416 × 100 cm) = 20.0 Ω
591.44 cm3 ft × 30.48 cm/ft = 91.44 cm
6Resistivity doublesρ = 2 × π × a × R, so if R doubles and a stays the same, ρ doubles
7NoSoil can vary across the site and with depth, moisture, temperature, and other conditions
8NoSoil resistivity supports review and planning. CP protection must be evaluated with proper measurements and criteria.

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