Engineering lesson
Why a calculated water saturation can exceed one
An Archie result above one is a diagnostic inconsistency under the supplied assumptions. Do not clip it to one before checking resistivity, porosity, units, temperature basis and model applicability.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Use the full-water resistivity threshold
For Sw = 1, Archie gives Rt = aRw/φ^m. If the supplied Rt is below that threshold, the mathematical result exceeds one.
With a = 1, Rw = 0.1 Ω·m, φ = 0.2 and m = 2, the threshold is 2.5 Ω·m. At Rt = 1 Ω·m and n = 2, Sw = √2.5 = 1.5811.
Investigate inputs before changing parameters
Check whether porosity was entered as 0.2 or 20, whether Rw matches formation temperature and salinity, and whether Rt represents the intended deep formation response.
Conductive clays, invasion and inappropriate a, m or n can also violate the clean-rock assumptions. A physically bounded saturation cannot rescue an unsuitable electrical model.
Preserve the invalid result in the audit
Clipping makes a plot look physical while hiding the magnitude and direction of the inconsistency. Retain the raw result and mark it out of range.
The related shaly-sand guide owns the broader model comparison. This page owns the diagnostic threshold and the sequence of checks before choosing a different model.
Explore the example
Replayed with the same equation record and calculation engine as /calculators/archie-water-saturation-from-resistivity-logs. Original synthetic inputs; engine outputs are saved in the downloadable record.
Select one of 5 recorded scenarios. Intermediate states are not interpolated.
True Formation Resistivity (ohm m) = 1. S_w (fraction): 1.58113883.
- S_w (fraction)
Worked example
Archie Water Saturation from Resistivity Logs
Fixed controls: a = 1 dimensionless; phi = 0.2 fraction; m = 2 dimensionless; R_w = 0.1 ohm m; n = 2 dimensionless. R_t is varied as shown.
Relation and unit basis
S_w = ((a / phi^m) * (R_w / R_t))^(1 / n)
a: Tortuosity Factor (dimensionless). phi: Porosity (fraction). m: Cementation Factor (dimensionless). R_w: Formation Water Resistivity (ohm m). R_t: True Formation Resistivity (ohm m). n: Saturation Exponent (dimensionless).
1. Define the baseline
True Formation Resistivity (ohm m) = 1
Fixed controls: a = 1 dimensionless; phi = 0.2 fraction; m = 2 dimensionless; R_w = 0.1 ohm m; n = 2 dimensionless. R_t is varied as shown.
2. Select the relation
S_w = ((a / phi^m) * (R_w / R_t))^(1 / n)
a: Tortuosity Factor (dimensionless). phi: Porosity (fraction). m: Cementation Factor (dimensionless). R_w: Formation Water Resistivity (ohm m). R_t: True Formation Resistivity (ohm m). n: Saturation Exponent (dimensionless).
3. Substitute the values
a = 1; phi = 0.2; m = 2; R_w = 0.1; R_t = 1; n = 2 → S_w = 1.58113883 fraction
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
S_w (fraction) = 1.58113883
The full-water threshold is Rt = 2.5 Ω·m. Sw above one remains visible as an input/model inconsistency, not a physical saturation.
Result and interpretation
S_w (fraction) = 1.58113883
The full-water threshold is Rt = 2.5 Ω·m. Sw above one remains visible as an input/model inconsistency, not a physical saturation.
Vary Rt through the full-water threshold. Values above 100% remain visible and explicitly invalid as physical saturations.
Exact scenario data
Replayed with the same equation record and calculation engine as /calculators/archie-water-saturation-from-resistivity-logs. Original synthetic inputs; engine outputs are saved in the downloadable record.
| True Formation Resistivity (ohm m) | S_w (fraction) |
|---|---|
| 1 | 1.581139 |
| 2 | 1.118034 |
| 2.5 | 1 |
| 5 | 0.7071068 |
| 10 | 0.5 |
Scroll the table to see all columns. The highlighted row is the worked baseline. Downloads retain full numerical precision.
Test your understanding
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. What is the full-water Rt threshold for Rw 0.08 Ω·m, φ 0.2, a 1 and m 2?
Show the worked answer
0.08/0.2² = 2 Ω·m. Smaller Rt produces Sw above one under these assumptions.
2. Reproduce the baseline S_w (fraction) without rounding intermediate values.
Show the worked answer
a = 1; phi = 0.2; m = 2; R_w = 0.1; R_t = 1; n = 2 → S_w = 1.58113883 fraction. S_w (fraction) = 1.58113883. The full-water threshold is Rt = 2.5 Ω·m. Sw above one remains visible as an input/model inconsistency, not a physical saturation.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
True Formation Resistivity (ohm m) changes from 1 to 10. Other controls remain as stated in the premise. S_w (fraction) changes from 1.58113883 to 0.5.
4. What would be needed to apply this example to a different physical system?
Show the worked answer
Replace the synthetic controls with compatible documented inputs and check the relation's assumptions. a: Tortuosity Factor (dimensionless). phi: Porosity (fraction). m: Cementation Factor (dimensionless). R_w: Formation Water Resistivity (ohm m). R_t: True Formation Resistivity (ohm m). n: Saturation Exponent (dimensionless). The full-water threshold is Rt = 2.5 Ω·m. Sw above one remains visible as an input/model inconsistency, not a physical saturation.
Before using the result
- Retain the unclipped diagnostic result.
- Check fraction versus percent inputs.
- Check temperature and conductive-mineral assumptions.
The full-water threshold is Rt = 2.5 Ω·m. Sw above one remains visible as an input/model inconsistency, not a physical saturation.
Continue learning
Start with the broader workflow guide
Sources and method
The explanation and synthetic examples are original PetroCalcHub teaching material. The references below support the definitions and model context. They do not supply field validation for the synthetic scenarios.