Original synthetic study
Two rocks with the same porosity and different Archie response
Two rocks with identical porosity can have different electrical formation factors and calculated saturations because Archie parameters reflect different pore structures and calibration.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Control porosity and resistivity
The study holds φ = 0.2, Rw and Rt fixed while varying cementation exponent m. Each scenario is a distinct synthetic parameterization, not a measured rock classification.
Formation factor a/φ^m changes even though void fraction does not. The resulting Sw difference follows through the same Archie equation.
Do not interpret m as permeability
Electrical tortuosity and connectivity influence m, while hydraulic permeability also depends strongly on pore-throat geometry. No unique permeability follows from one cementation exponent.
The study's controlled response is mathematical. Real rock comparisons need core calibration, water chemistry and compatible porosity definitions.
Use the result to guide data collection
If saturation is highly sensitive to m, obtaining an appropriate formation-factor calibration can matter more than retaining extra output decimal places.
The download includes the full parameter choices. It does not assign probability to them or claim one exponent is universally correct.
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.
Cementation Factor (dimensionless) = 2. S_w (fraction): 0.25.
- S_w (fraction)
Worked example
Archie Water Saturation from Resistivity Logs
Fixed controls: a = 1 dimensionless; phi = 0.2 fraction; R_w = 0.05 ohm m; R_t = 20 ohm m; n = 2 dimensionless. m 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
Cementation Factor (dimensionless) = 2
Fixed controls: a = 1 dimensionless; phi = 0.2 fraction; R_w = 0.05 ohm m; R_t = 20 ohm m; n = 2 dimensionless. m 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 = 0.05; R_t = 20; n = 2 → S_w = 0.25 fraction
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
S_w (fraction) = 0.25
Equal porosity does not fix formation factor when the calibrated cementation exponent changes.
Result and interpretation
S_w (fraction) = 0.25
Equal porosity does not fix formation factor when the calibrated cementation exponent changes.
Only m changes; porosity, a, n and both resistivities stay fixed.
Study data and reproduction
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.
| Cementation Factor (dimensionless) | S_w (fraction) |
|---|---|
| 1.5 | 0.1671851 |
| 1.75 | 0.2044414 |
| 2 | 0.25 |
| 2.25 | 0.3057111 |
| 2.5 | 0.3738372 |
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. Why can equal porosity fail to imply equal formation factor?
Show the worked answer
Formation factor also depends on a and m, which represent the calibrated electrical pore-network response.
2. Reproduce the baseline S_w (fraction) without rounding intermediate values.
Show the worked answer
a = 1; phi = 0.2; m = 2; R_w = 0.05; R_t = 20; n = 2 → S_w = 0.25 fraction. S_w (fraction) = 0.25. Equal porosity does not fix formation factor when the calibrated cementation exponent changes.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Cementation Factor (dimensionless) changes from 1.5 to 2.5. Other controls remain as stated in the premise. S_w (fraction) changes from 0.167185076 to 0.373837195.
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). Equal porosity does not fix formation factor when the calibrated cementation exponent changes.
Before using the result
- Hold porosity and resistivities fixed.
- Use calibrated exponents for actual rocks.
- Avoid inferring permeability from m alone.
Equal porosity does not fix formation factor when the calibrated cementation exponent changes.
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.