How Archie Exponents Change Water Saturation
Explore cementation exponent m and saturation exponent n with an Archie sensitivity table, worked example, pore diagram and interactive controls.
By PetroCalcHub Team | Updated 2026-10-23
Original diagram Interactive example Worked practice 4 technical sources
Engineering context
Changing the Archie cementation exponent m or saturation exponent n changes the inferred water saturation even when porosity and resistivity measurements stay fixed. That sensitivity is a property of the assumed electrical model. It does not, by itself, identify the correct exponent for a particular rock or prove that Archie is appropriate for the formation.
This lesson uses a synthetic baseline with a = 1, porosity = 0.20, formation-water resistivity Rw = 0.10 ohm m and true-formation resistivity Rt = 10 ohm m. At m = 2 and n = 2 the calculated water saturation is 0.50. The explorer changes only the exponents so that their mathematical effects remain visible rather than being hidden among simultaneous input changes.
Change one thing and observe the result
Fixed synthetic measurements: φ = 0.20, Rw = 0.10 ohm m, Rt = 10 ohm m and a = 1. Change only m and n.
Sw = 0.500000 as a fraction. This is a parameter sensitivity, not a calibrated rock estimate.
| m ↓ / n → | 1.8 | 2.0 | 2.2 |
|---|---|---|---|
| 1.8 | 38.71 | 42.57 | 46.00 |
| 2.0 | 46.29 | 50.00 | 53.25 |
| 2.2 | 55.36 | 58.73 | 61.64 |
Underlined cell: selected exponents. * Above 100%; physically inconsistent. The grid is a local sensitivity experiment, not a confidence interval.
Practical workflow
- 1
Check that porosity is a fraction and the resistivity values use compatible units and temperature bases.
- 2
Reproduce the baseline with a = 1, m = 2 and n = 2.
- 3
Change one exponent at a time, then inspect the two-parameter grid.
- 4
Use independent rock and fluid evidence to assess parameters and the suitability of an Archie model.
Read the two exponents separately
In the formation-factor relationship F = a/φ^m, the exponent m controls how the assumed electrical response changes with porosity. In the resistivity-index relationship I = Sw^(−n), the exponent n describes the saturation dependence. The two relationships are combined in the saturation expression shown here. Porosity and water saturation are fractions, while the compatible resistivity units cancel in the ratio.
Notation needs care when moving between references. The Kansas Geological Survey tutorial uses F to denote porosity in its displayed saturation equation; this lesson uses φ for porosity and F for formation factor. A symbol that looks familiar is not sufficient evidence that two inputs mean the same thing. Read the source's variable definition before reproducing its substitution or comparing a result with the calculator.
Why increasing m raises this baseline result
At porosity 0.20, raising a positive exponent reduces φ^m because porosity is less than one. With all other baseline inputs fixed, the denominator becomes smaller, the bracketed quantity grows, and calculated Sw rises. At n = 2, m values of 1.8, 2.0 and 2.2 give approximately 42.57%, 50.00% and 58.73% water saturation. Those are controlled synthetic substitutions, not recommended rock-property ranges.
The physical justification for m must come from suitable rock evidence and a compatible formation-factor model. Pore connectivity, rock fabric and conductive constituents can complicate a simple interpretation. Do not call a change in the computed answer a measured change in pore structure. The adjacent pore sketch is conceptual: it shows why a conducting water network matters, without claiming to derive an exponent from the drawing.
Why n changes the root of the saturation term
With the baseline m = 2, the quantity inside the brackets is 0.25. Raising n from 1.8 to 2.2 reduces the power 1/n. For a positive bracketed quantity below one, that produces a larger calculated Sw: approximately 46.29%, 50.00% and 53.25% for n = 1.8, 2.0 and 2.2. This reasoning depends on the bracketed value; it should not be generalized without checking the actual input state.
The effect of changing m and n together is not the sum of two fixed percentage increments. Each enters a different power operation, so use the full expression for each parameter pair. The sensitivity grid provides all nine calculations on one unchanged input basis. A practical comparison should likewise document the complete input set rather than comparing percentages copied from calculations with different porosity or water resistivity.
Treat an impossible saturation as information
For this synthetic input set, m = 3 and n = 2 produces Sw of about 1.1180, or 111.80%. A water-saturation fraction above one cannot describe the pore-volume fraction of water. The explorer preserves that numerical result and labels it outside the physical range. Replacing it with exactly 100% would hide the contradiction that the calculation is meant to reveal.
First check decimal fractions, resistivity units, temperature basis, measurement provenance and coefficient entry. Then assess the assumed electrical model. Archie-rock conditions and appropriate parameter support matter; conductive clay or a mismatched rock model cannot be repaired simply by moving sliders until the result looks familiar. Use the shaly-sand guide to review alternative model requirements, and keep sensitivity exploration separate from calibration against independent data.
Reproduce the 50% baseline before changing an exponent
Synthetic inputs: a = 1, φ = 0.20, Rw = 0.10 ohm m, Rt = 10 ohm m, m = 2 and n = 2. The resistivities are assumed compatible with the same interpretation basis.
- 1. Porosity powerφ^m = 0.20² = 0.04
Enter 0.20, not 20, when the input expects fractional porosity.
- 2. Resistivity ratioRw/Rt = 0.10/10 = 0.01
Both resistivities use ohm m, so this ratio is dimensionless.
- 3. Combine the termsa × (Rw/Rt)/φ^m = 1 × 0.01/0.04 = 0.25
This is the quantity to which the saturation-exponent root is applied.
- 4. Apply the rootSw = 0.25^(1/2) = 0.50 = 50%
Only the display converts the fraction to a percentage.
Result
The baseline is 50.00% water saturation. At fixed n = 2, changing m to 2.2 gives approximately 58.73%, an increase of about 8.73 percentage points.
Interpretation
The difference describes a controlled parameter sensitivity. It is not an uncertainty interval or a newly calibrated saturation estimate for measured rock.
Test your understanding
Keep m = n = 2 and all other baseline inputs fixed, but double Rt from 10 to 20 ohm m. What happens?
Show the worked answer
Sw = [0.10/(0.20² × 20)]^(1/2) = √0.125 = 0.353553, or 35.36%. The new value follows a separate resistivity sensitivity, not an exponent change.
Method selection guide
Match the calculation method to the physical question and the evidence available.
| Condition | Use | Why |
|---|---|---|
| Explain one parameter's effect | Change m or n separately | Keeping measurements fixed isolates the selected model assumption. |
| Choose parameters for a rock sample | Use independent electrical and rock evidence | A plausible calculated saturation does not identify unique coefficients. |
| Result exceeds one or contradicts geology | Audit inputs and model applicability | Silent clipping or arbitrary parameter tuning hides useful diagnostic evidence. |
Before using the result
- Use fractional porosity in the equation.
- Keep Rw and Rt units compatible.
- Check the temperature and interpretation basis of resistivity.
- Distinguish porosity φ from formation factor F.
- Record a, m and n with their provenance.
- Keep values outside the physical range visible during diagnosis.
Start with these calculators
These links keep the guide close to the working calculator flow.