Robert C. Ransom



What are Archie’s Basic Relationships

The Graphical Model

What is Meant by the Plot of Rt versus Swtϕt

Summary of Equations

Parallel Resistivity Equations Used in Resistivity Interpretation

What is the Formation Resistivity Factor

The m Exponents

How is Exponent n Related to Exponent m

The a Coefficient

The Saturation Evaluation

Challenging Well-Log Examples

Observations and Conclusions from Figure 10 about Exponent n

Are There Limitations to Archie's Relationships Developed in this Model?




Symbols Defined



All Figures

About the Author

Table of Retrievable Contents:


The Formation Resistivity Factor, F, is an intrinsic property of a porous insulating medium, related to the degree of efficiency or inefficiency for the electrolyte-filled paths to conduct electrical current through the medium. The formation factor pertains to and is intrinsic to the insulating medium only. It is independent of the electrical conductivity of the electrolyte in its pores. Any valid extraneous electrical conductivity that sometimes is seen to influence the value of the formation factor must be relegated to appropriate conductivity equations, such as Eq.(1a) and Eq.(1b), where conductivities can be accommodated by discrete terms.

An equation that demonstrates the purpose of the formation factor is the very basic resistance equation that converts resistivity to resistance:

                                    r ohms = ( ( L / A ) m / m2 ) (R ohms m2 /m )                             . . . (3 d )

where r is resistance, L is length, A is the cross-sectional area of a straight electrically-conductive path of length L, and R is the familiar resistivity. The term L / A in the equation is similar to a formation factor and describes a volume having length L and cross-sectional area A that is 100 percent occupied by a single, homogeneous, electrically-conductive medium, either an electrolyte or solid. This equation represents 100% efficiency in the conversion from resistivity to resistance, and the converse.

Figure 3 is similar to a figure used earlier by Ransom (1984, 1995). This figure represents a unit volume of insulating solid matter where each side of the cube has a length L equal to 1.0 meter. Remove 0.20 of the insulating cube from the center and fill the void with water of resistivity Rw . Electrical current now can pass from top to bottom of the cube through the water without deviation or interference. This form of electrical path through the insulating matter has the highest efficiency.

A measurable resistance implies a measurable resistivity, and resistivity implies conductivity. And, within this insulating cube, the only electrical conductivity occurs in water. Therefore, the measured resistance across the example cube is due to the resistance of the water. Eq.(3d) now can be written

r cube = ( r water ohms) = ( ( L /0.2 L2 ) m / m2 ) ( Rw ohms m2/ m )

If this cube were a unit volume of rock where all void volume was represented by 20% porosity, then, the relationship would be

                            rrock = Rwater = ( L / 0.2 L2 ) ( Rw ) = ( 1.0 / 0.2 L ) ( Rw )

The dimensional analysis of this relationship is

                                  ( ohms ) = ( m / m2 ) ( ohms m2 / m ) = ( ohms )

and the apparent formation factor is (1.0 / 0.2 L) with units of reciprocal meters (m-1 ). After dividing both sides of the equation by the unit volume represented by L / A, as observed in Eq.(3d), resistance becomes resistivity and the relationship becomes

( A / L ) r rock = R rock = Rt = ( A / L ) ( 1.0 / 0.2 L ) ( Rw )

Here, the dimensional analysis is

                                 ( m2 / m ) ( ohms ) = ( m2 / m ) ( 1 / m ) ( ohms m2 / m )

( ohms m2 / m ) = ( m2 / m2 ) ( ohms m2 / m ) = ( ohms m2 / m )

and the formation factor now is dimensionless and has become

                                 F = 1.0 / ( 0.2 ) = 1.0 / ϕt  = 1.0 / ϕtm

in its simplest form, where exponent m = 1.0.

In addition to showing the relationship between the formation factor and the void space in the rock, this exercise demonstrates that the absolute minimum value for exponent m or m1 is 1.0 at the highest possible electrical-path efficiency of 100%.

Theoretically, this degree of efficiency can be duplicated by the presence of an open fracture or other similar water-filled void aligned favorably with the electrical-survey current flow. Although the value of m might never reach 1.0 in practice, the presence of fractures and similar voids can and do reduce the value of m .

There is a fine distinction between the formation factor that is an intrinsic property of rock related to the shape of its voids, and the formation factor that is related to the shape of the water that occupies the voids. The intrinsic formation factor is directly related to the solid insulating framework of rock and how it shapes the electrically-conductive water volume when water saturation is 100%, and nothing else. This by definition is intrinsic. But, the size and shape of the conductive water volume also is influenced by the presence of insulating fluids, such as oil and gas, that can displace water and occupy part of that pore volume. Therefore, the distinction is made that the actual formation factor, that will be called Ft , used in calculations and derivations, will involve the term Swt .   At any saturation other than 100%, Ft is no longer intrinsic.

To carry this demonstration one step further, in the same rock where F = 1.0 / ϕt at 100% efficiency, if part is electrically inert, heterogeneous, porous, insulating rock framework, with a uniform distribution of constituents and porosity, and the remainder is formation water partly displaced by hydrocarbon, then the former volume of water ϕt now becomes Swtϕt and F now becomes 1.0 / ( Swtϕt ) for all values of water saturation and porosity. The fraction Swtϕt now has become the fractional cross-section of area in Figure 3 for all electrically-conductive water paths.

Not all water-filled electrically-effective pore paths are 100% efficient because the network of interstitial water within the rock framework can take on many different shapes and configurations imposed by the many and varied properties of the pore walls and rock framework. Where Swtϕt is the fractional cross-section area for all electrically-conductive water paths at their highest efficiency, as in a bundle of straight tubes, ( Swtϕt )m2 is an equivalent cross-sectional area resulting from all factors that impede the flow of electrical current or increase the resistance-to-flow through the rock. This interference to electrical-current flow is reflected in the steepness of the slopes of the tangential exponents m2 , m1 , and n in Figure 1. As a result, 1.0 / ( Swtϕt ) becomes 1.0 / ( Swtϕt )m2 to accommodate the varied geometries of pores and paths for all conditions of pore path inefficiency and interference, and the Formation Resistivity Factor once again becomes

                                             Ft = 1.0 / ( Swtϕt )m2                                                                from (3b)

the same as it was derived from the trigonometrics in triangle AEG of Figure 1.

The Formation Resistivity Factor is aptly named. Resistance r of the interstitial-water network, and therefore resistivity Rt of the rock, is a function of the size and geometric dimensions of configuration and shape imparted to the water volume within the network of interconnected pores. In one unit of total volume, size or cross-sectional area through which electrical current must flow is related to both porosity and water saturation as a fraction of that one unit area. Tortuous length, configuration, saturation distributions, and shape of the electrical pathway filled with water volume Swtϕt determine the efficiency or inefficiency of in-place water to conduct electrical current and, therefore, provide a value for exponent m2 .

In this conversion of formation water resistivity to pathway resistance, the greatest efficiency at any given value of Swtϕt occurs when exponent m or m1 equals 1.0 and m2 approaches 1.0. And, the greatest effectiveness in one unit of total volume occurs when both Swt and ϕt are 100% or 1.0. When that happens, Ft = 1.0 and r = R = Rw . At any given saturation, overall efficiency of water paths decreases as ϕt decreases and/or as m1 , m2 or n increases.

All factors that influence the parameters referred to in this discussion determine the efficiency for electrical-survey current to flow through the rock. These factors, through Archie’s parameters, ultimately determine the value given to the Formation Resistivity Factor. Next, we look at those factors.



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