- Now I want to talk about irreducible scatter. You would have noticed in Archie's data that there was quite a bit of scatter. As a young field engineer, I just presumed that scatter was due to experimental error. Well, that was a naive assumption. People do know how to make these measurements, and they make them very well and repeatably. So the scatter that we observe is part of the natural world. How can we understand it and where does it come from? So in this slide, we're showing the geometrical factor versus porosity for a model pore, and if you look at the vertical line, a constant porosity that has three red dots on it. This is a model for a unit cell made up of four quarter spheres with space between them. Imagine an electric field applied from right to left and current flowing through the brine in the pore space. So a maximum amount of current will flow when the pore is oriented favorably for that current flow. That would be the red dot at the top. If I rotate the unit cell an 1/8th of a turn, then it becomes more difficult for the current to go through, and the conductivity becomes less. If I rotate it a 1/4 of a turn, I get maximum resistance to current flow. Here we have the same pore with its core geometry varied only by the orientation, and we get three different values of conductivity or geometrical factor. Likewise, for the same geometrical factor we can have different porosities. So, this is showing that a small porosity oriented favorably will pass the same amount of current as a large porosity oriented obliquely to the applied electric field. So you just can't get around in rocks this kind of scatter. Now I want to take you back to high school physics and look at balls freely falling in a constant gravitational field. I do this because we don't really know the physics behind conductivity, porosity relation, and rocks. But we can make an analog model. And if we add scatter to the known solution to this model, then we may be able to illustrate what's going on with Archie rocks. So the equation is the equation for the displacement in the gravitational field, S. S0 is an initial displacement, V0 is an initial velocity, t is time, and g is the acceleration of gravity. Here we see a numerical experiment with balls freely falling in a gravitational field with scatter added. The amount of scatter that I added to the solution was to bring it into a visual conformity with the scatter in Archie's data. The time, which is on the x-axis, is chosen to be from .1 to .4 seconds, and that's to correspond to Archie's 10%-40% porosities. And the displacements are whatever they are in a gravitational field of 32 feet per second per second. The upper row of four slides is reciprocal displacement and I plotted that to simulate resistivity. And the lower row of four slides is displacement and I have plotted that to simulated conductivity. So we see that we've used all the different kinds of graph paper again and we have used eight different fitting functions, and each of the fitting functions gets approximately the same correlation coefficient. So, how would you pick one of these functions as more appropriate than any other function to predict displacement versus time? Notice the two slides in the red box, or the two graphs in the red box. They're duplicated here. Now this is a different numerical experiment. And this numerical experiment was chosen so that the coefficients of determination that R squared were roughly the same and pretty high. And so you can see on the left slide, the loglog presentation, that the fit is really very good. And if this is the only data that you had, you might stop there. On the other hand, on the right slide, that's where the underlying function is fit to the data, and you could notice that the coefficients determined from the data, 13.4 for g/2, is not 16, and 0.89 is not 0.08, or -0.48, and -0.16 is not -0.04, but at least there's some interpretation of those parameters. Looking beneath the left-hand slide, the fitting parameters, 0.04 and -2.4, you might scratch your head a long time trying to figure out what any of that has to do with the acceleration of gravity or your initial conditions. This is Archie's Nacatoch Sandstone Data plotted in the same way. So in the upper four panels, we see resistivity versus porosity. And in the lower four panels, we see conductivity versus porosity. Again, I've used every kind of graph paper to plot the data, and I have fit the data in conductivity and resistivity space with different functions all of which have comparable coefficients of determination. These two plots are again plots that were outlined in red for easy comparison. The one on the left corresponds to the loglog scale that Archie would have used. The one on the right corresponds to a linear scale, conductivity versus formation conductivity factor. And the thing that I want to point out here is that the fits in both cases are equally good, and one thing that you could notice about the fitting coefficients in Archie's equation is that the coefficient, 0.97, is almost equal to one, and the exponent is almost equal to two. I think that will be psychologically significant. And I'll mention that again later. I want to briefly mention that this geometrical factor theory that we're talking about is extensible to hydrocarbon-filled rocks. We don't, for purposes of time, want to go into this. It's discussed in detail in the papers that we cited. But for Archie rocks, we have on the left, the formation resistivity, pardon me, the resistivity index versus water saturation. And Archie discovered that this data, he had it from his precursors, fell on a straight line in loglog space. And if you've ever looked at this data for a single core plug it almost invariably falls really on a straight line. This proves that people know how to make this measurement without any error. If we plot the same data as a geometrical factor versus water saturation, then again it's fit with a straight line, but this is a linear function, not a power function. So, in summary and extension to water saturations less than equal to one, the 100% brine-saturated Archie rocks is given by this conductivity equation, the geometrical factor theory that we've been talking about. Extended to hydrocarbon-bearing Archie rocks, the amount of brine is the fractional volume of brine, phi S w, and the geometrical factors now are the product of the two geometrical factors that we had just talked about. So we still have a triple product of brine conductivity, amount of brine, and geometrical distribution of brine, which we express again as this triple product where beta is the brine fractional volume, E sub t is the product of E0 and lowercase e sub t.