090124-1540 EST
Some general comments.
To some extend you can consider a vertical rod driven into the ground to be an isotropic radiator.
What does isotropic mean? From dictionary.com
"Identical in all directions; invariant with respect to direction."
If we consider a long vertical rod and assume the resistivity of the soil to be uniform, then there is a uniform current density leaving the rod. This also means there is another imaginary large diameter conductive cylinder concentric and spaced a substantial distance from the rod. This will be an isotropic system and can be mathematically analyzed.
For each radial foot I move away from the rod the amount of soil in the 1 foot thick cylinder wall increases and thus the resistance of that cylinder becomes lower. I get more resistors of the same value in parallel as I move further from the center rod.
Maximum resistance is closest to the rod and becomes insignificant as you get further away. Therefore maximum voltage drop occurs close to the rod.
If you measure resistance between two rods spaced a substantial from one another, then you do not have an isotropic radiator but an approximation to one. You still see the majority of voltage drop close to the rods.
In my area If I measure the resistance between two 8 ft vertical rods about 30 ft apart I get in the range of 10 to 20 ohms. I make this resistance measurement from the voltage between the rods and the current thru them. My excitation is the output of a 28 V transformer. Before applying voltage I measure the rod to rod voltage to determine if there are stray currents. In the places I have done this there have been only a few millivolts and this is not a significant source of error.
If you were confronted with stray currents, then you need an excitation frequency that will allow you to eliminate these errors. You do not want to do a DC measurement because of polarization problems. However, with a Simpson 260 in the ohms position I can quickly switch between + and - and get results comparable to the AC measurement.
In the real world you will not have an isotropic field, but this type of analysis still gives you an estimate of what happens.
Next consider the resistance between two rods and what a resistance of 25 ohms might mean. Here is a real world problem you have to fight. Some machine tool manufacturers have told customers that they need to put a ground rod at a machine and disconnect the EGC from the machine. This is wrong and not safe.
Assume this was done and the resistance from the ground rod or electrode at the service entrance to this isolated rod at the machine is 25 ohms. Now short circuit a 120 V hot conductor to the machine chassis. The short circuit current will be 120/25 = 4.8 A. This will never trip a 20 A breaker, and the voltage on the chassis of the machine will be close to 120 V relative to the ground reference at the main panel. Next make the ground resistance between the rods 10 ohms. Still won't blow the breaker with just 12 A.
Next we will try a 1 ohm ground resistance and a 100 ft length of #12 copper for the hot wire. The wire resistance is 0.159 ohms. The current is 120/ 1.159 = 103.5 A. This will trip the breaker, but until it trips the voltage on the machine chassis is 103.5 V.
Note: that even a ground electrode resistance of 1 ohm results in a large voltage on the machine chassis.
If a #12 EGC had been connected to the machine and no auxiliary ground rod, then with a dead short circuit and an invariant 120 V source the current would have been 120/0.318 = 377 A and the machine chassis would be 60 V RMS with a peak of 170 V until the breaker trips, probably about 8 to 16 MS.
Lightning is a somewhat different story. Still consider a 1 ohm resistance to ground. Suppose a 10,000 A lightning current flows into this 1 ohm grounding electrode. Now there is a difference in potential between somewhere in the earth and the electrode of 10,000 V. Suppose there are no other connections in the building to earth and leakage resistance is 1000 ohms to earth. There would be less than 10 A flow into the building, and the amount would depend the fields from the building to earth.
But lightning bolts are rather high frequency signals and if you insert impedance in series with conductors into the building the amount of current entering the building can be greatly reduced. Additionally all paths on the outside of the building to and including the ground electrode should be low inductance, broad (wide) conductors to lower the high frequency impedance.
If internal to the building you can keep differential voltages between different parts of the building low during a lightning strike, then minimum damage occurs independent of what potential the building has relative to some place in the earth. This means keep lightning current out of the building, eliminate supplemental paths to ground, and keep impedances low within the building.
.
