How many rods do you actually have to drive? Give it the soil resistivity and the electrode you are thinking of building, and it returns the resistance to earth, the group efficiency you lose to rods crowding each other, the count or conductor length needed to hit your target, and the ground potential rise if you tell it the fault current. Rods, rings and grids — with the formula and its limits written out.
Soil
Measure it (Wenner). 100 is the usual stand-in for average moist soil.
Electrode
Barely matters — it sits inside a logarithm.
At least one rod length. Two to three is better.
Target and fault
From your code or your connection agreement.
For ground potential rise. Estimate it with the short-circuit calculator.
Result
What each extra rod actually buys
| Rods | Group Rn | R₁ ÷ n (ideal) | Efficiency |
|---|
The gap between the second and third column is what mutual coupling costs you. It widens as you add rods and as you crowd them together.
It turns a soil resistivity and an electrode geometry into a resistance to earth, then answers the question you actually have on site: will this pass, and if not, what do I add. It handles three geometries — a row of driven rods, a buried ring, and a grid or mesh — and it reverse-solves the rod count or the conductor length your target needs.
It does not tell you what your target is. That number comes from the wiring code, the earthing arrangement and, on a grid-connected job, the connection agreement. It is an input here, deliberately.
The λ values the group formula produces are the ones BS 7430 tabulates — 1.000 for two rods, 3.858 for ten — so the coupling term is not an invention. As a cross-check the whole rod model was run against Sunde's independent closed form for the same geometry; the two agree within 8 % across several thousand combinations of resistivity, length, diameter and spacing.
Current leaving a rod spreads through concentric shells of soil. The shell nearest the rod is thin and small, so its resistance is high; each shell further out has a larger area for the same current, so its resistance falls. Add them all up to infinity and the total converges — and nearly all of it accumulates in the first metre or so. Earth "far away" contributes almost nothing.
Two consequences fall straight out of that geometry, and both are worth the price of a rod:
It also explains the spacing rule. A second rod driven close to the first shares the shells the first one already owns, so it does not behave like an independent resistor in parallel. At a spacing of one rod length a second rod delivers roughly 85 % of what it would on its own; spread them two or three rod lengths apart and you get most of the rest back.
Three 3 m rods, 16 mm diameter, 3 m apart, in soil measured at 100 Ω·m.
Three rods buy you 14.11 Ω, not the 11.16 Ω you would get if they did not interact. To reach 5 Ω with this rod and this spacing you need 11 rods — at which point buying longer rods, or finding 6 m of better soil, is almost certainly cheaper than the trenching.
A 10 × 10 m earthing grid, 200 m of buried conductor including rods, 0.5 m deep, 100 Ω·m soil.
Read the area floor, not the resistance. This grid is at 4.56 Ω and the floor for its footprint is 4.06 Ω. That means no amount of extra conductor inside this 100 m² will get you below 4.06 Ω — you would have to lay thousands of metres to shave off half an ohm. If the spec says 1 Ω, more copper is not the answer; the grid has to get bigger, or you need deep rods into a better layer, or the soil needs treatment. That one number is usually what decides whether a substation earthing design is even feasible on the land available.
ρ dominates the answer and it is the input people are most tempted to guess. Salt content, moisture and temperature all move it, and the same ground can read several times higher at the end of a dry season than it does after rain. Frozen soil is close to an insulator.
Four pins in a line at equal spacing a, current through the outer pair, voltage across the inner pair. The reading averages the soil to a depth of roughly the spacing, so repeat it at 1, 2, 4 and 8 m: if ρ climbs with spacing there is something worse underneath and you should drive deep; if it falls, there is water or a conductive layer below and deep rods will pay. Designing a grid from the reading at 2 m when the grid works against the first 10 m is the classic mistake that ends in a non-compliant earth.
| Ground | Typical ρ (Ω·m) |
|---|---|
| Marshy ground, peat | 5 – 40 |
| Clay, wet | 8 – 50 |
| Clay, dry | 50 – 150 |
| Loam, garden soil | 50 – 200 |
| Sand, wet | 100 – 300 |
| Chalk | 100 – 500 |
| Gravel | 300 – 1000 |
| Sand, dry | 1000 – 3000 |
| Limestone | 500 – 10 000 |
| Rock (granite, slate) | 1000 and up |
Indicative ranges for planning only. Moisture alone spans orders of magnitude: mud around 1 Ω·m, wet soil near 10, moist near 100, dry near 1000, rock near 10 000. Measure it, and measure it in the dry season if the installation has to comply all year.