Earthing grid: touch voltage decides, not resistance
A 60x40 m substation grid solved per IEEE 80 in calculator #003 — it fails on touch voltage at a 10 m mesh, passes at 5 m, and every step is reproduced by hand.

An earthing grid is not designed to a resistance. It is designed to keep two voltages below what a person can survive — touch and step — and the grid resistance is only one term inside them. This guide runs a 132/11 kV substation grid through calculator #003, gets a fail, and fixes it. Every number is from that run.
What the tool computes
The IEEE Std 80-2013 sequence, in the order the standard sets it out:
- Tolerable voltages — E_touch and E_step for the body weight and clearing time you specify, with the surface-layer derating factor Cs when crushed rock is present;
- Grid current — the fault current actually flowing through the grid: I_G = D_f · Sf · If, where Sf is the split factor and D_f the decrement factor from the X/R ratio and fault duration;
- Grid resistance — Sverak or Schwarz, chosen automatically by geometry, with rods included and two-layer soil reduced to the resistivity each element sees;
- GPR = I_G · R_g, compared against E_touch as the early-exit test;
- Mesh and step voltages — E_m and E_s from the geometry factors Km, Ks, Ki, Kh and the buried length;
- Conductor sizing — the minimum cross-section for the fault energy, per material and joint type.
The point of the tool is that all six move together. Change the mesh spacing and you change L_C, which changes E_m, which changes the verdict — while R_g barely moves.
Worked example: Grid A, 132/11 kV substation
A 60 × 40 m grid, 2 400 m², buried 0.6 m, 12 mm copper conductor, 10 m × 10 m mesh, 12 perimeter rods 3 m long. Two-layer soil: 80 Ω·m to 1.5 m, 300 Ω·m below. Surface layer 100 mm of crushed rock at 2 500 Ω·m. Earth fault 10 kA for 0.5 s, split factor 0.6, X/R = 15. Body weight 70 kg.
Inputs


The geometry line is worth reading before the results: 580 m of buried conductor, 36 m of rods, 2 400 m² enclosed, 7 × 5 conductors. Those are the numbers E_m depends on.
The result — and it fails

| Quantity | Value | Allowed | Verdict |
|---|---|---|---|
| Grid current I_G | 6 280 A | — | D_f = 1.047, Sf = 0.6 |
| Grid resistance R_g | 0.857 Ω | — | Schwarz method |
| GPR | 5 383 V | — | above E_touch → full analysis required |
| Mesh voltage E_m | 1 091 V | 805 V | EXCEEDED — margin 0.74 |
| Step voltage E_s | 530 V | 2 552 V | OK — margin 4.82 |
| Conductor section | 113 mm² (12 mm) | 16.1 mm² min | OK |
Reading the failure
The step voltage passes by a factor of nearly five; the touch voltage fails. This is the normal outcome, not an unusual one. A person's feet are in series through the body for a step voltage — two foot resistances add — while for a touch voltage the hand-to-both-feet path halves that resistance. The tolerable step limit is therefore several times the touch limit, and a grid almost always fails on touch first. Designing on step voltage, or on "the grid looks dense enough", misses the governing case.
GPR of 5 383 V is not the failure. GPR only decides whether you must do the full analysis: if GPR were below E_touch, IEEE 80 lets you stop there. At 5 383 V against 805 V it is far above, so the mesh calculation governs. A high GPR is not itself a violation — it is a trigger.
Cs = 0.700 is doing real work. The 100 mm of crushed rock at 2 500 Ω·m raises the tolerable touch voltage substantially compared with bare soil. Remove the surface layer in the tool and the same grid fails harder — the crushed rock is a design element, not landscaping.
The conductor is not the problem. 16.1 mm² is the thermal minimum for this fault energy; the 12 mm conductor gives 113 mm². Sizing copper for mechanical robustness and corrosion, as is normal practice, leaves a large thermal margin — so a failed grid is almost never fixed by thicker copper.
Fixing it
The lever that matters is buried length inside the same area — that is what flattens the potential across each mesh. Tighten the mesh from 10 m to 5 m, and take the rods from 12 × 3 m to 20 × 4 m:

| 10 m mesh, 12 rods × 3 m | 5 m mesh, 20 rods × 4 m | |
|---|---|---|
| Buried conductor L_C | 580 m | 1 060 m |
| Rod length L_R | 36 m | 80 m |
| Grid resistance R_g | 0.857 Ω | 0.795 Ω |
| GPR | 5 383 V | 4 990 V |
| Mesh voltage E_m | 1 091 V | 600 V |
| Touch margin | 0.74 — fail | 1.34 — pass |
Look at what moved and what did not. Doubling the conductor length cut the mesh voltage by 45 %, from 1 091 V to 600 V. It cut the grid resistance by 7 %, from 0.857 Ω to 0.795 Ω. The design was rescued by potential grading, not by a lower resistance — and if you had spent the same copper chasing R_g, you would still be failing the touch check.
This is the practical lesson of IEEE 80 in one comparison: the resistance sets the GPR, but safety is decided by how uniformly the potential is graded inside the grid. A "good" earth resistance number on a specification sheet says almost nothing about whether the site is safe to stand on during a fault.
Hand check: Grid A, line by line
An earthing calculation has six or seven places where a factor can quietly go astray, so here is the whole chain of the run above reproduced by hand. Every number below is the arithmetic of IEEE Std 80-2013 as the calculator implements it, with the equation numbers it reports on screen.
1 — Surface layer and tolerable voltages (Eq. 27, 32, 33)
100 mm of crushed rock at 2 500 Ω·m over soil of 80 Ω·m, 0,5 s clearing time, 70 kg body (k = 0,157):
Cs = 1 - 0,09 (1 - rho/rho_s) / (2 hs + 0,09)
= 1 - 0,09 (1 - 80/2500) / (2 x 0,10 + 0,09) = 0,6996
E_touch = (1000 + 1,5 Cs rho_s) k / sqrt(ts)
= (1000 + 1,5 x 0,6996 x 2500) x 0,157 / sqrt(0,5) = 804,5 V
E_step = (1000 + 6,0 Cs rho_s) k / sqrt(ts)
= (1000 + 6,0 x 0,6996 x 2500) x 0,157 / sqrt(0,5) = 2 552,0 V
The factor of three between them is the whole reason a grid fails on touch voltage first: 1,5 body-to-two-feet in parallel against 6,0 foot-to-foot in series.
2 — Grid current (Eq. 79)
Ta = (X/R) / (2 pi f) = 15 / (2 pi x 50) = 0,04775 s
D_f = sqrt( 1 + (Ta/tf)(1 - e^(-2 tf/Ta)) )
= sqrt( 1 + (0,04775/0,5)(1 - e^(-20,94)) ) = 1,0467
I_G = D_f x Sf x If = 1,0467 x 0,6 x 10 000 = 6 280 A
3 — Grid resistance
The tool uses Schwarz here (grid plus rods, two-layer soil reduced per element) and reports R_g = 0,857 Ω. Two independent expressions bracket that number:
classical simplified: R = rho/L_T + (rho/4) sqrt(pi/A)
= 80/616 + 20 sqrt(pi/2400) = 0,8535 ohm
Sverak, uniform 80: R = rho [1/L_T + (1/sqrt(20A))(1 + 1/(1 + h sqrt(20/A)))] = 0,8412 ohm
Schwarz, two-layer (tool): 0,8571 ohm
Three formulations within 2 % of each other is what a correct implementation looks like. If they disagreed by tens of per cent, the geometry inputs — not the formula — would be the thing to check.
GPR = I_G R_g = 6 280 x 0,857 = 5 383 V >> E_touch 804,5 V
so the simplified GPR test does not clear the design and the full mesh analysis governs.
4 — Mesh and step voltage (Eq. 84 and following)
The geometry: 60 × 40 m, D = 10 m, so 7 conductors one way and 5 the other, L_C = 7 × 40 + 5 × 60 = 580 m, plus 12 rods × 3 m = 36 m.
n = n_a n_b = 5,80 x 1,01026 = 5,8595
K_h = sqrt(1 + h/h0) = sqrt(1 + 0,6/1,0) = 1,2649
K_m = 1/(2 pi) [ ln( D^2/(16 h d) + (D + 2h)^2/(8 D d) - h/(4 d) )
+ (K_ii/K_h) ln( 8/(pi (2n - 1)) ) ] = 0,91635
K_i = 0,644 + 0,148 n = 0,644 + 0,148 x 5,8595 = 1,5112
L_M = 637,6 m (buried conductor plus the rod contribution)
E_m = rho K_m K_i I_G / L_M = 80 x 0,91635 x 1,5112 x 6 280 / 637,6 = 1 091 V
E_s = rho K_s K_i I_G / L_S = 80 x 0,32493 x 1,5112 x 6 280 / 465,6 = 530 V
Verdict, hand-calculated: E_m = 1 091 V against E_touch = 805 V — the grid fails, with a margin of 805/1 091 = 0,737. The step voltage passes with 2 552/530 = 4,82.
5 — Conductor sizing (Eq. 37)
Hard-drawn copper, 97 % IACS: α_r = 0,00381 1/°C, K₀ = 242, ρ_r = 1,7774 µΩ·cm, TCAP = 3,42 J/(cm³·°C), fusing 1 084 °C, ambient 50 °C, 0,5 s:
A = I sqrt( tc alpha_r rho_r 10^4 / ( TCAP ln((K0 + Tm)/(K0 + Ta)) ) )
= 6,28 x sqrt( 0,5 x 0,00381 x 1,7774 x 10^4 / (3,42 x ln(1326/292)) ) = 16,06 mm2
against 113 mm² for the 12 mm conductor actually laid — the thermal check is not what sizes earthing copper, mechanical strength and corrosion allowance are.
The three answers side by side
| Quantity | Hand calculation | Calculator #003 |
|---|---|---|
| C_s | 0,6996 | 0,6996 |
| E_touch / E_step | 804,5 / 2 552,0 V | 804,5 / 2 552,0 V |
| D_f / I_G | 1,0467 / 6 280 A | 1,0467 / 6 280 A |
| R_g | 0,8535 Ω simplified, 0,8571 Ω Schwarz | 0,8571 Ω |
| GPR | 5 383 V | 5 383 V |
| E_m / E_s | 1 091 / 530 V | 1 091 / 530 V |
| Touch margin | 0,737 — fail | 0,737 — fail |
| A_min | 16,06 mm² | 16,06 mm² |
What a hand check like this is, and is not
It confirms the arithmetic and the factor chain: every intermediate the calculator shows can be reproduced from its inputs with the equations named on screen, and three independent expressions for R_g agree within 2 %. That catches the failure mode that actually bites in practice — a factor applied twice, a temperature or a time used in the wrong place, a geometry counted wrongly.
It is not a substitute for running your own soil resistivity measurement and your own fault-current study. The two inputs that move this result the most — the split factor S_f and the two-layer soil model — come from site data, and no software can validate them for you. Feed it a split factor of 1,0 when the real one is 0,6 and every voltage above rises by two thirds, perfectly correctly and completely wrongly.
What to check before trusting any grid result
- The split factor Sf. Here 0.6 — meaning 40 % of the fault current returns through overhead earth wires and cable screens rather than the soil. It comes straight off your fault-current study and the shielding arrangement, and taking Sf = 1.0 when it is really 0.5 doubles the grid current and can flip the verdict.
- The clearing time. Two times are used: tf for the decrement factor and thermal sizing, ts for the tolerable body voltages. Backup protection times, not primary, generally govern the body limits.
- The soil model. A two-layer model with 80 Ω·m over 300 Ω·m is not the same as a uniform 100 Ω·m. The grid sits in the upper layer, the rods reach into the lower — the tool reduces each to what that element actually sees (80 Ω·m for the grid, 234 Ω·m for the rods here).
- Where the fault current comes from. #002 Short-Circuit gives I″k1 at the substation — that is the input to this calculation, and the earth-fault current is often larger than the three-phase one at a Dyn11 transformer.
FREE and PRO: where the line runs
For #003 the division is between the numbers and the picture.
FREE — open access, no account:
- the full parametric IEEE 80 calculation, exactly as run above: tolerable touch and step voltages with the surface-layer factor, grid current with the decrement and split factors, grid resistance by Sverak or Schwarz, GPR, mesh and step voltages, conductor sizing and the verdict;
- the compact .docx report — project settings, the calculation steps with formulas, and the verdict.
PRO:
- the sketch editor on a 1 m grid, for grids that are not a plain rectangle (
sketch_editor); - the potential, touch and step voltage maps over the site (
potential_maps); - the full .docx report with the derivation, the layout drawing, the maps and the debug appendix (
full_report); - several grid scenarios in one project, so alternatives can be compared side by side (
multi_scenario); - the optimizer, which searches mesh spacing and rod arrangement against the margins (
optimizer).
The free tier is enough to prove a design safe or unsafe — the paid tier is what you need to show where on the site the voltage sits, and to hand a client a report with the drawing in it.
Where this sits next to the other tools
- #002 Short-Circuit supplies the earth-fault current and the X/R ratio this grid is designed against.
- #004 Cable Ampacity covers the cable screens that carry part of the return current, which is where the split factor comes from.