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Standard traction-elevator power formula + NEC Table 620.148 min read

Worked Example: Single Elevator Motor Power vs. a Four-Elevator Group Feeder Under NEC 620.14

One traction elevator's motor power from its rated load, speed, and counterweight balance, then the NEC Table 620.14 demand factor that lets a shared four-elevator feeder be sized well below the sum of all four running flat-out.

Scenario

Rated load per car1000 kg
Rated speed1.5 m/s
Counterweight balance factor0.5 (50% of rated load balanced)
Drive efficiency (mechanical x motor)0.7
Number of identical elevators on the shared feeder4
Feeder voltage / power factor415 V 3-phase / 0.85

Step-by-step calculation

Step 1: Find the net (unbalanced) load the motor must actually move

The counterweight balances a fraction of the rated load — the motor only has to move the remaining unbalanced portion at rated speed.

netLoad = ratedLoad x (1 - balanceFactor)
1000 x (1 - 0.5)
netLoad = 500 kg

Step 2: Compute one elevator's motor power

P(kW) = netLoad x 9.81 x speed / (1000 x efficiency)
500 x 9.81 x 1.5 / (1000 x 0.7)
singleUnitPowerKw = 10.51 kW

Step 3: Estimate the single unit's full-load current

FLA ≈ P(W) / (√3 x V x PF)
10,510.7 / (1.732 x 415 x 0.85)
singleUnitFlaApprox = 17.20 A

Step 4: Find the connected load for all 4 elevators (before any demand factor)

connectedLoad = singleUnitPower x numElevators
10.51 x 4
connectedLoadKw = 42.04 kW — this is what the feeder would need if all 4 ran at full power simultaneously with zero diversity

Step 5: Apply the NEC Table 620.14 demand factor for 4 elevators on one feeder

Table 620.14 reflects the reality that not every elevator in a group runs at full duty simultaneously — the demand factor drops as more elevators share a feeder.

demandLoad = connectedLoad x demandFactor(n)
42.04 x 0.85 (the Table 620.14 factor for exactly 4 elevators)
demandLoadKw = 35.74 kW, demandLoadFlaApprox = 58.49 A

Result summary

CheckRequirementActualStatus
Single unit motor powern/a (sizing basis)10.51 kW✓ PASS
Connected load (4 units, no diversity)n/a (informational)42.04 kW✓ PASS
NEC 620.14 demand factor (n=4)n/a (code table lookup)0.85✓ PASS
Feeder demand loadn/a (sizing basis)35.74 kW / 58.49 A✓ PASS
Each elevator's motor draws 10.51 kW to move its 500 kg net (unbalanced) load at 1.5 m/s. Four of them would connect to 42.04 kW with zero diversity, but NEC Table 620.14's 0.85 demand factor for a 4-elevator group brings the feeder design load down to 35.74 kW (58.49 A) — about 15% smaller than sizing for all four running flat-out at once.

Key insight: NEC 620.14's demand factor exists because elevator group operation is inherently non-simultaneous by design — a dispatch system rarely runs every car in a bank at full duty at the same instant, so sizing the shared feeder for 100% connected load on every unit would be a real but avoidable oversizing. The demand factor gets smaller as more elevators share a feeder (down to 0.72 for 10 or more), reflecting that duty-cycle overlap becomes statistically less likely to hit 100% concurrently as the group grows.

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Frequently asked questions

Does the demand factor apply to each elevator's own branch circuit, or only to the shared feeder?

Only to the shared feeder — each individual elevator's own branch circuit and motor protection still has to be sized for that elevator's own full-load current with no diversity applied, since any single car can and does run at full duty at any moment. The demand factor is specifically a feeder-level allowance for the statistical unlikelihood of every car in the group being at full duty at exactly the same instant.

Why does a higher counterweight balance factor reduce motor power?

The counterweight's whole purpose is to offset most of the car's weight so the motor only has to supply the net difference — a higher balance factor (closer to 1.0) leaves less unbalanced load for the motor to lift, directly reducing both motor power and current, though in practice the balance factor is chosen based on expected average passenger loading, not maximized purely to minimize motor size.

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