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Street lighting scheme design procedure

Short answer

Street lighting design: select average illuminance E_av, lamp lumens F, CU, MF. Spacing S = F×CU×MF/(E_av×W), ensure S/h~3‑4, check uniformity by point method. Example: 9 m wide road, 12 lux, 100W LED (12,000 lm), CU 0.45, MF 0.8, h=9 m gives S=35 m, 30 poles, uniformity 0.52 acceptable.

With an example, explain the procedure for street lighting scheme design.

Street lighting design means choosing lamps, mounting height, spacing and arrangement so that a road gets the required average illumination with good uniformity and low glare at minimum cost.

Design procedure

  1. Classify the road (main, secondary, residential) and traffic; select the required average illumination E_av and uniformity (E_min/E_av ≥ 0.4).
  2. Note road data: width W, length, number of lanes, median, surface type.
  3. Select lamp and luminaire: HPSV or LED; cut-off/semi-cut-off luminaire; lumen output F.
  4. Choose mounting height h: 7.5–9 m for minor roads, 10–12 m for major roads.
  5. Choose arrangement:
    • Single side: W ≤ h
    • Staggered: W up to 1.5 h
    • Opposite / central twin-arm: wider roads or dual carriageways
  6. Find spacing (lumen method): S = F × CU × MF / (E_av × W); keep S/h ≈ 3–4.
  7. Check minimum and uniformity by the point method midway between poles: E = I h / d³.
  8. Number of poles: N = L/S + 1.
  9. Electrical design: balance lamps on three phases, size cables for ≤ 5% voltage drop, provide a feeder pillar, MCBs and photocell/timer control.
  10. Cost estimate and maintenance plan.

Example

A 1 km road, 9 m wide, needs E_av = 12 lux. Use 100 W LED lanterns of 12 000 lm, CU = 0.45, MF = 0.8, h = 9 m, single-side.

S = F·CU·MF / (E·W)
  = 12 000 × 0.45 × 0.8 / (12 × 9)
  = 40 m
S/h = 40/9 = 4.4  > 4  → take S = 35 m
E_av = 12 000×0.45×0.8 / (35×9) = 13.7 lux
Poles = 1000/35 = 28.6 → 29 spans, 30 poles

Check midway point (x = 17.5 m). Street lanterns concentrate their light at about 60–75° from the downward vertical; from the luminaire's photometric data take I ≈ 3000 cd towards this point:

d² = 9² + 17.5² = 387.25,  d³ = 7620.6
E (one lamp) = 3000 × 9 / 7620.6 = 3.54 lux
E_min (2 lamps) ≈ 7.1 lux
Uniformity = 7.1 / 13.7 ≈ 0.52 ≥ 0.4  ✓

The design is acceptable. (If the luminaire gave only about 1500 cd in this direction, E_min would be 3.5 lux and uniformity 0.26; the spacing would then have to be reduced, or a staggered arrangement or a wider (batwing) distribution used, and the check repeated.)

Load: 30 × 100 W = 3 kW, about 10 lamps per phase.

✓ Answer checked by Claude Opus 5.5 (high effort)

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