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Flood lighting a 50 × 16 m building

Short answer

Total lumens = 100×800×1.5×1.2/0.5 = 288,000 lm. Assuming 1000 W tungsten-halogen (20 lm/W): 15 projectors. Beam spread ≈ 35°. Alternative: 400 W metal-halide (80 lm/W) yields 9 projectors.

A building elevation is 50m x 16m is to be illuminated by flood lighting projectors situated at 25 meters away. If the illumination is 100 lumen/m², coefficient of utilization is 0.5, depreciation factor 1.5, waste light factor 1.2, estimate the number and size of projectors. Also find out angle of spread.

Given: building face 50 m × 16 m, projectors 25 m away, E = 100 lm/m² (lux), CU = 0.5, depreciation factor DF = 1.5, waste light factor WLF = 1.2.

Total lumens required

Area A   = 50 × 16 = 800 m²
Total lm = E × A × DF × WLF / CU
         = 100 × 800 × 1.5 × 1.2 / 0.5
         = 288 000 lm

Number and size of projectors

Size is not given, so assume 1000 W tungsten-halogen floodlight projectors with efficacy about 20 lm/W (standard textbook assumption):

Lumens per lamp = 1000 × 20 = 20 000 lm
N = 288 000 / 20 000 = 14.4 → 15 projectors

(If 400 W metal-halide lamps of 80 lm/W = 32 000 lm are used instead, N = 9.)

Angle of spread

With projectors 25 m away, aimed at the middle of the face (8 m high), the vertical beam must cover 16 m:

θ = 2 tan⁻¹(8 / 25) = 35.5°

(If projectors are at ground level aimed upward, the face subtends tan⁻¹(16/25) = 32.6°.) So projectors with a medium beam of about 35° are chosen. Horizontally the 15 projectors are spread along the 50 m frontage so that their beams overlap.

Answer: total 288 000 lm; 15 projectors of 1000 W (20 lm/W assumed); beam spread ≈ 35°.

✓ Answer checked by Claude Opus 5.5 (high effort)

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