- 2073 Bhadra · 2+2+4 marks
Waste light factor and floodlight calculation
Short answer
Waste light factor (WLF) increases required lumens to compensate for light spilling beyond edges (typical 1.2–1.5). It prevents a 15–35% illumination shortfall. For a 30 m × 12 m facade needing 120 lux: F = 120 × 360 × 1.3 × 1.2 / 0.4 = 168,480 lm; using 250 W lamps (20,000 lm) yields 9 projectors.
What is Waste Light Factor? Why is it necessary to be considered in floodlighting design? With a suitable example, explain the process of floodlight calculation?
Waste light factor
Waste light factor (WLF) is the factor by which the required lumens are increased to allow for light from the projectors that spills beyond the edges of the surface (or passes between parts of it) and so does no useful work. Usual values: 1.2 for regular, flat surfaces and 1.5 for irregular objects such as statues and towers.
Why it must be considered
- Beams are circular or oval, while buildings are rectangular; overlapping beams at the edges always spill some light.
- Without WLF, the actual illumination on the surface would be about 15–35% lower than designed, making the facade look dull and patchy.
- It gives a realistic number and wattage of projectors and avoids redesign after installation.
Floodlight calculation process
- Decide the illumination level E (lux) from the surface colour and surroundings.
- Find the area A of the surface.
- Choose CU (or beam factor), depreciation factor DF and WLF.
- Total lumens: F = E × A × DF × WLF / CU.
- Choose lamp wattage and efficacy; lamp lumens = W × η.
- Number of projectors N = F / lamp lumens (round up).
- Choose the beam spread from the distance: θ = 2 tan⁻¹(half-height / distance), then arrange and aim.
Example: A building front 30 m × 12 m, in a medium-bright area, needs 120 lux. Projectors at 15 m; CU = 0.4, DF = 1.3, WLF = 1.2; 250 W metal-halide lamps of 80 lm/W.
A = 30 × 12 = 360 m²
F = 120 × 360 × 1.3 × 1.2 / 0.4
= 168 480 lm
Lamp lm = 250 × 80 = 20 000 lm
N = 168 480 / 20 000 = 8.4 → 9
θ = 2 tan⁻¹(6 / 15) = 43.6° (wide beam)
So 9 wide-beam 250 W projectors are used. Without WLF (= 1.0), F would be only 140 400 lm (N = 7.02 → 8), one projector fewer, and the surface would be under-lit.
✓ Answer checked by Claude Opus 5.5 (high effort)
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