--- title: "Floodlighting a 25 × 10 m monument wall" subject: "Industrial Power Distribution and Illumination" chapter: 8 key: "semester_6/industrial_power_distribution_and_illumination:floodlight-red-wall-monument" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/floodlight-red-wall-monument site: BCT Engineering --- # Floodlighting a 25 × 10 m monument wall Industrial Power Distribution and Illumination · Chapter 8: Outdoor Lighting Design - Asked in: 2080 Chaitra, 8 marks - Video: [ILLUMINATION ENGINEERING - PART - 14 - THREE PROBLEMS IN FLOODLIGHTING DESING](https://www.youtube.com/watch?v=kZA2zZFlxRg) — zenmurali Sachithanandam - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/floodlight-red-wall-monument - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-8 ## Question A vertical red wall of size 25 m x 10 m (High) of the monument is to be illuminated by a floodlight from a distance of 30m by fittings planted in the ground. Find the number of fittings/lamps, their arrangement and wattage of lamps if the Average Illumination required is 120 lux, Beam Factor is 0.5, the Maintenance Factor is 0.625, the waste factor is 1.2, Lamp Efficiency is 116 lumen/Watt and Beam Spread (vertical) Angle is 12.5°. ## Short answer 8 fittings in 2 vertical tiers × 4 horizontal positions. Each lamp ~124 W → use 150 W. Total lumens required 115,200 lm. Beam spread 12.5° covers 6.57 m at 30 m. Arrange 4 ground positions 6.25 m apart, each with two projectors aimed at upper and lower halves. ## Answer **Given:** wall 25 m × 10 m, distance d = 30 m, E = 120 lux, beam factor (BF) = 0.5, maintenance factor (MF) = 0.625, waste light factor (WLF) = 1.2, lamp efficacy = 116 lm/W, vertical beam spread = 12.5°. #### Step 1: Total lamp lumens required ```text Area A = 25 × 10 = 250 m² Total lm = E × A × WLF / (BF × MF) = 120 × 250 × 1.2 / (0.5 × 0.625) = 36 000 / 0.3125 = 115 200 lm ``` #### Step 2: Coverage of one beam (decides arrangement) At 30 m, a beam of total spread 12.5° covers a length of ```text D = 2 × d × tan(12.5°/2) = 2 × 30 × tan 6.25° = 6.57 m ``` - **Vertically:** the wall is 10 m high; the wall subtends tan⁻¹(10/30) = 18.4° at ground-level fittings, which is more than 12.5°. One beam (6.57 m) cannot cover 10 m, so **2 tiers** of aiming are needed (one row aimed at the lower half, one at the upper half). - **Horizontally:** 25 / 6.57 = 3.8 → **4 beams** across the wall (assuming a symmetrical beam). So number of fittings = 2 × 4 = **8**. #### Step 3: Lamp wattage ```text Lumens per lamp = 115 200 / 8 = 14 400 lm Wattage = 14 400 / 116 = 124.1 W ``` Choose the next standard size: **150 W** lamps (e.g. 150 W metal-halide or LED floodlight). This gives a little extra margin (8 × 150 × 116 = 139 200 lm). #### Arrangement ```text 25 m wall (10 m high) ┌──────┬──────┬──────┬──────┐ │ U1 │ U2 │ U3 │ U4 │ upper aim ├──────┼──────┼──────┼──────┤ │ L1 │ L2 │ L3 │ L4 │ lower aim └──────┴──────┴──────┴──────┘ ↑ 4 ground positions, 6.25 m apart, 30 m in front; 2 fittings per position ``` The fittings are placed at 4 points about 6.25 m apart, 30 m in front of the wall, with two fittings at each point – one aimed at the upper half and one at the lower half. Beams overlap slightly for uniformity. **Answer: 8 fittings (2 rows × 4), each about 124 W → use 150 W lamps; total lamp lumens required 115 200 lm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expression gives total lamp lumens when beam factor and maintenance factor are given? A. E x A x WLF / (BF x MF) = 115200 lm B. E x A x BF x MF / WLF C. E x A / (BF x MF x WLF) D. E x A x WLF x BF x MF Answer: A. 120 x 250 x 1.2/(0.5 x 0.625) = 115200 lm. (Key point: Total lumens) 2. How wide is a 12.5° beam at 30 m? A. About 6.57 m B. About 13 m C. About 3.3 m D. About 30 m Answer: A. D = 2 x 30 x tan(6.25°) = 6.57 m. (Key point: Beam coverage) 3. Why are 8 fittings used? A. Wall is 10 m high, beyond 6.57 m, so 2 tiers; 25/6.57 gives 4 across; 2 x 4 = 8 B. Total lumens divided by 14400 C. One for each 3 m of wall D. Because 8 is a standard number Answer: A. One beam cannot cover the 10 m height, so two tiers are needed, and four beams cover the 25 m width. (Key point: Number of fittings) 4. Each lamp must give 14400 lm at 116 lm/W. What is chosen? A. 124 W calculated, next standard 150 W used B. 14400 W C. 116 W exactly D. 1000 W Answer: A. 14400/116 = 124.1 W, so the next standard size of 150 W is selected. 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