---
title: "Outdoor lighting, flood and street calculation"
subject: "Industrial Power Distribution and Illumination"
chapter: 8
key: "semester_6/industrial_power_distribution_and_illumination:outdoor-flood-street-calculation"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/outdoor-flood-street-calculation
site: BCT Engineering
---

# Outdoor lighting, flood and street calculation

Industrial Power Distribution and Illumination · Chapter 8: Outdoor Lighting Design

- Asked in: 2072 Asoj, 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/outdoor-flood-street-calculation
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-8

## Question

What is outdoor lighting? Explain about calculation for flood lighting and street lighting.

## Short answer

Outdoor lighting is for roads, facades etc. Floodlight calculation: total lumens = E×A×DF×WLF/CU; e.g. 20m×10m face, 150 lux, CU 0.5, DF 1.3, WLF 1.2, 150W LED (16,500 lm) yields 6 fixtures. Street lighting lumen method: spacing S = (F×CU×MF)/(E_av×W); with 150W HPSV (15,000 lm), 8m road, E_av 15 lux, CU 0.4, MF 0.8 gives S≈35 m.

## Answer

**Outdoor lighting** is artificial lighting of open areas – roads, building facades, monuments, sports grounds, yards and parking areas – for safety, security, night work, advertising and decoration. The two main calculations are for **floodlighting** and **street lighting**.

#### Floodlighting calculation
Floodlighting means lighting a large surface with projectors giving concentrated beams.
1. Select illumination E (lux) based on the surface reflectance and surroundings.
2. Area A = length × height of the surface.
3. Total lumens required:
```text
F = E × A × DF × WLF / CU
```
   where CU = coefficient of utilisation (beam lumens/lamp lumens, 0.3–0.6), DF = depreciation factor (1.3–1.5), WLF = waste light factor (1.2–1.5).
4. Number of projectors N = F / (lamp watts × efficacy).
5. Beam spread θ = 2 tan⁻¹(half-height / distance).

**Example:** 20 m × 10 m face, E = 150 lux, CU = 0.5, DF = 1.3, WLF = 1.2, 150 W LED floodlights of 110 lm/W.
```text
F = 150 × 200 × 1.3 × 1.2 / 0.5 = 93 600 lm
Lamp lm = 150 × 110 = 16 500 lm
N = 93 600 / 16 500 = 5.7 → 6 floodlights
```

#### Street lighting calculation
**(a) Lumen method (average illumination):**
```text
E_av = F × CU × MF / (S × W)
S    = F × CU × MF / (E_av × W)
```
F = lamp lumens, S = pole spacing, W = road width, MF = maintenance factor.

**(b) Point-by-point method (minimum illumination):** for a lamp of intensity I at height h, a point at horizontal distance x gets
```text
E = I cos θ / d² = I h / (h² + x²)^1.5
```
Add contributions of adjacent lamps (usually the two nearest) midway between poles to check minimum and uniformity.

**Example:** Road 8 m wide, 150 W HPSV lamps (15 000 lm), CU = 0.4, MF = 0.8, E_av = 15 lux, h = 9 m.
```text
S = 15 000 × 0.4 × 0.8 / (15 × 8) = 40 m
S/h = 40/9 = 4.4  → too high; take S = 35 m
E_av = 15 000×0.4×0.8/(35×8) = 17.1 lux
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is outdoor lighting?
   A. Lighting using only natural daylight in open areas
   B. Lighting inside industrial workshops
   C. Lighting only used for advertising
   D. Artificial lighting of open areas like roads, facades and yards for safety, security and night work
   Answer: D. It covers roads, monuments, sports grounds and yards for safety, security, work, advertising and decoration. (Key point: Outdoor lighting)

2. In the flood example (20 m x 10 m, 150 lux, CU 0.5, DF 1.3, WLF 1.2), what is the total lumen requirement?
   A. 93 600 lm
   B. 46 800 lm
   C. 187 200 lm
   D. 30 000 lm
   Answer: A. 150 x 200 x 1.3 x 1.2 / 0.5 = 93 600 lm, giving 5.7, i.e. 6 floodlights. (Key point: Floodlight lumens)

3. Which expression gives pole spacing in the lumen method?
   A. S = E_av x W / (F x CU x MF)
   B. S = F x E_av / (CU x W)
   C. S = F x CU x MF / (E_av x W)
   D. S = I h / (h^2 + x^2)^1.5
   Answer: C. Rearranging E_av = F CU MF/(S W) gives S = F CU MF/(E_av W). (Key point: Street lumen method)

4. What does the point-by-point method do?
   A. Gives the average illumination over the road
   B. Finds illumination at a point (e.g. midway) from I h/(h^2+x^2)^1.5, adding the nearest lamps, to check the minimum
   C. Calculates cable size for the lamps
   D. Finds the total lumens for floodlights
   Answer: B. It checks minimum illumination and uniformity, which the lumen method cannot. (Key point: Point-by-point method)

## Related in Industrial Power Distribution and Illumination

- [Outdoor lighting purposes and street lighting](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aoutdoor-lighting-street-methods) — Industrial Power Distribution and Illumination, chapter 8, asked 2×
- [Factors for street and flood lighting design](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aflood-street-design-factors) — Industrial Power Distribution and Illumination, chapter 8, asked 2×
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- [Ten factors for flood, street lighting, luminaires](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aten-factors-flood-street-luminaires) — Industrial Power Distribution and Illumination, chapter 8
- [Street lighting a 2 km road with sodium lamps](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Astreet-lighting-sodium-2km) — Industrial Power Distribution and Illumination, chapter 8
- [Street lighting scheme design procedure](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Astreet-lighting-design-procedure) — Industrial Power Distribution and Illumination, chapter 8

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