---
title: "Point to point method of lighting calculation"
subject: "Industrial Power Distribution and Illumination"
chapter: 7
key: "semester_6/industrial_power_distribution_and_illumination:point-to-point-method"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/point-to-point-method
site: BCT Engineering
---

# Point to point method of lighting calculation

Industrial Power Distribution and Illumination · Chapter 7: Illumination Design Principles

- Asked in: 2068 Bhadra (old course), 4 marks
- Video: [Calculator Techniques - Illumination Engineering| Point by Point Method (Three-Dimensional Space)](https://www.youtube.com/watch?v=6LxpFMUXC1E) — ENGR. CALAQUE REVIEW EXCELLENCE 
- 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/point-to-point-method
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7

## Question

Explain point to point method of lighting calculation.

## Short answer

It calculates illuminance at a specific point using the inverse square law and Lambert's cosine law: E = (I/d²) cos θ. For horizontal plane, E_H = I h/d³. It is used for street, flood, and local lighting where point values are needed, ignoring reflected light.

## Answer

The **point-to-point method** (also called the point-by-point or inverse-square-law method) calculates the illuminance at a particular point due to one or more lamps from their luminous intensity, using the **inverse square law** and **Lambert's cosine law**. It is used where the illumination at specific points matters: street lighting, flood lighting, local lighting and checking uniformity.

#### Principle

For a lamp of intensity I (cd) in the direction of point P, at distance d from P, with θ the angle between the light ray and the normal to the surface:

```text
E = (I / d²) × cos θ        lux

On a horizontal plane, lamp at height h,
point at horizontal distance x:
  d = √(h² + x²),  cos θ = h/d
  E_H = I h / d³ = (I / h²) cos³θ
On a vertical plane:
  E_V = I x / d³ = (I / h²) cos²θ sin θ
```

```text
   Lamp L
     |\
   h |  \ d
     |θ   \
     +-----P   (horizontal plane)
        x
```

#### Steps

1. Find the intensity I of each lamp towards P from the polar curve of the luminaire.
2. Find d and θ from the geometry.
3. Calculate E from each lamp; **add the contributions** of all lamps at that point.
4. Repeat for several points to check the minimum, maximum and uniformity.

Example: I = 400 cd, h = 2 m: directly below, E = 400/2² = 100 lux; at x = 2 m, d = 2.83 m, E = 400 × 2/2.83³ = 35.4 lux.

Limitation: it ignores light reflected from walls and ceiling, so the lumen method is used instead for general indoor lighting.

## Check yourself

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

1. Which two laws are the basis of the point-to-point method?
   A. Ohm's law and Kirchhoff's law
   B. Inverse square law and Lambert cosine law
   C. Faraday's law and Lenz's law
   D. Snell's law and Coulomb's law
   Answer: B. Illuminance E = (I/d^2) cos(theta) combines the inverse square and cosine laws. (Key point: Laws used)

2. For a lamp of intensity I at height h, what is the illuminance directly below?
   A. I x h^2
   B. I / h
   C. I / h^2
   D. I / (2h)
   Answer: C. Directly below, d = h and theta = 0, so E = I/h^2. For I = 400 cd and h = 2 m this is 100 lux. (Key point: Horizontal illuminance)

3. What is true of the method when several lamps light a point?
   A. Only the nearest lamp counts and reflections are included
   B. Contributions of all lamps are added, but reflected light from walls and ceiling is ignored
   C. Contributions are multiplied
   D. Average of the lamps is taken and reflection is fully included
   Answer: B. E from each lamp is added at the point; because reflected light is ignored, indoor general lighting uses the lumen method. (Key point: Multiple lamps and limitation)

4. In the example I = 400 cd, h = 2 m, what is E at x = 2 m?
   A. 100 lux
   B. 35.4 lux
   C. 70.7 lux
   D. 50 lux
   Answer: B. d = 2.83 m, so E = I h/d^3 = 400 x 2/2.83^3 = 35.4 lux. (Key point: Off-axis value)

## Related in Industrial Power Distribution and Illumination

- [Outdoor lighting, flood and street calculation](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aoutdoor-flood-street-calculation) — Industrial Power Distribution and Illumination, chapter 8
- [Workshop hall lighting and power distribution](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aworkshop-hall-lighting-power) — Industrial Power Distribution and Illumination, chapter 7

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