CCTV Essentials – Formulas & Examples

 

CCTV Essentials – Formulas & Examples

🧩 1. Field of View (FOV)

Formula:

$$FOV=\frac{SensorSize\times Distance}{FocalLength}$$

• Sensor Size: width of the camera sensor (mm).

• Distance: distance from camera to target (m).

• Focal Length: lens focal length (mm).

Example: Sensor size = 6 mm, focal length = 12 mm, distance = 10 m.

$$FOV=\frac{6\times 10}{12}=5m$$

→ The camera covers 5 m width at 10 m distance.

🔧 2. Resolution Requirement

Formula:

$$PixelsperMeter(PPM)=\frac{HorizontalResolution}{SceneWidth}$$

Guidelines:

• Recognition: ≥ 125 PPM

• Identification: ≥ 250 PPM

Example: Camera resolution = 1920 pixels, scene width = 8 m.

$$PPM=\frac{1920}{8}=240$$

→ Suitable for identification.

📊 3. Storage Calculation

Formula:

$$Storage=Bitrate\times Time\times NumberofCameras$$

Example: Bitrate = 4 Mbps, time = 24 hrs, cameras = 10.

$$Storage=4\times 3600\times 24\times 10=3.45TB\mathrm{/}day$$

🏗️ 4. Bandwidth Requirement

Formula:

$$Bandwidth=Bitrate\times NumberofCameras$$

Example: Bitrate = 2 Mbps, cameras = 20.

$$Bandwidth=2\times 20=40Mbps$$

→ Network must support at least 40 Mbps.

🧪 5. Illumination Requirement

Formula:

$$Lux=\frac{Lumens}{Area}$$

Example: Lamp output = 2000 lumens, area = 20 m².

$$Lux=\frac{2000}{20}=100lux$$

→ Adequate for general surveillance.

📑 Quick Reference Table

Parameter	Formula	Example Result
FOV	(Sensor × Distance) ÷ Focal Length	5 m
PPM	Resolution ÷ Scene Width	240 PPM
Storage	Bitrate × Time × Cameras	3.45 TB/day
Bandwidth	Bitrate × Cameras	40 Mbps
Illumination	Lumens ÷ Area	100 lux

CCTV design depends on field of view, resolution, storage, bandwidth, and illumination. Using these formulas ensures proper coverage, identification capability, and reliable system performance.