Guides · CCTV design
How to calculate CCTV field of view, scene width and pixel density
Three numbers decide whether a CCTV camera does the job: how wide it sees (field of view), how wide the scene is at a given distance (scene width), and how many pixels land on each metre of that scene (pixel density, in px/m). Here are the formulas, with worked examples.
The short version
Scene width = 2 × distance × tan(field of view ÷ 2)
Pixel density = horizontal pixels ÷ scene width
1. Find the horizontal field of view
Use the datasheet figure if there is one. Most manufacturers publish the horizontal field of view (HFOV) in degrees, often as a range for a varifocal lens. That figure already allows for the real sensor size and the lens, so it's the most reliable number to use.
If you only have the focal length and sensor width, calculate it:
For example, a 4 mm lens on a sensor 6.4 mm wide gives 2 × arctan(6.4 ÷ 8) ≈ 77°.
Be careful with sensor sizes like "1/2.8-inch". They are a naming convention, not a measurement, so look up the actual width in millimetres. Wide-angle lenses also tend to differ from the calculated figure because of distortion, which is another reason to prefer the published HFOV.
2. Work out the scene width at your distance
Take the AXIS Q1728-LE at its widest setting: a published HFOV of 108°. At 10 m the scene is 2 × 10 × tan(54°) ≈ 27.5 m wide.
3. Turn that into pixels per metre
The Q1728-LE is 3840 pixels wide, so at 10 m that's 3840 ÷ 27.5 ≈ 140 px/m.
Compare the answer with the EN 62676-4 DORI levels to see what the camera is good for at that distance:
| Level | px/m | What you get |
|---|---|---|
| Detect | 25 | A person is there |
| Observe | 62 | What they're doing and wearing |
| Recognise | 125 | Someone who knows them can say who it is |
| Identify | 250 | A stranger can be identified |
At 140 px/m, the wide setting at 10 m is enough to recognise someone, but not to identify them.
4. How far can it identify someone?
Turn the formula round to find the distance at which a camera still meets a level:
For the Q1728-LE at 108°, identification (250 px/m) holds out to about:
Zoom the same camera in to its 42° tele setting and the identification distance grows to about 20 m, but the scene at 10 m shrinks to about 7.7 m wide. That trade-off, coverage against detail, is the whole game in CCTV layout.
What the formulas leave out
- Mounting height and tilt. A camera 3 m up looking at someone 8 m away is really about 8.5 m from them, and tilting the camera changes the shape of what it covers on the ground.
- The edges of the picture. Pixel density is highest along the centre of the view and drops towards the sides.
- Walls, doors and obstacles. The maths assumes a clear view, which most rooms don't give you.
- Light and compression. Low light, glare and heavy compression all reduce the detail you actually get.
Skip the calculator
Plot-LV FOV does these sums for every camera on your own floor plan, allowing for mounting height, tilt and walls, and shows where each one meets the DORI level you need. It's free and runs in your browser.
Looking for a specific camera? The camera library has field of view and DORI distances worked out for each model.