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CCTV

CCTV Lens & FOV

Enter the distance to the target, the sensor, the focal length and the horizontal resolution and get the HFOV, the scene width, the px/m and the DORI level (EN 62676-4) reached, with the maximum distance for detection, observation, recognition and identification.

m

Active-area width of the sensor, in millimetres.

mm

Image width in pixels. With MP the width is estimated from a 16:9 aspect ratio.

Examples
redzilla.cl — cctv
 
Field of view (HFOV)
horizontal angle
Density at the distance
pixels per metre
DORI level reached
at the entered distance
HFOV
Scene
px/m

Maximum distance per DORI level

LevelThresholdMax. distanceStatus at D

Status at D shows whether the threshold of each level is met at the entered distance.

How it is calculated · FOV, px/m and DORI (EN 62676-4)

1. Horizontal field of view: HFOV = 2·atan(sensorW / (2·f)) · 180/π, with the sensor width and focal length in mm.

2. Scene width at distance D: width = D · sensorW / f (in metres, with D in metres).

3. Density: px/m = horizRes / scene_width. More pixels on the target means more detail.

4. DORI thresholds (EN 62676-4, px/m): Detection 25, Observation 63, Recognition 125, Identification 250.

5. Maximum distance for a level: Dmax = horizRes · f / (threshold · sensorW). Up to that distance the required density is kept.

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How it works

The calculator solves the optics of an IP camera from four inputs: distance to the target, sensor width (from 1/4 inch to 1/1.8), lens focal length and horizontal resolution. The horizontal field of view comes from HFOV = 2·atan(sensorW / (2·f)), the scene width at distance D is D · sensorW / f, and the detail density is px/m = horizontal resolution / scene width.

That density is compared against the DORI thresholds of EN 62676-4: Detection 25 px/m, Observation 63, Recognition 125 and Identification 250. The tool reports the level achieved at the entered distance and the maximum distance for each level using Dmax = resolution × focal / (threshold × sensorW), which helps choose lens and resolution before installing.

Example: a 1080p camera with a 4 mm lens — can it identify at 10 m?

  1. 1/2.8 inch sensor (5.0 mm wide) and 4 mm focal: HFOV = 2·atan(5/8) ≈ 64°.
  2. Scene at 10 m: 10 × 5 / 4 = 12.5 m wide → density 1920 / 12.5 ≈ 154 px/m.
  3. At 154 px/m you reach Recognition (≥125), but not Identification (≥250).
  4. Identification only up to 1920 × 4 / (250 × 5) ≈ 6.1 m: identifying at 10 m requires a longer lens or more resolution.

Frequently asked questions

What does DORI mean in CCTV?
It is the EN 62676-4 scale that defines how much detail each task needs: Detection (25 px/m, noticing someone is there), Observation (63 px/m, seeing what they do), Recognition (125 px/m, recognizing a known person) and Identification (250 px/m, identifying a stranger with evidentiary value).
At how many meters can a 1080p camera identify someone?
It depends on lens and sensor. With a 1/2.8 (5 mm) sensor and a 4 mm lens, a 1080p camera identifies (250 px/m) only up to about 6 m; an 8 mm lens reaches about 12 m, and 12 mm about 18 m. Extra reach always costs angle: the FOV narrows as focal length increases.
What is the difference between a 2.8 mm and a 6 mm lens?
The 2.8 mm gives a very wide angle (over 80° with a typical sensor) to cover rooms or nearby entrances, but pixel density drops quickly with distance. The 6 mm narrows the field to about 45° and concentrates pixels farther away: same camera, nearly double the range for each DORI level.
Do more megapixels always mean more range?
Yes in terms of density: doubling the horizontal resolution doubles the maximum distance of every DORI level with the same lens. In practice lighting also matters (small high-resolution sensors perform worse at night), along with focus and compression, so the px/m figure is a necessary condition, not a sufficient one.
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