CCTV Lens Calculator

How the CCTV lens calculation works

Everything comes from the sensor width s, the focal length f, the horizontal pixels P and the distance D (in mm inside the formulas):

The angle is at infinity focus, the way manufacturers publish it, and the width uses the thin lens, as in our field-of-view calculator; at surveillance distances it is practically s × D ÷ f. The maximum distance for each level is the one that gives exactly that density: D = f + P × f ÷ (s × px/m). The reverse, "what focal length do I need?", solves for the focal length: f = D × s ÷ (width + s).

If the resolution does not match the sensor's shape (a 16:9 stream on a 4:3 sensor), the largest centred rectangle with the stream's shape is used: full width and cropped height, or the other way round. That way the vertical angle does not come from sensor height that is not being recorded.

DORI distances under EN 62676-4

The standard measures detail as horizontal pixels per metre on the person, based on a 16 cm face. The calculator flags the level you reach at your distance and the maximum distance for each one:

• 2014 (DORI): detection 25, observation 62.5, recognition 125, identification 250 px/m.
• 2025: Overview 20, Outline 40, Discern 80, Perceive 125, Characterize 250, Validate 500, Scrutinize 1500 px/m.

Switch the units to feet and every length and density converts exactly (1 ft = 0.3048 m): 250 px/m is 76.2 px/ft; Axis prints 76.

CCTV sensor formats: why 1/2.8″ is not 1/2.8 of an inch

The fractional-inch name comes from old camera tubes and is not a measurement: two "1/2.8″" sensors can have different areas. So the named sensors use the real recording area = pixels × unit cell from the maker's datasheet:

• Sony IMX327 (1/2.8″, 2 MP): 1920 × 2.9 µm = 5.57 × 3.13 mm.
• Sony IMX415 (1/2.8″, 8 MP): 3840 × 1.45 µm = 5.57 × 3.13 mm, the same width with four times the pixels.
• Sony IMX335 (1/2.8″, 5 MP 4:3): 2592 × 2.0 µm = 5.18 × 3.89 mm.
• OmniVision OS04C10 (1/2.9″, 4 MP): 2688 × 1.998 µm = 5.37 × 3.04 mm.
• Sony IMX334 (1/1.8″, 8 MP): 3840 × 2.0 µm = 7.68 × 4.32 mm.

The "classic" 1/3″, 1/2″ and 2/3″ (4:3) are the traditional tube table, handy for unnamed sensors. If you know your sensor's pixel size, use "Custom" with pixels × pixel size.

Example: 2 MP camera with a 2.8 mm vs a 12 mm lens

1/2.8″ sensor (IMX327, 5.57 mm), 1920 × 1080, person at 10 m:

• 2.8 mm: 89.7° × 58.4°; at 10 m the scene is 19.9 m wide → 96.6 px/m, observation. It identifies up to 3.86 m and detects up to 38.6 m.
• 12 mm: 26.1° × 14.9°; at 10 m it is 4.63 m wide → 414.3 px/m, identification. It identifies up to 16.6 m and detects up to 166 m.

The datasheet of a 2 MP 1/2.8″ Dahua with a 2.8 mm lens gives DORI 38.6 / 15.4 / 7.7 / 3.9 m, the same as here (38.6 / 15.5 / 7.73 / 3.86 m) apart from rounding; we do not know which sensor it uses, so the maths uses the IMX327 area. Its angle, though, is 102°: with "By datasheet angle" at 102° the scene at 10 m is 24.7 m wide and the average density drops to 77.7 px/m. The difference is wide-angle distortion: the centre keeps the detail and the edges lose it.