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CCTV DORI Distance Calculator

Pick resolution, lens and sensor; get the field of view and the distances to detect, observe, recognise and identify (px/m per IEC 62676-4).

Camera and lens

Common lenses
mm

For a varifocal lens (e.g. 2.8–12 mm) enter the setting you will actually use.

Check at a distance

m

DORI distances

Levelpx/m neededMax distanceScene width
DetectYou can tell a person or vehicle is there2538.6 m76.8 m
ObserveClothing colour and what the person is doing62.515.4 m30.7 m
RecogniseSomeone who knows the person can say who it is1257.7 m15.4 m
IdentifyThe face is clear enough to establish identity2503.9 m7.7 m

Results

Identify up to

3.9m

Horizontal FOV

89.7°

Detect up to

38.6m

At 10 m

Pixel density

97px/m

Scene width

19.9m

Level reached: Observe

Reading number plates needs about Identify level (≥ 250 px/m) where the vehicle passes, a narrow lens and a fast shutter. One camera should not try to cover a wide area and plates at the same time.
These figures assume good conditions. IR range, lighting, video compression, rain, fog and dust all reduce real results. At night, check that the IR actually reaches the distance you need.
How we calculate

1. Horizontal FOV = 2 × atan(sensor width ÷ (2 × focal length)). If you enter the datasheet HFOV, we use it directly.

2. Scene width at distance d = 2 × d × tan(HFOV ÷ 2). Pixel density px/m = horizontal pixels ÷ scene width.

3. DORI levels per IEC 62676-4: Detect 25, Observe 62.5, Recognise 125, Identify 250 px/m. Maximum distance d = horizontal pixels ÷ (px/m × 2 × tan(HFOV ÷ 2)).

4. Sensor widths used (16:9; pixel pitch × horizontal pixels of a common sensor of that size): 1/2.9" = 5.38 mm (2.8 µm × 1920), 1/2.8" = 5.57 mm (2.9 µm × 1920), 1/2.7" = 5.76 mm (3.0 µm × 1920), 1/1.8" = 7.68 mm (2.0 µm × 3840). 1/3" = 4.8 mm is the standard 4:3 width. Individual camera models can differ slightly.

5. The formula assumes a distortion-free lens. Real wide-angle lenses usually have a wider HFOV than the formula gives, so use the datasheet HFOV when you have it.

These results are estimates to help with planning and budgeting. Our engineers will verify everything on site before final design and equipment proposal.

The same rules of thumb our engineers use for a first design. Good enough for budgeting; the final numbers come from a site survey.

How to use it and read the result

What DORI means

DORI is four levels of detail, defined in IEC 62676-4 by how many pixels fall on each metre of the scene at the target. The calculator uses these four values.

Level Pixels per metre What you can see 2MP, 2.8 mm, 1/2.8" 2MP, 6 mm, 1/2.8"
Detect 25 px/m A person or vehicle is there 38.6 m 82.7 m
Observe 62.5 px/m Clothing colour and what the person is doing 15.4 m 33.1 m
Recognise 125 px/m Someone who knows the person can say who it is 7.7 m 16.5 m
Identify 250 px/m The face is clear enough to establish identity 3.9 m 8.3 m

Most disputes after an incident are about Identify: the camera "saw" the person, but only at Detect or Observe level.

What to enter and where to find each number

  • Resolution. Only the horizontal pixel count matters: 2MP is 1920, 4MP is 2560, 5MP is 2592 or 2880 depending on the sensor, 8MP (4K) is 3840. Find it on the datasheet or the camera's video settings page, or choose "Custom".
  • Lens focal length. Printed on the lens or the label, e.g. 2.8 mm or 4 mm. For a varifocal lens (2.8–12 mm) enter the setting you will actually use, not the range.
  • Sensor size. On the datasheet, e.g. 1/2.8" CMOS. The calculator uses these active widths: 1/3" = 4.8 mm, 1/2.9" = 5.38 mm, 1/2.8" = 5.57 mm, 1/2.7" = 5.76 mm, 1/1.8" = 7.68 mm. Each 16:9 width is pixel pitch × horizontal pixels of a common sensor of that size (1/2.8" = 2.9 µm × 1920); 1/3" is the standard 4:3 width. Individual models can differ slightly; use "Custom width" if the datasheet gives it.
  • "I know the HFOV from the datasheet". Tick this and type the horizontal field of view instead of lens and sensor. It is more accurate, especially for 2.8 mm lenses with barrel distortion.
  • Distance. The straight line from the camera to where the face or plate will be, not the distance along the ground. A camera high on a pole is further from a face than it looks.

How the calculator works it out

  1. Horizontal field of view = 2 × atan(sensor width ÷ (2 × focal length)), or the datasheet HFOV if you entered it.
  2. Scene width at distance d = 2 × d × tan(HFOV ÷ 2). Pixel density = horizontal pixels ÷ scene width.
  3. Maximum distance for each level = horizontal pixels ÷ (px/m × 2 × tan(HFOV ÷ 2)).

The results show "Identify up to" as the big number, then the HFOV, "Detect up to", and at your chosen distance the pixel density, the level reached and the scene width. Above 100° HFOV the tool warns that the image edges bend and carry less detail than the centre. There is no safety margin built in: the figures assume good light, a sharp focus and a lens without distortion. Treat them as the best case.

Worked example: faces at a warehouse gate

A warehouse near Vientiane wants faces of people walking through the gate, about 8 m from the camera. The installer proposed a 2MP camera with a 2.8 mm lens on a 1/2.8" sensor.

  • HFOV: 2 × atan(5.57 ÷ 5.6) = 89.7°. Width factor 5.57 ÷ 2.8 = 1.989.
  • Identify: 1920 ÷ (250 × 1.989) = 1920 ÷ 497.3 = 3.86 m, shown as 3.9 m.
  • At 8 m: scene width 8 × 1.989 = 15.9 m, density 1920 ÷ 15.9 = 121 px/m. That is Observe, below Recognise (125). You see someone in a blue shirt, not who it is.

Change only the lens to 6 mm:

  • HFOV: 2 × atan(5.57 ÷ 12) = 49.8°. Width factor 5.57 ÷ 6 = 0.928.
  • Identify: 1920 ÷ (250 × 0.928) = 1920 ÷ 232.1 = 8.27 m, shown as 8.3 m.
  • At 8 m: scene width 8 × 0.928 = 7.43 m, density 1920 ÷ 7.43 = 259 px/m, so Identify. Second check: 250 px/m × 7.43 m = 1,857 pixels needed, under the 1,920 available.

The 6 mm camera only covers 7.4 m of width, so it watches the gate while a separate wide camera covers the yard. A 4MP camera with a 6 mm lens on a 1/2.7" sensor would reach 2560 ÷ (250 × 0.96) = 10.7 m for Identify.

Now the HFOV option. If the 2.8 mm camera's datasheet says 105° rather than the 89.7° from the formula, the width factor becomes 2 × tan 52.5° = 2.606, and Identify drops to 1920 ÷ (250 × 2.606) = 2.9 m. The wide-angle warning also appears.

Common mistakes we see on Lao sites

  • 2.8 mm lenses everywhere because they show the most area. They give overview, not identity.
  • One 4K camera expected to watch a whole yard and read faces at the far fence.
  • Taking the "IR 30 m" on the box as the distance at which faces are visible. IR range is about light, not detail.
  • Cameras mounted too high at an entrance, so the recording shows the tops of heads.
  • A gate facing the afternoon sun with no WDR, turning faces into silhouettes.
  • A low bitrate set to save disk, which smears faces even when the pixel density is enough. Check storage with the CCTV storage calculator instead of cutting quality.

Night, weather and when a calculator is not enough

These are daylight, best-case figures. At night the IR has to reach the distance you need, and IR light often reflects off rain, insects and dust. Rainy-season mornings bring fog, mine sites bring dust on the dome, and video compression removes fine detail. Reading number plates needs about Identify level (≥ 250 px/m) where the vehicle passes, a narrow lens and a fast shutter; one camera should not try to do both area coverage and plates. That is a number plate recognition job.

For perimeters, car parks and gates, we walk the site, mark where faces and plates must be captured, and test a camera at the real distance, day and night, before the full order. The CCTV resolution and lens guide explains the trade-offs, and design and installation is part of our IP CCTV work.

Next step

Send us on WhatsApp a screenshot of your calculator result, a photo of the camera's model label or datasheet, and a phone photo taken from where the camera will be mounted, with the distance to the gate or door. We will tell you which lens gives Identify at that point.

FAQ

How far can a 2MP camera with a 2.8 mm lens see a face?

With a 1/2.8" sensor, Identify level (250 px/m) reaches only about 3.9 m: 1920 ÷ (250 × 5.57 ÷ 2.8). Recognise reaches about 7.7 m and Detect about 38.6 m. If the datasheet HFOV is wider than the formula gives, the distances are shorter still. A 2.8 mm lens is for an overview of a room or yard, not for a gate where you need faces.

How many pixels per metre do I need to identify a face?

IEC 62676-4 sets Identify at 250 px/m at the target, Recognise at 125, Observe at 62.5 and Detect at 25. You get the figure by dividing the horizontal pixels by the scene width at that distance. For example, 1920 pixels across a 7.43 m wide scene gives 259 px/m, which passes Identify. Real footage also loses detail to poor light and compression, so keep some margin.

Should I choose a 2.8 mm or a 6 mm CCTV lens?

It depends on the job. A 2.8 mm lens on a 1/2.8" sensor sees about 90°: wide area, little detail. A 6 mm lens sees about 50°, narrower, but a 2MP camera then reaches Identify level to about 8.3 m. Most sites need both: wide cameras for overview and longer lenses at gates, counters and vehicle entrances.

Is the IR distance on a camera box the same as its face-recognition distance?

No. The IR figure says how far the infrared light reaches, not how much detail you get. A 2MP camera with a 2.8 mm lens labelled IR 30 m may show that someone is there at 30 m, but Identify level ends around 3.9 m. At night detail drops further because of noise and compression. Use the DORI calculator, then test at night on site.

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Send us a floor plan or a few photos of the site

We’ll come back with a few questions, then a first design. The first consultation is free, in Lao, Thai or English, on WhatsApp or our Facebook page.