Horizontal, vertical and diagonal FOV, IFOV and real-world coverage at range — from sensor size, pixel pitch and focal length.
Field of view. For each axis, the angle the sensor sees:
Scene coverage. How much real-world width, height and diagonal the frame spans at the set distance:
IFOV (central). The exact rectilinear angular width of one pixel at the centre of the field:
Angular sampling. The average angular size per pixel across the whole image (FOV ÷ pixel count). This is close to, but not identical with, the central IFOV — it is an image-average, shown here in µrad/px.
Model limits. Rectilinear (pinhole) projection. Real lenses vary across the field and add their own MTF and chromatic effects — treat this as first-order geometry.
Projection: Rectilinear · Lens distortion: Not modelled · Atmosphere: Not modelled / simplified
Optical MTF: Not modelled · Stabilization: Not modelled · Ground: Flat horizontal plane
This calculator works out the angular field of view (FOV) of a camera from its sensor size and lens focal length, then projects that angle onto a scene at a chosen distance to give real-world coverage width and height. It also returns the single-pixel angular sampling — both the central instantaneous field of view (IFOV) and the frame-average angular sampling — which together tell you how much ground each pixel sees.
It is a first-order geometric model intended for sizing a sensor-and-lens pairing early in a design: choosing a focal length for a required coverage, checking whether a given camera resolves enough detail at range, or comparing candidate sensors on a like-for-like basis. For a fixed sensor, a shorter focal length widens the field and coarsens angular sampling; a longer focal length narrows the field and samples the scene more finely.
Angular FOV along any axis follows directly from the right-triangle geometry of a pinhole (rectilinear) lens:
A 1/2.8″ sensor (5.57 × 3.13 mm, 1920 × 1080 px, 2.9 µm pitch) with a 50 mm lens, imaging at 1 km:
HFOV = 2·arctan(5.57 ÷ (2·50)) = 6.37° Scene width = 2·1000·tan(6.37°/2) = 111 m IFOV = 2·arctan(0.0029 ÷ (2·50)) = 58 µrad (≈ 5.8 cm at 1 km)So this configuration frames a 111 m-wide scene at 1 km, with each pixel subtending about 5.8 cm on the ground — enough to detect a vehicle, but not to read markings on it.
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