1/2 inch crop factor and field of view calculator
A 1/2 inch sensor measures 6.4 by 4.8 mm, a diagonal of 8.00 mm. Against the full-frame diagonal of 43.27 mm that is a crop factor of 5.408.
The diagonal divided by this gives the acceptable blur circle. 1500 is the usual modern figure.
- Full-frame equivalent focal length
- 189.29
- Equivalent aperture for depth of field
- 15.143
- Sensor diagonal
- 8.000
- Sensor area
- 30.72 mm2
- Full frame gathers this many times more light
- 28.125
- Horizontal field of view
- 10.448 degrees
- Vertical field of view
- 7.845 degrees
- Diagonal field of view
- 13.040 degrees
- Hyperfocal distance
- 82 066
- Depth of field
- 217
- Frame width at the subject
- 0.549
Show the working
- Focal length = 35
- Aperture f-number = 2.8
- Subject distance = 3
- Circle of confusion divisor = 1500
- sensor_w = 6.4 = 6.4
- sensor_h = 4.8 = 4.8
- Sensor diagonal = hypot(6.4, 4.8) = 8
- Crop factor = 8 > 0 ? 43.266615305567875 / 8 : 0 = 5.408 33
- Full-frame equivalent focal length = 35 * 5.408 33 = 189.291
- Equivalent aperture for depth of field = 2.8 * 5.408 33 = 15.1433
- Sensor area = 6.4 * 4.8 = 30.72
- Full frame gathers this many times more light = 30.72 > 0 ? 864 / 30.72 : 0 = 28.125
- Horizontal field of view = deg(2 * atan(6.4 / (2 * 35))) = 10.4479
- Vertical field of view = deg(2 * atan(4.8 / (2 * 35))) = 7.845 43
- Diagonal field of view = deg(2 * atan(8 / (2 * 35))) = 13.0396
- coc = 8 / 1500 = 0.005 333 33
- Hyperfocal distance = 35^2 / (2.8 * 0.005 333 33) + 35 = 82 066.3
- near = 3*1000 * 82 066.3 / (82 066.3 + (3*1000 - 35)) = 2895.39
- far = 3*1000 < 82 066.3 ? 3*1000 * 82 066.3 / (82 066.3 - (3*1000 - 35)) : 1e9 = 3112.45
- Depth of field = 3112.45 >= 1e9 ? 1e9 : 3112.45 - 2895.39 = 217.059
- Frame width at the subject = 35 > 0 ? 3 * 6.4 / 35 : 0 = 0.548 571
Values are shown to six significant figures here; the arithmetic itself runs on exact fractions and is only rounded where a number is printed.
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The formula
- crop factor = full-frame diagonal / sensor diagonal
- field of view = 2 arctan(sensor dimension / 2 focal length)
- hyperfocal distance = f^2 / (N c) + f
- equivalent aperture = aperture x crop factor, for depth of field and total light
Crop factor changes the framing, not the lens. A 50 mm lens is always a 50 mm lens; it just sees a smaller rectangle.
Equivalent aperture matters for depth of field and total light gathered, not for exposure. An f/2.8 lens meters as f/2.8 on any format.
The circle of confusion is a judgement about acceptable sharpness at a viewing size, not a physical constant.
Equivalent focal lengths on 1/2 inch
| Lens | Full-frame equivalent |
|---|---|
| 12 mm | 64.9 mm |
| 16 mm | 86.5 mm |
| 24 mm | 129.8 mm |
| 35 mm | 189.3 mm |
| 50 mm | 270.4 mm |
| 85 mm | 459.7 mm |
| 135 mm | 730.1 mm |
| 200 mm | 1081.7 mm |
Frequently asked
What is the crop factor of a 1/2 inch sensor?
5.408, calculated from its 8.00 mm diagonal against full frame at 43.27 mm.
Does a crop sensor make a lens longer?
No. The lens projects the same image; the smaller sensor simply records a smaller part of it, which looks like a longer lens.