1/2.3 inch crop factor and field of view calculator
A 1/2.3 inch sensor measures 6.17 by 4.55 mm, a diagonal of 7.67 mm. Against the full-frame diagonal of 43.27 mm that is a crop factor of 5.644. The classic compact camera and action camera size.
The diagonal divided by this gives the acceptable blur circle. 1500 is the usual modern figure.
- Full-frame equivalent focal length
- 197.53
- Equivalent aperture for depth of field
- 15.803
- Sensor diagonal
- 7.666
- Sensor area
- 28.07 mm2
- Full frame gathers this many times more light
- 30.776
- Horizontal field of view
- 10.074 degrees
- Vertical field of view
- 7.438 degrees
- Diagonal field of view
- 12.500 degrees
- Hyperfocal distance
- 85 637
- Depth of field
- 208
- Frame width at the subject
- 0.529
Show the working
- Focal length = 35
- Aperture f-number = 2.8
- Subject distance = 3
- Circle of confusion divisor = 1500
- sensor_w = 6.17 = 6.17
- sensor_h = 4.55 = 4.55
- Sensor diagonal = hypot(6.17, 4.55) = 7.666 25
- Crop factor = 7.666 25 > 0 ? 43.266615305567875 / 7.666 25 : 0 = 5.643 78
- Full-frame equivalent focal length = 35 * 5.643 78 = 197.532
- Equivalent aperture for depth of field = 2.8 * 5.643 78 = 15.8026
- Sensor area = 6.17 * 4.55 = 28.0735
- Full frame gathers this many times more light = 28.0735 > 0 ? 864 / 28.0735 : 0 = 30.7764
- Horizontal field of view = deg(2 * atan(6.17 / (2 * 35))) = 10.0744
- Vertical field of view = deg(2 * atan(4.55 / (2 * 35))) = 7.437 99
- Diagonal field of view = deg(2 * atan(7.666 25 / (2 * 35))) = 12.5
- coc = 7.666 25 / 1500 = 0.005 110 83
- Hyperfocal distance = 35^2 / (2.8 * 0.005 110 83) + 35 = 85 637.5
- near = 3*1000 * 85 637.5 / (85 637.5 + (3*1000 - 35)) = 2899.61
- far = 3*1000 < 85 637.5 ? 3*1000 * 85 637.5 / (85 637.5 - (3*1000 - 35)) : 1e9 = 3107.59
- Depth of field = 3107.59 >= 1e9 ? 1e9 : 3107.59 - 2899.61 = 207.985
- Frame width at the subject = 35 > 0 ? 3 * 6.17 / 35 : 0 = 0.528 857
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.
The classic compact camera and action camera size.
Equivalent focal lengths on 1/2.3 inch
| Lens | Full-frame equivalent |
|---|---|
| 12 mm | 67.7 mm |
| 16 mm | 90.3 mm |
| 24 mm | 135.5 mm |
| 35 mm | 197.5 mm |
| 50 mm | 282.2 mm |
| 85 mm | 479.7 mm |
| 135 mm | 761.9 mm |
| 200 mm | 1128.8 mm |
Frequently asked
What is the crop factor of a 1/2.3 inch sensor?
5.644, calculated from its 7.67 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.