1/1.28 inch crop factor and field of view calculator
A 1/1.28 inch sensor measures 9.8 by 7.3 mm, a diagonal of 12.22 mm. Against the full-frame diagonal of 43.27 mm that is a crop factor of 3.541. A large phone sensor, as used in flagship main cameras.
The diagonal divided by this gives the acceptable blur circle. 1500 is the usual modern figure.
- Full-frame equivalent focal length
- 123.92
- Equivalent aperture for depth of field
- 9.914
- Sensor diagonal
- 12.220
- Sensor area
- 71.54 mm2
- Full frame gathers this many times more light
- 12.077
- Horizontal field of view
- 15.939 degrees
- Vertical field of view
- 11.907 degrees
- Diagonal field of view
- 19.805 degrees
- Hyperfocal distance
- 53 738
- Depth of field
- 332
- Frame width at the subject
- 0.840
Show the working
- Focal length = 35
- Aperture f-number = 2.8
- Subject distance = 3
- Circle of confusion divisor = 1500
- sensor_w = 9.8 = 9.8
- sensor_h = 7.3 = 7.3
- Sensor diagonal = hypot(9.8, 7.3) = 12.2201
- Crop factor = 12.2201 > 0 ? 43.266615305567875 / 12.2201 : 0 = 3.540 62
- Full-frame equivalent focal length = 35 * 3.540 62 = 123.922
- Equivalent aperture for depth of field = 2.8 * 3.540 62 = 9.913 74
- Sensor area = 9.8 * 7.3 = 71.54
- Full frame gathers this many times more light = 71.54 > 0 ? 864 / 71.54 : 0 = 12.0772
- Horizontal field of view = deg(2 * atan(9.8 / (2 * 35))) = 15.9392
- Vertical field of view = deg(2 * atan(7.3 / (2 * 35))) = 11.9072
- Diagonal field of view = deg(2 * atan(12.2201 / (2 * 35))) = 19.8049
- coc = 12.2201 / 1500 = 0.008 146 71
- Hyperfocal distance = 35^2 / (2.8 * 0.008 146 71) + 35 = 53 737.7
- near = 3*1000 * 53 737.7 / (53 737.7 + (3*1000 - 35)) = 2843.13
- far = 3*1000 < 53 737.7 ? 3*1000 * 53 737.7 / (53 737.7 - (3*1000 - 35)) : 1e9 = 3175.19
- Depth of field = 3175.19 >= 1e9 ? 1e9 : 3175.19 - 2843.13 = 332.064
- Frame width at the subject = 35 > 0 ? 3 * 9.8 / 35 : 0 = 0.84
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.
A large phone sensor, as used in flagship main cameras.
Equivalent focal lengths on 1/1.28 inch
| Lens | Full-frame equivalent |
|---|---|
| 12 mm | 42.5 mm |
| 16 mm | 56.6 mm |
| 24 mm | 85.0 mm |
| 35 mm | 123.9 mm |
| 50 mm | 177.0 mm |
| 85 mm | 301.0 mm |
| 135 mm | 478.0 mm |
| 200 mm | 708.1 mm |
Frequently asked
What is the crop factor of a 1/1.28 inch sensor?
3.541, calculated from its 12.22 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.