1/1.33 inch crop factor and field of view calculator
A 1/1.33 inch sensor measures 9.6 by 7.2 mm, a diagonal of 12.00 mm. Against the full-frame diagonal of 43.27 mm that is a crop factor of 3.606.
The diagonal divided by this gives the acceptable blur circle. 1500 is the usual modern figure.
- Full-frame equivalent focal length
- 126.19
- Equivalent aperture for depth of field
- 10.096
- Sensor diagonal
- 12.000
- Sensor area
- 69.12 mm2
- Full frame gathers this many times more light
- 12.500
- Horizontal field of view
- 15.618 degrees
- Vertical field of view
- 11.745 degrees
- Diagonal field of view
- 19.455 degrees
- Hyperfocal distance
- 54 723
- Depth of field
- 326
- Frame width at the subject
- 0.823
Show the working
- Focal length = 35
- Aperture f-number = 2.8
- Subject distance = 3
- Circle of confusion divisor = 1500
- sensor_w = 9.6 = 9.6
- sensor_h = 7.2 = 7.2
- Sensor diagonal = hypot(9.6, 7.2) = 12
- Crop factor = 12 > 0 ? 43.266615305567875 / 12 : 0 = 3.605 55
- Full-frame equivalent focal length = 35 * 3.605 55 = 126.194
- Equivalent aperture for depth of field = 2.8 * 3.605 55 = 10.0955
- Sensor area = 9.6 * 7.2 = 69.12
- Full frame gathers this many times more light = 69.12 > 0 ? 864 / 69.12 : 0 = 12.5
- Horizontal field of view = deg(2 * atan(9.6 / (2 * 35))) = 15.618
- Vertical field of view = deg(2 * atan(7.2 / (2 * 35))) = 11.7453
- Diagonal field of view = deg(2 * atan(12 / (2 * 35))) = 19.4552
- coc = 12 / 1500 = 0.008
- Hyperfocal distance = 35^2 / (2.8 * 0.008) + 35 = 54 722.5
- near = 3*1000 * 54 722.5 / (54 722.5 + (3*1000 - 35)) = 2845.81
- far = 3*1000 < 54 722.5 ? 3*1000 * 54 722.5 / (54 722.5 - (3*1000 - 35)) : 1e9 = 3171.86
- Depth of field = 3171.86 >= 1e9 ? 1e9 : 3171.86 - 2845.81 = 326.052
- Frame width at the subject = 35 > 0 ? 3 * 9.6 / 35 : 0 = 0.822 857
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.
Equivalent focal lengths on 1/1.33 inch
| Lens | Full-frame equivalent |
|---|---|
| 12 mm | 43.3 mm |
| 16 mm | 57.7 mm |
| 24 mm | 86.5 mm |
| 35 mm | 126.2 mm |
| 50 mm | 180.3 mm |
| 85 mm | 306.5 mm |
| 135 mm | 486.7 mm |
| 200 mm | 721.1 mm |
Frequently asked
What is the crop factor of a 1/1.33 inch sensor?
3.606, calculated from its 12.00 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.