1 inch crop factor and field of view calculator
A 1 inch sensor measures 13.2 by 8.8 mm, a diagonal of 15.86 mm. Against the full-frame diagonal of 43.27 mm that is a crop factor of 2.727. The "1 inch" name comes from vidicon tube diameters and has nothing to do with any dimension of the sensor.
The diagonal divided by this gives the acceptable blur circle. 1500 is the usual modern figure.
- Full-frame equivalent focal length
- 95.45
- Equivalent aperture for depth of field
- 7.636
- Sensor diagonal
- 15.864
- Sensor area
- 116.16 mm2
- Full frame gathers this many times more light
- 7.438
- Horizontal field of view
- 21.358 degrees
- Vertical field of view
- 14.331 degrees
- Diagonal field of view
- 25.539 degrees
- Hyperfocal distance
- 41 401
- Depth of field
- 432
- Frame width at the subject
- 1.131
Show the working
- Focal length = 35
- Aperture f-number = 2.8
- Subject distance = 3
- Circle of confusion divisor = 1500
- sensor_w = 13.2 = 13.2
- sensor_h = 8.8 = 8.8
- Sensor diagonal = hypot(13.2, 8.8) = 15.8644
- Crop factor = 15.8644 > 0 ? 43.266615305567875 / 15.8644 : 0 = 2.727 27
- Full-frame equivalent focal length = 35 * 2.727 27 = 95.4545
- Equivalent aperture for depth of field = 2.8 * 2.727 27 = 7.636 36
- Sensor area = 13.2 * 8.8 = 116.16
- Full frame gathers this many times more light = 116.16 > 0 ? 864 / 116.16 : 0 = 7.438 02
- Horizontal field of view = deg(2 * atan(13.2 / (2 * 35))) = 21.3579
- Vertical field of view = deg(2 * atan(8.8 / (2 * 35))) = 14.3306
- Diagonal field of view = deg(2 * atan(15.8644 / (2 * 35))) = 25.539
- coc = 15.8644 / 1500 = 0.010 576 3
- Hyperfocal distance = 35^2 / (2.8 * 0.010 576 3) + 35 = 41 401.1
- near = 3*1000 * 41 401.1 / (41 401.1 + (3*1000 - 35)) = 2799.51
- far = 3*1000 < 41 401.1 ? 3*1000 * 41 401.1 / (41 401.1 - (3*1000 - 35)) : 1e9 = 3231.42
- Depth of field = 3231.42 >= 1e9 ? 1e9 : 3231.42 - 2799.51 = 431.914
- Frame width at the subject = 35 > 0 ? 3 * 13.2 / 35 : 0 = 1.131 43
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 "1 inch" name comes from vidicon tube diameters and has nothing to do with any dimension of the sensor.
Equivalent focal lengths on 1 inch
| Lens | Full-frame equivalent |
|---|---|
| 12 mm | 32.7 mm |
| 16 mm | 43.6 mm |
| 24 mm | 65.5 mm |
| 35 mm | 95.5 mm |
| 50 mm | 136.4 mm |
| 85 mm | 231.8 mm |
| 135 mm | 368.2 mm |
| 200 mm | 545.5 mm |
Frequently asked
What is the crop factor of a 1 inch sensor?
2.727, calculated from its 15.86 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.