1/1.7 inch crop factor and field of view calculator
A 1/1.7 inch sensor measures 7.6 by 5.7 mm, a diagonal of 9.50 mm. Against the full-frame diagonal of 43.27 mm that is a crop factor of 4.554.
The diagonal divided by this gives the acceptable blur circle. 1500 is the usual modern figure.
- Full-frame equivalent focal length
- 159.40
- Equivalent aperture for depth of field
- 12.752
- Sensor diagonal
- 9.500
- Sensor area
- 43.32 mm2
- Full frame gathers this many times more light
- 19.945
- Horizontal field of view
- 12.393 degrees
- Vertical field of view
- 9.310 degrees
- Diagonal field of view
- 15.457 degrees
- Hyperfocal distance
- 69 114
- Depth of field
- 258
- Frame width at the subject
- 0.651
Show the working
- Focal length = 35
- Aperture f-number = 2.8
- Subject distance = 3
- Circle of confusion divisor = 1500
- sensor_w = 7.6 = 7.6
- sensor_h = 5.7 = 5.7
- Sensor diagonal = hypot(7.6, 5.7) = 9.5
- Crop factor = 9.5 > 0 ? 43.266615305567875 / 9.5 : 0 = 4.554 38
- Full-frame equivalent focal length = 35 * 4.554 38 = 159.403
- Equivalent aperture for depth of field = 2.8 * 4.554 38 = 12.7523
- Sensor area = 7.6 * 5.7 = 43.32
- Full frame gathers this many times more light = 43.32 > 0 ? 864 / 43.32 : 0 = 19.9446
- Horizontal field of view = deg(2 * atan(7.6 / (2 * 35))) = 12.3928
- Vertical field of view = deg(2 * atan(5.7 / (2 * 35))) = 9.310 49
- Diagonal field of view = deg(2 * atan(9.5 / (2 * 35))) = 15.4573
- coc = 9.5 / 1500 = 0.006 333 33
- Hyperfocal distance = 35^2 / (2.8 * 0.006 333 33) + 35 = 69 113.9
- near = 3*1000 * 69 113.9 / (69 113.9 + (3*1000 - 35)) = 2876.59
- far = 3*1000 < 69 113.9 ? 3*1000 * 69 113.9 / (69 113.9 - (3*1000 - 35)) : 1e9 = 3134.47
- Depth of field = 3134.47 >= 1e9 ? 1e9 : 3134.47 - 2876.59 = 257.876
- Frame width at the subject = 35 > 0 ? 3 * 7.6 / 35 : 0 = 0.651 429
Values are shown to six significant figures here; the arithmetic itself runs on exact fractions and is only rounded where a number is printed.
Scenarios & local history
Compare a saved scenario
Recent calculations
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 1/1.7 inch
| Lens | Full-frame equivalent |
|---|---|
| 12 mm | 54.7 mm |
| 16 mm | 72.9 mm |
| 24 mm | 109.3 mm |
| 35 mm | 159.4 mm |
| 50 mm | 227.7 mm |
| 85 mm | 387.1 mm |
| 135 mm | 614.8 mm |
| 200 mm | 910.9 mm |
Frequently asked
What is the crop factor of a 1/1.7 inch sensor?
4.554, calculated from its 9.50 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.