How large does a lens image circle need to be for tilt and shift movements?
Asked 1/26/2016
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I'm adapting full-frame lenses to an APS-C camera and experimenting with tilt. A full-frame lens can still cover the smaller sensor when tilted a bit, which made me wonder: how large does a lens's image circle need to be to support tilt and/or shift without vignetting? Is there a simple way to estimate the required image circle for a given sensor and amount of movement? Also, are tilt-shift lenses expensive mainly because they need a larger image circle, or because of other design complexities too?
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
10y ago
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How large does the image circle have to be to get it to project correctly on the image sensor?
This part of the answer deals with the shift of the lens only. The answer for the tilt is much more complicated (i.e., I haven't cranked out the maths).
In order for the image to be projected onto the sensor, without any clipping at the corners of the sensor, regardless of the orientation of the lens with respect to the camera, the amount of shift is limited by the camera's sensor's diagonal distance. Thus, for the following parameters:
- Sensor height h (in mm);
- Sensor width w (in mm);
- Maximum shift distance s (in mm);
the minimum image circle diameter D (in mm) is given by
D = 2 s + √(h² + w²)
Alternately, for a lens designed for a particular camera format (h0 for height of the sensor the lens was designed for, w0 for the width of the sensor the lens was designed for), the maximum amount the lens can be shifted (smax) on a smaller sensor with height h and width w is:
smax = [√(h0 + w0) - √(h + w)] / 2
By taking into consideration the heights and widths of both the designed-for sensor and the sensor of the camera the lens will be used on, any difference in the aspect ratio of the two cameras is handled (i.e., it doesn't matter).
Your A600 has a sensor with h = 15.6 mm and w = 23.5 mm, so the diagonal of your sensor (the part under the square root) is 28.2 mm. If you need 10 mm shift in any direction, then the image circle of the lens must be at least 28.2 mm + 2 * 10 mm = 48.2 mm.
A non-shift lens designed for a full-frame sensor (36 mm × 24 mm) has a minimum image circle of 43.3 mm. Its image circle extends ~ 7.5 mm farther than the corners of your camera's sensor, so you can shift up to 7.5 mm in any direction (i.e., along the sensor's diagonal) without clipping.
Related, are tilt-shifts expensive because of their larger image circle (and thus cost of glass/etc) or is there more than meets the eye?
Assuming the price of a lens is strictly proportional to the construction cost of the lens, that's probably most of the reason why they are more expensive. Especially when you consider that there is no autofocus mechanism in a tilt-shift lens, so that cost & design constraint is eliminated. But since the market for such lens is much smaller than non-movement prime lenses, there is probably also some "rarity" or small-market pricing going on as well.
For really expensive tilt-shift lenses, check out Schneider Optics PC-TS lenses. Their tilt and shift mechanics are completely internal to the lens – there are rings on the lens to control the movements. Their 90mm ƒ/4.5 and 50mm ƒ/2.8 tilt-shift lenses are around $4,000, and have 95mm front filter threads. But that's nothing compared to the $10,000 28mm ƒ/4.5 with 120mm filter threads (!!).
Originally by user11924. Source · Licensed CC BY-SA 4.0
user11924
10y ago
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The required image circle depends on the sensor size and on how much shift and/or tilt you want.
For a normal lens with no movements, the image circle only needs to cover the sensor diagonal.
For shift alone, a useful minimum is:
D = 2s + √(w² + h²)
where D is image-circle diameter, s is maximum shift, and w/h are sensor width and height. In other words, the image circle must cover the sensor diagonal plus the extra off-center movement.
Tilt is harder to reduce to one simple formula, because the needed coverage depends on tilt angle and geometry. More tilt generally requires a larger usable image circle. If you combine tilt and shift, the geometry becomes more complex still, so there isn’t really a single one-size-fits-all equation.
That’s why APS-C can often tolerate more movement from a full-frame lens than a full-frame sensor can: the smaller sensor uses less of the image circle.
As for cost: yes, a larger image circle is part of why tilt-shift lenses are expensive, but not the only reason. They also require more complex optical and mechanical design to maintain image quality while the lens is moved off-axis.
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