I've been using my friend's Otus 55 on my Sony A6000 and I'm having a blast tilting that lens. With a smaller sensor APS-C, I haven't had any problems tilting a Full Frame lens and keeping the image circle covering the whole sensor.


How large does the image circle have to be to get it to project correctly on the image sensor? 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?


4 Answers 4


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

Image circle dimensions

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 (!!).

  • 1
    \$\begingroup\$ Shouldn't the first equation read D = 2s + \sqrt{h^2+w^2} ("+" instead of "*")? The shift should just add to the minimum needed image circle, or do I miss something? \$\endgroup\$
    – Chris
    Jan 27, 2016 at 10:15
  • \$\begingroup\$ @Chris oops, thanks. I fixed it now. Noticed you're a TeXie. Oh how I wish LaTeX math were enabled for this site... =/ \$\endgroup\$
    – scottbb
    Jan 27, 2016 at 13:40
  • 1
    \$\begingroup\$ Is there something we can do about getting MathJax support on this site? I saw it enabled at mathematics.sx and it is impressive. IMO all sites with scientific and/or technical background should have it enabled. \$\endgroup\$
    – Chris
    Feb 1, 2016 at 8:36
  • \$\begingroup\$ @Chris Probably will never happen (per Jeff Atwood). The SE devs are keen to keep SE sites load as quickly as possible. MathJax adds a lot to the average page weight & load latency. \$\endgroup\$
    – scottbb
    Feb 1, 2016 at 20:33
  • \$\begingroup\$ @scottbb Hm, from the link I don't get the point, there are contradictory statements. The original statement from Jeff is from 2010, and the world has moved on. More processing power, faster internet, and many of the assessments state that it is not a heavy dependency, it does not add that much to the overall data & load latency, at least not if it is not used on the specific page. Maybe this has to be triggered one more time, after 6 years. But who am I to make such suggestions. Hm, OK, I'm a user desperate about the fact that the web is still sooooooooooo ugly. Especially concerning math ;-) \$\endgroup\$
    – Chris
    Feb 1, 2016 at 20:56

The answer would depend upon the degree of tilt and shift one wishes to effect. An image circle needs to be larger to accomplish 12º of tilt than to accomplish 8º of tilt on the same sensor. Likewise, a small shift movement requires a smaller image circle than a larger shift movement does.

With a traditional lens the center of the lens' optical axis is aimed perpendicularly at the center of the film/sensor. The usable image circle need only be large enough that its diameter is equal to or greater than the length of the diagonal of the sensor or film frame.

With a shift movement the perpendicular aspect is maintained, but the point at which the optical axis intersects the film/sensor is moved away from the center. The further one moves from the center of the film/sensor, the larger the image circle needs to be to continue to cover the edge of the sensor on the "far" side.

With a tilt movement it gets a little more complicated. Because the optical center of the lens is no longer perpendicular to the film/sensor, the image circle becomes an image oval. The short dimension of the oval is the same width as the diameter of the former image circle, but the longer axis of the oval is now greater. The center of the optical axis, however, may or may not be still pointed at the center of the film/sensor. It all depends on the design of the lens and where the point of rotation of the optical axis of the lens is centered.

As with a shift movement, when tilting a lens the more distance the center of the optical axis is moved from the center of the film/sensor the larger the image circle needs to be to accommodate the move. But since the edge of the image cast by the lens along the long axis (also the direction of movement) is now narrower at the apex as it intersects the non-perpendicular plane of the film/sensor, with a tilt movement it takes an even larger image circle cast by the lens to allow a move of the optical center the same distance from film/sensor center as would be required by a shift movement.

Note also that a movement of a specific distance along the short axis of the film/sensor requires more additional image circle than the same size movement along the long axis of the film/sensor. Movements the same distance made at angles in between the long and short axis of the film/sensor require an image circle size between those two extremes (some tilt/shift lenses are capable of such movements, others are not).

Combining both shift and tilt movements at the same time may compound the degree of additional image circle required if the two movements are in the same direction. But they may also serve to cancel the relative movement created by each of the optical axis in relation to the center of the film/sensor if the shift movement is in the opposite direction as the tilt movement. It all depends on the relative angles of each movement.

  • \$\begingroup\$ I know you're using oval in a casual way, but a more precise term would be ellipse. \$\endgroup\$
    – Caleb
    Jan 27, 2016 at 15:00
  • \$\begingroup\$ An ellipse is a specific type of oval. All ellipses are ovals, but not all ovals are ellipses. \$\endgroup\$
    – Michael C
    Jan 27, 2016 at 21:42
  • \$\begingroup\$ Like I said, ellipse is the more precise term. The light coming from the lens is shaped like a cone, and tilting the lens causes the focal plane to cut the cone at an angle, thus forming an ellipse. It's a really minor point, I know, but ellipse gives a better sense of the shape of the image formed by the lens. \$\endgroup\$
    – Caleb
    Jan 27, 2016 at 21:50
  • \$\begingroup\$ Ellipses are a subset of ovals. Therefore it is an oval. The description is accurate, if not as precise as you wish. The entire answer is intended to be easily readable in non-technical layman's terms, thus the more generic oval was intentionally selected. There is a much more technical answer here as well. \$\endgroup\$
    – Michael C
    Jan 27, 2016 at 21:57

If you only consider tilt or shift alone but not the possibility of both simultaneously then you omit considering instances where both can be used together to reduce the required diameter of the lens.

Shift only

You'll need a large lens to do that.

Shifted and Tilted

A smaller lens can be used where both tilting and shifting are employed.

Providing a simple one size fits all equation is nearly impossible because of the variety of equipment to do tilting and shifting combined with the number of lens available. Usually it is possible to adjust the image projected from the lens to fully cover large format film, this is much easier with a comparatively small digital sensor.

Cambo ACTUS-XL-35

With so much to adjust you'll need to read the above book (or the calculator linked below) and use the equations to devise a calculator for your setup. Vignette is uncommon (due to going up a step in lens size for your format - EG: Medium Format lens on FF sensor, FF lens on APS-C or shoot crop mode) and the bellows can be adjusted to increase (or decrease) the image circle diameter.

A Scheimpflug Calculator is really needed to do all the calculations simultaneously without error. There are a few calculators on this Webpage that provide the answer in Aperture.

For a mathematical background you can visit Wikipedia's Scheimpflug Principle Webpage or Harold M. Merklinger's Website where you can download his Shareware Book called "Focusing the View Camera" (see Chapter 4 - Optical Principles). Harold would prefer people to download the book for free and pay if they like it rather than copy it (or provide direct links to the .PDF).

With a high enough resolution sensor you can have a bit of it vignetted and still crop a large resolution image out of the remainder.

A lot of cropping can be done to a more than 100MP image and still leave a substantially sized image. Perfect coverage of the sensor is neither practical in all cases nor as important as tilting and shifting the required amount to produce the desired results with minimal (preferably no) cropping.

Here is a wonderful interactive Scheimpflug Calculator from OptiWiki. This calculator allows choosing the sensor size (and thus the size of the image circle). The interactive graphic keeps the optical axis in the center of the lens and maps the edges of the sensor to the wanted object plane.

  • \$\begingroup\$ Qualitatively, the pictures in your answer speak volumes. However, with regards to the specific question about the size of the image circle, the linked pages and calculators don't really refer to image circle size. But good answer, nonetheless. \$\endgroup\$
    – scottbb
    Nov 11, 2017 at 21:22
  • \$\begingroup\$ @scottbb - I added a link to an interactive calculator. I am looking for better calculators and for an ultimate one that would calculate the ellipse using all possible variables. What's good about the above calculator is that the image circle size (sensor size) is sticky (assumes that once adjusted it won't be continuously updated) and allows you to quickly set up the scene and see the results of changes graphically. Works well with a Mobile Touchscreen also. My search continues. \$\endgroup\$
    – Rob
    Nov 12, 2017 at 0:05

This extends scottbb's answer for shift by allowing a one-dimensional tilt. The general solution requires much more extensive knowledge of projective geometry and optics than I have, but it shouldn't be impossible to generalize the ideas here presented, in principle.

Tilt (1D)

Assume that the (rectangular) sensor/frame lies in the xy-plane with its centre coinciding with the origin (x,y) = (0,0). Let the sensor/frame width and height be 2w and 2h, respectively:

Sensor/frame diagram

Let the lens reside in the third, z-dimension at a focal length f from the sensor/film plane, transmitting a right rectangular cone of light of aperture 2φ degrees and tilted by θ degrees from the normal:

Lens tilt diagram

The cone intersects the sensor/film plane at the angle θ and produces an image ellipse whose major axis is of length Y1 + Y2 (see above figure). Using the law of sines, it is possible to figure out Y1 and Y2:

Equation for Y1

Equation for Y2

Since ellipses are symmetric, the centre of this ellipse must lie at the point (0, (Y2 - Y1)/2). Its major semi-axis is given by (Y1 + Y2)/2. Its minor semi-axis is

Equation for Y0

(i.e. the special case of θ = 0 degrees tilt). From the shifted ellipse equation, any point (x,y) on this ellipse must satisfy

Ellipse equation, no shift

Shift (2D)

A lens shift of Δx and Δy units in the x and y-dimensions, respectively, corresponds to a shift of the ellipse centre by the amounts Δx and Δy. The above equation becomes:

Ellipse equation, with shift

Maximum sensor/frame size

The largest sensor/frame that can be positioned at the origin of the xy-plane to fit in the image ellipse is characterized by

enter image description here

The solution is indeterminate since both w and h are free to vary. But if we know the width-to-height ratio of the sensor/frame (e.g. 3:2), we can express one of w and h in terms of the other and find the unique solution. Suppose for example w = (3/2)h (35mm aspect ratio). Substituting into the above equation and solving for h gives the unique largest 3:2 aspect ratio sensor/frame that the ellipse will accommodate.

Minimum un-tilted image circle radius

Conversely, given a sensor/frame extending from x = -w to x = w and from y = -h to y = h, a lens tilt of θ degrees, and a lens shift of amounts Δx and Δy, the smallest radius Y0 of the image circle projected by the un-tilted lens sufficient to guarantee that the sensor/frame is covered by the image ellipse of the tilted lens, can be recovered from the above equation by solving for Y0.


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