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I am considering the x1d II, and already have Canon tilt shift lenses, 24 and 45mm, with image circles of 67 and 58mm, respectively.

By using, say, a Novoflex adapter, how many mm of shift would I lose with the x1d I or II sensor?

And what would the corresponding focal length be, approximatively?

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The Hasselblad X1D II has a 43.8mm x 32.9mm sensor. That computes to a diagonal of just under 55mm and an aspect ratio of 4:3 or 1.333:1.

With the two lenses in question, and assuming you want to preserve infinity focus by using the lenses' designed registration distance of 44mm:

  • You'd give up pretty much the possibility of any shift with the TS-E 45mm f/2.8 as you'd only have 1.5mm to spare at each corner of the frame. Most lenses are designed to allow about that much room between the absolute edge of the image circle and the usable portion of the image circle.
  • You'd have a bit more room to play with using the TS-E 24mm f/3.5 L II (be sure and confirm that the measurement you're using of 67mm image circle applies to whichever of the two TS-E 24mm variants you're considering). You'd nominally have about 4.5mm on each corner to play with (4.5mm + 1.5mm = 6mm x 2 = 12mm), but that is in the direction of the sensor's diagonal. If you want to move in a direction that is horizontal with respect to the long side of the sensor, You'd be able to move about 4.8mm in either direction. Moving vertically with respect to the long side of the sensor, you'd be limited to about 6.7mm of movement.

I've taken the liberty of using scottbb's diagram with the addition of the usable portion of the image circle and superimposed light blue rectangles where the image circle would be over the sensor at the limits of usable movement.

Please note: The ratio of the sensor diagonal to the diameter of the image circle in the diagram is approximately 1.414:1, which would mean an image circle of about 78mm if the sensor diagonal is 55mm. (The sensor's aspect ratio is also 3:2 (1.5:1) instead of 4:3 (1.33:1), and the differences between Sw and Sh would be less with a more square sensor). Image circles of 58mm and 67mm, respectively, would be much smaller in relation to the sensor.

enter image description here

And what would the corresponding focal length be, approximately?

The focal length of either lens will not change, they'll still be 24mm and 45mm lenses. The larger sensor size will yield a wider angle of view than the same lenses would give on a 35mm/FF camera. Your "crop factor" would be roughly 0.8X, so using the 43.8 x 32.9 mm sensor would give similar diagonal angles of view as 19mm and 35mm lenses would give using a 35mm/FF camera. Keep in mind that the aspect ratio is different. The larger sensor is 4:3 or 1.333:1, while the smaller 35mm/FF sensor is 3:2 or 1.5:1, so you'd get slightly more vertical coverage and slightly less horizontal coverage with the Hasselblad after applying the "crop factor" than you would get with a 35mm/FF camera.

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  • thanks a lot for the info. Movements are a bit too constrained for me. Do you know better alternative for using tilt shift with hasselblad or fuji gfx 50?
    – Bob
    Dec 1 '19 at 7:30
  • @Bob have a look at the Cambo Actus view camera. They have bayonet mounts for the GFX and lens plates for Canon EOS. You don't need a tilt-shift lens to use the camera but TS lenses should work just fine.
    – MrUpsidown
    Dec 1 '19 at 21:32
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    @MrUpsidown Fuji X-mount is APS-C, not MF. GFX mount is the mount for Fuji's MF cameras.
    – Michael C
    Dec 2 '19 at 0:06
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    Oh... my bad... Ok. So no option for Canon lenses on the Actus G which is the one with the GFX mount. Makes sense. In other words, you could only use Canon TS lenses on an Actus B and one of the listed and compatible mirrorless bodies (which doesn't include the GFX 50).
    – MrUpsidown
    Dec 2 '19 at 0:13
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    Yeah, they're all APS-C or smaller except for the Sony E-mount, which has a short enough registration distance of only 16mm. The 20mm registration distance of the EOS R makes it incompatible with Canon EF lenses on the Actus B.
    – Michael C
    Dec 2 '19 at 0:22
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Here is the geometry to determine how much shift you theoretically have:

Geometry of sensor and image circle to calculate shift

  • C is the diameter of the image circle projected by the lens;
  • h is the height of the image sensor in landscape orientation (i.e., the short edge);
  • w is the width of the image sensor in landscape orientation (i.e., the long edge);
  • sh is the amount of shift (+/-) along the height dimension of the sensor (i.e., vertical shift in landscape orientation);
  • sw is the amount of shift (+/-) along the width dimension of the sensor (i.e., horizontal shift in landscape orientation).

According to Pythagoras's Theorem, the square of the diagonal of a right triangle is equal to the sum of the squares of other legs of the triangle. The two equations, one for the shift along the height (sh), and one for the shift along the width (sw), are:

Pythagorean equations for height and width shift

Solving the equations for sh and sw, respectively:

enter image description here

The X1D sensor's height and width are h = 32.9mm and w = 43.8mm. Plugging into the equations, we can see that

  • With the 24mm lens's image circle of C = 67mm, the available shift should be
    • sh = ±8.9mm
    • sw = ±7.3mm
  • for the 45mm lens with an image circle of C = 58mm,
    • sh = ±2.5mm
    • sw = ±2.0mm
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  • This assumes all of the image circle is considered "usable". With the vast majority of lenses, the edge of their image circle is not a well defined line, but a graduated transition from light to dark. For lenses intended for 135 format (FF) cameras, it's usually about 1.5mm wide on each side of the image circle.
    – Michael C
    Dec 6 '19 at 20:47
  • @MichaelC Only my final calculations assume "precise" image circle values, and that's just based on the info given in OP. That's why I gave the formulas, so people can plug in whatever value for C they want or need.
    – scottbb ♦
    Dec 6 '19 at 22:10
  • Even with that taken into account, I think your calculated values are double what they should be. They're the total amount of shift possible, from +x/2 to -x/2.
    – Michael C
    Dec 7 '19 at 6:31
  • There's a much simpler way to draw it. Place the image circle diameter (D) along the sensor diagonal (d). The excess diameter on each end of the image circle is (D - d)/2. This dimension is also the length of the diagonal formed by a right triangle with sides Sh and Sw. Then just solve for the 3:4:5 triangles at each end (since the sensor is a 3:4 sensor it's pretty easy).
    – Michael C
    Dec 7 '19 at 6:37
  • @MichaelC re: 2x values: I don't think so. If you plug in Sh and Sw into the Pythagorean formulas, it works. So my algebra is correct. That only leaves whether or not my geometry is correct, and I'm confident it is. Which leads me to...
    – scottbb ♦
    Dec 7 '19 at 18:43

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