Maximum Vertical Shift needed?

[Thierry Schreiner]

Hi to all,

First of all, let me apologize in advance for the maybe stupid, but definitely weird question I am going to ask.

I hope someone out there can tell me a formula or share with me his/her experience.

I am planning to do night photography of industrial buildings / factories.

Apart from the long exposure times, I may encounter windy conditions. That’s why I decided to build a rigid body camera with fixed focus at hyperfocal distance but allowing shift control for perspective correction, a bit in the style of the former “Orbitar” by Burke & James.

The lens that I will be using will be the Sironar W / Sinaron WS 210mm.

Now the question: How much maximum shift do I need?

With the aforementioned lens, maximum vertical shift capability in the 8x10 format will be 29mm, whereas in 5x7 it will be 94mm. (horizontal format at infinity at f22)

Depending on what I need, I will decide to go either the 8x10 or the 5x7 way.

Is there a formula to calculate the amount of shift needed, taken probably into consideration the focal length of the lens, the distance from the subject, the height of the building, etc.

I hope someone can help me with my dilemma.

Thank you in advance and best regards from France.

Thierry

[ki6mf]

If you will use a ground glass back get the kind with a 45 degree angle in each corner. Then do your shifts and as long as you can see a round portion of the lens the image circle will cover the amount of shift. Open up your aperture to it up to F 4.5/5.6 like when you focus. This method you don't have to worry about how much shift you have. I like to leave a yellow filter on the lens and as long as i can see yellow I avoid vignetting.

[Glenn Thoreson]

Since there are no tall buildings around these parts, I rarely need those movements. When I do, I just go by what the image on the glass looks like. It's not hard to tell when things are vertical and the coverage is right. Simple is my motto. :D

[mikec]

Hi Thierry,
I am not sure that a formula is possible due to differing lens formulas. As pointed out previously, the image circle (attenuated by any falloff) will be the limting factor.

Lenses of the same focal length may have different image circles. IIRC these are usually expressed at f22.

Michael Gudzinowicz compiled a list of lenses with their image circles which you can find on http://www.largeformatphotography.info/lenseslist.html .

Simple math will give you the shift range.

[Armin Seeholzer]

Best you make a changing back so you can use 5x7 and or 8x10!
But the 210 is on 8x10 around a 28mm on 35mm Film so it is quit wide.
On 5x7 its around a 35mm lens on 35 mm Film so not so wide!
If you always can go so far away as needed for the 5x7 then this would be the better choice in my opinion.
But I'm a very wide guy so my choice would be even a shorter lens.

Cheers Armin

[cjbroadbent]

The 210mm is a good lens for 8x10 in those conditions. I would mock it up with a borrowed 8x10. When that is working to the limit, take measurements and make the bird-box. Don't screw the front on until you have checked focus outdoors ( a far away street lamp at night). If you plan to stop down, make a shim to get somewhere close to hyperfocal.

[Jim Noel]

A sliding front like those on a Deardorff will probably be an asset to you .

[Leonard Evens]

Consider the height Y of the building above the level of the lens. (In most cases, that will be only slightly less than the height of the building above ground level, so you can use that height instead.)

To get the height y in the film plane of the image of that building height, multiply Y by the magnification M. In symbols,

y = M x Y

To find M, divide the distance of the lens to the building, converted to mm, by the focal length of the lens. also in mm, subtract one from that ratio, and finally take the reciprocal of the result. In symbols, if D is the distance to the building and f is the focal length,

M = 1/((D/f) -1)

Let h be the vertical height of the frame, in mm. This will depend on the format of the film, and whether you are in portrait or landscape orientation.

The the necessary vertical rise is the image height y less half the frame height. In symbols

rise = y - h/2

Of course, as others have suggested, the rise necessary to include the entire building in the frame may take you outside the circle of coverage of the lens.

If R is the radius of the circle of coverage (half the diameter of the circle of coverage), and w is the width of the frame (which again will depend on the format and the orientation), then the maximum vertical rise (assuming no shift) which keeps you within the circle of coverage is

sqrt[R^2 - (w/2)^2] - h/2

It is remotely possible I made some error in my calculations, so if this doesn't check out, let me know.

[srbphoto]

Leonard...you did make one mistake,

M = 1/((D/f) -1) should be M = 1/((D/f) -1²)

Hope this helps

Scott

[Leonard Evens]

Leonard...you did make one mistake,

M = 1/((D/f) -1) should be M = 1/((D/f) -1²)

Hope this helps

Scott

I hope this is a joke. Of course, you could have put 1^N where N is any number, including infinite cardinals.

[Robert Hughes]

M = 1/((D/f) -1²)


Clearly, the answer is 42. What was the question?