Trying to work out DSLR vs 5x7 or 6x17 shift (rise) equivalency

[gerald.d]

Hi all -

I hope this is the right forum to ask this, and that someone here can help me because my brain has got a bit frazzled trying to work it all out.

I'm trying to work out whether a shot I've taken with a Canon 17mm TS-E on a FF DSLR can be replicated on the kit that I'm awaiting delivery on.

I'm getting a Walker Titan XL 5x7 with a back to take the Canham 6x17, along with a couple of lenses. The lens in question for this shot would be the Schneider 72XL.

Thanks to the wonderful resources on this site, I've managed to get some way through the theory, but am not really sure of where it's got me...

The shot in question is in portrait, and required around 8mm of vertical shift/rise (I can't recall exactly, but am going from the location of the horizon in the photo). Here's the shot:

http://dxbae.com/old_images/Burj/IMG_7985_1k.JPG

To my advantage, I've got more FoV to play with, because I understand the 72XL is equivalent to a 15mm on FF DSLR. However, the 72XL is pretty tight on 5x7 and doesn't provide much room for movements.

Assuming the camera back is in portrait...

I understand from the lens tables on this site that on 5x7 I'd have 10.44mm of rise to play with, and on 6x17 I'd have a more substantial 24.48mm available.

A couple of questions - the first rather basic and no doubt very easy to answer:

When the tables indicate 24.48mm of rise, that means that in total, I have double that movement (i.e. i have 24.48mm of rise, and 24.48mm of fall available), yes?

Secondly, would I be able to get the building fully in shot and still keep the camera back perfectly vertical? I'm not fussed about aspect ratio or losing width here - all I'm interested in is the building. Am I correct in assuming that if on 24x36mm format in portrait a shot required 8mm of rise, then on 6x17 it would need ((8/36)*170)=38mm of rise assuming the lenses had the same FoV?

Obviously I'll find out for myself in a few weeks when everything arrives and I set it all up for real. My gut feel is that this is going to be very close, and that if I have understood the relative movements between the different formats properly, then the extra FoV of the Schneider 72XL over the Canon 17mm TS-E might just wing it for me.

If anyone can actually work it out in advance though, it would be nice to know :)

Kind regards,

Gerald.

[engl]

When the tables indicate 24.48mm of rise, that means that in total, I have double that movement (i.e. i have 24.48mm of rise, and 24.48mm of fall available), yes?


Yes, as long as you are talking about the same standard. You can't do 24mm front rise and 24mm back fall.

As for the shot in question, you can start by figuring out the field of view needed for the shot you took. Drawing the sensor within the image circle makes it easy to see that the radius of the needed image circle is sqrt(12^2 + 26^2)=28.6mm with 8mm rise in portrait orientation, making the needed image circle diameter 57.3mm. The field of view of a 17mm lens with a 57.3mm image circle is 119 degrees.

In other words, the 72XL is not quite wide enough for the same shot. You'd need the 47XL (120 degrees) to get the same shot, but you'd need to crop your negative to get it.

(someone please check my maths :) )

[Jack Dahlgren]

Hi all -

I hope this is the right forum to ask this, and that someone here can help me because my brain has got a bit frazzled trying to work it all out.

I'm trying to work out whether a shot I've taken with a Canon 17mm TS-E on a FF DSLR can be replicated on the kit that I'm awaiting delivery on.

I'm getting a Walker Titan XL 5x7 with a back to take the Canham 6x17, along with a couple of lenses. The lens in question for this shot would be the Schneider 72XL.

Thanks to the wonderful resources on this site, I've managed to get some way through the theory, but am not really sure of where it's got me...

The shot in question is in portrait, and required around 8mm of vertical shift/rise (I can't recall exactly, but am going from the location of the horizon in the photo). Here's the shot:

http://dxbae.com/old_images/Burj/IMG_7985_1k.JPG

To my advantage, I've got more FoV to play with, because I understand the 72XL is equivalent to a 15mm on FF DSLR. However, the 72XL is pretty tight on 5x7 and doesn't provide much room for movements.

Assuming the camera back is in portrait...

I understand from the lens tables on this site that on 5x7 I'd have 10.44mm of rise to play with, and on 6x17 I'd have a more substantial 24.48mm available.

A couple of questions - the first rather basic and no doubt very easy to answer:

When the tables indicate 24.48mm of rise, that means that in total, I have double that movement (i.e. i have 24.48mm of rise, and 24.48mm of fall available), yes?

Secondly, would I be able to get the building fully in shot and still keep the camera back perfectly vertical? I'm not fussed about aspect ratio or losing width here - all I'm interested in is the building. Am I correct in assuming that if on 24x36mm format in portrait a shot required 8mm of rise, then on 6x17 it would need ((8/36)*170)=38mm of rise assuming the lenses had the same FoV?

Obviously I'll find out for myself in a few weeks when everything arrives and I set it all up for real. My gut feel is that this is going to be very close, and that if I have understood the relative movements between the different formats properly, then the extra FoV of the Schneider 72XL over the Canon 17mm TS-E might just wing it for me.

If anyone can actually work it out in advance though, it would be nice to know :)

Kind regards,

Gerald.

Can you move back a few steps?

If you know the height of the building and your distance from the building as well as the angle of coverage of the lens you should be able to figure it out.

-Jack

[gerald.d]

Can you move back a few steps?

If you know the height of the building and your distance from the building as well as the angle of coverage of the lens you should be able to figure it out.

-Jack

Thanks Jack -

I know the height of the building (828 metres), but what I don't know is how high the shooting position is off the ground, but I suppose I could estimate it. I guess I could also work out the distance from the building using Google Earth.

engl -

Sorry, but you lost me at "it easy to see that the radius of the needed image circle is sqrt(12^2 + 26^2)=28.6mm with 8mm rise in portrait orientation". If you do have the time to explain that bit in a bit more detail, it would be apprecaited.

Kind regards to you both,

Gerald.

[engl]

I made a quick sketch showing the image circle, the unshifted image sensor in portrait orientation (36x24mm big, gray in the sketch) and the shifted image sensor (black in the sketch). Drawn like this, the radius of the *needed* image circle is the hypotenuse of the triangle in the drawing. The two legs of triangle are known, one is 12mm, the other is 18+8mm.

The diameter is two times the radius.