17mm TSE + aps-c

[WorkonSunday]

Hi all,

I must admit i have only spent 30mins searching on google, but i will ask anyway

I understand that a 17mm TSE without shift will give 17mm on FF. And if use on aps-c i get 17x1.6=27mm eqv on FF. But thats without shifting.

On FF, if i do maximum shift horizontally on left and right, and stitch the images in post, i get a field of view around 10.5mm eqv. In theory, the shift will allow aps-c bodies to do the same. Afterall, the lens just project the same image, it's only the sensor thats determining how much of that image can be captured. However, in practice i know there may be other mechanical limitation on how much i can shift with the lens (the knob and slide wont let u go further?). So my question is, do you know if i can still get the full 10.5mm horizontal coverage if i use something like 7D2 with 17mm TSE? Thanks.

eDIT: or word it differently, if i use 7D2 with 17TSE and shift to the most far left (or right), will i see the black edge where the image circle ends? Thanks

 

[J.R.]

The image circle of the 17mm TSE is huge and even with full shift, covers the FF sensor easily. There is no reason why the crop sensor will not be covered in its entirety.

 

[Loswr]

You'll get the full coverage, but you'll still have the 1.6x crop sensor such that your fully shifted FoV will approximate 17mm on FF.

An EF-S 10-18mm would get you there cheaper...

 

[StudentOfLight]

You will not see the edge of the image circle on Full frame or on APS-C.

Here is a slide I created which might help explain.

The circle represents the lens' image circle.
1 represents full shift left
2 represents neutral position
3 represents full shift right

FYI, I call shifting in the direction of the long end of frame "like-shifting" and shifting in the direction of the short end of the frame "cross-shifting" these are just my own terms.

 

[rfdesigner]

Hi all,

I must admit i have only spent 30mins searching on google, but i will ask anyway

I understand that a 17mm TSE without shift will give 17mm on FF. And if use on aps-c i get 17x1.6=27mm eqv on FF. But thats without shifting.

On FF, if i do maximum shift horizontally on left and right, and stitch the images in post, i get a field of view around 10.5mm eqv. In theory, the shift will allow aps-c bodies to do the same. Afterall, the lens just project the same image, it's only the sensor thats determining how much of that image can be captured. However, in practice i know there may be other mechanical limitation on how much i can shift with the lens (the knob and slide wont let u go further?). So my question is, do you know if i can still get the full 10.5mm horizontal coverage if i use something like 7D2 with 17mm TSE? Thanks.

eDIT: or word it differently, if i use 7D2 with 17TSE and shift to the most far left (or right), will i see the black edge where the image circle ends? Thanks

think about it this way.

Imagine holding the lens still and moving the sensor under it. If you have a FF sensor, the sensor is moved entirely within the huge image circle of the TSE17, at no point does the sensor extend beyond the image circle.

Now relace the FF sensor for a smaller one. It's now further from the edge of the image circle so no, no black edge.. (I don't think the mirror box vignetts from very wide angles, but you can check that)

However you could now need 4 or 5 images to cover the full shift as you have a smaller sensor, and so the difference between crop and FF in terms of FOV will be reduced compared to the normal 1.6:1 ratio.

But the 10mm Wide angles zooms will get you roughly that FOV in one hit.

IMHO the three ways forward are: a 10-something EF-S zoom, the TSE17+FFcamera, or a panoramic tripod head.. which guarentees everything lines up but is much cheaper than the TSE+FF but should yield better resolution than the 10-something zoom.

It all comes down to what you are trying to do.

Also concider renting if you really aren't sure.