Image circle increment wrt to decrease in subject distance

[sweat100]

I got myself a Darlot Petzval 10.5" to use for the 8x10. I understand that the image circle for Petzval design is approximately its focal length, thus 10.5" image circle is not enough for 8x10. Since I am using this lens for portraits mainly, how close should my subject distance be in order to cover the entire full frame 8x10 (without any movements). I calculated that I need the image circle of at least 12.8" to cover the full frame. Is there any formula for this?

[ic-racer]

You can use trigonometry and the 'thin lens equation.' I find it easier to just mount the lens, focus and then check the corners. Do you have clipped corners on your ground glass?

[jb7]

I got myself a Darlot Petzval 10.5" to use for the 8x10. I understand that the image circle for Petzval design is approximately its focal length, thus 10.5" image circle is not enough for 8x10. Since I am using this lens for portraits mainly, how close should my subject distance be in order to cover the entire full frame 8x10 (without any movements). I calculated that I need the image circle of at least 12.8" to cover the full frame. Is there any formula for this?

This is a Voigtlander 11" f/3.5 ish Petzval on 8x10- you might need to get even a bit closer than I should have-

http://farm3.staticflickr.com/2794/4536965857_726c07c762_z.jpg (http://www.flickr.com/photos/joseph-jb7/4536965857/)
Genevieve jb7 Voigtlander2 (http://www.flickr.com/photos/joseph-jb7/4536965857/) by joseph - jb7 (http://www.flickr.com/people/joseph-jb7/), on Flickr

[jumanji]

Hey jb7, did you coat your Voigtlander :cool:?

To the question of the OP: The best solution is testing it out. Each petzval lens, though same FL, has slightly (some cases it's very!) different IC.

Indeed there is a formula. We have: 1/f = 1/d + 1/s (1), with f: FL, d: distance from the GG to the focal plane, s: distance from focal plane to the subject. So at 1:1 the IC will be twice at infinity. Assuming that the IC at inf of your lens is 10.5", same as FL. At 1:1 the IC will be 21", same as the bellows extension. Use uniform formula, the bellows extension needed to have IC 12.8" is same 12.8". Use formula (1) again, the distance from the subject to the focal plane (consider it is the front standard of the camera) will be approx 58" = 4.8 ft = 1.5m. So you should keep your subject closer than 1.5m

Sorry if my explanation sounds stupid :p. My english is bad. I could do it better if I have a paper and a pen and you are next to me.

[sweat100]

Thanks! I followed your guide and got the answer. At least its a rough estimate to get things right. I did not realise that there is a linear relationship for IC and distance from lens to film plane. Guess that was the missing link for me!


Hey jb7, did you coat your Voigtlander :cool:?

To the question of the OP: The best solution is testing it out. Each petzval lens, though same FL, has slightly (some cases it's very!) different IC.

Indeed there is a formula. We have: 1/f = 1/d + 1/s (1), with f: FL, d: distance from the GG to the focal plane, s: distance from focal plane to the subject. So at 1:1 the IC will be twice at infinity. Assuming that the IC at inf of your lens is 10.5", same as FL. At 1:1 the IC will be 21", same as the bellows extension. Use uniform formula, the bellows extension needed to have IC 12.8" is same 12.8". Use formula (1) again, the distance from the subject to the focal plane (consider it is the front standard of the camera) will be approx 58" = 4.8 ft = 1.5m. So you should keep your subject closer than 1.5m

Sorry if my explanation sounds stupid :p. My english is bad. I could do it better if I have a paper and a pen and you are next to me.

[sweat100]

You can use trigonometry and the 'thin lens equation.' I find it easier to just mount the lens, focus and then check the corners. Do you have clipped corners on your ground glass?

Yup, the corners of my ground glass is clipped and I do not understand why is that so.

[Doremus Scudder]

Yup, the corners of my ground glass is clipped and I do not understand why is that so.

So air can easily get in and out of the bellows when opening and closing the camera and, most importantly, so you can focus on the aerial image at the corner and/or check to see if you can see the lens aperture through the cut-out. If you can see all of the aperture when viewing from the corner, you are not vignetting; if you see a thin ellipse, you will likely get some fall-off; nothing at all means dark corners (all this at taking aperture, of course).

Best,

Doremus