Question du Jour
So Andreas Feininger used a 40 inch telephoto lens with a 4 x 5 inch camera to do many of his famous shots of New York in the 40’s, 50’s.
Here’s a link to the poster I had on my wall for many years of Fifth Avenue Lunchtime.
So if you convert the lens to mm, you get 1016 mm.
Fifth Avenue does rise a bit from where he took this photo, but I think he had a platform to get higher.
My thinking was that a normal lens on a 4×5 camera is 150mm. So 1016 / 150 means this was roughly 7x normal.
So from that, if normal on a cropped sensor is about 40mm, then 7x 40mm = 280mm.
There are a number of ways to get to this number but they wind up about the same.
So the frame is going to be similar with a 280mm lens on a 40D, but will the “sense of compression” be the same?
I think it’s sort of the same question as what happens when you put a 20mm full-frame lens on a cropped sensor. Yes, you multiply it by 1.6 and get 32mm. But the relationship between foreground and background is still the same as with a 20mm lens on a full frame camera. It’s just cropped.
So, according to my theory, to get the same compression as Feininger got, you’d still need a 1000mm lens (more or less).
One of my blog readers is sure that I’m wrong about this. What do you think?
Hopefully I’m wrong.
Comments
Comment from Patrick Cooney
Time: June 10, 2008, 9:22 pm
“Hopefully I’m wrong.” Yes.
Perspective is determined solely by the distance from the lens to the two objects being compared.
Consider two 1.8m models standing nearly in line and facing the lens. If one is 5 m away and the other is 10 m away from the lens, the image of the farther model will be half as high on the film as that of the closer model. Move back from these two models until one is 100 m distant and the other is 105 m distant and the height of the image on the film of the more distant model will be 95% of that of the closer one. That’s perspective.
All that the lens focal length and sensor height determine are the fraction of the frame height occupied by the near model.
Comment from dave
Time: June 11, 2008, 7:15 am
Patrick, not to beat a dead horse, but isn’t the lens effecting the size relationship between the near and far objects. In other words, if the near object is twice the frame size of the far object with a given focal length, won’t that relationship be the same no matter what the crop factor if the distance from the lens to the objects is kept constant?
Put another way. You put a 20mm lens on a full frame camera and the near object is twice as high in the frame as the far object. Won’t you have the exact same proportion when the lens is stuck on a cropped sensor (though the objects will take up more frame size).
That’s the relationship (perspective) that I’m interested in.
Comment from Patrick Cooney
Time: June 11, 2008, 4:33 pm
Dave, first, we should assure our readers that no horses have been injured in the course of this discussion ;.}
The short responses to each sentence in your reply are no, yes, yes, yes, yes. So you are definitely on the right track. The big bad ideas here are “crop factor” and its unmentioned evil twin, “effective focal length”. Let’s see if I can banish these bad actors with a story.
Knowing that you love images of old New York, your friends pool their life saving and buy you the newly invented time machine. So worrisome is this business about focal length and perspective that you grab your cameras and lenses and immediately project yourself back to the very time and spot where Andreas Feininger is using his 40 inch telephoto lens with his 4 x 5 inch camera to make “Fifth Avenue Lunchtime”. You put your 1000 mm focal length lens on your full-frame 35 mm camera and point it right at the same scene. You look in your viewfinder and what do you see?
(please pause here to reflect and answer)
YES! You see exactly the same perspective as Feininger! So you are right and so Cooney is … also right. Huh? How can that be?
Looking again in your viewfinder, you realize that you see only a 1″ wide by 1.5″ high rectangle cut out of the center of Feininger’s image. Bummer! You wanted that great lampost as well as the crowd and the awnings. So what do you do?
With no time to change position, you switch to a shorter focal length lens. But which one? For the same vertical angle of view as Feininger has, you quickly realize that his 40″ lens is 8 times the long dimension of his negative, so you grab your 8 x 36mm = 288 mm lens. Now you have the same vertical angular image size as Feininger AND the same perspective (since you are still the same distance from everything in the scene).
Wow! What a story, and not a mention of “crop factor” or “effective focal length”.
Comment from dave
Time: June 11, 2008, 5:02 pm
Patrick — exactly. And as you say - no horses have been injured during this discussion 
Will let you know when I get the long telephoto — though I’ll probably just rent one for a weekend.
Comment from Jon Maxson
Time: June 10, 2008, 1:35 pm
Try it. Take your two lenses and standing in the same spot take two pictures of something far away and delightfully photogenic. (Stick your pole out the window and down the street?) Load them up and crop them down so they show the same far away area. I have a guess but I really don’t have a clue what the result will be. Lurking and enjoying your work.