Given a 600mm lens and a distance of 2040 mm, how much magnification do we get ?
M = F / (D-F)
where M = magnification ratio, D = distance, F = focal length
F = 600mm
D = 2040mm
M = 600 / (2040-600)
M= 600 / 1440
Magnification = 41%
Given 11x14 inch film, your subject size will be 27 by 34 inches in size: head and shoulders if the camera can shoot in vertical orientation.
Quick question. I've read this statement a number of times, and it confuses me. Assuming we have a 12" and a 16" lens, both giving a life-size repro at 1:1. For the 12" lens we'd need 24" of bellows draw, and for the 16", we need 32". Using the formula for bellows extension compensation - (bellows draw / focal length)^2 , both of these lenses focused at 1:1 would have the same compensation factor (4), right? Therefore, there should be no difference with a wider lens in regard to the amount of compensation, at the same magnification ratio.Originally Posted by Monty McCutchen
I dunno about 11X14, but I gotta take exception to what John said. The 450 Fuji has huge coverage and absolutely nothing
is "soft" off-center on 8x10 film size. Such results might be from film sagging in the holder. It sure ain't the lens' fault! It's
a spectacularly sharp lens clear across the field of coverage. The 600 is nearly as good, just has a tad more vibration risk due
to the bigger shutter. Both are utter overkill in terms of detail rendition. But it would seem to my way of seeing 600 would be better suited to portrait use than 450 on 11X14, namely, it would get you in closer. I think "normal" focal lengths are way too wide for this kind of application. Might be a little harder to focus a long lens if your arms are short however!
If I plug in numbers for a 35mm camera, an 85mm lens gets you about the same magnification. That doesn't seem right to me.
Looking at some of my negs... It might have been the varying distances of the buildings I was shooting in the urban landscapes that caused this effect.
As a quick rule of thumb, 450mm is normal length on 11x14, and 600mm is portrait length: 3/2 x normal. On a 35mm camera, 3/2 normal length is 75mm - but that's a crude approximation.
We need to be careful when comparing formats of different aspect ratio. See Rui Salgueiro's field of view calculator. I always consider horizontal rather than diagonal measurement.
Christopher's calculation of 2040 mm distance is correct.
1/F = 1/D + 1/B
where F = focal length, D = distance, B = Bellows Draw
F = 600mm, B = 850mm
1/600 = 1/D + 1/850
1/D = 1/600 - 1/850
1/D = 0.0017 - 0.0012
1/D = 0.0005
D = 2000mm, which is exactly what Christopher got, with greater precision than my calculation.
You got the wrong guy!! My Literature degree precludes me from answering this! I'm mathematically hopeless. All I know for my wet plate images when I use that 12 inch lens and the necessary two feet of bellows draw and then switch to longer lenses the Dallmeyer 3A for example with the same f-stop my exposure are longer. I look forward to hearing the math from others but that has been my experiential knowledge base that I mentioned above.
Monty
Thanks for checking me. It's been a while since I did any amount of optical math.
These calculations won't work well for 35mm type lenses. These equations are for "thin lenses" which have the front and the rear node in the same place. In reality, this is almost never the case, though. View camera lenses aren't usually that far from a thin lens but a 35mm type lens is, as are tele design lenses for view cameras. Since 35mm cameras don't usually use a bellows to focus and since the flange focal depth must always be the same, they are designed so the lens nodes are in strange places. The formula should tell you the right numbers but interpreting them can be very difficult.
Bryan, you're right. If you're shooting identical magnification with two different focal lengths, the bellows compensation will be the same. The likelihood of the longer lens simply being slower is pretty good, though.
-Chris
Okay, that's what I thought. I've seen it stated that larger formats inherently have more bellows compensation by virtue of the longer lenses, which seems completely wrong to me. Assuming, as Ken mentioned, a standard prime lens, non-tele design, for the given Field of View and magnification amount (which will be the same regardless of 4x5 or 20x24, using equivalent lenses), bellows compensation should stay constant.
This concept of longer extension changing the FoV or whatever might be the cause of this statement??
That one is true. Imagine a normal lens focused at infinity. It will be throwing something like a 45 degree cone of light out the back and onto the film. As you focus closer, the lens is still throwing that same angular cone of light (we know this because a lens gives greater room for movements as we focus closer than infinity) but the film, being further away, is using less and less of that cone. Just picture that the angle between the edges of the film and the lens in back of the lens is roughly the same as the angle of view in front of the lens.
-Chris
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