Why no FF equiv. to EF-S 17-55/2.8?

Started Jan 11, 2013 | Discussions thread
David Naylor
Contributing MemberPosts: 602Gear list
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Re: On "aperture"
In reply to schmegg, Jan 12, 2013

I wake up this morning (in Sweden) and find that this whole thread has gone the way of the dodo.

As usual, semantics and different mental models has caused an endless debate where no-one understands anything anyone is trying to say. Thanks to those of you who take the time to repeatedly explain things clearly and precisely.

Anyway, I'd like to try to summarize for my own understanding. My example is based on the idea that I would switch lenses when switching to FF, in order to get the same angle of view.

Is the following correct?

Given the same print/output size, same angle of view (say 50 mm on crop and 80mm on FF) and the same f-ratio (f/2.8 on both), same shutterspeed and same ISO setting; FF will have a 2.56x sensor size advantage over crop, thereby having a 2.56x total light advantage. This means FF will have a theoretical signal to noise advantage of 1.36 stops. The image taken on FF will have shallower depth of field (equal to shooting 50mm @ f/1.75 on crop).

What I don't understand is how image resolution relates to signal-to-noise ratio. For instance, if I compare the 5D III and the 7D, does the 5D have *both* a 1.36 stop total light advantage as well as x amount more image resolution? In other words, could I increase ISO by 1.36 stops on the 5D and still have more image detail than from the 7D?

 David Naylor's gear list:David Naylor's gear list
Canon EOS 7D Canon EF-S 17-55mm f/2.8 IS USM Canon EF 100mm f/2.8 Macro USM Tokina AT-X Pro 11-16mm f/2.8 DX Sigma 50-150mm F2.8 EX DC APO OS HSM +1 more
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