Hey,
I've tried to think of how to explain this, but I can't with any degree of sophistication, so here goes.
Because the 5D is FF, does it give a consistent and measurable difference in aperture?
I.E. Because it is 60% bigger than the S35 sensor, can we take the f stop and divide it 1.6 to get the full frame equivalent f-stop?
Say I take my newly ordered 24-70 2.8 = because of the FF sensor, does that 2.8 give an equivalent aperture, and so dof, of 2.8/1.6 = 1.75, lets say 1.8?
Or does it still remain a 2.8, but the sensor perceives that 2.8 image in a different way.
Am I explaining myself here because I think I feel I'm not properly enunciating, so let me try a real world example.
If I put, say, my 50mm f1.4 on a 7D and on a 5D and make all settings equal - how many stops worth brighter would the 5D image be because it's a FF sensor and does this brightness bring with it an equivalent DoF relationship?
To use my maths above - b/c the FF sensor is 60% bigger, does that 50mm 1.4 actually have the brightness of a 1.4/1.6 = f0.9 ish lens with the same shallow depth of field of a 0.9 lens.
Or is it the other way round - are all these settings based on FF cameras and everything with a crop suffers a loss in light accordingly - so a 7D with it's 1.6 crop from full frame suffers a 1.6 times light loss - so a 2.8 becomes 2.8*1.6 = f4.5?
Or am I just getting this all wrong?
Sorry if this is confusing, I'm trying to understand this from a technical perspective and not coming up with an answer! Thanks!
I've tried to think of how to explain this, but I can't with any degree of sophistication, so here goes.
Because the 5D is FF, does it give a consistent and measurable difference in aperture?
I.E. Because it is 60% bigger than the S35 sensor, can we take the f stop and divide it 1.6 to get the full frame equivalent f-stop?
Say I take my newly ordered 24-70 2.8 = because of the FF sensor, does that 2.8 give an equivalent aperture, and so dof, of 2.8/1.6 = 1.75, lets say 1.8?
Or does it still remain a 2.8, but the sensor perceives that 2.8 image in a different way.
Am I explaining myself here because I think I feel I'm not properly enunciating, so let me try a real world example.
If I put, say, my 50mm f1.4 on a 7D and on a 5D and make all settings equal - how many stops worth brighter would the 5D image be because it's a FF sensor and does this brightness bring with it an equivalent DoF relationship?
To use my maths above - b/c the FF sensor is 60% bigger, does that 50mm 1.4 actually have the brightness of a 1.4/1.6 = f0.9 ish lens with the same shallow depth of field of a 0.9 lens.
Or is it the other way round - are all these settings based on FF cameras and everything with a crop suffers a loss in light accordingly - so a 7D with it's 1.6 crop from full frame suffers a 1.6 times light loss - so a 2.8 becomes 2.8*1.6 = f4.5?
Or am I just getting this all wrong?
Sorry if this is confusing, I'm trying to understand this from a technical perspective and not coming up with an answer! Thanks!
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