« on: August 28, 2014, 10:18:07 PM »
Here is a way of calculating the effective extra reach or resolving power of a crop body versus FF, which will amuse the geeks among us.
Measure the MTF of a lens on the crop (= MTFcrop) and the same lens on the FF (= MTFff). The ratio of the MTFs, MTFcrop/MTFff, gives the relative resolving power of the bodies with that lens. However, the crop body can be placed 1.6x further away to give the same field of view. Therefore, the true effective relative resolving power, R, is given by:
R = 1.6x MTFcrop/MTFff.
Photozone lists measured MTFs for a set of lenses on the 5DII and 50D. I calculated their ratios for the Canon 200mm f/2.8 II, 85mm f/1.2 II and 35mm f/2 at wide apertures below the DLA. MTFcrop/MTFff is very close to 0.726 in all cases.
This gives R for 50D/5DII = 1.16.
So the effective extra reach is 16%. (Based on the ratio of their pixel sizes, a value of 36% is expected.
The dpreview widget gives values for the 5DIII and 7D only for a few lenses. I did the same calculations with the Tamron 150-600mm (between 150-400mm), the Canon 200-400mm and the Sigma 50mm f/1.4 A at wider apertures below the DLA. In all cases, MTFcrop/MTFff is close to 0.742.
This gives R for 7D/5DIII = 1.19.
So, the effective extra reach is 19%. (Based on the ratio of their pixel sizes, a value of 45% is expected).
There are always arguments about using MTFs quantitatively, but I think in this particular calculation it is reasonably valid to use them. It fits in reasonably well with experience - Jon has shown there is better resolving power in photos of the moon with the 7D, but it doesn't look 45% better. And my own experience is that the 7D and 70D aren't much better than the 5DIII, certainly not 1.6x.
I assume you mean you're dividing the line widths per picture height (LW/PH) at the center when you say you're dividing the MTFs. If so, how do you figure that you can divide the APS-C by FF LW/PH values when the max LW/PH stated in the charts are different for the two sensors?