Forum: Pentax DSLR Discussion
05-23-2009, 06:44 PM
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So, not that I actually care about the tiny fraction of a difference we're talking about here, but I just noticed that in this wikipedia entry APS-C - Wikipedia, the free encyclopedia
the K20D is listed as having a crop factor of 1.54× rather than the 1.53× listed for all of the other Pentax dSLRs.
Sure enough, the specs give the K20D as having a
23.4×15.6mm sensor rather than the older 23.5×15.7. So, that gives 1.5385× instead of 1.5319× horizontally.
But then, there's the issue of _effective_ pixels — every dSLR uses some of the edge of the sensor for determining black level, demosaicing the edges, etc. So, I wonder if despite the fractionally smaller sensor chip itself the actual imaging area is identical. I notice that the K10D uses 10.2m of 10.75m actual pixels (5.39% excess) whereas the K20D uses 14.5m of 15.1m (4.14% in excess). Since the fraction is somewhat lower, I speculate that the actual image area is virtually identical.
In fact, assuming that the above percentage of pixels is removed from each, that gives the actual image size for both as about 22.9×15.3 — or a "real" crop factor of 1.57×.
(I put "real" in quotes because typical prints crop the edge of 35mm full-frame anyway, and because surely the same issue affects full-frame dSLRs, which also have approximately 5% more photosites than effective pixels.)
It's all really pedantic and doesn't seem even slightly important. Still, I'm curious. Anyone know the answer?
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