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07/28/2005 01:05:49 PM · #1 |
I'm going over some lens decisions in my mind and was looking at the MTF charts for various Canon lenses. My question is this. Should I ignore the portion of the chart past a certain mm from the center of the lens? (ie. past 15mm on the x axis, or so) The digital sensor in the 300D is, of course, smaller than 35mm film so my sensor should never see the true "edge" of the lens right?
So at what point on the X-axis does a MTF chart become irrelevant for a digital SLR (with the 300D specifically in mind)? |
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07/29/2005 03:54:29 AM · #2 |
Past approx 13.6mm is irrelevant. The MTF charts for the EF-S Canon lenses stop at 13mm because of the sensor size.
Good old Pythagoras...
The sensors are 22.7mm x 15.1mm in a 300D. Therefore the horizontal distance from centre to edge is 11.35mm and vertical is 7.55mm. Square both of them and add them together. Then square root the result. You get 13.63mm so it's 13.63mm from the centre of the CMOS to the corner.
Hope that's a help - if not, then the answer's essentially 13.6mm!
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07/29/2005 05:15:03 AM · #3 |
Richard's math is correct; he's just missing the "divide by two" after taking the square root :-)
Remember that good lenses are likely to far outlast your camera body; consider your lenses a long-term investment. If you feel you may ever move up to a camera with a larger sensor, consider that in the purchase decision. Also, remember that Canon's posted MTF graphs are theoretical, based on a software model of the lens design, and not on testing of an actual unit. As such, they represent what the lens *can* achieve, not necessarily what it *will* achieve. That said, I find them useful and a decent indicator of performance, though they really only tell you about sharpness and contrast.
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07/29/2005 05:27:29 AM · #4 |
PhotoDo, although no longer updated, wraps up real-world MTF results into a single score, which is very useful in evaluating the quality of older lenses, and extrapolating for newer lenses.
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07/29/2005 06:49:46 AM · #5 |
Originally posted by kirbic:
Richard's math is correct; he's just missing the "divide by two" after taking the square root :-) |
No, I divided the sensor by two at the very beginning to describe the horizontal and vertical distances from the centre of the sensor to the edges. Thus basically constructing my Pythagoras triangle within a quarter of the sensor.
GCSE's were useful for something - hoorah! |
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