antihero
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taken from
http://photo.net/bboard/q-and-a-fetch-msg?msg_id=007YEP
Peter Groen, Feb 29, 2004; 01:39 p.m.
Ofcourse it is YES. Think logically people. The shaking of the camera is of course amplified because of the multiplier factor.
As an example: let's assume you make a picture using a Canon 1Ds and a Canon 10D (1.6 multiplier), both with the same lens:
You shake both camera's 1 mm up and down.
For the 1Ds, this means that the picture is being shaken for 1mm of the 23.8 mm of the sensor's height (sensor is 35.8 x 23.8 mm). That means that the picture is being blurred for 1/23.8th part of it's total height: 4.2%
For the 10D, this means that the picture is being shaken for 1mm of the 15.1 mm of the sensor's height (sensor is 22.7 mm x 15.1 mm). That means that the picture is being blurred for 1/15.1th part of it's total height: 6.6%
Ok, which picture will be 'blurrier'? That of the 10D of course, and exactly 1.6 times blurrier than the picture taken with the 1Ds.
So, yes, you should multiply the 1/focal length rule with the multiplier for digital camera's...
well....using (my) logical thinking, the crop factor is exactly that. It CROPS the viewed image, it doesn't magnify it. The fact that the image is cropped only gives the effect of it being magnified.
So if you took a blurred shot on a full frame camera, would you not in fact see LESS blur if the image had a x1.6 crop (only because part of the image has been cropped out)? Or have I completely got the wrong end of the stick?