Thread: S9000
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Old Oct 11, 2005, 6:04 AM   #476
proton
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read this interesting theory, it explains why RAW is 18mp and JPG is 9mp on the S9500! (source:Fuji forum on dpreview)

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In JPEG the camera delivers 9 million pixels. In RAW, it's 17,743,872 pixels (17,7MP). Why this difference? A photosite is not a pixel.

There are 9 million fotosites on the SUPER CCD. There are organized diagonally, which gives an advantage (the fotosites can be larger and thus more sensitive, which leads to a higer ISO capabilities). The downside is: some kind of a translation has to be made to make the pixels fall into a vertical/horizontal organized grid. So the pixels in between the diagonal photosites are interpolated (from the info of their neighbours). Saving a RAW file from the s9500 saves this picture. Saving a picture in jpeg saves it in a recalculated 9 MP

normal CCD grid:
O O O O O O O O
O O O O O O O O
O O O O O O O O
O O O O O O O O

total: 32 photosites = 32 pixels

super ccd grid:
O O O O O O O O
O O O O O O O O
O O O O O O O O
O O O O O O O O

total: 32 more sensitive photosites

Raw interpolated pixels are "+", photo sites are "o"
o+o+o+o+o+o+o+o+
+o+o+o+o+o+o+o+o
o+o+o+o+o+o+o+o+
+o+o+o+o+o+o+o+O

total: 32 more sensitive photosites= 64 pixels on the same surface as the 32 pixels of a normal ccd

beware: a photosite is the fysical spot on a CCD that captures light. A pixel is the smallest part of an image. In cameras an "effective pixel" equals a photosite. But the 9mp of the s9500 are recalculated from a 17,7MP Raw file.

so why not interpolate a normal ccd then? Because the photosites are less sensitive due to their positioning on the CCD. That is exactly why all competitors struggle with noisy cameras.

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more discussion can be found here:

http://forums.dpreview.com/forums/re...hread=15357537

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