D70 resolution: 3008x2000 - what's the 8 all about

Discussion in 'Digital SLR' started by paul, Jan 1, 2005.

  1. paul

    John Francis Guest

    The full sensor imaging area is 3040x2024 (6,152,960 pixels).
    John Francis, Jan 2, 2005
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  2. Alan Browne wrote:
    If you save it as JPEG that will normally produce an extra loss compared
    to the original image. However, if you have a JPEG which is exactly a
    multiple of 16 in its X and Y dimension it can be losslessly rotated by 90

    David J Taylor, Jan 2, 2005
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  3. paul

    John Francis Guest

    Yes you have. If you don't save it as a JPG, you've lost the space savings.
    If you do re-save it as a JPG, you're going to re-run the compression algorithm.
    That's not being run on the same (original, uncompressed) image data that was
    the input to the first JPG compression, so it's extremely unlikely to come up
    with the same final result. That means you've lost some more image quality.
    The same holds true even if you don't rotate the image; you can still end up
    with a loss of quality just by re-saving a JPG image with no editing changes.
    That's why some image editing software checks for this case, and will simply
    copy the input file to the output in that situation.
    John Francis, Jan 2, 2005
  4. paul

    Alan Browne Guest

    I think I see what you're saying, though I don't know why truncation should
    affect the lossiness as the image is edited as an image not as a JPG. (JPG
    compression is a function of storage, not editing).

    I also assume that at its least lossy setting JPG is lossless (which is not
    quite true, but close enough).

    In terms of a RAW image it's a non-issue.

    So, I believe the reason the engineers chose 3008:2000 is simply for the usual
    computer design reasons, not image related reasons. This also follows with the
    other saveable image sizes Maxxum 7d/Nikon d70: (2256x1496, 2240x1488,
    1504x1000) ... 1496 and 1000 do not divide by 16, but all the dimensions
    mentioned divide by 8.

    Alan Browne, Jan 2, 2005
  5. paul

    Jeremy Nixon Guest

    It matters for JPEGs, not for images in general. JPEGs can only be
    rotated losslessly (without decompressing and then recompressing the
    data) if the dimensions are divisible by 16.
    Jeremy Nixon, Jan 2, 2005
  6. paul

    Jeremy Nixon Guest

    Yes, you have, if you saved it as a JPEG. You have re-compressed the image.
    If you load a JPEG image, do something to it, and then save it again as a
    JPEG, that is a lossy operation due to the compression.

    However, a JPEG image with dimensions divisible by 16 can be rotated without
    decompressing and recompressing the image at all -- that is, without ever
    loading it into an image editor in the first place, requiring no further
    step of re-saving as a JPEG and thus recompressing the data.
    Jeremy Nixon, Jan 2, 2005
  7. paul

    Alan Browne Guest

    I'm beginning to understand that through the force of people saying so and not
    because I've looked at the mechanics.

    You bring up 'uncompressing and recompressing' and that really is it... a JPG
    image is only such when stored. When it is 'in the camera' prior to storage or
    in photoshop it is not JPG... it is 'image'. So it can be rotated without loss
    regardless of dimensions. It's all semantics, I suppose.

    Alan Browne, Jan 2, 2005
  8. paul

    John Bean Guest

    Lossless rotation is done on the JPEG data itself, not on a decoded image.
    The JPEG data is not decoded/rotated/re-encoded as it must be for any JPEG
    whose dimensions are not multiples of 16. That's why it's lossless.
    Not so. As I said earlier the Pentax *istDS uses 3008x2008 for raw images
    but 3008x2000 for JPEG, for the reasons I've stated. This was a change from
    the older *istD, which used 3008x2008 for both - much to the annoyance of
    some users.
    John Bean, Jan 2, 2005
  9. paul

    Jeremy Nixon Guest

    The normal case is that someone takes a bunch of pictures, loads them onto his
    computer, looks at a screen full of thumbnails, and clicks a "rotate" button
    to make the vertical compositions vertical. If the saved images are JPEGs,
    this can be done with no recompression loss if the image dimensions are
    divisible by 16.

    If you don't use JPEG, this doesn't matter to you.
    Jeremy Nixon, Jan 2, 2005
  10. paul

    C J Campbell Guest

    There are actually 6.24 million pixels on the sensor, of which 3008x2000 are
    used in the max size JPG or in raw NEF files, which Nikon rounds to 6.1
    million pixels.
    C J Campbell, Jan 2, 2005
  11. paul

    Will D. Guest

    Anyone ask to see Roland's badge number?


    Will D.
    Will D., Jan 3, 2005
  12. Yes, I read that somewhere myself, but I still don't see how 3008 x 2000 =
    6.016 million pixels "rounds" to 6.1 million pixels. Seems to me it
    rounds to 6.0 million pixels.

    Can you clarify further?


    Merritt Mullen, Jan 3, 2005
    Merritt Mullen, Jan 3, 2005
  14. paul

    C J Campbell Guest

    Nope. No doubt some lawyer will sue the camera companies for false
    advertising, claiming to represent all us consumers as a class. As part of
    the settlement, he will be paid millions of dollars, and we will get a
    coupon for $10 off on our next purchase of a camera. It will be hailed as "a
    great victory for consumers." At least, that is what happened to the
    manufacturers of disk drives and computer screens.
    C J Campbell, Jan 3, 2005
  15. Hehe ...

    Roland Karlsson, Jan 3, 2005
  16. paul

    Ken Tough Guest

    You're right, it's because of compressing, but no you don't understand
    it; it has nothing to do with semantics.

    I'll have a stab at explaining it:

    The original image comes out of the camera in JPG. That is lossy
    compression with effects; it has artifacts. For illustration, let's
    say that horizontal high-contrast areas cause a shadow in the lighter

    You open that image (uncompressing it) in your viewer, and edit the
    pixels. What you're looking at has loss and artifacts. You do a
    pixel-for-pixel rotation (also duly rotating the artifacts). The
    result in your editing window, as you point out, is exactly the
    same as the original rotated.

    Now, you save that edited image as JPG. The saving process compresses,
    and introduces new artifacts due to the new horizontal high-contrast
    areas (as well as the previous, rotated artifacts of course). This
    compression loses more info. When you open the newly formed JPG,
    it no longer looks exactly like the original, rotated. It's a lossless
    rotation only so long as you don't save it in JPG.

    On the other hand, lossless rotation is a mathematical operation on
    the original JPG, effectively transforming it so that it uncompresses
    into a rotated version. There is no additional loss, no additional

    Hope that helps?
    Ken Tough, Jan 10, 2005
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