PS printing in 8 or 16 bits

Discussion in 'Photoshop' started by 8or16, Aug 18, 2004.

  1. 8or16

    8or16 Guest

    When printing a 16-bit depth file from PS (CS or other versions), is it
    converted to 8-bits first before sending it to a printer (Epson
    8or16, Aug 18, 2004
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  2. 8or16

    Xalinai Guest

    It is.

    It is also reduced to 8bit when you view it on the monitor.

    Xalinai, Aug 18, 2004
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  3. Ink jet printers can only reproduce about 65 distinct shades of a given
    color, not even the 255 that an 8 bit image would indicate. So there is
    no point is sending a 16 bit image to the printer.
    16 bit is useful for doing large changes in contrast or color without
    worrying about posterization.
    Take a look at the tips section of my web site for some relevant
    Robert Feinman, Aug 18, 2004
  4. 8or16

    JPS Guest

    In message <>,
    Perhaps, but many printer dots usually are used to make one image pixel,
    with dithering, so far more than 65 colors are perceived in the print.
    There's no point in converting it to 8 bit, just to print it, either.
    JPS, Aug 20, 2004
  5. 8or16

    Xalinai Guest

    Even with the multitude of dots used for shading, there are no more
    than 65 shades of <inkcolor> possible with current inkjet printers. So
    65^<number of inks> is the number of colors the printer could create if
    overprinting would not result in something near black too.

    More shades require more space - resulting in lower ppi.
    You do not need to convert manually. Conversion to printer compatible
    ranges is done automatically, even if you still work in 16bit mode
    (maybe PS needs a little more memory and time when printing 16bit

    Xalinai, Aug 21, 2004
  6. 8or16

    Chris Cox Guest

    Yes - because there are no APIs for printing more than 8 bits/channel.

    Chris Cox, Aug 22, 2004
  7. 8or16

    Thomas Fors Guest

    Where does this number, 65, come from?

    I can print 256 patches from paper white to full cyan ink on my Epson
    2200, measure them with the eye-one spectro and get distinct XYZ,
    L*a*b* or spectral reflectance values for each patch.

    Is there some deltaE limit on what is considered a distinct shade?

    Thomas Fors, Aug 23, 2004
  8. It comes from info found on Epson's web site. The reason for the limited
    number of values has to do with the size of the square used to create
    the various densities. For an 8x8 square you can get 65 distinct values.
    If you print at 1440 dpi then an 8x8 square reduces the effective
    resolution to 180 dpi, for example. Each of these 8x8 squares represents
    one pixel in the original image in the simplest case.
    Modern printers get more shades by using more colors of inks and some
    also use variable size dots.
    This is the standard tradeoff between smoothness of gradiation and
    resolution that all halftoning printing systems face.
    It's possible that the 2200 by using a light cyan get generate values
    "in between" the standard ones and thus get 128 values. It may also
    use a bigger than 8x8 grid thus allowing more values. A careful
    study of their technical specs should reveal the details.

    Whatever the value is (even 256) it is still much less than what can
    be obtained with 16 bits and thus the advice to shift to 8 bits for
    output still is valid.
    I have some discussions of this and other 16 bit matters on my web
    site. Just follow the tips link on the home page.
    Robert Feinman, Aug 24, 2004
  9. 8or16

    Thomas Fors Guest

    Robert, thank you for explaining that. If you have a link to the info
    on the Epson site, I would appreciate it.

    Given that at 1440dpi, an 8x8 square of ink droplets represents one
    pixel in a 180ppi image. That 8x8 cell gives 64 possible ink droplet
    locations. In a four-color printer, if you count up how many
    combinations of ink droplets can populate that 8x8 area, you get
    864497. This number goes up approximately exponentially as you add
    inks to the printer:

    4 color 864497
    5 color 12103009
    6 color 143218993
    7 color 1473109697

    As soon as you get to the 6 color printer, there are more possible
    combinations of ink droplets possible in that 8x8 square than an 8-bit
    RGB pixel can represent.

    Of course, these numbers assume the perfect printer that makes the
    perfect ink droplets. In reality there are many other factors that
    affect how these ink droplets combine spectrally to create a visible
    color. Mechanical dot gain, optical dot gain, Kubelka-Munk theory.

    I suspect that 65 is a number that is over-simplified and outdated.
    As you point out, variable dot size in the newer printers is one
    factor that will affect this. Also, I believe the newer printers can
    place ink droplets over previously placed ones which changes things

    So, it seems there might be something to gain from newer printers by
    driving them with 16-bit data. In practice, I don't know how much of
    that could be realized. I do know that ImagePrint can drive the 2200
    with 16-bit data. I'll have to run some tests to see if I can notice
    a difference.

    Thomas Fors, Aug 25, 2004
  10. 8or16

    Xalinai Guest

    The number of dot combinations is irrelevant for the number of shades.
    If you use a 8x8 square you can cover from zero to 64 of the squares so
    the intensity of ink in this 8x8 area varies in 65 steps.

    If you take in account, that the size of an ink drop is larger than the
    positioning precision (=dpi value) you understand that if you fill more
    than 50% of the possible positions in the 8x8 square you already cover
    the full area.
    There are 64 ways to place one dot in an 8x8 area but all of them are
    the same shade.
    The same is true for the 64*63 ways to place two dots, and so on.
    65 is the number of possible shades in an 8x8 square and will never
    change. The arrangement of the 32 dots for a 50% fill does not affect
    the fact that it is a 50% fill.
    Variable dot size is irrelevant for resolution as the smallest ink drop
    is still bigger than the (virtual) square it should color.

    Overprinting has been part of the calculation but its effect is not too
    big as mixing colors leads to black
    To resolve for 16 bits you would need a 256x256dot square or get a ppi
    value that is 1/256th of the mechanical dpi.

    Xalinai, Aug 25, 2004
  11. 8or16

    Thomas Fors Guest


    Thank you for taking the time to explain this. I was thinking only in
    terms of combinations of the various colored ink droplets and ignoring
    the fact that the 8x8 cell is used to create an effective halftone
    dot. I believe I understand now.

    Thomas Fors, Aug 26, 2004
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