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I just purchased the Canon MG8220 all-in-one photo printer. To get the most accuracy with my photo prints, do I need to calibrate my monitor to match the printer? I am running OSX Lion on a 2011 iMac 27", I shoot RAW with a Nikon D90 mostly using my 17-55/ 2.8 and using Aperture for color correction.

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Actually, in theory, no. Calibration of the two devices is independent. If you calibrate just the printer, perfectly-adjusted images may look wrong on the monitor but will print just fine. You could use images corrected by someone else somewhere (which you download), or you could calibrate "by wire", just looking at histograms and numeric values rather than trusting your eyes, probably in combination with making test prints.

This may sound a bit silly (because, in practice, you really want to calibrate your monitor too), but is actually an important concept to understand in any case, because the range of colors and the degree of contrast which can be displayed on monitors and which can be printed with various inks and various papers differs significantly. Your monitor can show colors which can't be printed, and you can print colors which can't be displayed on screen.

There's a nice 3D diagram of overlap between an Apple LCD and an Epson printer in this article by Keith Cooper, which also covers this whole topic in more depth.

Color management can help you visualize what the results will be, but if it's critical, nothing beats making test prints. If you have a higher-end wide-gamut monitor, you will be able to display almost everything that can be printed with most inks on most paper. Then, the challenge becomes making sure that the colors outside of that space aren't used — or at least, aren't critical. When you go to make your print, these will be "clipped" to in-gamut colors; that is, they'll be automatically reassigned to something the printer software determines to be close enough.

An alternative is to just work in sRGB, which is a constrained color space designed to both display well-enough and print well-enough on most equipment. You lose out on anything that can't be represented by either — and on a little more just to be safe. That sounds bad, but it's actually a big enough space that if absolutely perfect color isn't top on your list of important factors, you might not even notice.

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  • \$\begingroup\$ Nice and informative! :) \$\endgroup\$
    – R4D4
    Commented Dec 18, 2011 at 16:04
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You should calibrate both monitor and printer to match the color space you're using. If they'd only match each other, your photos would still look wrong when you use some other device to output them (such as printing service or sharing on Internet).

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    \$\begingroup\$ I don't think this is quite right. When you calibrate a device, you build a mapping of its own peculiar and unique rendering of color — its native color space. Having done that, color-aware software/libraries can map to and from any other color space — either that of another device, or standard color spaces like Adobe RGB. These standard spaces aren't (usually) used as native spaces – instead, you save a file in that space, and when it's displayed on a different color-managed device, the software on that side knows where to start from (and then maps from that to the "new" native device). \$\endgroup\$
    – mattdm
    Commented Dec 18, 2011 at 15:57
  • \$\begingroup\$ I don't even know of a good way to calibrate a monitor and printer together and not to a standard — I guess by eye? \$\endgroup\$
    – mattdm
    Commented Dec 18, 2011 at 16:48
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YES! - I used the ColorMunki system.

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    \$\begingroup\$ Perhaps you could describe your experience with ColorMunki in a bit more depth? \$\endgroup\$
    – Imre
    Commented Dec 18, 2011 at 12:20
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    \$\begingroup\$ ColorMunki Photo does two independent calibrations: one for the display device, and one for the printer. They profiles are not interrelated in any way (except, perhaps, by any quirky inaccuracies of the ColorMunki photospectrometer unit you own). End-to-end results match well because both profiles are accurate, not because they are matched to each other. \$\endgroup\$
    – user2719
    Commented Dec 19, 2011 at 13:19

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