Will desaturated colors use less ink?

If I print a full-color document or spreadsheet, will the printer use less ink if the colors used are less intense? If I desaturate the color without actually approaching grayscale, I would assume that the printer will either use less color ink or will start substituting (cheaper) black ink.

If that is indeed the case, is the effect significant and linear with reduced saturation / decreased opacity? Does it depend on the type of printer?


Yes, in most instances, reducing the saturation of a given color (particularly bright colors) will result in less ink/toner being needed to produce the printed piece. The result can be directly linear when using CMYK swatch values on a CMYK printer. That is to say, 50% opacity of solid (100%) yellow, or a screen of 50% yellow at 100% opacity will result in half of the ink coverage, thus half of the ink used.
However, many printers have automatic desaturation features that kick in when the ink coverage exceeds a specific amount. For example, rich blacks like registration black, which is defined as 100% each of CMYK, will often be reduced to save resources as well as to prevent the ink from rubbing or flaking off.
Therefore, desaturating registration black will not necessarily result in a direct linear savings in ink, since some savings were being employed by default. Aside from that, desaturating shades of gray will not likely make much of a difference unless you redefine the swatches to exclude CMY and only use black ink, which is usually cheaper than colored ink or toner.

To be clear, this applies to medium to light colors. Dark or strong colors, on the other hand, may have the opposite effect, especially if you are working in RGB as your source gamut.
For example, full RGB red (255,0,0), if desaturated to grayscale (77,77,77), then converted to CMYK will result in C:65, M: 58, Y:57, K:37, which is 216% ink coverage. That same red, if printed “as-is” in CMYK would be C: 0, M: 100, Y: 100, K: 0. In that case, full red uses 16% less ink than its desaturated analog.

Source : Link , Question Author : Lilienthal , Answer Author : Kromster

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