Re: CMYKOG profiles in PosterPrint RIP?
Re: CMYKOG profiles in PosterPrint RIP?
- Subject: Re: CMYKOG profiles in PosterPrint RIP?
- From: Robert L Krawitz <email@hidden>
- Date: Sun, 4 Jan 2004 18:51:32 -0500
From: email@hidden
Date: Sun, 4 Jan 2004 18:05:48 EST
In a message dated 1/4/04 1:44:03 PM, email@hidden writes:
> Does anyone have experience how to accurately profile
> CMYK+Orange+Green inks for large format inkjet printing specially
> with the Ergosoft PosterPrint RIP ColorProf-Module?
I did quite a bit of testing with this combination a while
back... it works, but for most of the uses I was testing did not
add nearly as much gamut to the photo color range as it did to the
vector color (spot color) range. For six color printers the loss of
light Cyan and Magenta for the addition of Orange and Green cost
more in image quality then it added in gamut. With current finer
dot sizes, or with eight color and higher devices, this would be
less of an issue. It also complicates the profiling process
considerably.
This is going to be a mainstream issue in the near future (and already
is in Japan). The Epson Stylus Photo R800 adds blue and red inks to
the CMYK; rather than using light cyan and magenta inks, it uses 1.5
pl droplets; it's supposed to be out in February. This is roughly
equivalent to a 4.5 pl 6-color printer, except that you could use a
lot more black ink.
I'm thinking about how to handle this in Gimp-Print. The option that
seems most obvious to me is to offer a choice between conventional
4-color and CMYKRB, and if the user selects CMYKRB, to perform an
operation analogous to GCR to generate red and blue if there's a large
component of blue (C+M) or red (M+Y) left over, after subtracting out
virtual (CMY) black (real black is removed from CMY in the GCR
process).
So for example if the CMYK components are:
C = .8
M = .7
Y = .2
K = .1
I would subtract the virtual black -- min(C, M, Y) -- from the CMY
components to yield
C = .6
M = .5
Y = 0
K is ignored
This leaves a blue component of .5 and a red component of 0 (note that
with this procedure there can never be both a blue and a red
component; that would be magenta). After determining the B or R
components, of course, the virtual black has to be added back in.
In principle, we could then have the following separation
C = .3
M = .2
Y = .2
K = .1
R = .0
B = .5
In practice, the "blue component reduction" would be a curve of some
kind, and would be only partial, so I might wind up with
C = .5
M = .4
Y = .2
K = .1
R = .0
B = .3
--
Robert Krawitz <email@hidden>
Tall Clubs International --
http://www.tall.org/ or 1-888-IM-TALL-2
Member of the League for Programming Freedom -- mail email@hidden
Project lead for Gimp Print --
http://gimp-print.sourceforge.net
"Linux doesn't dictate how I work, I dictate how Linux works."
--Eric Crampton
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