7

In my document I'm using the colour lightgray which looks nice on screen, but is barely decipherable when printed (with a BW-laser-printer). Using the xcolor pkg:

  • Would it make sense to try and convert lightgray with \convertcolorspec to a cmyk equivalent for more legible printing results?
  • Would this necessitate choosing a darker colour, or could a cmyk 1:1 equiv. of lightgray be chosen that looks exactly like the original lightgray on screen ?
  • If so, how to code it?

A pretty plain MWE:

\documentclass{article}

\usepackage{lipsum}

\usepackage{xcolor}

\begin{document}


\lipsum[6]

{\textcolor{lightgray} {\lipsum[3]}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%% Some more or less random snippets from the xcolor manual %%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \extractcolorspec{lightgray}\tmpa
% \expandafter\convertcolorspec\tmpa{cmyk}\tmpb
% \expandafter\convertcolorspec\tmpa{RGB}\tmpc

% {\textcolor{\extractcolorspec{lightgray}{\tmpa\expandafter\convertcolorspec\tmpa{cmyk}\tmpb}}{\lipsum[2]}}

% \convertcolorspec

%%%%%%%%%%%%%%%%%%%%%%%%%
% Colors via svgnames option

% LightSlateGray
% LightSlateGrey

% Silver
%%%%%%%%%%%%%%%%%%%%%%%%%

% - enhanced color definition syntax to allow for target-model specific color parameters, e.g., \definecolor {red}{rgb/cmyk}{1,0,0/0,1,1,0}, facilitating the usage of tailor-made colors both for displays and printers; 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\end{document}

See also here.

closed as off-topic by Joseph Wright Aug 4 '14 at 12:04

  • This question does not fall within the scope of TeX, LaTeX or related typesetting systems as defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

Browse other questions tagged or ask your own question.