## Abstract

A digital computer was programed to calculate the concentrations required to produce a given transparent color by mixing soluble dyes. The computation is based on Beer’s law calculations at 65 wavelengths across the visible spectrum. The color to be formulated is specified in terms of CIE tristimulus values derived from instrumental measurement. The computer formulation technique was tested by making up to computed formulas a series of mixtures of dyes in a solvent and in an acrylic resin. The correspondence was good between the measured colors of the mixtures and those for which the formulations were calculated, as indicated by color differences averaging about 1 NBS unit (Adams chromatic value formula, normalized as in ASTM Method D 1482-57T).

© 1960 Optical Society of America

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### Equations (4)

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(1)
$${T}_{\mathrm{\lambda}}=\text{exp}({c}_{1}{D}_{1\mathrm{\lambda}}+{c}_{2}{D}_{2\mathrm{\lambda}}+{c}_{3}{D}_{3\mathrm{\lambda}}),$$
(2)
$$X=\sum _{\mathrm{\lambda}=1}^{65}{T}_{\mathrm{\lambda}}{E}_{\mathrm{\lambda}}{\overline{x}}_{\mathrm{\lambda}},$$
(3)
$$\begin{array}{l}\frac{\partial X}{\partial {c}_{i}}=\sum _{\mathrm{\lambda}=1}^{65}{E}_{\mathrm{\lambda}}{\overline{x}}_{\mathrm{\lambda}}{D}_{i\mathrm{\lambda}}\hspace{0.17em}\text{exp}({c}_{1}{D}_{1\mathrm{\lambda}}+{c}_{2}{D}_{2\mathrm{\lambda}}+{c}_{3}{D}_{3\mathrm{\lambda}}),\\ \frac{\partial Y}{\partial {c}_{i}}=\sum _{\mathrm{\lambda}=1}^{65}{E}_{\mathrm{\lambda}}{\overline{y}}_{\mathrm{\lambda}}{D}_{i\mathrm{\lambda}}\hspace{0.17em}\text{exp}({c}_{1}{D}_{1\mathrm{\lambda}}+{c}_{2}{D}_{2\mathrm{\lambda}}+{c}_{3}{D}_{3\mathrm{\lambda}}),\\ \frac{\partial Z}{\partial {c}_{i}}=\sum _{\mathrm{\lambda}=1}^{65}{E}_{\mathrm{\lambda}}{\overline{z}}_{\mathrm{\lambda}}{D}_{i\mathrm{\lambda}}\hspace{0.17em}\text{exp}({c}_{1}{D}_{1\mathrm{\lambda}}+{c}_{2}{D}_{2\mathrm{\lambda}}+{c}_{3}{D}_{3\mathrm{\lambda}}),\end{array}$$
(4)
$$\begin{array}{ll}\hfill \mathrm{\Delta}X& =\partial X/\partial {c}_{1}\mathrm{\Delta}{c}_{1}+\partial X/\partial {c}_{2}\mathrm{\Delta}{c}_{2}+\partial X/\partial {c}_{3}\mathrm{\Delta}{c}_{3},\\ \hfill \mathrm{\Delta}Y& =\partial Y/\partial {c}_{1}\mathrm{\Delta}{c}_{1}+\partial Y/\partial {c}_{2}\mathrm{\Delta}{c}_{2}+\partial Y/\partial {c}_{3}\mathrm{\Delta}{c}_{3},\\ \hfill \mathrm{\Delta}Z& =\partial Z/\partial {c}_{1}\mathrm{\Delta}{c}_{1}+\partial Z/\partial {c}_{2}\mathrm{\Delta}{c}_{2}+\partial Z/\partial {c}_{3}\mathrm{\Delta}{c}_{3}.\end{array}$$