Commit 34457f5a authored by Turnhout, M.C. van's avatar Turnhout, M.C. van
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typesetting

parent d06e16b1
......@@ -87,7 +87,7 @@ The consequence so far is that interpreting RGB values in terms of `amounts of v
For starters: twice as much pigment does not mean `twice as high a pixel value', on the contrary. More pigment means more absorption and thus \textsl{less} (transmitted) light. Furthermore, when 10\,\% of the light is absorbed, intensity is 90\,\%, and when 20\,\%, twice as much, light is absorbed intensity is 80\,\%: there is a factor 2, but that is not apparent from (the fraction of) the intensity values.
To investigate the effects of `amounts of pigments' on recorded RGB values, we first have to switch to subtractive colour mixing or `absorption space'. The RGB (nominalized) intensity values for yellow are $y_I = [1 1 0]$. Thus, to produce yellow with a pigment we have remove all light but the blue: the RGB \textsl{absorption} values are $y_A = [0 0 1]$.
To investigate the effects of `amounts of pigments' on recorded RGB values, we first have to switch to subtractive colour mixing or `absorption space'. The RGB (nominalized) intensity values for yellow are $y_I = \begin{bmatrix}1 & 1& 0\end{bmatrix}$. Thus, to produce yellow with a pigment we have remove all light but the blue: the RGB \textsl{absorption} values are $y_A = \begin{bmatrix}0 & 0 & 1\end{bmatrix}$.
In other words: we have to switch from `transmission space' (Intensity, RGB) to `absorption space'.
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