Commit 847dfc37 by Turnhout, M.C. van

 ... ... @@ -152,13 +152,58 @@ We therefore did not deconvolve with the actual colours (absorptions) in the fla \end{figure} \section{Two colour flag analysis} In this section, we are only interested in deconvolving the yellow and green from the flag (figure \ref{flagJA}) and we will not treat black' as an input colour, or as a colour to be deconvolved (in yellow and green). However, since colour deconvolution in RGB \textsl{has} to be done with \textsl{three} input colours' (matrix $\mat{K}$ needs to be 3 by 3), we'll have come up with an alternative third colour'. \subsection{Adding a third perpendicular' absorption column} Our calculations are done with the $\mat{K}$ matrix with (normalised) absorption columns. For computational reasons it is best when the angles between the three absorption columns in that matrix are all 90\,\degree . To find the angle between two columns, you can take the inverse cosine of the inner dot product' of the two columns (equation \ref{angledot}). So the absorption angle between the flags yellow dye and green dye is \begin{align} \varphi_{yg} & = \acos \left(\colt{\hat{k}}_y \cdot \col{\hat{k}}_g\right) \notag \\ & = \acos \left(k_{y_1} \cdot k_{g_1} + k_{y_2} \cdot k_{g_2} + k_{y_3} \cdot k_{g_3} \right) \notag \\ & = \acos \left( 0.0007\cdot 0.9631 + 0.0357\cdot 0.0860 + 0.9994 \cdot 0.2549\right) \notag \\ & = \acos\left(0.2585\right) \notag \\ & = 75\,\degree \end{align} Similarly, we can find for the absorption angle between the flags yellow dye and black dye, and between the flags green dye and black dye: \begin{align} \varphi_{yb} & = 53\,\degree\\ \varphi_{gb} & = 41\,\degree \end{align} For our current purposes, we cannot help the suboptimal angle between the flags yellow dye and green dye: that is a given. But since we now get to pick our third complementary' dye $c$ ourselves, we can make both angles $\varphi_{yc}$ and $\varphi_{yc}$ equal to 90\,\degree . \section{Two colour flag analysis} To do so, we need to calculate the normalised cross product of the normalised yellow dye and green dye absorption columns (equation \ref{anglecross}): \begin{align} \col{\hat{A}}_c & = \col{\hat{k}}_y \times \col{\hat{k}}_g = \begin{bmatrix*}[r] -0.0769 \\ 0.9623 \\ -0.0343\end{bmatrix*}\\ \col{\hat{k}}_c & = \frac{\col{\hat{A}}_c }{\abs{\col{\hat{A}}_c }} = \begin{bmatrix*}[r] -0.0796 \\ 0.9962\\ -0.0355\end{bmatrix*} \end{align} Note the negative absorption coefficients: this column represents a dye' that produces, instead of absorbs, red and blue. As a result the RGB' values for this dye (normalised amount of 1) come out as: \col{\hat{R}}_c = 256\cdot \mathrm{e}^{\begin{bmatrix*}[r] 0.0796 \\ -0.9962\\ 0.0355\end{bmatrix*} \cdot 1} - 1 = \begin{bmatrix} 277\\ 95 \\ 265 \end{bmatrix} In other words: this third dye colour' is not a real colour' and cannot be (properly) represented in RGB space. \subsection{Deconvolution} \subsection{Reconstruction} \subsection{Adding a third `Ruifrok' absorption column} \col{\hat{A}}_c = \begin{bmatrix} \sqrt{1 - k_{y_1}^2 - k_{g_1}^2} \\[1em] \sqrt{1 - k_{y_2}^2 - k_{g_2}^2} \\[1em] \sqrt{1 - k_{y_3}^2 - k_{g_3}^2}\end{bmatrix},\, \col{\hat{A}}_c \leq 0 \subsection{Deconvolution} \subsection{Reconstruction} \section{Two blending flag colours analysis} \section{Two blending flag colours analysis} \ No newline at end of file \nocite{Haub2015} \ No newline at end of file