### update for new figure/values

parent 8d6db9f9
 ... ... @@ -190,8 +190,8 @@ For our current purposes, we cannot help the suboptimal angle between the flags 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*} \label{Apcross}\\ \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*} \col{\hat{A}}_c = \col{\hat{k}}_y \times \col{\hat{k}}_g & = \begin{bmatrix*}[r] -0.0769 \\ 0.9623 \\ -0.0343\end{bmatrix*} \label{Apcross}\\ \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: \begin{equation} ... ... @@ -225,10 +225,9 @@ And solve again for the amounts of dye in the flag (equation \ref{solveap}). \noindent The results for the yellow and green parts in the flag are the same as in the three colour deconvolution (figure \ref{flagJA2pas}): the yellow stripes have $\hat{a}_y = \abs{\col{\hat{A}}_y} = 5.5487$ of yellow and nothing else, and the green patches have $\hat{a}_g = \abs{\col{\hat{A}}_g} = 5.7575$ of green and nothing else. The black patches our now made up of a combination of all three dyes: yellow, green and our third complementary dye'. According to the deconvolution, black' is the result of a combination of 75\,\% of the flags yellow dye (figure \ref{flagJA2pay}), 100\,\% of the flags green dye (figure \ref{flagJA2pag}), and 16\,\% of our third perpendicular dye'.\\ The black patches our now made up of a combination of all three dyes: yellow, green and our third complementary dye'. According to the deconvolution, black' is the result of a combination of 75\,\% of the flags yellow dye (figure \ref{flagJA2pay}), 100\,\% of the flags green dye (figure \ref{flagJA2pag}), and about 5 units' of our third perpendicular dye'. \noindent Note that the total absorption' (length) of our complementary perpendicular column is large (30-ish) compared to those of the yellow and green (5.5-ish). That is because this column was calculated from the yellow and green absorption columns \textsl{before} normalisation. When calculated from the normalised columns (as in equations \ref{anglecross} and \ref{Apcross}), the length is $\abs{\col{\hat{A}}_c} = 0.9660$. Calculated from the columns before normalisation, $\frac{\abs{\col{\hat{A}}_c}}{\abs{\col{\hat{A}}_y}\cdot\abs{\col{\hat{A}}_g}} = 0.9660$. And since the total absorption' (length) of our third column is $\abs{\col{\hat{A}}_c} = 0.9660$, black' contains roughly about as much of the flags normalised yellow dye (4.2) and normalised green dye (5.8) as of our complementary perpendicular dye' (4.9). \subsection{Reconstruction} ... ... @@ -248,9 +247,9 @@ It is instructive to reconstruct the single dye and combined dye RGB images as b The black patches in the flag has 75\,\% of the yellow dye of the stripes (figure \ref{flagJA2pay}). Less dye means less absorption, so the yellow for the black patches in the flag comes out lighter' (figure \ref{flagJA2py}, RGB = 252, 281, 3). The black in the flag patches are as green' as the green patches (figures \ref{flagJA2pag}, \ref{flagJA2pg}), and they contain a certain amount of our complementary perpendicular' dye (figures \ref{flagJA2pac}, \ref{flagJA2pc}). Note that these RGB values have been scaled to make them representable in RGB, the actual RGB colour for this image does not exist. The black in the flag patches are as green' as the green patches (figures \ref{flagJA2pag}, \ref{flagJA2pg}), and they contain a certain amount of our complementary perpendicular' dye (figures \ref{flagJA2pac}, \ref{flagJA2pc}). Note that these RGB values have been scaled by $\frac{255}{300}$ to make them representable in RGB, the actual RGB colour for this image does not exist. Also note, finally, that the combination of 75\,\% yellow, 100\,\% green and 16\,\% of our third dye, \textsl{does} result in proper black'. And it should be obvious from figure \ref{flagJA2precon} that to add the effects of (absorbing) dyes, you cannot add RGB values'. Also note, finally, that the combination of 75\,\% yellow, 100\,\% green and 5 units of our third dye, \textsl{does} result in proper black'. And it should be obvious from figure \ref{flagJA2precon} that to add the effects of (absorbing) dyes, you cannot add RGB values'. \subsection{Adding a third `Ruifrok' absorption column} ... ...
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