Update on Overleaf.
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parent
ed9f7d0977
commit
161272bcd3
2 changed files with 10 additions and 10 deletions
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@ -5,9 +5,9 @@
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\foreach \x in {0,...,\raynum} {
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\foreach \x in {0,...,\raynum} {
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\ifthenelse{\x < 10 \OR \x>50}
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\ifthenelse{\x < 10 \OR \x>50}
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{
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{
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\draw[color = kit-blue100] (\x /\raynum * 6 ,0) -- (3,4);
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\draw[color = blue] (\x /\raynum * 6 ,0) -- (3,4);
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}{
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}{
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\draw[color = kit-green100] (\x /\raynum * 6 ,0) -- (3,4);
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\draw[color = green] (\x /\raynum * 6 ,0) -- (3,4);
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}
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}
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}
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}
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@ -17,13 +17,13 @@
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] (1.49,1) -- (4.51,1) -- (5.01,0) -- (0.99,0) -- cycle;
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] (1.49,1) -- (4.51,1) -- (5.01,0) -- (0.99,0) -- cycle;
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\draw [color=kit-blue100] (0,0) -- node[below]{wall} (6,0) ; % wall
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\draw [color=blue] (0,0) -- node[below]{wall} (6,0) ; % wall
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\draw [color=kit-green100](1.5,1) -- node[below]{blocker} (4.5,1); %blocker
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\draw [color=green](1.5,1) -- node[below]{blocker} (4.5,1); %blocker
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%cam
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%cam
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\node[
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\node[
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circle,
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circle,
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fill = kit-gray70,
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fill = black,
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minimum width = 1mm,
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minimum width = 1mm,
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outer sep = 2mm
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outer sep = 2mm
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] (camera) at (3,4) {};
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] (camera) at (3,4) {};
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@ -37,17 +37,17 @@
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\fill[
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\fill[
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kit-green100
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green
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] (7,2) -- (9,2) -- (9,1.7) -- (7,1.7) -- cycle;
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] (7,2) -- (9,2) -- (9,1.7) -- (7,1.7) -- cycle;
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\draw[kit-blue100] (10,1) coordinate (blue_ray_isec) -- (7,4) coordinate (orig);
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\draw[blue] (10,1) coordinate (blue_ray_isec) -- (7,4) coordinate (orig);
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\draw[kit-green100] (8.7,2) coordinate (green_ray_isec) -- (7,4);
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\draw[green] (8.7,2) coordinate (green_ray_isec) -- (7,4);
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%line
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%line
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\draw (mag.north east) -- (7,2);
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\draw (mag.north east) -- (7,2);
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%magnifier box
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%magnifier box
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\draw (7,4) -- (10,4) -- (10,1) -- (7,1) -- cycle;
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\draw (5,4) -- (10,4) -- (10,1) -- (5,1) -- cycle;
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\pic [
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\pic [
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draw,
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draw,
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@ -78,7 +78,7 @@ Differentiable ray tracing is the task of calculating the gradient of this proce
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The geometry term $g(x,x^\prime)$ in equation \ref{eq:rendering_integral} is the main problem when it comes do differentiating the rendering integral. This term is 1 iff $x$ is visible from $x^\prime$, 0 otherwise.
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The geometry term $g(x,x^\prime)$ in equation \ref{eq:rendering_integral} is the main problem when it comes do differentiating the rendering integral. This term is 1 iff $x$ is visible from $x^\prime$, 0 otherwise.
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\begin{figure}
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\begin{figure}
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\centering
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\centering
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\include{presentation/diagrams/edge_sampling.tex}
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\include{handout/diagrams/diagramm_occlusion}
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\caption{Caption}
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\caption{Caption}
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\label{fig:enter-label}
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\label{fig:enter-label}
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\end{figure}
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\end{figure}
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