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7

Our TikZers getting lazy I guess. This can't get away without a TikZ answer :P Same idea but using decorations (over and over again). \documentclass[tikz]{standalone} \usetikzlibrary{decorations.pathmorphing,decorations.markings,calc} \begin{document} \begin{tikzpicture} \shadedraw[thick,top color=gray!10,bottom color=gray, postaction=decorate, ...


16

Here's an attempt in Metapost. Using the direction x of y macro to find the required angle of reflection. Here's what you get with r=0: And with r=0.33: prologues := 3; outputtemplate := "%j%c.eps"; beginfig(1); path base, ray[]; u = 5mm; r=0.33; base = (-6u,0) for x=-5.8u step 0.2u until 5.8u: -- (x,r*normaldeviate) endfor -- (6u,0); draw ...


2

As a work around, you can write (i mod 2)=1 instead of odd i. This behaves as expected for both positive and negative integers but not quite for decimals. To get the advertised behaviour of the odd primitive you need: (round i mod 2)=1.



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