# Tag Info

## New answers tagged coordinates

4

The problem here is that using \draw[d_arrow] (nuc.west)++(0, 0.25) -- (smps.east)++(0, -0.25); you are not adding/subtracting 0.25 from the coordinates without affecting the path. You are moving the pencil, as the following simple example shows (look where the line ends in both cases and where the arrow tip results): ...

0

It seems that the source of your problem is scaling. It scales coordinates of center of rotation twice. Once when you define (A1), second for rotation (you can see this by explicit notation {90:(0,1.91)}) \begin{tikzpicture} \tikzstyle{subj} = [circle, minimum width=0.2mm, draw,inner sep=1mm] \coordinate (A0) at (0,0); \coordinate (B0) at (0,5); ...

0

Ok, I got it. That question helped a lot: Why such rotate in tikz doesn't work? Now my code enriched with loop is like this and works perfectly: \documentclass[tikz]{standalone} \usetikzlibrary{shapes.arrows,chains,positioning,intersections,decorations.text} \usetikzlibrary{calc} \usepackage{ifthen} \begin{document} % \center ...

1

Like this? \documentclass[12pt]{article} \pagestyle{empty} \usepackage{tikz} \begin{document} \center \begin{tikzpicture}[scale=5] \coordinate (A0) at (0,0); \node (A) at (A0) {A}; \coordinate (B) at (0,1); \foreach \k in {0,90,180,270} { \node (B\k) at ([rotate around={\k:(A0)}]B) {B\k};; } \end{tikzpicture} \end{document}

2

Here an approach using the evaluate option of the foreach loop. Using this you can create the whole plot in one loop and pre-calculate all needed values. The coordinates on the plot are saved via coord-\x and can be reused. All the calculations are done via evaluate. I think the code is quite self-explanatory. If you still got questions, feel free to ask. ...

3

Another example using the math library. This illustrates the use of the r operator which is a post-fix operator which converts the preceding number to degrees. Also, points and derivatives are stored as coordinate macros. \documentclass[tikz,border=5]{standalone} \usetikzlibrary{math} \tikzmath{ coordinate \c, \d; integer \n; \n = 0; for \t in {0, ...

8

Here is a solution where: (f-\t) is a point on the curve, (f'-\t) is the associate tangent vector. \documentclass[border=5pt]{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} \draw[ultra thin,color=lightgray] (-4,-2) grid (7,5); % coordinate grid \draw[<->] (-4.5,0) -- (7.5,0) node[right] {$x$}; % x-axis ...

6

Except for the missing gnuplot file, this works: \documentclass[border=5pt]{standalone} \usepackage{tikz} \usepackage{pgfplots} \pgfplotsset{compat=1.12} \usetikzlibrary{arrows,backgrounds,patterns,shapes.geometric,calc,positioning} \tikzset{small dot/.style={fill=black,circle,outer sep=8pt,scale=0.25}} \begin{document} \begin{tikzpicture}[scale=0.8] ...

2

I tried to show it visually why and let me know if you need details. 2*==> denotes the result of the one that doesn't work. !==> is the working one. Though both of them are working properly as seen from the paths. They are just not equivalent syntax. Especially, (B)+(D) is the sum relative to the origin, not to each other. Hence the one that doesn't ...

0

I find that part of the manual a bit out of place as it looks like it is from T.Tantau's beamer days. Anyway here is, in my opinion, more structured (TikZic ?!) version of that using pgfkeys \documentclass[tikz]{standalone} \makeatletter \tikzset{mishm@sh/.cd, r/.store in=\myr, a/.store in=\mya, b/.store in=\myb } \tikzdeclarecoordinatesystem{mesh}{% ...

Top 50 recent answers are included