[Tex/LaTex] TiKZ: How to define new 2D canvas

3dtikz-pgf

I have the code below to draw a isometric 3D figure. Il would like to draw something in the "rz" plane defined by \theta=30.

Would it be possible to define a new canvas that would NOT be the xy xz or yz plane?
Something like "canvas is rz plane at angle=30" in cylindrical coordinates?

I see this question is related to this post:
TikZ: How to draw an isometric drawing in tikz.
But I did not managed to use it properly.

 \documentclass[14pt,a4paper]{article}
 \usepackage{tikz}
 \usetikzlibrary{arrows,3d,calc}

 \begin{document}

 \begin{center}
 \begin{tikzpicture}
 \draw[line width=0.3mm,->,>={latex}](0,0,0)--(8,0,0)node[right]{\emph{y}};
 \draw[line width=0.3mm,->,>={latex}](0,0,0)--(0,6,0)node[above]{\emph{z}};
 \draw[line width=0.3mm,->,>={latex}](0,0,0)--(0,0,8)node[below left=-3pt]{\emph{x}};
 \draw[line width=0.7mm,-stealth](0,0)--(2,0)node[pos=0.9, above]{$\vec{u}_y$};
 \draw[line width=0.7mm,-stealth](0,0)--(0,2)node[pos=0.8, left]{$\vec{u}_z$};
 \draw[line width=0.7mm,-stealth](0,0,0)--(0,0,2)node[pos=0.8, above left=-3pt]{$\vec{u}_x$};
 \node at (0,0,0) [left]{$O$};
 \draw[thick,->,>={stealth}](0,0)--(6,0,4)node[below]{$M'$}node[pos=0.6,above,sloped]{$r$};
  \draw[dashed,thick](6,0,4)--(6,4,4);
  \draw[dotted] (0,0,0) grid (7,5.5,0);
  \begin{scope}[canvas is yz plane at x=0,]
 \draw[dotted] (0,0) grid (5.5,7.5);
 \end{scope}
 \begin{scope}[canvas is xz plane at y=0,]
 \draw[dotted] (0,0) grid (7.5,7.5);
 \end{scope}
 \draw[line width=0.7mm,-stealth](6,4,4)--({6+3/sqrt(2)},4,{4+2/sqrt(2)})node[pos=0.6, below]{$\vec{u}_r$};
 \draw[line width=0.7mm,-stealth](6,4,4)--($(6,4,4)+({2/sqrt(2)},0,{-3/sqrt(2)})$)node[pos=0.8, below]{$\vec{u}_\theta$};
 \draw[line width=0.7mm,-stealth](6,4,4)--(6,6,4)node[pos=0.8, right]{$\vec{u}_z$};
 \draw[thick,->,>={stealth}](0,0)--(6,4,4)node[below=5pt,circle,fill=white,inner sep=1pt]{$M$};
 \end{tikzpicture}
 \end{center} 

 \end{document}

Best Answer

Here is a solution adapted from the link that you gave, TikZ: How to draw an isometric drawing in tikz. Add the following lines to your preamble:

\makeatletter
\tikzoption{canvas is plane}[]{\@setOxy#1}
\def\@setOxy O(#1,#2,#3)x(#4,#5,#6)y(#7,#8,#9)%
  {\def\tikz@plane@origin{\pgfpointxyz{#1}{#2}{#3}}%
   \def\tikz@plane@x{\pgfpointxyz{#4}{#5}{#6}}%
   \def\tikz@plane@y{\pgfpointxyz{#7}{#8}{#9}}%
   \tikz@canvas@is@plane
  }
\makeatother

They define a new tikz option to be used as

canvas is plane={O(#1,#2,#3)x(#4,#5,#6)y(#7,#8,#9)}

where (#1,#2,#3) are the coordinates of the new origin and the start of the new x- and y-unit-vector, (#4,#5,#6) are the end coordinates of the new x-unit-vector and (#7,#8,#9) are the end coordinates of the new y-unit-vector. This option subsumes the canvas is .. plane options of tikz.

Here is an example:

\documentclass[border=2mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{3d}
\makeatletter
\tikzoption{canvas is plane}[]{\@setOxy#1}
\def\@setOxy O(#1,#2,#3)x(#4,#5,#6)y(#7,#8,#9)%
  {\def\tikz@plane@origin{\pgfpointxyz{#1}{#2}{#3}}%
   \def\tikz@plane@x{\pgfpointxyz{#4}{#5}{#6}}%
   \def\tikz@plane@y{\pgfpointxyz{#7}{#8}{#9}}%
   \tikz@canvas@is@plane
  }
\makeatother 
\begin{document}
\begin{tikzpicture}
\draw[thick,->] (0,0,0) -- (5,0,0) node[right]{$x$};
\draw[thick,->] (0,0,0) -- (0,5,0) node[above]{$y$};
\draw[thick,->] (0,0,0) -- (0,0,5) node[below left]{$z$};

% canvas is yz plane at x=1
\begin{scope}[canvas is plane={O(1,0,0)x(1,1,0)y(1,0,1)},blue!70]
\draw[->] (0,0) -- (2,0) node[right] {$x$};
\draw[->] (0,0) -- (0,2) node[right] {$y$};
\draw[fill] (1,1) rectangle (2,2);
\end{scope}

% canvas is xz plane at y=4
\begin{scope}[canvas is plane={O(0,4,0)x(1,4,0)y(0,4,1)},green!70]
\draw[->] (0,0) -- (2,0) node[right] {$x$};
\draw[->] (0,0) -- (0,2) node[right] {$y$};
\draw[fill] (1,1) rectangle (2,2);
\end{scope}

% canvas is yz plane at x=4, tilted 45 deg
\begin{scope}[canvas is plane={O(4,0,0)x(3.3,0.7,0)y(4,0,1)},red!70]
\draw[->] (0,0) -- (2,0) node[right] {$x$};
\draw[->] (0,0) -- (0,2) node[right] {$y$};
\draw[fill] (1,1) rectangle (2,2);
\end{scope}

\end{tikzpicture}
\end{document}

The following code adds a blue rectangle to your sample image, in the rz-plane.

\begin{scope}[canvas is plane={O(6,4,4)x({6+3/sqrt(2)},4,{4+2/sqrt(2)})y(6,6,4)}]
\draw[fill=blue] (0.1,0.1) rectangle (0.5,0.5);
\end{scope}

\documentclass[a4paper]{article}
 \usepackage{tikz}
 \usetikzlibrary{arrows,3d,calc}
 \makeatletter
 \tikzoption{canvas is plane}[]{\@setOxy#1}
 \def\@setOxy O(#1,#2,#3)x(#4,#5,#6)y(#7,#8,#9)%
   {\def\tikz@plane@origin{\pgfpointxyz{#1}{#2}{#3}}%
    \def\tikz@plane@x{\pgfpointxyz{#4}{#5}{#6}}%
    \def\tikz@plane@y{\pgfpointxyz{#7}{#8}{#9}}%
    \tikz@canvas@is@plane
   }
 \makeatother  
 \begin{document}

 \begin{center}
 \begin{tikzpicture}
 \draw[line width=0.3mm,->,>={latex}](0,0,0)--(8,0,0)node[right]{\emph{y}};
 \draw[line width=0.3mm,->,>={latex}](0,0,0)--(0,6,0)node[above]{\emph{z}};
 \draw[line width=0.3mm,->,>={latex}](0,0,0)--(0,0,8)node[below left=-3pt]{\emph{x}};
 \draw[line width=0.7mm,-stealth](0,0)--(2,0)node[pos=0.9, above]{$\vec{u}_y$};
 \draw[line width=0.7mm,-stealth](0,0)--(0,2)node[pos=0.8, left]{$\vec{u}_z$};
 \draw[line width=0.7mm,-stealth](0,0,0)--(0,0,2)node[pos=0.8, above left=-3pt]{$\vec{u}_x$};
 \node at (0,0,0) [left]{$O$};
 \draw[thick,->,>={stealth}](0,0)--(6,0,4)node[below]{$M'$}node[pos=0.6,above,sloped]{$r$};
  \draw[dashed,thick](6,0,4)--(6,4,4);
  \draw[dotted] (0,0,0) grid (7,5.5,0);
  \begin{scope}[canvas is yz plane at x=0,]
 \draw[dotted] (0,0) grid (5.5,7.5);
 \end{scope}
 \begin{scope}[canvas is xz plane at y=0,]
 \draw[dotted] (0,0) grid (7.5,7.5);
 \end{scope}
 \draw[line width=0.7mm,-stealth](6,4,4)--({6+3/sqrt(2)},4,{4+2/sqrt(2)})node[pos=0.6, below]{$\vec{u}_r$};
 \draw[line width=0.7mm,-stealth](6,4,4)--($(6,4,4)+({2/sqrt(2)},0,{-3/sqrt(2)})$)node[pos=0.8, below]{$\vec{u}_\theta$};
 \draw[line width=0.7mm,-stealth](6,4,4)--(6,6,4)node[pos=0.8, right]{$\vec{u}_z$};
 \draw[thick,->,>={stealth}](0,0)--(6,4,4)node[below=5pt,circle,fill=white,inner sep=1pt]{$M$};
 \begin{scope}[canvas is plane={O(6,4,4)x({6+3/sqrt(2)},4,{4+2/sqrt(2)})y(6,6,4)}]
 \draw[fill=blue] (0.1,0.1) rectangle (0.5,0.5);
 \end{scope}
 \end{tikzpicture}
 \end{center} 

 \end{document}

Related Solutions

[Tex/LaTex] TikZ: Cropping the Bounding Box

The code adds some completely useless invisible (or rather white) stuff. The lines

\clip(0pt,403pt) -- (389.957pt,403pt) -- (389.957pt,99.6166pt) -- (0pt,99.6166pt) -- (0pt,403pt);
\color[rgb]{1,1,1}
\fill(3.76406pt,399.236pt) -- (380.923pt,399.236pt) -- (380.923pt,253.19pt) -- (3.76406pt,253.19pt) -- (3.76406pt,399.236pt);
\fill(53.4497pt,394.719pt) -- (374.901pt,394.719pt) -- (374.901pt,289.325pt) -- (53.4497pt,289.325pt) -- (53.4497pt,394.719pt);

draw a white background that is larger than the actual picture. TikZ sees that and thinks it is part of the picture. Simply removing/uncommenting these lines removes most of the whitespace.

Near the end of the first scope,

\color[rgb]{1,1,1}
\fill(3.76406pt,249.426pt) -- (386.193pt,249.426pt) -- (386.193pt,103.381pt) -- (3.76406pt,103.381pt) -- (3.76406pt,249.426pt);

does the same. Additionally (near the end of the second scope),

\pgftext[center, base, at={\pgfpoint{220.95pt}{106.392pt}}]{\sffamily\fontsize{9}{0}\selectfont{\textbf{ }}}

adds a blank node below the picture, again enlarging the bounding box.

Removing all those lines gives a tight bounding box.

As far as I know, TikZ cannot do the cropping for you, as it can't know whether the white stuff is intentional or not (there might for example be a dark background behind the image so that white is visible).

[Tex/LaTex] Tikz new shape : how to access/define anchors

Using Spike's approach is for sure a possibility, but the shape to be defined is quite simple, so I think there's a simplest way. Credit should go to Jake's answer How to have a cross out rectangle in tikz?, because here I just adapted things.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}

\tikzset{
        orange square/.style={draw=red,ultra thick, fill=orange,%
        minimum width=1cm, minimum height=1cm},%
        orange circle/.style={circle,draw=red,ultra thick, fill=orange,minimum width=1cm},%
        orange triangle/.style={regular polygon, regular polygon sides=3,%
        draw=red,ultra thick, fill=orange,minimum height=1cm},%
        orange rectangle/.style={rectangle,draw=red,ultra thick,%
        fill=orange,minimum height=1cm,minimum width=0.5cm,append after command={                       
            (\tikzlastnode.north east) edge[draw=red,ultra thick,shorten >=1.75\pgflinewidth,
                shorten <=1.75\pgflinewidth,](\tikzlastnode.south west)
            }
        },%
}

\begin{document}
\begin{tikzpicture}[scale=1.75,transform shape]
\node[orange square] (a) at (0,0){};
\node[orange circle] (b) at (2,0){};
\node[orange triangle] (c) at (4,0){};
\node[orange rectangle] (d) at (6,0){};
\draw[->,>=stealth] (a.south) -- (b.west) (b.north) -- (c.north)-- (d.center);
\end{tikzpicture}
\end{document}

Result:

enter image description here

The edit of the question was very useful to better understand the request. In this case perhaps it is better to define a new shape, but that's Spike's domain: I provide you an alternative. Ok, it could be an alternative if one is used to the calc library.

The basic hint is that regular polygons define also corner anchors so, by taking them as reference, it is possible to access every position without defining a new shape.

The revised example:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric,calc}

\tikzset{
        orange square/.style={regular polygon, regular polygon sides=4,%
        draw=red,ultra thick, fill=orange,minimum width=1.5cm},%
        orange triangle/.style={regular polygon, regular polygon sides=3,%
        draw=red,ultra thick, fill=orange,minimum height=1.5cm},%
        }

\begin{document}

\begin{tikzpicture}[scale=1.75,transform shape]
\node[orange square,] (e) at (0,-2){};
\node[orange triangle] (f) at (4,-2.165){};

% Display the anchors
\foreach \anchor/\placement in {corner 1/above, corner 2/above, corner 3/below, corner 4/below}
\draw[shift=(e.\anchor)] plot[mark=x] coordinates{(0,0)}node[\placement] {\tiny\texttt{(e.\anchor)}};

\foreach \anchor/\placement in {corner 1/above, corner 2/below, corner 3/below}
\draw[shift=(f.\anchor)] plot[mark=x] coordinates{(0,0)}node[\placement] {\tiny\texttt{(f.\anchor)}};

% Drawing the arrows    
\draw[-latex] ($(e.corner 3)!0.5!(e.corner 4)-(0,1)$)--($(e.corner 3)!0.5!(e.corner 4)$);
\draw[-latex] ($(e.corner 1)!0.2!(e.corner 4)$)--($(f.corner 1)!0.25!(f.corner 2)$);
\draw[-latex] ($(e.corner 1)!0.5!(e.corner 4)$)--($(f.corner 1)!0.525!(f.corner 2)$);
\draw[-latex] ($(e.corner 1)!0.8!(e.corner 4)$)--($(f.corner 1)!0.8!(f.corner 2)$);
\end{tikzpicture}
\end{document}