Using the graphicx package, I can make any arbitrary box rotated using the \rotatebox{...}{...} command. Is there a comparable command in any package that lets me shear a box, i.e., producing a kind of general slanting?


With a slightly more recent pdfTeX than in David's answer, you can more directly do affine transforms using \pdfsetmatrix. I don't claim to know anything about this, but here is roughly what graphicx does under the hood in \rotatebox (with a different matrix, of course).

        \pdfsetmatrix{1 0 #1 1}%
\slantbox{Hello, world!}

\slantbox[-2]{Hello, world!}
\slantbox[-1]{Hello, world!}
\slantbox[-.8]{Hello, world!}
\slantbox[-.6]{Hello, world!}
\slantbox[-.4]{Hello, world!}
\slantbox[-.2]{Hello, world!}

\slantbox[.2]{Hello, world!}
\slantbox[.4]{Hello, world!}
\slantbox[.6]{Hello, world!}
\slantbox[.8]{Hello, world!}
\slantbox[1]{Hello, world!}
\slantbox[2]{Hello, world!}


enter image description here

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    Do we want this for LaTeX3? It should be pretty easy to code as I've already got the basics sorted. – Joseph Wright Jul 13 '12 at 17:10
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    @Seamus: yes, the four numbers 1, 0, #1, and 1 are the entries in the 2×2 matrix defining the linear transformation. David's answer with \pdfliteral allows arbitrary affine transformation (so the 4 coefficients for the linear part, and 2 coefficients for the offset, I think). – Bruno Le Floch Jul 13 '12 at 17:47
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    @Joseph: I was thinking of it (I didn't upvote your comment, so at least one other person would like that), but I'm not sure whether we should give \box_linear_transform:Nnnnn for arbitrary parameters, or \box_vertical_shear:Nn and \box_horizontal_shear:Nn, or... – Bruno Le Floch Jul 13 '12 at 17:59
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    @BrunoLeFloch If we provide a mechanism here, we should have the 'general case' shear, something like \box_shear:Nnn or even as you say a totally-general transformation (\box_affine:Nnnnn?). One for LaTeX-L – Joseph Wright Jul 13 '12 at 18:40
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    Indeed I was thinking of \pdfliteral. I've confused your answer with David's probably. But maybe it is possible to use directly that command in your case too. Great answer anyway and I hope it goes into the L3 magic. – percusse Jul 14 '12 at 17:48

Aha, TikZability opportunity!

\begin{scope}[cm={1,0,1,1,(0,0)}] % Sets the coordinate trafo matrix entries.
\node[transform shape] at (0,0) {ABC};
\node[transform shape] at (3,0) {Hello};
\node[transform shape] at (2,2) {World};
\node[transform shape] at (1,2) {FOObar?};

You can put into nodes instead of boxes (with less risk :P).

enter image description here

  • 'transform shape' is not needed – Black Mild Jun 2 '19 at 15:23

enter image description here

You can mess with the coordinate matrix, but at your own risk...



ABC\pdfliteral{ q 2 0.1 0.6 .4 0 0 cm}\rlap{XYZ}\pdfliteral{ Q}

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    q (\pdfsave) and Q (\pdfrestore) should be used at the same place. Otherwise the coordinate system of TeX and its output device get indeed messed up. – Heiko Oberdiek Jun 19 '14 at 15:36
  • @HeikoOberdiek ah yes I wrote some driver graphics back end files once, clearly I forgot everything, I'll add an rlap.... – David Carlisle Jun 19 '14 at 15:49

A little late, but perhaps useful. An easier solution is the use of xslant and yslant in the node options.

Example 1

    \node[yslant=0.5, draw] (1) {Latex};
    \node[xslant=0.5, draw, anchor=south] at ([yshift=10]1.north) {Latex};

Result 1

Result example 1

Then I wanted to use it for a 3D view of a box, where on one plane I project an external *.png image.

Example 2


\newcommand{\w}{4 cm} % width of the box
\newcommand{\dep}{1 cm} % depth of the box
\newcommand{\h}{0.7 cm} % height of the box

    % Front side of the box
    \node[minimum width = \w, minimum height=\h, fill=gray, outer sep =0] (front) {};
        \pgfset{minimum width=\w, minimum height= \dep, outer sep = 0}
        % Image on top of the box
        \node[anchor=south, inner sep =0, xslant=1, outer sep = 0] (img) at (front.north) 
            {\includegraphics[width=\w, height=\dep]{Example_image.png}}; 
    % Side of the box
    \node[anchor=west, yslant=1, minimum height =\h, minimum width=\dep,
          inner sep=0, outer sep=0, fill=black!70] at (front.east) {};

Result 2

Result example 2

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    "For every problem, there is a TikZ-attribute". This should be the higher-rated TikZ-answer, w.r.t. to it's simplicity. – Bubaya Mar 29 '20 at 8:01
  • I completely agree @Bubaya ;) – Roald Jun 4 '20 at 19:17

Shear transforms can be decomposed into scalings and rotations.

% \hshearbox{vertical_prescale_times_shearfactor}{one_divide_by_shearfactor}{content}
% an initial vertical downscale is often necessary for a 3d projection
% \vshearbox{horizontal_prescale_times_shearfactor}{one_divide_by_shearfactor}{content}
% an initial horizontal downscale is often necessary for a 3d projection

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