Why does rotating a circle alter its bounding box?

Question

Why does rotating a tikzpicture with a circle alter the height and width of the bounding box, and is there a way to not have this effect?

The MWE below yields:

Code:

\documentclass{article}
\usepackage{tikz}
\usepackage{xstring}
\newcommand*{\DrawBoundingBox}[1][]{%
    \draw [red]
    ([shift={(-5pt,-5pt)}]current bounding box.south west)
    rectangle
    ([shift={(5pt,+5pt)}]current bounding box.north east);
% https://tex.stackexchange.com/questions/418499/
  %         align-tikzpictures-at-internal-bounding-box-with-text-below-it
  \coordinate (X) at (current bounding box.south);
  \tikzset{baseline={(X)}} % X is the alignment point
\IfStrEq{#1}{}{}{%
    \node [below, anchor=north,  align=center,
        baseline=0pt, thin, shift only, solid,
    ]
        at (current bounding box.south)
        {#1\strut};
}%

}
\newcommand*{\MyCircle}[2][]{%
    %% #1 = tikz picture options
    %% #2 = text
    \begin{tikzpicture}
        \draw [fill=yellow!20, draw=black, #1] (0,0) circle (1.0cm);
        \DrawBoundingBox[#2]
    \end{tikzpicture}%
}
\begin{document}
\section{Height Changes}
\noindent
    \MyCircle[fill=green!25]{rotate=0}~%
    \MyCircle[fill=cyan!40, rotate=20]{rotate=20}~%
    \MyCircle[fill=orange!40, rotate=60]{rotate=60}%
\section{Width Changes}
\par\noindent\MyCircle[fill=green!25]{rotate=0}%
\par\noindent\MyCircle[fill=cyan!40, rotate=20]{rotate=20}%
\end{document}

The control points of the Bézier curves are taken into account and a circle is composed of 4 Bézier curves. — Paul Gaborit, Apr 13 '18 at 05:15
@PaulGaborit PDF has has a low level circle command, no ? Why TikZ is using Bézier approximation here ? — Kpym, Apr 13 '18 at 13:35
@PaulGaborit really ? PS has arc command, and I thought that PDF has it too ... strange. — Kpym, Apr 13 '18 at 14:20
@Kpym Internally, the arc command uses Bézier curves (at least in ghostscript). ;-) — Paul Gaborit, Apr 13 '18 at 14:50
@PaulGaborit I know that the final rendering is often done in Bézier curves, but if we deffer this to the renderer we can treat it at higher level as arcs and circles (with the appropriate bounding box). In any case it looks strange to calculate the bounding box of a circle as if it is Bézier curve, even if we use them for rendering. — Kpym, Apr 13 '18 at 16:09
@PaulGaborit In any case the way TikZ calculate the bounding box of a Bézier curve (as convex envelope of the control points) is over simplistic ;) — Kpym, Apr 13 '18 at 16:10
@PaulGaborit: I'm curious, why are you providing the answer in the comments section? You have a very good answer posted below also. — Lightness Races in Orbit, Apr 13 '18 at 17:40
@LightnessRacesinOrbit See Why do people answer in comments? (in the present case: too little time) — Paul Gaborit, Apr 13 '18 at 22:12
@PaulGaborit: Interesting. The equivalent meta post on other SE sites almost invariably declare that this is a terrible idea that violates the very purpose of the network, including explaining that "too little time" is an insufficiently strong reason to violate the SE model. — Lightness Races in Orbit, Apr 14 '18 at 17:38

score 37 · Accepted Answer · answered Apr 13 '18 at 05:36

A circle is drawn by four Bézier curves. The control points of the Bézier curves are taken into account to compute the bounding box:

\documentclass[tikz]{standalone}
\usetikzlibrary{decorations.pathreplacing}
\begin{document}
\begin{tikzpicture}[>=stealth, every node/.style={midway, sloped, font=\tiny},
  decoration={show path construction,
    moveto code={},
    lineto code={},
    curveto code={
      \fill[red] (\tikzinputsegmentsupporta) circle(.5pt);
      \fill[red] (\tikzinputsegmentsupportb) circle(.5pt);
      \fill[blue] (\tikzinputsegmentfirst) circle(.5pt);
      \fill[blue] (\tikzinputsegmentlast) circle(.5pt);
      \draw[blue,->] (\tikzinputsegmentfirst) -- (\tikzinputsegmentsupporta);
      \draw[blue,->] (\tikzinputsegmentlast) -- (\tikzinputsegmentsupportb);
      \draw [black] (\tikzinputsegmentfirst) .. controls
      (\tikzinputsegmentsupporta) and (\tikzinputsegmentsupportb)
      ..(\tikzinputsegmentlast);
    },
    closepath code={},
  }]
  \draw [help lines] grid (6,3);
  \begin{scope}[local bounding box=lbb]
    \path [decorate,rotate around={0:(1.5,1.5)}] (1.5,1.5) circle(1);
    \draw[red] (lbb.north west) rectangle (lbb.south east);
  \end{scope}
  \begin{scope}[local bounding box=lbb]
    \path [decorate,rotate around={30:(4.5,1.5)}] (4.5,1.5) circle(1);
    \draw[red] (lbb.north west) rectangle (lbb.south east);
  \end{scope}
\end{tikzpicture}
\end{document}

Congratulations!! I was just looking up show path constructions. — , Apr 13 '18 at 05:47
Is it a bug, that Bézier control points are taken into account for the bounding box? If not, do you know the reason behind this decision? — T. Verron, Apr 13 '18 at 09:12
@T.Verron It's not a bug (see p.124, pgfmanual, v3.0.1a)! The reason is undoubtedly efficiency. To work around this feature, you can try my answer to question tikz bounding box / cropping: too much space for curves. — Paul Gaborit, Apr 13 '18 at 09:41
@PaulGaborit and Marmot congratulations for excellent code and Grill for excellent answer. +1 — Sebastiano, Apr 13 '18 at 10:11

score 18 · Answer 2 · 2018-04-13T05:45:30.730

I guess that you are using not the appropriate means to rotate the circles. If you use transform canvas instead, none of the issues arises. UPDATE I forgot to put transform canvas to the third circle.

\documentclass{article}
\usepackage{tikz}
\usepackage{xstring}

\newcommand*{\DrawBoundingBox}[1][]{%
    \draw [red]
    ([shift={(-5pt,-5pt)}]current bounding box.south west)
    rectangle
    ([shift={(5pt,+5pt)}]current bounding box.north east);

  % https://tex.stackexchange.com/questions/418499/
  %         align-tikzpictures-at-internal-bounding-box-with-text-below-it
  \coordinate (X) at (current bounding box.south);
  \tikzset{baseline={(X)}} % X is the alignment point

    \IfStrEq{#1}{}{}{% 
        \node [below, anchor=north,  align=center, 
            baseline=0pt, thin, shift only, solid,
        ] 
            at (current bounding box.south)
            {#1\strut};
    }%
}

\newcommand*{\MyCircle}[2][]{%
    %% #1 = tikz picture options
    %% #2 = text
    \begin{tikzpicture}
        \draw [fill=yellow!20, draw=black, #1] (0,0) circle (1.0cm);
        \draw [#1] (-1,0) -- (1,0); % added to see that the transformatio does
                                    % something
        \DrawBoundingBox[#2]
    \end{tikzpicture}%
}

\begin{document}
\section{No Height Changes}
\noindent
    \MyCircle[fill=green!25]{rotate=0}~%
    \MyCircle[fill=cyan!40, transform canvas={rotate=20}]{rotate=20}~%
    \MyCircle[fill=orange!40, transform canvas={rotate=60}]{rotate=60}%

\section{No Width Changes}
\par\noindent\MyCircle[fill=green!25]{rotate=0}%
\par\noindent\MyCircle[fill=cyan!40, transform canvas={rotate=20}]{rotate=20}%

\end{document}

(I added a line in order to show that this transformation does indeed rotate the circles, which is hard to see otherwise ;-)

So, what is the difference between rotate=x and transform canvas={rotate=x} and when would one want to use rotate=x instead of transform canvas={rotate=x}? — Peter Grill, Apr 13 '18 at 05:08
Also, it appears as if the heights and widths are still not exactly the same. — Peter Grill, Apr 13 '18 at 05:11
@PeterGrill The height change is because I forgot to put transform canvas in the third circle. — , Apr 13 '18 at 05:42
+1 people have been trying to square the circle a long time. — David Carlisle, Apr 13 '18 at 07:51
transform canvas doesn't update the bounding box that's why. But usually other funny things happen if you don't use it carefully. — percusse, Apr 13 '18 at 09:07

Why does rotating a circle alter its bounding box?

Code:

2 Answers2

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