Consider following example:

\draw (0,0) arc (0:90:1 and 2);
\filldraw (0,0) circle (2pt);
\filldraw[red] (-.1,.9) circle (2pt);

enter image description here

Given a curve and an origin (A in the example), I would like to place a point on the curve at a given arclength from the origin (B at an arclength 1 in the example).

How can I achieve this? I found this question and this question, but do not see how to specify the arclength.

1 Answer 1


This can be done (approximately) with decorations.markings, which accepts lengths (as well as percentages) for its position.

enter image description here

All markings in the image are at 1cm.

Define a style arclen that takes a length as its argument. Then \draw[arclen=1cm] <path> will make the mark at 1cm from the start.

\usetikzlibrary {decorations.markings}

\tikzset{arclen/.style={decoration={markings, mark=at position #1 with{\filldraw[red] (0,0) circle (2pt);}},

\draw[gray!50, very thin](0,-1) grid (-2,2);
\draw[arclen=1cm] (0,0) -- (0,2);
\draw[arclen=1cm] (0,0) arc (0:90:1 and 2);
\draw[arclen=1cm] (0,0) arc (0:180:1);
\draw[arclen=1cm] (0,0) arc (0:270:.5);
\draw[arclen=1cm] (0,0) arc (0:330:.4);
\draw[arclen=1cm] (0,0) arc (0:330:.3);
  • +1. Really interesting, especially if we know that the length may be defined by an integral not expressed by elementary functions. May 21 at 1:31
  • 3
    @PrzemysławScherwentke the “curve” is really a sequence of line segments. So there’s no integral involved. Just a sum. I suspect that if you tried this on a segment with large curvature you would see some inaccuracies. But I haven’t looked at the raw code to see how tikz calculates the number of points.
    – Sandy G
    May 21 at 2:16
  • This is so nice, I didn't know it, thanks Sandy. +1
    – SebGlav
    May 21 at 7:46
  • 1
    @SebGlav: Glad I could show you something new for a change. It's usually the other way around.
    – Sandy G
    May 21 at 14:57

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