Adapting KnotDiagramData to form more visually pleasing diagram

Question

Graphics[KnotData[{5, 1}, "KnotDiagramData"]] gives

but I would like to create diagrams like this

remeniscent of Celtic knot designs (which I made by adapting the above Mathematica image in Illustrator). Is there any way to do this in Mathematica directly?

Hey Martin: It's not Mathematica, but check out the KnotPlot site for everything knotical. — KennyColnago, Nov 14 '15 at 16:52

score 19 · Accepted Answer · edited Apr 13 '17 at 12:55

19

You can use a common 3D trick with double boundary (1, 2, 3):

fun = KnotData[{5, 1}, "SpaceCurve"];
ParametricPlot3D[fun[t], {t, 0, 2 Pi}, PlotPoints -> 100, MaxRecursion -> 1, 
   Axes -> False, Boxed -> False, ViewPoint -> {0, 0, ∞}, 
   ViewVertical -> {0, -1, 0}, PlotRange -> All, 
   Lighting -> {{"Ambient", White}}] /. 
  Line[pts_] :> {White, Tube[pts, 0.1], FaceForm[Opacity[0], Black], 
    Tube[pts, 0.12]}

edited Apr 13 '17 at 12:55

Community

1

answered Nov 14 '15 at 14:23

ybeltukov

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Nice solution thanks! :) One minor poit - mine is showing up slightly differently to youur - ie with contour lines - I an on v.10.2 - can you relpicate this? – martin Nov 15 '15 at 15:49
1

@martin It is a bug of versions 10.0 – 10.2, I advise to upgrade to 10.3. – ybeltukov Nov 15 '15 at 16:10
ok will do - thanks again :) – martin Nov 15 '15 at 16:13

score 4 · Answer 2 · answered Nov 14 '15 at 14:10

This is only a first idea how one could proceed

KnotData[{5, 1}, "Name"]

"Solomon seal knot"

fun = KnotData["Solomon seal knot", "SpaceCurve"];

par =
 ParametricPlot3D[fun[t], {t, 0, 2 Pi},
  Axes -> False,
  Boxed -> False,
  ViewPoint -> Above,
  ImageSize -> Medium,
  PlotStyle -> Thickness[0.05]]

ColorNegate @ EdgeDetect @ Image @ par

Adapting KnotDiagramData to form more visually pleasing diagram

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