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Brunnian 5 links

Posted 9 years ago

Hello, Can anyone tell me how to get Mathematica to draw a Brunnian 5 links. Many thanks.

POSTED BY: ali cherad
11 Replies
Posted 9 years ago

Yes but you forgot, the precedence order property, in temporal space, on its own the red loop always occupies the first order, with any other color, red is always in the 2nd order of precedence (time-wise), this (timing) in my model is most important, and the most obvious thing, but strange YOU have not seen !

POSTED BY: Updating Name

I do not entirely get the question either. Do you mean 5 links or 5 intersections of a single link - because that is a thing too:

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

enter image description here

POSTED BY: Sam Carrettie
Posted 9 years ago

Hi Krause .Yes that is (Mandala) what I am using (I googled it) but my question is how to get Mathematica to draw the links so that I am able to change some parameters and alter the shape

POSTED BY: ali cherad

Do a redraw of a redraw of a redraw

(* Take 5 from http://katlas.org/wiki/File:6Loops-Brunnian-link.png *) 
Graphics3D[{{Yellow, Specularity[White, 70], 
   Tube[{{0, 12, 0}, {23, 12, 0}, {23, -12, 0}, {0, -12, 0}, {0, 12, 
      0}}, 1/3]},
  {Green, Specularity[White, 70], 
   Tube[{{1, 9, 0}, {22, 9, 0}, {22, -9, 0}, {1, -9, 0}, {1, 9, 0}}, 
    1/3]},
  {Blue, Specularity[White, 70], 
   Tube[{{2, 6, 0}, {21, 6, 0}, {21, -6, 0}, {2, -6, 0}, {2, 6, 0}}, 
    1/3]},
  {Cyan, Specularity[White, 70], 
   Tube[{{3, 3, 0}, {20, 3, 0}, {20, -3, 0}, {3, -3, 0}, {3, 3, 0}}, 
    1/3]},
  {Red, Specularity[White, 40], 
   Tube[BSplineCurve[{{23/2, 2, 0}, {19, 2, 0}, (* 
      über 1 *) {19, -3, 2},
      (* über 2, 3, 
      4 *) {19, -6, 2}, {19, -9, 2}, {19, -12, 2}, {19, -15, 0},
      (* unter 4, über 3 *) {18, -15, 0}, {18, -12, -2}, {18, -9, 2},
      (* unter 3, 
      unter 4 *) {18, -15/2, 0}, {17, -15/2, 
       0}, {17, -9, -2}, {17, -12, -2}, {17, -15, 0},
      (* über 4, unter 3, 
      über 2 *) {16, -15, 0}, {16, -12, 2}, {16, -9, -2}, {16, -6, 
       2}, {16, -9/2, 0}, {15, -9/2, 0},
      (* unter 2, unter 3, 
      über 4 *) {15, -6, -2}, {15, -9, -2}, {15, -12, 2}, {15, -15, 
       0}, {14, -15, 0},
      (* unter 4, 
      unter 3 *) {14, -12, -2}, {14, -9, -2}, {14, -15/2, 
       0}, {13, -15/2, 0},
      (* über 3, 
      unter 4 *) {13, -9, 2}, {13, -12, -2}, {13, -15, 0}, {12, -15, 
       0},
      (* über 4, über 3, unter 2, 
      über 1 *) {12, -12, 2}, {12, -9, 2}, {12, -6, -2}, {12, -3, 
       2}, {12, -3/2, 0}, {11, -3/2, 0},
      (* unter 1, unter 2, über 3, 
      über 4 *) {11, -3, -2}, {11, -6, -2}, {11, -9, 2}, {11, -12, 
       2}, {11, -15, 0}, {10, -15, 0},
      (* unter 4, 
      über 3 *){10, -12, -2}, {10, -9, 2}, {10, -15/2, 0}, {9, -15/2, 
       0},
      (* unter 3, 
      unter 4 *) {9, -9, -2}, {9, -12, -2}, {9, -15, 0}, {8, -15, 0},
      (* über 4, unter 3, 
      unter 2 *){8, -12, 2}, {8, -9, -2}, {8, -6, -2}, {8, -9/2, 
       0}, {7, -9/2, 0},
      (* über 2,unter 3, 
      über 4 *) {7, -6, 2}, {7, -9, -2}, {7, -12, 2}, {7, -15, 
       0}, {6, -15, 0},
      (* unter 4, 
      unter 3 *) {6, -12, -2}, {6, -9, -2}, {6, -15/2, 0}, {5, -15/2, 
       0},
      (* über 3, 
      unter 4 *) {5, -9, 2}, {5, -12, -2}, {5, -15, 0}, {4, -15, 0},
      (* über 4,über 3, über 2, 
      unter 1 *) {4, -12, 2}, {4, -9, 2}, {4, -6, 2}, {4, -3, -2}, {4,
        2, 2}, {23/2, 2, 0}
      }], 1/3]}
  }, ViewPoint -> Top, BoxRatios -> Automatic, Background -> Black, 
 Boxed -> False]

enter image description here

POSTED BY: Udo Krause
Posted 9 years ago

Hi Krause Thanks for the code. In your last graphics however the blues the green and yellow are not linked (only through he red). If I undo the red but only the red, all other loops will be trivial (loose). My situation is that every loop plays the same role, if I undo any loop (no only red) all will become undone!

POSTED BY: ali cherad

My situation is that every loop plays the same role, if I undo any loop all will become undone!

Exactly that is the case: If one removes the red link, the other 4 links are free, this is obvious and you have seen it too.

But what happens, if one removes the most inner (the cyan) rectangular link? Now one can weave out the read one (do it in your mind) and the other three are then free. The same holds for every of the rectangular links. So they have all the same topological quality.

POSTED BY: Udo Krause
Posted 9 years ago

Hi Sam I meant 5 links (not 5 intersections)

POSTED BY: ali cherad

So basically the model is not mathematically defined

So it's not a model but rather a description. Let's try it again: Is the following (see Borromean Rings) the intended representation to your description?

enter image description here

POSTED BY: Udo Krause

But note that the Discordian "mandala" is not actually Brunnian. An easy way to see that is to look at the folllowing sublink and notice that yellow lies atop red, which lies atop green, which lies atop yellow, etc., so these three are the Borromean rings (and are thus linked, which must not be the case if the original link is Brunnian). enter image description here

POSTED BY: Rick Mabry

Something like this

enter image description here

this one

... serves as an illustration of a symmetric Brunnian link of order 5.

So do you have a model of a Brunnian 5 link?

POSTED BY: Udo Krause
Posted 9 years ago

Hello Thanks for replying. Well yes I think i have a model. I am doing research in sociology and my model is to do with bureaucratic administration in the sense that there are 5 perspectives to be considered when a decision has to be made. However if one perspective is trivialized then the whole structure is functionally undone. So basically the model is not mathematically defined but I just want to use the graphics because they are very representative of my model.

POSTED BY: ali cherad
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