thank you for sharing, I'm going to print up a few!

On Sun, Apr 9, 2017 at 5:03 AM, Mr C Camacho chris@bedroomcoders.co.uk
wrote:

Does work with fn=3 you have to change the fudge multiple tho

For three sides it would be 1.33333

It needs changing in two places

It could be calculated if $fn was assigned from a "settings" variable called SIDES

Where the multiple would be 1+(1/SIDES)  .... off the top of my head...

Three sides does look very effective but you will have to rotate the whole knot on the x and y axis slightly for it to sit on the base

Hope this helps
Chris

On 10 April 2017 01:55:31 BST, MichaelAtOz oz.at.michael@gmail.com wrote:

--
Sent from my Android device with K-9 Mail. Please excuse my brevity.

okay he's for n sides while retaining its mobius quality - I will stop
playing with this now!

----snip-----

STEP=4; // fine in most cases
SCALE=14; // of overall knot
// three SIDES will need the whole model rotating to sit flat on base
SIDES=3; // number of sides in the profile 3 to 16 (unless printed very
large 16 should be plenty for printing)

$fn=SIDES; // use to select number of faces in the path profile

// calculates position in path a=angle s=scale
function knotAng(a,s) =
[(sin(a)+(2sin(2a)))s,
(cos(a)-(2cos(2a)))s,
sin(3a)(s*1.5)];

// from wiki

// Find the unitary vector with direction v. Fails if v=[0,0,0].
function unit(v) = norm(v)>0 ? v/norm(v) : undef;
// Find the transpose of a rectangular matrix
function transpose(m) = // m is any rectangular matrix of objects
[ for(j=[0:len(m[0])-1]) [ for(i=[0:len(m)-1]) m[i][j] ] ];
// The identity matrix with dimension n
function identity(n) = [for(i=[0:n-1]) [for(j=[0:n-1]) i==j ? 1 : 0] ];

// computes the rotation with minimum angle that brings a to b
// the code fails if a and b are opposed to each other
function rotate_from_to(a,b) =
let( axis = unit(cross(a,b)) )
axis*axis >= 0.99 ?
transpose([unit(b), axis, cross(axis, unit(b))]) *
[unit(a), axis, cross(axis, unit(a))] :
identity(3);

for (a=[0:STEP:360]) {
v = knotAng(a,SCALE) - knotAng(a+STEP,SCALE);
v2 = knotAng(a+STEP,SCALE) - knotAng(a+STEP+STEP,SCALE);
hull() {
translate(knotAng(a,SCALE))
multmatrix(rotate_from_to([0,0,1],v))
rotate(a*(1+(1.0/SIDES)))  // fudge
cylinder(r=SCALE/2,h=0.1,center=true);
translate(knotAng(a+STEP,SCALE))
multmatrix(rotate_from_to([0,0,1],v2))
rotate((a*(1+(1.0/SIDES)))+STEP) // fudge
cylinder(r=SCALE/2,h=0.1,center=true);
}
}

//base
translate([0,0,-1.35*(SCALE1.5)]) cylinder(r=3SCALE,h=2,$fn=128);

On 10/04/17 01:55, MichaelAtOz wrote:

nice catch, thats defo a coding error! fortunately its tiny enough to be
well outside the resolution of most 3d printers....

On 10/04/17 19:32, Ronaldo Persiano wrote:

On 10/04/17 19:32, Ronaldo Persiano wrote:

how do you mean? sounds interesting...

codifies wrote

Yes, it is indeed. You may want to take a look on it in  wikipedia
https://en.wikipedia.org/wiki/Torus_knot  . The particular knot you have
made is a knot(2,3):

http://forum.openscad.org/file/n21173/Knot2_3.png
​
In general in a knot(p,q), p is the number of turns the cord does around the
torus and q the number of times the cord goes through the torus hole. I
don't know how to change your definition of the knot path to include p and
q. I have used another definition:

where:

d is the cord section radius
R is the torus major radius
r  is the torus minor radius

The control of the sweep twist is important to get a nice form. Your fudge
does exactly that. In the image above I have used another twist value such
that one face of the swept cord is facing the torus surface along all the
path. Certainly there no Moebius strip with that twist.

If you are interested on that I can provide you my changes to your code with
the twist control.

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