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.

--
View this message in context: http://forum.openscad.org/simple-Celtic-knot-tp21141p21173.html
Sent from the OpenSCAD mailing list archive at Nabble.com.

I posted (here) a newer version where any value of SIDES creates the
correct twist

@Ronaldo I'd like to see it too, could you post it here for everyone ?

On 11/04/17 04:55, Neon22 wrote:

2017-04-11 5:16 GMT-03:00 Mr C Camacho chris@bedroomcoders.co.uk:

Sorry for the delay in answering you. I have been struggling with
trigonometric stuff trying to understand the specific parametrization of
the knot 2,3 you have used in your celtic knot. It produces a far better
shape the the parametrization I have been using. The effort was
rewarding... at least for me. :)

I will consider here the celtic knot parametrization (CKP) in the following
form without the scale parameters (they can be applied in the calling code):

function knot1(a) =  // CKP parametrization for knot 2,3
[(sin(a)+(2sin(2a))),
(cos(a)-(2cos(2a))),
sin(3*a)];

and I will compare it with the traditional parametrization for knots p,q:

function knot2(a,p,q,R,r) =
[ (rcos(qa) + R)cos(pa),
(rcos(qa) + R)sin(pa),
rsin(qa) ];

This parametrization may be rewritten to be clear that it is a knot on a
circular torus:

function knot3(a,p,q,R,r) =
let( t = [(rcos(qa) + R), 0, rsin(qa) ] ) // circle in XZ
[ [ cos(pa), sin(pa), 0],
[-sin(pa), cos(pa), 0],  // rotation in Z of p*a degrees
[0,0,1] ] * t;

This parametrization is a rotation in Z with an angle p*a of a point 't'
that also depends on 'a'. The expression of point t is itself a
parametrization of a circle of radius 'r' on the XZ plane centered at the
point [ R,0,0 ]. For 0<= a <= 360, the parametrization t does q turns on
the circle and the rotation does p turns around Z axis. The net result will
be a p,q knot on the surface of a circular torus with major radius 'R' and
minor radius 'r'. The following image shows the knot 2,3 (also known as
trefoil) computed with knot3():

​

In the image, a partial torus appears transparent and its transversal
section highlighted in red. Note that the knot cord section center is on
the torus surface.

However, the shape of the knot generated by knot1() is far better specially
in the section it goes thought the torus hole.

​

Observe that the torus section of knot1() parametrization is not a circle
anymore. If we get a good parametrization of this section we may rewrite
knot1() in the form:

function knot1a(a) =
let( t = section_parametrization(3a) )
[ [ cos(2a), sin(2a), 0],
[-sin(2a), cos(2a), 0],  // rotation in Z of 2a degrees
[0,0,1] ] * t;

and then substitute p for 2 and q for 3 to get a general knot p,q with the
good shape of knot1().

Here is where the trigonometric stuffs are handy. After many failed trials
I have found the proper parametrization behind knot1().

function knot1b(a) =
let( t = [ sin(3a), cos(3a) - 2, sin(3a) ] ) // section
parametrization
[ [ cos(2a), -sin(2a), 0],
[ sin(2a),  cos(2a), 0],  // rotation in z of 2a degrees
[0,0,1] ] *
t;

This formulation is exactly equivalent to knot1() but it is now evident the
p,q numbers in it. If we compare knot1b() with knot3 it is also evident
that the section parametrization here is not a vertical section of the
torus anymore. In fact, if we sweep just the section t, we get:

​
The torus transversal section is a slanted ellipsis!

Now, introducing p, q, R and r in the knot1b() we get our general form of
knot1():

function knot4(a,p,q,R,r) =
let( t = [ sin(qa+alpha), cos(qa+beta) - R, rsin(qa) ] ) //
elliptical slanted section
[ [ cos(pa), -sin(pa), 0],
[ sin(pa),  cos(pa), 0],  // rotation in z of p*a degrees
[0,0,1] ] * t;

With this formulation we may explore other knots like the cinquefoil (3,5)
and its "dual" 5,3 :

​
​

Along of my multiple attempts to find knot4() expressions I have discovered
new possibilities to define interesting parametrizations of p,q knots, like
for instance the following interwoven basket:

​
This is a knot 5,8 we get with simple changes in knot4() introducing two
more shape parameters:

function knot5(a,p,q,R,r,alpha,beta) =
let( t = [ sin(qa+alpha), cos(qa+beta) - R, rsin(qa) ] ) //
elliptical slanted transversal section
[ [ cos(pa), -sin(pa), 0],
[ sin(pa),  cos(pa), 0],  // rotation in z of p*a degrees
[0,0,1] ] * t;

The phase parameters alpha and beta are angles that produce very different
torus shapes. In the last basket-like knot I have used alpha=beta=60. In
the following, we have p=7, q=11, alpha=80 and beta=50 with a different look

​
​
This same technique may be introduced in the definition of the circular
knot3():

function knot6(a,p,q,R,r) =
let( t = [(rcos(qa+alpha) + R), 0, rsin(qa) ] ) // circle in xz
[ [ cos(pa), sin(pa), 0],
[-sin(pa), cos(pa), 0],  // rotation in z of p*a degrees
[0,0,1] ] * t;

to achieve other basket shapes:

​

with the same parameters (7,11,80,50).

All this open up infinite possibilities to shape torus knots: just define a
suitable expression of t in knot5() !

That was a rather long message. I will leave the sweep aspects to another
time.

Michael,

To make my discussion clearer I have dropped all scales from knotAng(). To
apply the my knot functions to @codifies code you should recode knotAng()
as:

p = 7;
q = 11;
alpha = 80;
beta =50;
function knotAng(a,s) =
[[s,0,0],[0,s,0],[0,0,s*1.5]]*knot4(a,p,q,2,1);

where you may use almost any of my knot functions instead of knot4(). Chose
the values of p, q, alpha and beta at your will.

The second @codifies code possibly work for any case even with SIDES=3, but
I haven't tested this. For higher values of p and/or q it is advisable to
reduce the cylinder radius and the STEP size. In my examples I have used
larger values for SIDES to get a rounded cord and less apparent twists.

I don't know what you mean by:
"I can see how it gets a face pointing to the centre."

2017-04-13 0:10 GMT-03:00 MichaelAtOz oz.at.michael@gmail.com: