Cylindrical Skew?
python code is here:
od=150
id=75
n=16
t=3
sl_ang=50
sl_h=(od/2-id/2)*tan(d2r(sl_ang))
a=c2t3(circle(od/2,s=n+1))
b=q_rot([f'z{360/len(a)/2}'],c2t3(circle(od/2,s=n+1)))
h=l_len(a[:2])/2
sol_1=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=flip(c2t3(circle(od/2,s=n+1)))
b=flip(q_rot([f'z{-360/len(a)/2}'],c2t3(circle(od/2,s=n+1))))
h=l_len(a[:2])/2
sol_2=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=c2t3(circle(id/2,s=n+1))
b=q_rot([f'z{360/len(a)/2}'],c2t3(circle(id/2,s=n+1)))
h=l_len(a[:2])/2
sol_3=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=flip(c2t3(circle(id/2,s=n+1)))
b=flip(q_rot([f'z{-360/len(a)/2}'],c2t3(circle(id/2,s=n+1))))
h=l_len(a[:2])/2
sol_4=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=[seg(p)[:-1] for p in cpo(sol_1)]
b=[seg(p)[:-1] for p in translate([0,0,sl_h],cpo(sol_3))]
sol_5=array([a,b]).transpose(1,2,0,3,4).tolist()
a=[seg(p)[:-1] for p in cpo(sol_2)]
b=[seg(p)[:-1] for p in translate([0,0,sl_h],cpo(sol_4))]
sol_6=array([a,b]).transpose(1,2,0,3,4).tolist()
with open('trial.scad','w+') as f:
f.write(f'''
include<dependencies2.scad>
for(p={sol_5})for(p1=p)
let(
a=slice_sol(p1,10)
)
for(i=[1:len(a)-1])
hull(){{
p_line3d(a[i-1],{t/2},1);
p_line3d(a[i],{t/2},1);
}}
for(p={sol_6})for(p1=p)
let(
a=slice_sol(p1,10)
)
for(i=[1:len(a)-1])
hull(){{
p_line3d(a[i-1],{t/2},1);
p_line3d(a[i],{t/2},1);
}}
''')
Apart from above you would need file dependencies2.scad, a library I have
written , which is a much smaller version of dependencies.scad
Also attached revised trial.scad
On Sun, 3 Mar 2024 at 22:36, azdle azdle@azdle.net wrote:
Yeah! That looks like pretty much exactly what I'm going for.
Any chance you can share the scad file and the python too?
On Sun, Mar 3 2024 at 10:14:47 PM +0530, Sanjeev Prabhakar
sprabhakar2006@gmail.com wrote:
I do all the calculations in python and render results in openSCAD
and use native features of openSCAD like intersection, difference etc
to draw shapes.
and it is perfectly parametric in python
Is this the shape?
Wonderful, thanks!
On Sun, Mar 3 2024 at 10:41:33 PM +0530, Sanjeev Prabhakar
sprabhakar2006@gmail.com wrote:
python code is here:
od=150
id=75
n=16
t=3
sl_ang=50
sl_h=(od/2-id/2)*tan(d2r(sl_ang))
a=c2t3(circle(od/2,s=n+1))
b=q_rot([f'z{360/len(a)/2}'],c2t3(circle(od/2,s=n+1)))
h=l_len(a[:2])/2
sol_1=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=flip(c2t3(circle(od/2,s=n+1)))
b=flip(q_rot([f'z{-360/len(a)/2}'],c2t3(circle(od/2,s=n+1))))
h=l_len(a[:2])/2
sol_2=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=c2t3(circle(id/2,s=n+1))
b=q_rot([f'z{360/len(a)/2}'],c2t3(circle(id/2,s=n+1)))
h=l_len(a[:2])/2
sol_3=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=flip(c2t3(circle(id/2,s=n+1)))
b=flip(q_rot([f'z{-360/len(a)/2}'],c2t3(circle(id/2,s=n+1))))
h=l_len(a[:2])/2
sol_4=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=[seg(p)[:-1] for p in cpo(sol_1)]
b=[seg(p)[:-1] for p in translate([0,0,sl_h],cpo(sol_3))]
sol_5=array([a,b]).transpose(1,2,0,3,4).tolist()
a=[seg(p)[:-1] for p in cpo(sol_2)]
b=[seg(p)[:-1] for p in translate([0,0,sl_h],cpo(sol_4))]
sol_6=array([a,b]).transpose(1,2,0,3,4).tolist()
with open('trial.scad','w+') as f:
f.write(f'''
include<dependencies2.scad>
for(p={sol_5})for(p1=p)
let(
a=slice_sol(p1,10)
)
for(i=[1:len(a)-1])
hull(){{
p_line3d(a[i-1],{t/2},1);
p_line3d(a[i],{t/2},1);
}}
for(p={sol_6})for(p1=p)
let(
a=slice_sol(p1,10)
)
for(i=[1:len(a)-1])
hull(){{
p_line3d(a[i-1],{t/2},1);
p_line3d(a[i],{t/2},1);
}}
''')
Apart from above you would need file dependencies2.scad, a library I
have written , which is a much smaller version of dependencies.scad
Also attached revised trial.scad
On Sun, 3 Mar 2024 at 22:36, azdle azdle@azdle.net wrote:
Yeah! That looks like pretty much exactly what I'm going for.
Any chance you can share the scad file and the python too?
On Sun, Mar 3 2024 at 10:14:47 PM +0530, Sanjeev Prabhakar
sprabhakar2006@gmail.com wrote:
I do all the calculations in python and render results in openSCAD
and use native features of openSCAD like intersection, difference
etc
to draw shapes.
and it is perfectly parametric in python
Is this the shape?
sorry i missed
you would also need to import another python library written by me:
openscad1.py
it is available here:
https://github.com/sprabhakar2006/openSCAD/blob/main/openscad1.py
it is a little complicated initially
On Sun, 3 Mar 2024 at 22:41, Sanjeev Prabhakar sprabhakar2006@gmail.com
wrote:
python code is here:
od=150
id=75
n=16
t=3
sl_ang=50
sl_h=(od/2-id/2)*tan(d2r(sl_ang))
a=c2t3(circle(od/2,s=n+1))
b=q_rot([f'z{360/len(a)/2}'],c2t3(circle(od/2,s=n+1)))
h=l_len(a[:2])/2
sol_1=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=flip(c2t3(circle(od/2,s=n+1)))
b=flip(q_rot([f'z{-360/len(a)/2}'],c2t3(circle(od/2,s=n+1))))
h=l_len(a[:2])/2
sol_2=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=c2t3(circle(id/2,s=n+1))
b=q_rot([f'z{360/len(a)/2}'],c2t3(circle(id/2,s=n+1)))
h=l_len(a[:2])/2
sol_3=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=flip(c2t3(circle(id/2,s=n+1)))
b=flip(q_rot([f'z{-360/len(a)/2}'],c2t3(circle(id/2,s=n+1))))
h=l_len(a[:2])/2
sol_4=[translate([0,0,ih],a) if i%2==0
else translate([0,0,ih],b)
for i in range(5)]
a=[seg(p)[:-1] for p in cpo(sol_1)]
b=[seg(p)[:-1] for p in translate([0,0,sl_h],cpo(sol_3))]
sol_5=array([a,b]).transpose(1,2,0,3,4).tolist()
a=[seg(p)[:-1] for p in cpo(sol_2)]
b=[seg(p)[:-1] for p in translate([0,0,sl_h],cpo(sol_4))]
sol_6=array([a,b]).transpose(1,2,0,3,4).tolist()
with open('trial.scad','w+') as f:
f.write(f'''
include<dependencies2.scad>
for(p={sol_5})for(p1=p)
let(
a=slice_sol(p1,10)
)
for(i=[1:len(a)-1])
hull(){{
p_line3d(a[i-1],{t/2},1);
p_line3d(a[i],{t/2},1);
}}
for(p={sol_6})for(p1=p)
let(
a=slice_sol(p1,10)
)
for(i=[1:len(a)-1])
hull(){{
p_line3d(a[i-1],{t/2},1);
p_line3d(a[i],{t/2},1);
}}
''')
Apart from above you would need file dependencies2.scad, a library I have
written , which is a much smaller version of dependencies.scad
Also attached revised trial.scad
On Sun, 3 Mar 2024 at 22:36, azdle azdle@azdle.net wrote:
Yeah! That looks like pretty much exactly what I'm going for.
Any chance you can share the scad file and the python too?
On Sun, Mar 3 2024 at 10:14:47 PM +0530, Sanjeev Prabhakar
sprabhakar2006@gmail.com wrote:
I do all the calculations in python and render results in openSCAD
and use native features of openSCAD like intersection, difference etc
to draw shapes.
and it is perfectly parametric in python
Is this the shape?
Is the final thing supposed to be cylindrical?
If so, maybe make it oversized and then intersect to make it curved.
Here's my take on this in library-free OpenSCAD:
wall = 3;
cells_around = 16;
cells_down = 3;
down_ang = 50;
body_od = 150;
body_id = 75;
circ = body_od * PI;
cell_side = circ / cells_around / sqrt(2);
cell_l = (cells_down + 1) * cell_side;
cell_h = circ/cells_around;
module xflip_copy() {children(); mirror([1,0,0]) children();}
module zxskew(a) {multmatrix([[1,0,0,0],[0,1,0,0],[tan(a),0,1,0],[0,0,0,1]]) children();}
for (zi = [0:1:cells_down-2]) translate([0,0,cell_hzi]) {
xflip_copy() {
for (ri = [0:1:cells_around-1]) rotate(360ri/cells_around) {
difference() {
zxskew(-down_ang) {
linear_extrude(height=cell_h, center=true, twist=-360/cells_around, convexity=4) {
rotate(-180/cells_around) {
translate([body_od/4,0,0]) {
square([body_od/2,wall], center=true);
}
}
}
}
cylinder(d=body_id, h=10*cell_h, center=true);
}
}
}
}

- Revar
