How to make this parametric code?
Rather than include the brute-force code I generated, it's easier to describe
it and provide the results.
http://forum.openscad.org/file/n18352/curve.jpg
The primary dimensions in this drawing are the vertical and horizontal
segments ending at the blue handles. In this specific image, the vertical is
75 mm and the horizontal is 12.7 mm.
The arc center is shown as the blue handle to the right. I used AutoCAD to
generate a circle with two points at the ends of the lines and an arbitrary
third location to provide a "pleasing" curve shown in the image. Doing so
provided a center point reference to use in OpenSCAD.
The circle was extruded as a torus using the information from the drawing.
It didn't fit perfectly for reasons unknown to me, but I then manually
adjusted the figures until I had the desired result, hence the "brute-force"
description.
In order to eyeball the dimensions, I dropped a cylinder at the top of the
curve and at the bottom and adjusted the torus diameter and location until
it looked close. That's just not good coding.
http://forum.openscad.org/file/n18352/curvedduct.png
The flange at the bottom can be ignored. A cylinder is generated at the
origin and the torus is subtracted. The curved object so created is then
subtracted from the same object scaled up in x and y creating an open duct
with a suitable curve as shown.
I performed a number of searches but lack the correct description for my
searches to locate the proper formulae. I'm not certain that a correct
formula exists but am hopeful that one can be created.
Having only the two points makes me believe that it would result in an
ambiguous solution. The other constraint is that the circle should be
tangent to the top-most point, to keep the curve to the right of the
vertical line in all cases. Perhaps that constraint reduces or removes the
ambiguity?
Given the two points, the tangent restriction and the need to know an arc's
(circle's) center, can this process be written in parametric form?
I am open to other methods of creating such a duct from two reference
points, but that seems a tough one in itself.
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Yes the fact that it is a tangent at the top will define a unique circle
with two points. You haven't drawn it as a tangent in your 2D sketch
though. The center needs to be at the same height as your top point, i.e.
75mm, and equidistant from your top point and your bottom point. You can
then use the equation of a circle x^2 + y^2 = r^2 to find the radius, which
is also the x coordinate of the centre.
At your bottom point (r - 12.7)^2 + 75^2 = r^2. Here is code to solve for r
and draw the curve:
yc = 75;
x0 = 12.7;
function sqr(x) = x * x;
xc = (sqr(yc) + sqr(x0)) / (2 * x0);
echo(xc);
difference() {
square([x0, yc]);
translate([xc,yc]) circle(xc, $fn = 360);
}
On 10 September 2016 at 00:49, fred_dot_u fred_dot_u@yahoo.com wrote:
Rather than include the brute-force code I generated, it's easier to
describe
it and provide the results.
http://forum.openscad.org/file/n18352/curve.jpg
The primary dimensions in this drawing are the vertical and horizontal
segments ending at the blue handles. In this specific image, the vertical
is
75 mm and the horizontal is 12.7 mm.
The arc center is shown as the blue handle to the right. I used AutoCAD to
generate a circle with two points at the ends of the lines and an arbitrary
third location to provide a "pleasing" curve shown in the image. Doing so
provided a center point reference to use in OpenSCAD.
The circle was extruded as a torus using the information from the drawing.
It didn't fit perfectly for reasons unknown to me, but I then manually
adjusted the figures until I had the desired result, hence the
"brute-force"
description.
In order to eyeball the dimensions, I dropped a cylinder at the top of the
curve and at the bottom and adjusted the torus diameter and location until
it looked close. That's just not good coding.
http://forum.openscad.org/file/n18352/curvedduct.png
The flange at the bottom can be ignored. A cylinder is generated at the
origin and the torus is subtracted. The curved object so created is then
subtracted from the same object scaled up in x and y creating an open duct
with a suitable curve as shown.
I performed a number of searches but lack the correct description for my
searches to locate the proper formulae. I'm not certain that a correct
formula exists but am hopeful that one can be created.
Having only the two points makes me believe that it would result in an
ambiguous solution. The other constraint is that the circle should be
tangent to the top-most point, to keep the curve to the right of the
vertical line in all cases. Perhaps that constraint reduces or removes the
ambiguity?
Given the two points, the tangent restriction and the need to know an arc's
(circle's) center, can this process be written in parametric form?
I am open to other methods of creating such a duct from two reference
points, but that seems a tough one in itself.
--
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to-make-this-parametric-code-tp18352.html
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That's a beautiful simplification. Rather than playing with a torus and
subtracting it from a cylinder, I've taken your code and rotate_extruded it
for the desired shape.
I've yet to figure out how to create a consistent wall thickness after the
rotation, or perhaps I should be doing so before? Hmm, the "after" method
is giving me fits, but the "before" method may be a good bit simplere.
Thank you for your contribution. This will make future ducts so much easier
to create.
--
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To get a consistent wall thickness the inner wall needs to be part of a
circle with the same centre as the outer wall but a bigger radius.
$fn = 360;
yc = 75;
x0 = 12.7;
wall = 3;
or = 20;
ir = or - wall;
function sqr(x) = x * x;
xc = (sqr(yc) + sqr(x0)) / (2 * x0);
rotate_extrude()
translate([ir, 0])
intersection() {
difference() {
square([x0 + wall, yc]);
translate([xc + wall, yc]) circle(xc);
}
translate([xc + wall, yc]) circle(xc + wall);
}
On 10 September 2016 at 18:46, fred_dot_u fred_dot_u@yahoo.com wrote:
That's a beautiful simplification. Rather than playing with a torus and
subtracting it from a cylinder, I've taken your code and rotate_extruded it
for the desired shape.
I've yet to figure out how to create a consistent wall thickness after the
rotation, or perhaps I should be doing so before? Hmm, the "after" method
is giving me fits, but the "before" method may be a good bit simplere.
Thank you for your contribution. This will make future ducts so much easier
to create.
--
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to-make-this-parametric-code-tp18352p18356.html
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Not surprisingly, your code is far simpler and orders of magnitude more
elegant than mine. I used to be able to think better, but I was much younger
too.
Thanks again for your assistance.
--
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What you may want with this particular shape is to have walls with a
constant horizontal cross section designed for a set number of extruder
passes to avoid weird things with infilling tiny gaps. To do that, you
would just rotate extrude the same curve translated closer to the origin.
On Sun, Sep 11, 2016 at 3:38 AM, nop head nop.head@gmail.com wrote:
To get a consistent wall thickness the inner wall needs to be part of a
circle with the same centre as the outer wall but a bigger radius.
$fn = 360;
yc = 75;
x0 = 12.7;
wall = 3;
or = 20;
ir = or - wall;
function sqr(x) = x * x;
xc = (sqr(yc) + sqr(x0)) / (2 * x0);
rotate_extrude()
translate([ir, 0])
intersection() {
difference() {
square([x0 + wall, yc]);
translate([xc + wall, yc]) circle(xc);
}
translate([xc + wall, yc]) circle(xc + wall);
}
On 10 September 2016 at 18:46, fred_dot_u fred_dot_u@yahoo.com wrote:
That's a beautiful simplification. Rather than playing with a torus and
subtracting it from a cylinder, I've taken your code and rotate_extruded
it
for the desired shape.
I've yet to figure out how to create a consistent wall thickness after the
rotation, or perhaps I should be doing so before? Hmm, the "after" method
is giving me fits, but the "before" method may be a good bit simplere.
Thank you for your contribution. This will make future ducts so much
easier
to create.
--
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to-make-this-parametric-code-tp18352p18356.html
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I've seen a video in which the narrator specifies exactly what you describe.
The extrusion width should be evenly divided into the wall thickness for the
best result on thin walls. Simplify3D and perhaps other slicers will "play
back" the toolpath and allow one to observe the best combination of
thickness and extrusion.
I'd expect that a well-calculated path would reduce the print time by quite
a large factor.
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Another thought appeared recently in the light bulb that popped up above my
head. What other curves can be used to create a similar profile? The
objective would be to have no constriction from the smaller ID but any
variation from that location to the larger ID is acceptable.
How would one place and calculate a parabolic curve, or is such a curve
impractical given the above restrictions? Hyperbolas? or is it hyperbolae?
Google says either one is acceptable.
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Parabolas give a nice curve and require a very simple code:
s = 75/2; // choose any value between 0 and top_point[1]
small_radius = 30; // the radius at top
top_point = [small_radius, 75];
bottom_point = [12.7+small_radius, 0];
middle_point = [small_radius, s];
np = 10;
// this parabola will be tangent to the segment |top_point, middle_point|
at the top_point
// and to the segment |middle_point, bottom_point| at the bottom_point
function parabola(np) =
[ for(i=[0:np]) let( u=i/np )
top_point*(1-u)(1-u) + middle_point2*(1-u)u +
bottom_pointu*u ];
rotate_extrude()
polygon(concat([ [0,top_point[1]] ], parabola(np=np), [[0,0]] ));
translate([0, 0, top_point[1]+50])
cylinder(r=small_radius, h=100, center=true);
However, I guess the offset of a parabola is not a parabola.
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If you have AutoCad there is simple to do this. In AutoCad create a DXF file
of the profile. Then use the DXF import function within OpenSCAD. Note
that OpenSCAD only recognizes ARC and LINE entities from a DXF file.
http://forum.openscad.org/file/n18370/acad_arc.jpg
http://forum.openscad.org/file/n18370/arc3.jpg
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