I'm struggling how to approximate the shape of the sails on this windmill:
[image: image.png]
The edges AC and BC could probably be defined as a catenary, but beyond
that I'm struggling to come up with anything. Suggestions?
Dan
yes, but but A-B is a straight!
so you can
On Fri, Jun 28, 2024 at 11:49 AM Dan Perry via Discuss <
discuss@lists.openscad.org> wrote:
I'm struggling how to approximate the shape of the sails on this windmill:
[image: image.png]
The edges AC and BC could probably be defined as a catenary, but beyond
that I'm struggling to come up with anything. Suggestions?
Dan
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
The catenary is from A to the midpoint of BC.
On Fri, Jun 28, 2024 at 6:13 AM Guenther Sohler via Discuss <
discuss@lists.openscad.org> wrote:
yes, but but A-B is a straight!
so you can
The OP said "approximate". I would think that a catenary would be more of
an "exact" solution.
IMO, an approximation would simply be a large radius arc-segment.
On Fri, Jun 28, 2024 at 11:00 AM David Phillip Oster via Discuss <
discuss@lists.openscad.org> wrote:
The catenary is from A to the midpoint of BC.
On Fri, Jun 28, 2024 at 6:13 AM Guenther Sohler via Discuss <
discuss@lists.openscad.org> wrote:
yes, but but A-B is a straight!
so you can
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
I would agree on this, but as far as I remember from long ago in
Engineering School, the catenary only describes the shape of the rope if:
It's density is homogeneous (which is almost always implied, when not
stated otherwise),
we are in an homogeneous gravitational field (which is even more
implied, if we don't have an awful long rope and the extremes are very
far away).
the rope is subject to its own weight only, which in this case doesn't
seem to apply.
On 28/06/2024 18:11, Joe H via Discuss wrote:
The OP said "approximate". I would think that a catenary would be
more of an "exact" solution.
IMO, an approximation would simply be a large radius arc-segment.
On Fri, Jun 28, 2024 at 11:00 AM David Phillip Oster via Discuss
discuss@lists.openscad.org wrote:
The catenary is from A to the midpoint of BC.
On Fri, Jun 28, 2024 at 6:13 AM Guenther Sohler via Discuss
<discuss@lists.openscad.org> wrote:
yes, but but A-B is a straight!
so you can
* define the catenary(chain- curve) in 2D space
* linear_extrude it to 3D space
* use multmatrix to skew it on the way from AB to C
* mirror-copy the result as you probably only get half of the
riangle
_______________________________________________
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
OpenSCAD mailing list
To unsubscribe send an email todiscuss-leave@lists.openscad.org
I think if the wind is uniform across the sail hitting it straight on it
would have pretty much the same effect as gravity on a rope. The catenary
formula comes from balancing tension along the rope with gravity and
integrating. There is a function in NopSCADlib to make one:
https://github.com/nophead/NopSCADlib/tree/master?tab=readme-ov-file#catenary
On Fri, 28 Jun 2024 at 17:36, Andreas Croci via Discuss <
discuss@lists.openscad.org> wrote:
I would agree on this, but as far as I remember from long ago in
Engineering School, the catenary only describes the shape of the rope if:
It's density is homogeneous (which is almost always implied, when not
stated otherwise),
we are in an homogeneous gravitational field (which is even more
implied, if we don't have an awful long rope and the extremes are very far
away).
the rope is subject to its own weight only, which in this case doesn't
seem to apply.
On 28/06/2024 18:11, Joe H via Discuss wrote:
The OP said "approximate". I would think that a catenary would be more of
an "exact" solution.
IMO, an approximation would simply be a large radius arc-segment.
On Fri, Jun 28, 2024 at 11:00 AM David Phillip Oster via Discuss <
discuss@lists.openscad.org> wrote:
The catenary is from A to the midpoint of BC.
On Fri, Jun 28, 2024 at 6:13 AM Guenther Sohler via Discuss <
discuss@lists.openscad.org> wrote:
yes, but but A-B is a straight!
so you can
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
looks to me a shape like this:
[image: Screenshot 2024-06-28 at 10.54.30 PM.png]
On Fri, 28 Jun 2024 at 15:20, Dan Perry via Discuss <
discuss@lists.openscad.org> wrote:
I'm struggling how to approximate the shape of the sails on this windmill:
[image: image.png]
The edges AC and BC could probably be defined as a catenary, but beyond
that I'm struggling to come up with anything. Suggestions?
Dan
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
I found a solution:
[image: image.png]
I used the bezier functions in BOSL2. Adjusting the center_z and edge_z
parameters is a quick way to tweak the shape of the sail. My solution to
turn the zero thickness bezier patch into a 1 unit thick shape is
inelegant, hopefully the BOSL2 team can explain a better way.
include <BOSL2/std.scad>
include <BOSL2/beziers.scad>
$fs = 0.2;
$fa = 1.0;
center_z = 20;
edge_z = 8;
patch = [
[[0, 50, 0], [25, 33.3, edge_z], [50, 16.7, edge_z], [75, 0, 0]],
[[0, 0, 0], [25, 0, center_z], [50, 0, center_z-5], [75, 0, 0]],
[[0, -50, 0], [25, -33.3, edge_z], [50, -16.7, edge_z], [75, 0, 0]]
];
// debug_bezier_patches(patches=[patch], size=1, showcps=false);
vnf = bezier_vnf([
patch,
up(1, patch),
]);
difference() {
hull() vnf_polyhedron(vnf);
translate([0, 0, -1]) scale([1.01, 1.01, 1.00]) hull() vnf_polyhedron
(vnf);
}
On Fri, Jun 28, 2024 at 10:49 AM Dan Perry dan.p3rry@gmail.com wrote:
I'm struggling how to approximate the shape of the sails on this windmill:
[image: image.png]
The edges AC and BC could probably be defined as a catenary, but beyond
that I'm struggling to come up with anything. Suggestions?
Dan