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Adrian Mariano

Fri, Aug 5, 2022 10:38 PM

I don't have a specific need for a circumscribing elliptical polygon with
equal sides. I was writing code to make ellipses and it seemed like equal
side approximations would be nice, so I solved the case for inscribed, but
couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
that this discussion reminded me of.

On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:

A long discussion but I have to ask why do you need an elliptical polygon
with equal tangential sides?

On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:

You think that a circumscribing polygon with equal length segments does
not exist?? I'm skeptical of this claim.

For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.

So for example, I have segments lengths of:

ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]

Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
[image: image.png]

The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.

On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:

I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.

On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:

I think you need to be a little careful about defining what you want to
do. If you want to approximate an ellipse with points that are distributed
uniformly over the ellipse, that could be done by approximating the arc
length of the ellipse to find the necessary points---e.g. advance distance
d along the ellipse at each step. But if you want to approximate an
ellipse with a polygon with equal length sides, that approach does not do
it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.

On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:

I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)

Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.

On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:

Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.

On Tue, Aug 2, 2022 at 10:56 PM Father Horton fatherhorton@gmail.com
wrote:

Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).

On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:

Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse

On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:

In openscad drawing a ellipse is not too difficult
Refer following code

function to draw an ellipse with semi-major and semi-minor axis
"r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);

function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;

On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:

On 8/2/2022 4:46 PM, Brian Allen wrote:

I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);

Tip: use scale() rather than resize(); it previews faster in
complicated cases.

It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.

You're taking a circle and stretching it; the result of that
stretching is an ellipse.

I'm not 100% sure - maybe one of the real math wizards can chime
in - but I think the shape you are asking for would not be an ellipse.

I believe that an ellipse is fully defined by its major (long)
and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.

You can make something else, but you can't make it using this
technique.

Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.

You can use a Bézier function.

Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves

DIY, here's a function that I cobbled together some years ago:

// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);

function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];

// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];

It is that final function bez() that I actually call, e.g.

polygon(bez([0,0], [0,4], [5,1], [7,0], 20));

often concatenating the results of several bez() calls together.

Note that the four coordinates correspond to the coordinates of
the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.

Note that Bézier functions are not the only mathematical curves.
(For instance, they cannot exactly match an arc of a circle.) I'm sure
there are numerous other functions that also produce curves.

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To unsubscribe send an email to discuss-leave@lists.openscad.org

I don't have a specific need for a circumscribing elliptical polygon with equal sides. I was writing code to make ellipses and it seemed like equal side approximations would be nice, so I solved the case for inscribed, but couldn't solve the circumscribed case. So it's an unsolved puzzle I've got that this discussion reminded me of. On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote: > A long discussion but I have to ask why do you need an elliptical polygon > with equal tangential sides? > > On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote: > >> You think that a circumscribing polygon with equal length segments does >> not exist?? I'm skeptical of this claim. >> >> For circumscribed, it's not so hard. You start with a uniform-in-angle >> point distribution, compute the segment lengths, compare to the mean >> segment length to get an error, and then compute an adjustment based on the >> error. It converges quickly. >> >> So for example, I have segments lengths of: >> >> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, >> 5.57297] >> >> Maybe not exactly equal, since tolerance is 1e-9, but surely good enough. >> [image: image.png] >> >> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm >> should exist for the circumscribed case, but as I said, my efforts to find >> it have failed. >> >> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar < >> sprabhakar2006@gmail.com> wrote: >> >>> I think circumscribed polygon segments cannot be equal lengths. >>> inscribed ones may be approximately equal, but would be a very tough >>> problem to solve. >>> I tried something. >>> >>> >>> >>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote: >>> >>>> I think you need to be a little careful about defining what you want to >>>> do. If you want to approximate an ellipse with points that are distributed >>>> uniformly over the ellipse, that could be done by approximating the arc >>>> length of the ellipse to find the necessary points---e.g. advance distance >>>> d along the ellipse at each step. But if you want to approximate an >>>> ellipse with a polygon with equal length sides, that approach does not do >>>> it, because the length of the segments will not be constant as the span >>>> sections of the ellipse with differing curvature. And since the perimeter >>>> of the desired polygon is unknown, you can't just step it out analogously. >>>> The circle doesn't present this challenge due to having constant >>>> curvature. >>>> >>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com> >>>> wrote: >>>> >>>>> I think our emails crossed. I was addressing how to approximate an >>>>> ellipse with constant-length segments in general, not how to solve your >>>>> problem. When I had to do something similar, I used polar coordinates and >>>>> stepped up the value of theta until the distance from the previous point >>>>> was correct. (In the constant-length version, I'd keep a running total of >>>>> the maximum desired length so that round-off errors don't accumulate.) >>>>> >>>>> Figuring out the circumscribed polygon is, as you note, a greater >>>>> challenge. I can visualize how to do it for n = 4, but it doesn't >>>>> extrapolate at all. >>>>> >>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu> >>>>> wrote: >>>>> >>>>>> Eh, there's no such thing as "brute force" in this situation. Step >>>>>> through *what* with sufficiently small steps? Of course an iterative >>>>>> algorithm is necessary. This is true for most interesting problems. No >>>>>> big deal. Show me an iterative algorithm that converges to the >>>>>> circumscribing polygon. When I tried to devise such an algorithm I >>>>>> couldn't find one that would converge. Writing an algorithm that would >>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note >>>>>> also that computing arc length of the ellipse doesn't really help here. >>>>>> You can do that by solving an elliptical integral, or summing up small >>>>>> segments. But what's the connection between arc length of the ellipse and >>>>>> a circumscribing polygon with equal segment lengths? If there's a >>>>>> connection, I don't see it. >>>>>> >>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com> >>>>>> wrote: >>>>>> >>>>>>> Except by brute force (step through it with sufficiently small >>>>>>> steps), it’s not possible because there’s no closed form expression for the >>>>>>> distance between two points on a general ellipse (as opposed to special >>>>>>> cases like a circle). >>>>>>> >>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu> >>>>>>> wrote: >>>>>>> >>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N >>>>>>>> segments that are all the same length that circumscribes a perfect ellipse >>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not >>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon >>>>>>>> with all segments the same length---though it's not easy---but the >>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every >>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment >>>>>>>> is tangent to the ellipse >>>>>>>> >>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar < >>>>>>>> sprabhakar2006@gmail.com> wrote: >>>>>>>> >>>>>>>>> In openscad drawing a ellipse is not too difficult >>>>>>>>> Refer following code >>>>>>>>> >>>>>>>>> function to draw an ellipse with semi-major and semi-minor axis >>>>>>>>> "r1" and "r2" respectively and with center "cp" and number of segment "s" >>>>>>>>> / >>>>>>>>> // example: >>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30); >>>>>>>>> >>>>>>>>> function ellipse(r1,r2,cp,s=30)= >>>>>>>>> let( >>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]] >>>>>>>>> )sec; >>>>>>>>> >>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, < >>>>>>>>> openscad@jordan.maileater.net> wrote: >>>>>>>>> >>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote: >>>>>>>>>> >>>>>>>>>> I created an oval using: >>>>>>>>>> inMm = 25.4; >>>>>>>>>> fullW = 35.5 * inMm; >>>>>>>>>> fullL = 54.5 * inMm; >>>>>>>>>> resize([fullW, fullL, 1]) >>>>>>>>>> circle(d=fullW); >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> Tip: use scale() rather than resize(); it previews faster in >>>>>>>>>> complicated cases. >>>>>>>>>> >>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world. >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> You're taking a circle and stretching it; the result of that >>>>>>>>>> stretching is an ellipse. >>>>>>>>>> >>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can chime >>>>>>>>>> in - but I think the shape you are asking for would not be an ellipse. >>>>>>>>>> >>>>>>>>>> I believe that an ellipse is fully defined by its major (long) >>>>>>>>>> and minor (short) axis dimensions; the other parts of the curve >>>>>>>>>> mathematically flow from those two measurements. >>>>>>>>>> >>>>>>>>>> You can make something else, but you can't make it using this >>>>>>>>>> technique. >>>>>>>>>> >>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed. >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> You can use a Bézier function. >>>>>>>>>> >>>>>>>>>> Using BOSL2: >>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves >>>>>>>>>> >>>>>>>>>> DIY, here's a function that I cobbled together some years ago: >>>>>>>>>> >>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443 >>>>>>>>>> // but that yielded either single points or a raft of triangles; >>>>>>>>>> // this yields a vector of points that you can then concatenate >>>>>>>>>> // with other pieces to form a single polygon. >>>>>>>>>> // If we were really clever, I think it would be possible to >>>>>>>>>> // automatically space the output points based on how linear >>>>>>>>>> // the curve is at that point. But right now I'm not that clever. >>>>>>>>>> function BEZ03(u) = pow((1-u), 3); >>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2)); >>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u); >>>>>>>>>> function BEZ33(u) = pow(u,3); >>>>>>>>>> >>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [ >>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0], >>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]]; >>>>>>>>>> >>>>>>>>>> // p0 - start point >>>>>>>>>> // p1 - control point 1, line departs p0 headed this way >>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way >>>>>>>>>> // p3 - end point >>>>>>>>>> // segs - number of segments >>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [ >>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs) >>>>>>>>>> ]; >>>>>>>>>> >>>>>>>>>> It is that final function bez() that I actually call, e.g. >>>>>>>>>> >>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20)); >>>>>>>>>> >>>>>>>>>> often concatenating the results of several bez() calls together. >>>>>>>>>> >>>>>>>>>> Note that the four coordinates correspond to the coordinates of >>>>>>>>>> the control points in a GUI drawing program: the start point, control >>>>>>>>>> point associated with the start, the control point associated with the end, >>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an >>>>>>>>>> object in a drawing program and then reading off the coordinates of the >>>>>>>>>> various control points. >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> Note that Bézier functions are not the only mathematical curves. >>>>>>>>>> (For instance, they cannot exactly match an arc of a circle.) I'm sure >>>>>>>>>> there are numerous other functions that also produce curves. >>>>>>>>>> _______________________________________________ >>>>>>>>>> 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 >>>>>>>> >>>>>>> _______________________________________________ >>>>>>> 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 >>>>> >>>> _______________________________________________ >>>> 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 >> > _______________________________________________ > OpenSCAD mailing list > To unsubscribe send an email to discuss-leave@lists.openscad.org >

BC

Bob Carlson

Fri, Aug 5, 2022 11:15 PM

Seems like I remember Nop Head talking about using circumscribed holes for screws and even rotating the polygons to produce a more even hole.

If an ellipse is the nominal shape, the using a circumscribed polygon for a hole and an inscribed polygon for a solid might make sense.

Btw, can someone confirm my intuition that a circle scaled in one dimension is not an ellipse?

-Bob
Tucson AZ

On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:

I don't have a specific need for a circumscribing elliptical polygon with equal sides. I was writing code to make ellipses and it seemed like equal side approximations would be nice, so I solved the case for inscribed, but couldn't solve the circumscribed case. So it's an unsolved puzzle I've got that this discussion reminded me of.

On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com mailto:nop.head@gmail.com> wrote:
A long discussion but I have to ask why do you need an elliptical polygon with equal tangential sides?

On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu mailto:avm4@cornell.edu> wrote:
You think that a circumscribing polygon with equal length segments does not exist?? I'm skeptical of this claim.

For circumscribed, it's not so hard. You start with a uniform-in-angle point distribution, compute the segment lengths, compare to the mean segment length to get an error, and then compute an adjustment based on the error. It converges quickly.

So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.

<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm should exist for the circumscribed case, but as I said, my efforts to find it have failed.

On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com mailto:sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough problem to solve.
I tried something.

On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu mailto:avm4@cornell.edu> wrote:
I think you need to be a little careful about defining what you want to do. If you want to approximate an ellipse with points that are distributed uniformly over the ellipse, that could be done by approximating the arc length of the ellipse to find the necessary points---e.g. advance distance d along the ellipse at each step. But if you want to approximate an ellipse with a polygon with equal length sides, that approach does not do it, because the length of the segments will not be constant as the span sections of the ellipse with differing curvature. And since the perimeter of the desired polygon is unknown, you can't just step it out analogously. The circle doesn't present this challenge due to having constant curvature.

On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com mailto:fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to approximate an ellipse with constant-length segments in general, not how to solve your problem. When I had to do something similar, I used polar coordinates and stepped up the value of theta until the distance from the previous point was correct. (In the constant-length version, I'd keep a running total of the maximum desired length so that round-off errors don't accumulate.)

Figuring out the circumscribed polygon is, as you note, a greater challenge. I can visualize how to do it for n = 4, but it doesn't extrapolate at all.

On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu mailto:avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this situation. Step through what with sufficiently small steps? Of course an iterative algorithm is necessary. This is true for most interesting problems. No big deal. Show me an iterative algorithm that converges to the circumscribing polygon. When I tried to devise such an algorithm I couldn't find one that would converge. Writing an algorithm that would converge to the inscribed polygon seemed to be straight forward. Note also that computing arc length of the ellipse doesn't really help here. You can do that by solving an elliptical integral, or summing up small segments. But what's the connection between arc length of the ellipse and a circumscribing polygon with equal segment lengths? If there's a connection, I don't see it.

On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com mailto:fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small steps), it’s not possible because there’s no closed form expression for the distance between two points on a general ellipse (as opposed to special cases like a circle).

On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu mailto:avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments that are all the same length that circumscribes a perfect ellipse with given semiaxes a and b. I tried to solve this problem and could not find a solution. I can draw an inscribed N segment "ellipse" polygon with all segments the same length---though it's not easy---but the circumscribed case eludes me. I think of the inscribed case as, every vertex is located on the ellipse and the circumscribed case, every segment is tangent to the ellipse

On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com mailto:sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code

function to draw an ellipse with semi-major and semi-minor axis "r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);

function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;

On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net mailto:openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:

I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);

Tip: use scale() rather than resize(); it previews faster in complicated cases.

It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.

You're taking a circle and stretching it; the result of that stretching is an ellipse.

I'm not 100% sure - maybe one of the real math wizards can chime in - but I think the shape you are asking for would not be an ellipse.

I believe that an ellipse is fully defined by its major (long) and minor (short) axis dimensions; the other parts of the curve mathematically flow from those two measurements.

You can make something else, but you can't make it using this technique.

Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.

You can use a Bézier function.

Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves

DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443 https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);

function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];

// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.

polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.

Note that the four coordinates correspond to the coordinates of the control points in a GUI drawing program: the start point, control point associated with the start, the control point associated with the end, and the end point. I have sometimes created OpenSCAD objects by tracing an object in a drawing program and then reading off the coordinates of the various control points.

Note that Bézier functions are not the only mathematical curves. (For instance, they cannot exactly match an arc of a circle.) I'm sure there are numerous other functions that also produce curves.

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Seems like I remember Nop Head talking about using circumscribed holes for screws and even rotating the polygons to produce a more even hole. If an ellipse is the nominal shape, the using a circumscribed polygon for a hole and an inscribed polygon for a solid might make sense. Btw, can someone confirm my intuition that a circle scaled in one dimension is not an ellipse? -Bob Tucson AZ On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote: I don't have a specific need for a circumscribing elliptical polygon with equal sides. I was writing code to make ellipses and it seemed like equal side approximations would be nice, so I solved the case for inscribed, but couldn't solve the circumscribed case. So it's an unsolved puzzle I've got that this discussion reminded me of. On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com <mailto:nop.head@gmail.com>> wrote: A long discussion but I have to ask why do you need an elliptical polygon with equal tangential sides? On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu <mailto:avm4@cornell.edu>> wrote: You think that a circumscribing polygon with equal length segments does not exist?? I'm skeptical of this claim. For circumscribed, it's not so hard. You start with a uniform-in-angle point distribution, compute the segment lengths, compare to the mean segment length to get an error, and then compute an adjustment based on the error. It converges quickly. So for example, I have segments lengths of: ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297] Maybe not exactly equal, since tolerance is 1e-9, but surely good enough. <image.png> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm should exist for the circumscribed case, but as I said, my efforts to find it have failed. On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com <mailto:sprabhakar2006@gmail.com>> wrote: I think circumscribed polygon segments cannot be equal lengths. inscribed ones may be approximately equal, but would be a very tough problem to solve. I tried something. On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu <mailto:avm4@cornell.edu>> wrote: I think you need to be a little careful about defining what you want to do. If you want to approximate an ellipse with points that are distributed uniformly over the ellipse, that could be done by approximating the arc length of the ellipse to find the necessary points---e.g. advance distance d along the ellipse at each step. But if you want to approximate an ellipse with a polygon with equal length sides, that approach does not do it, because the length of the segments will not be constant as the span sections of the ellipse with differing curvature. And since the perimeter of the desired polygon is unknown, you can't just step it out analogously. The circle doesn't present this challenge due to having constant curvature. On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com <mailto:fatherhorton@gmail.com>> wrote: I think our emails crossed. I was addressing how to approximate an ellipse with constant-length segments in general, not how to solve your problem. When I had to do something similar, I used polar coordinates and stepped up the value of theta until the distance from the previous point was correct. (In the constant-length version, I'd keep a running total of the maximum desired length so that round-off errors don't accumulate.) Figuring out the circumscribed polygon is, as you note, a greater challenge. I can visualize how to do it for n = 4, but it doesn't extrapolate at all. On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu <mailto:avm4@cornell.edu>> wrote: Eh, there's no such thing as "brute force" in this situation. Step through *what* with sufficiently small steps? Of course an iterative algorithm is necessary. This is true for most interesting problems. No big deal. Show me an iterative algorithm that converges to the circumscribing polygon. When I tried to devise such an algorithm I couldn't find one that would converge. Writing an algorithm that would converge to the *inscribed* polygon seemed to be straight forward. Note also that computing arc length of the ellipse doesn't really help here. You can do that by solving an elliptical integral, or summing up small segments. But what's the connection between arc length of the ellipse and a circumscribing polygon with equal segment lengths? If there's a connection, I don't see it. On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com <mailto:fatherhorton@gmail.com>> wrote: Except by brute force (step through it with sufficiently small steps), it’s not possible because there’s no closed form expression for the distance between two points on a general ellipse (as opposed to special cases like a circle). On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu <mailto:avm4@cornell.edu>> wrote: Here's a challenge for you: draw an "ellipse" polygon with N segments that are all the same length that circumscribes a perfect ellipse with given semiaxes a and b. I tried to solve this problem and could not find a solution. I can draw an *inscribed* N segment "ellipse" polygon with all segments the same length---though it's not easy---but the circumscribed case eludes me. I think of the inscribed case as, every vertex is located on the ellipse and the circumscribed case, every segment is tangent to the ellipse On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com <mailto:sprabhakar2006@gmail.com>> wrote: In openscad drawing a ellipse is not too difficult Refer following code function to draw an ellipse with semi-major and semi-minor axis "r1" and "r2" respectively and with center "cp" and number of segment "s" / // example: // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30); function ellipse(r1,r2,cp,s=30)= let( sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]] )sec; On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net <mailto:openscad@jordan.maileater.net>> wrote: On 8/2/2022 4:46 PM, Brian Allen wrote: > I created an oval using: > inMm = 25.4; > fullW = 35.5 * inMm; > fullL = 54.5 * inMm; > resize([fullW, fullL, 1]) > circle(d=fullW); Tip: use scale() rather than resize(); it previews faster in complicated cases. > It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world. You're taking a circle and stretching it; the result of that stretching is an ellipse. I'm not 100% sure - maybe one of the real math wizards can chime in - but I think the shape you are asking for would not be an ellipse. I believe that an ellipse is fully defined by its major (long) and minor (short) axis dimensions; the other parts of the curve mathematically flow from those two measurements. You can make something else, but you can't make it using this technique. > Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed. You can use a Bézier function. Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves <https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves> DIY, here's a function that I cobbled together some years ago: // Bezier functions from https://www.thingiverse.com/thing:8443 <https://www.thingiverse.com/thing:8443> // but that yielded either single points or a raft of triangles; // this yields a vector of points that you can then concatenate // with other pieces to form a single polygon. // If we were really clever, I think it would be possible to // automatically space the output points based on how linear // the curve is at that point. But right now I'm not that clever. function BEZ03(u) = pow((1-u), 3); function BEZ13(u) = 3*u*(pow((1-u),2)); function BEZ23(u) = 3*(pow(u,2))*(1-u); function BEZ33(u) = pow(u,3); function PointAlongBez4(p0, p1, p2, p3, u) = [ BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0], BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]]; // p0 - start point // p1 - control point 1, line departs p0 headed this way // p2 - control point 2, line arrives at p3 from this way // p3 - end point // segs - number of segments function bez(p0, p1, p2, p3, segs) = [ for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs) ]; It is that final function bez() that I actually call, e.g. polygon(bez([0,0], [0,4], [5,1], [7,0], 20)); often concatenating the results of several bez() calls together. Note that the four coordinates correspond to the coordinates of the control points in a GUI drawing program: the start point, control point associated with the start, the control point associated with the end, and the end point. I have sometimes created OpenSCAD objects by tracing an object in a drawing program and then reading off the coordinates of the various control points. Note that Bézier functions are not the only mathematical curves. (For instance, they cannot exactly match an arc of a circle.) I'm sure there are numerous other functions that also produce curves. _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org> _______________________________________________ OpenSCAD mailing list To unsubscribe send an email to discuss-leave@lists.openscad.org

AM

Adrian Mariano

Fri, Aug 5, 2022 11:19 PM

Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.

On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:

Seems like I remember Nop Head talking about using circumscribed holes for
screws and even rotating the polygons to produce a more even hole.

If an ellipse is the nominal shape, the using a circumscribed polygon for
a hole and an inscribed polygon for a solid might make sense.

Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?

-Bob
Tucson AZ

On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:

I don't have a specific need for a circumscribing elliptical polygon with
equal sides. I was writing code to make ellipses and it seemed like equal
side approximations would be nice, so I solved the case for inscribed, but
couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
that this discussion reminded me of.

On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:

A long discussion but I have to ask why do you need an elliptical polygon
with equal tangential sides?

On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:

You think that a circumscribing polygon with equal length segments does
not exist?? I'm skeptical of this claim.

For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.

So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]

Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>

The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.

On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:

I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.

On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:

I think you need to be a little careful about defining what you want
to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.

On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:

I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)

Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.

On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:

Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.

On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote: