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AM

Adrian Mariano

Mon, Aug 8, 2022 1:32 AM

Father Horton,

I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.

It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)

On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com wrote:

How about this to construct N equal length segments around an ellipse with
axes a and b:

Select an initial segment length estimate (e.g., Ramanujan's formula /
N) = L
Construct a half-segment from (0, b) to (0, L/2).
Find a line tangent to the ellipse and passing through (0, L/2).
Find the point on the tangent line that is L away from (0, L/2).
Repeat (3) substituting the new point for (0, L/2). until you have N
segments.

If everything worked perfectly, the last point found would be (0, -L/2).
If there's a gap, increase L and try again. If there's overlap, decrease L
and try again.

On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:

Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.

[image: Uniformly_sampled_ellipsis.png]

Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:

Circumscribed means tangents to points on the ellipse. But there are no
arc lengths (equal or unequal) in this problem, nor equally spaced angles.
The angles of each segment will be different. But the lengths of the
segments are not the same as the arc lengths, which is why arc length does
not matter. Here's a solution to the N=6 polygon sort of inspired by the
idea Jordan suggested, where I adjust the first segment until the second
segment has the right length. The rest of the solution is then obtained
using symmetry, which means there is only a single unknown that can simply
be solved for. But this doesn't generalize to higher N.

[image: image.png]
include<BOSL2/std.scad>

// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;

// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));

// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;

// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];

err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);

x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));

stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);

// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));

On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:

Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .

On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:

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:

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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Father Horton, I'm having trouble understanding your proposed method. A segment from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then there won't be a tangent. It also sounds like your idea is somewhat similar to Jordan's idea: basically start at one side and work around the ellipse and see if it works. If not, make a correction. The problem with this idea is that there are two parameters for the first segment, so how do you make a correction? (The two parameters are the location of the endpoint of the first segment and its length.) On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com> wrote: > How about this to construct N equal length segments around an ellipse with > axes a and b: > > 1) Select an initial segment length estimate (e.g., Ramanujan's formula / > N) = L > 2) Construct a half-segment from (0, b) to (0, L/2). > 3) Find a line tangent to the ellipse and passing through (0, L/2). > 3) Find the point on the tangent line that is L away from (0, L/2). > 4) Repeat (3) substituting the new point for (0, L/2). until you have N > segments. > > If everything worked perfectly, the last point found would be (0, -L/2). > If there's a gap, increase L and try again. If there's overlap, decrease L > and try again. > > On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <rcmpersiano@gmail.com> > wrote: > >> Adrian, why to use a (either inscribed or circumscribed) polygon with >> equal side as an approximation of an ellipsis? It doesn't seem to be a good >> approximation of it. I get a better approximation by taking an angle >> uniform sample of the curve in the parameter space. >> >> [image: Uniformly_sampled_ellipsis.png] >> >> >> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu> >> escreveu: >> >>> Circumscribed means tangents to points on the ellipse. But there are no >>> arc lengths (equal or unequal) in this problem, nor equally spaced angles. >>> The angles of each segment will be different. But the lengths of the >>> segments are not the same as the arc lengths, which is why arc length does >>> not matter. Here's a solution to the N=6 polygon sort of inspired by the >>> idea Jordan suggested, where I adjust the first segment until the second >>> segment has the right length. The rest of the solution is then obtained >>> using symmetry, which means there is only a single unknown that can simply >>> be solved for. But this doesn't generalize to higher N. >>> >>> [image: image.png] >>> include<BOSL2/std.scad> >>> >>> // ellipse semiaxes >>> // If a is much smaller than b it will fail >>> a=5; >>> b=2; >>> >>> // Find point on ellipse at specified x coordinate >>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2)); >>> >>> >>> // Find tangent line to ellipse at point (u,w) >>> // and determine the x value where that line has >>> // given y value. >>> function findx(a,b,u,w,y) = >>> let( >>> A = -u*b^2/w/a^2, >>> B = w+(u*b)^2/w/a^2 >>> ) >>> (y-B)/A; >>> >>> >>> // First segment if tangent point is at x=u >>> function first_seg(a,b,u) = >>> let( >>> w = ellipse_pt(a,b,u), >>> x1 = findx(a,b,u,w,0), >>> x2 = findx(a,b,u,w,b) >>> ) >>> [[x1,0],[x2,b]]; >>> >>> >>> err = function (u) let(pt = first_seg(a,b,u)) >>> 2*pt[1].x - norm(pt[0]-pt[1]); >>> >>> x = root_find(err,.2*a,.9*a); >>> first = first_seg(a,b,x); >>> top = concat(first,reverse(xflip(first))); >>> poly = concat(top, reverse(yflip(list_tail(top)))); >>> >>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true); >>> color("red")stroke(poly,width=.05,closed=true); >>> >>> // Verify equality of segment lengths >>> echo(path_segment_lengths(poly,closed=true)); >>> >>> >>> >>> >>> >>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar < >>> sprabhakar2006@gmail.com> wrote: >>> >>>> Circumscribed means tangents from points on ellipse >>>> How can the tangents from equally spaced arc lengths or equally spaced >>>> angles be equal for an ellipse >>>> You need to change the arc lengths dynamically and also accurately >>>> estimate the segment length which fits the ellipse . >>>> >>>> >>>> >>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote: >>>> >>>>> 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: >>>>>>>>>>>> >>>>>>>>>>>>> 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 >>>>>>> >>>>>> _______________________________________________ >>>>>> 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 >

FH

Father Horton

Mon, Aug 8, 2022 1:52 AM

I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
That's always outside the ellipse.

I will look at the second problem as soon as I figure out how to calculate
a tangent line to an ellipse containing a point not on the ellipse. Google
is not helping much.

On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu wrote:

Father Horton,

I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.

It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)

On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:

How about this to construct N equal length segments around an ellipse
with axes a and b:

Select an initial segment length estimate (e.g., Ramanujan's formula /
N) = L
Construct a half-segment from (0, b) to (0, L/2).
Find a line tangent to the ellipse and passing through (0, L/2).
Find the point on the tangent line that is L away from (0, L/2).
Repeat (3) substituting the new point for (0, L/2). until you have N
segments.

If everything worked perfectly, the last point found would be (0, -L/2).
If there's a gap, increase L and try again. If there's overlap, decrease L
and try again.

On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:

Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.

[image: Uniformly_sampled_ellipsis.png]

Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:

Circumscribed means tangents to points on the ellipse. But there are
no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.