Rounded Polygon

adrianv <[hidden email]> wrote:

I'm a little puzzled.  Maybe you are using the word degree differently than
I think?  I was thinking a degree 3 Bezier is a cubic polynomial and has 4
control points.  It can be set to zero curvature at both ends but has only
one degree of freedom, which just controls the size of the curve.  That
isn't enough.  It appears to me that degree 4, with 5 control points, is
sufficient.  Yes, you do need to put p2 on the vertex.  This doesn't seem to
cause any undesirable restriction, and it is definitely possible to satisfy
the zero curvature condition.   One parameter remains that controls the
degree of curvature, as I noted in my previous message.  Having fewer
parameters is often better than more, so I'm not sure about what the value
of more degrees of freedom is of going to the case with 6 control points
(which is a 5th degree polynomial).  I did experiment a bit with your code
and you can get some rather sharp cornered results.  It seemed like r1
usually had little effect. 

You are right regarding the degrees. I wrongly said 5 instead of 4 and 6 instead of 5. A degree n Bezier curve has n+1 control points.

However I disagree that you could meet the zero curvature constraint at both end with degree 3 curves. To have a zero curvature at the beginning of the arc, the first 3 control points should co-linear. The same is true for the ending point. If we require zero curvature at both ends all four control points should be co-linear and the arc reduces to a straight segment.

Degree 4 is in fact the minimal degree for zero curvature at both ends for a non-straight line Bezier arc. However degree 5 solution give us not only more degrees of freedom but a better arc shape as will see bellow. Degree 4 solution produces an arc with a greater curvature variation and a more pointed arc. Degree 5 shape is able to produce arcs that resemble circular arcs.

I have revised my codes in order to generate alternatively both solutions. The two missed functions and line() module are defined as before.

q = [ [0,0], [30,0], [30,30], [0,30] ];

color("blue")   roundedPoly(q, 20, 4, 1/3);

color("yellow") roundedPoly(q, 20, 4, 2/3);

color("red")    roundedPoly(q, 20, 4,   1);

translate([35,0,0]) {

  color("blue")   roundedPoly(q, 20, 5, 1/3, 17/32);

  color("yellow") roundedPoly(q, 20, 5, 2/3, 17/32);

  color("red")    roundedPoly(q, 20, 5,   1, 17/32);

  color("green")  translate([15,15,-1]) circle(15);

}  

  

module roundedPoly(p, n=20, degree=4, r=1/4, r0=2/3, r1=17/32) 

{

  assert(len(p)>2, "polygonal has less than 3 points");

  assert(r>0 && r<=1, "r out of range");

  assert(r0>0 && r<=1, "r0 out of range");

  assert(degree==4 || degree==5, "degree should be 4 or 5");

  assert(degree==5 || (r1>0 && r1<=1), "r1 out of range");

  l = len(p);

  q = [for(i=[0:l-1]) 

         let( p0 = (1-r/2)*p[i] + r*p[(i+l-1)%l]/2 ,

              p1 = p[i],

              p2 = (1-r/2)*p[i] + r*p[(i+1)%l]/2 )

         each BZeroCurvature(p0,p1,p2,n,degree,r0,r1) ];

  line(q, closed=true,w=0.2);

}  

  

// a Bezier arc of degree 4 or 5 from p0 to p2, tangent to

// [p0,p1] at p0, tangent to [p1,p2) at p2 and

// with zero curvature at p0 and p2

// n  - # of points in the arc

// r0 - form factor in the interval [0,1] 

// r1 - form factor in the interval [0,1] (for degree=5 only)

function BZeroCurvature(p0,p1,p2,n=20,degree=4,r0=2/3,r1=1/2) =

  assert(r0>0 && r0<=1, "r0 out of range")

  assert(degree==4 || degree==5, "degree should be 4 or 5")

  assert(degree==5 || (r1>0 && r1<=1), "r1 out of range")

  let( r1 = degree==4 ? 1 : r1,

       p  = [ p0, 

              p0 + r0*(p1-p0)*r1,

              each degree==4 ?

                    [p1] :

                    [p0 + r0*(p1-p0), p2 + r0*(p1-p2)], 

              p2 + r0*(p1-p2)*r1,

              p2 ] )

  BezierCurve(p,n);

And got the following image:

At left we have the arcs of degree 4 and at right the degree 5 alternative for similar form factors. With r=1, which means that the straight lines between arcs reduce to a point, drawn here with red lines, the degree 4 solution keeps the middle point of each arc much nearer to the polygon vertex. With degree 5 on the other hand it is possible to get a good approximation of a circular arc with r1~= 17/32, a value I got empirically.

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Ronaldo wrote
> adrianv &lt;

> avm4@

> &gt; wrote:
>
>
> However I disagree that you could meet the zero curvature constraint at
> both end with degree 3 curves. To have a zero curvature at the beginning
> of
> the arc, the first 3 control points should co-linear. The same is true for
> the ending point. If we require zero curvature at both ends all four
> control points should be co-linear and the arc reduces to a straight
> segment.

No it doesn't.  I posted graphs a few messages back showing zero curvature
order 3 beziers, and I had directly verified that the curvature was indeed
zero by calculating the derivatives both symbolically and numerically, so I
am reasonably sure it was correct.   With p0=p3 set to the endpoints of the
arc you then set p1=p2 equal to the point of the corner.  You achieve the
co-linearity condition in a degenerate fashion, since p1 and p2 are on both
lines. 


> Degree 4 is in fact the minimal degree for zero curvature at both ends for
> a non-straight line Bezier arc. However degree 5 solution give us not only
> more degrees of freedom but a better arc shape as will see bellow. Degree
> 4
> solution produces an arc with a greater curvature variation and a more
> pointed arc. Degree 5 shape is able to produce arcs that resemble circular
> arcs.

It appears that your degree 4 code can also generate a nearly circular arc:
just set r1 very small, like .01 and you get something visually
indistinguishable from a circle.  In fact, it looks very similar to your
empirically determined 17/32 value.  I wonder what are the parameters for
the order 5 bezier that reduce it to the order 4.

On the left is order 4 with r1=0.01 and on the right, order 5 with r1=17/32. 

<http://forum.openscad.org/file/t2477/round2.png>





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No it doesn't.  I posted graphs a few messages back showing zero curvature
order 3 beziers, and I had directly verified that the curvature was indeed
zero by calculating the derivatives both symbolically and numerically, so I
am reasonably sure it was correct.   With p0=p3 set to the endpoints of the
arc you then set p1=p2 equal to the point of the corner.  You achieve the
co-linearity condition in a degenerate fashion, since p1 and p2 are on both
lines.

You are right. Putting both intermediate control points at the corner will meet the conditions of zero curvature at the two ends of a degree 3 Bezier arc. I could not imagine that solution before. And certainly this is a very constrained alternative. 

 

It appears that your degree 4 code can also generate a nearly circular arc:
just set r1 very small, like .01 and you get something visually
indistinguishable from a circle.  In fact, it looks very similar to your
empirically determined 17/32 value.  I wonder what are the parameters for
the order 5 bezier that reduce it to the order 4.

I have confirmed your statement about the approximation of circular arcs by degree 4 curves. It seems even better than the degree 5 arc with r0=17/32.

I have applied the degree elevation formulas to deduce algebraically the conditions r0 and r1 should satisfy in order the degree 4 and degree 5 curves be equal. From my math exercise, we should have:

degree 5 with  r0 = 3/5  and r1=0 will be equivalent to

degree 4 with r0=0

I also found that r0 may be set to 0 for degree 4 and r1 may be set to 0 for degree 5 but r0 should not be 0 for degree 5 because the arc degenerate into a line segment. I have changed the assert conditions in the my functions accordingly.

I agree that rounding corners with degree 4 Bezier arcs give us enough flexibility and there is no point to use greater degree. In my functions, the parameter r of roundedPoly() controls how much the corner is rounded like your proposed d parameter. That parameter sets the end points relatively to the polygon edge length instead of an absolute value. The problem with this solution is that the arc will not be symmetrical when the two edges meeting a corner have different length. But the alternative of a absolute value d requires a validation to avoid that two arcs rounding the two corners of an edge intercept.

Finally, I have checked that my code works well even with not convex polygons as can be seen in the image.

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On 15/03/2019 14:07, Ronaldo Persiano wrote:

Besides, curvature continuity is a requeriment on the design of a road track. Any discontinuity on the track is prone to an accident because a sudden wheel turn is needed to keep the vehicle on track when you cross the discontinuity.

Railways used to be designed with straight and curves, and it made them uncomfortable and restricted speeds, and ditto "loop the loop" on roller coasters as the sudden changes in acceleration (high jerk) caused neck injuries.

Then along came what I still call clothoid curves, but wikipedia prefers Euler Spirals.

https://en.wikipedia.org/wiki/Euler_spiral

They are used to link straights to circles on railways, rollercoasters, and much more.

https://en.wikipedia.org/wiki/Track_transition_curve

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