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I was in a rush when I replied last time and didn't have this to hand. Second
installment.
If you want a Bezier Spline approximation to an arc of a circle subtending angle
theta at the centre, the four control points are
(1,0), (1,a), (cos(theta) + asin(theta),sin(theta) 
acos(theta)),(cos(theta),sin(theta))
where a = (8/3)(sin(theta/2)  sin(theta)/2)/(1cos(theta))
As mentioned before, this is tangent to the circle at theta/2 as well as the
endpoints and outside the circle otherwise. For theta = pi/2 the defect is
0.03% of the radius. For theta = pi it is 1.8%, still not bad, but for theta =
5pi/4 it is an unusable 7.6%.
Tweaking a to have the spline cross the circle twice reduces the error only to
about 70% of the value given and is probably not worth it.
If anyone wants code for a single segment Bezier Spline sphere sweep, I will
post it again, since the last time was years ago.
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On 20190122 11:23 AM (4), JimT wrote:
> If you want a Bezier Spline approximation to an arc of a circle subtending angle
> theta at the centre, the four control points are
>
> (1,0), (1,a), (cos(theta) + asin(theta),sin(theta) 
> acos(theta)),(cos(theta),sin(theta))
>
> where a = (8/3)(sin(theta/2)  sin(theta)/2)/(1cos(theta))
Hmmm. This is the formula for 'a' that I used for the ring shank cross
section in GemCuts:
#declare Gem__fn_Bezier_arc = function (x)
{ (8 * cos (x / 2)  4  4 * cos (x)) / (3 * sin (x))
}
Presumably, they are somehow the same formula, though I haven't figured
out how to reduce one to the other.
> As mentioned before, this is tangent to the circle at theta/2 as well as the
> endpoints and outside the circle otherwise. For theta = pi/2 the defect is
> 0.03% of the radius. For theta = pi it is 1.8%, still not bad, but for theta =
> 5pi/4 it is an unusable 7.6%.
My notes say that the curve deviates markedly from a circle if x > 90
degrees (pi / 2), though I did not quantify the error.
> Tweaking a to have the spline cross the circle twice reduces the error only to
> about 70% of the value given and is probably not worth it.
For large angles I just use 2 segments.
I've actually tried multiple crossings to approximate a quadrant of a
superquadric ellipse, but it required an arbitrary amount of fudging. I
do not have a general formula for this. I also tried using two 45
degree segments, but the result was horrible.
> If anyone wants code for a single segment Bezier Spline sphere sweep, I will
> post it again, since the last time was years ago.
There are two Object Collection modules that already do this for
multiple segments: PointArrays and SphereSweep.
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Cousin Ricky <ric### [at] yahoocom> wrote:
> On 20190122 11:23 AM (4), JimT wrote:
> > If you want a Bezier Spline approximation to an arc of a circle subtending angle
> > theta at the centre, the four control points are
> >
> > (1,0), (1,a), (cos(theta) + asin(theta),sin(theta) 
> > acos(theta)),(cos(theta),sin(theta))
> >
> > where a = (8/3)(sin(theta/2)  sin(theta)/2)/(1cos(theta))
>
> Hmmm. This is the formula for 'a' that I used for the ring shank cross
> section in GemCuts:
>
> #declare Gem__fn_Bezier_arc = function (x)
> { (8 * cos (x / 2)  4  4 * cos (x)) / (3 * sin (x))
> }
>
Sorry I'm a month late. I didn't spot your post. And you probably won't spot
this.
I plotted them in Matlab to convince myself they were the same. Then I
multiplied top and bottom by sin(x/2)/cos(x/2). 1+cos(x) is 2cos^2(x/2) and
1cos(x) is 2sin^2(x/2).
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On 20190220 10:06 AM (4), JimT wrote:
> Cousin Ricky <ric### [at] yahoocom> wrote:
>> On 20190122 11:23 AM (4), JimT wrote:
>>> If you want a Bezier Spline approximation to an arc of a circle subtending angle
>>> theta at the centre, the four control points are
>>>
>>> (1,0), (1,a), (cos(theta) + asin(theta),sin(theta) 
>>> acos(theta)),(cos(theta),sin(theta))
>>>
>>> where a = (8/3)(sin(theta/2)  sin(theta)/2)/(1cos(theta))
>>
>> Hmmm. This is the formula for 'a' that I used for the ring shank cross
>> section in GemCuts:
>>
>> #declare Gem__fn_Bezier_arc = function (x)
>> { (8 * cos (x / 2)  4  4 * cos (x)) / (3 * sin (x))
>> }
>>
> Sorry I'm a month late. I didn't spot your post. And you probably won't spot
> this.
>
> I plotted them in Matlab to convince myself they were the same. Then I
> multiplied top and bottom by sin(x/2)/cos(x/2). 1+cos(x) is 2cos^2(x/2) and
> 1cos(x) is 2sin^2(x/2).
Good to know. That's more trig than I can remember.
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