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There is nurbscurve with four control points.
For an example: (1,1.5,0.5),(0.5,0.5,1.2),(0.8,1,0.3),(2,0.1,0.8)
It is necessary to create a tube along a curve, using two bicubic patches
Radius tube = R
The mathematics is necessary :) :) :)
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Am 22.02.2015 um 14:07 schrieb LanuHum:
> Hi!
> There is nurbscurve with four control points.
> For an example: (1,1.5,0.5),(0.5,0.5,1.2),(0.8,1,0.3),(2,0.1,0.8)
> It is necessary to create a tube along a curve, using two bicubic patches
> Radius tube = R
> The mathematics is necessary :) :) :)
Note that the mathematics say that strictly speaking this is impossible:
With bicubic patches you can neither create a tube with perfectly
circular crosssection (because a cubic spline in 2d space never forms a
perfect circular arc), nor can you use them for any curved tubular
structure with constant crosssection (because the parallel curve of a
cubic spline in 2d space is never another cubic spline unless they both
are linear).
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clipka <ano### [at] anonymousorg> wrote:
> Am 22.02.2015 um 14:07 schrieb LanuHum:
> > Hi!
> > There is nurbscurve with four control points.
> > For an example: (1,1.5,0.5),(0.5,0.5,1.2),(0.8,1,0.3),(2,0.1,0.8)
> > It is necessary to create a tube along a curve, using two bicubic patches
> > Radius tube = R
> > The mathematics is necessary :) :) :)
>
> Note that the mathematics say that strictly speaking this is impossible:
> With bicubic patches you can neither create a tube with perfectly
> circular crosssection (because a cubic spline in 2d space never forms a
> perfect circular arc), nor can you use them for any curved tubular
> structure with constant crosssection (because the parallel curve of a
> cubic spline in 2d space is never another cubic spline unless they both
> are linear).
Not necessarily. Approximate.
I am going to use rectangular triangles for calculation
But, I don't know how it is effective.
Therefore asked masters.
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