Initializing spline data
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On 15/04/2014 at 11:55, xxxxxxxx wrote:
So in my development of my python plugin I have the following code.
self.sd = c4d.SplineData()
self.sd.MakeLinearSplineBezier(3)self.sd.SetRound(1.0)
self.sd.SetRange(xmin=0, xmax=1, xsteps=.1, ymin=0, ymax=1, ysteps=.1)
self.sd.SetKnot(index=0, vPos= c4d.Vector(0,0, 0))
self.sd.SetKnot(index=1, vPos= c4d.Vector(.50,1, 0))
self.sd.SetKnot(index=2, vPos= c4d.Vector(1,0, 0))This creates a linear graph representation, I am looking more for a graph representation that looks like
y= 2 - (-x)^2
I guess more of a curve.
Any help would be much appreciated.
Jimmy -
On 16/04/2014 at 00:49, xxxxxxxx wrote:
You need to set the tangents for each knot to get a curve.
Steve
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On 16/04/2014 at 08:01, xxxxxxxx wrote:
Steve,
Thank you, can you elaborate a bit more? I'm still a bit lost as to how to do that? Would you be so kind as to supply a code example perhaps?
Thank you! -
On 16/04/2014 at 08:27, xxxxxxxx wrote:
So I modified the code in this fashion to make the y= 2 - (-x)^2 curve.
self.sd = c4d.SplineData()
self.sd.MakeLinearSplineBezier(3)self.sd.SetRound(1.0)
self.sd.SetRange(xmin=0, xmax=1, xsteps=.1, ymin=0, ymax=1, ysteps=.1)
self.sd.SetKnot(index=0, vPos= c4d.Vector(0,0, 0))
self.sd.SetKnot(index=1, vPos= c4d.Vector(.50,1, 0),vTangentLeft=c4d.Vector(-.10, 0, 0), vTangentRight=c4d.Vector(.10, 0, 0))
self.sd.SetKnot(index=2, vPos= c4d.Vector(1,0, 0))