# SOLVEDHow to get spline offset by spline point index ？

hi，
How to get spline offset(0-1) by spline point index ? i want know offset between points to caculate the value points should be.

Hi Mike, thanks for writing us.

With regard to your question, if I properly understood you, you can use the SplineHelp Class and its SplineHelp::GetPosition() method to retrieve the spline's points value given a certain offset.

Something like this:

import c4d

# Main function
def main():
if op is None:
return

realSpline = op.GetRealSpline()
if realSpline is None:
return

splineH = c4d.utils.SplineHelp()
if splineH is None:
return

splineH.InitSpline(realSpline)

samples = 10
for sample in range(0, samples):
offset = float(sample) / samples
print splineH.GetPosition(offset)

# Execute main()
if __name__=='__main__':
main()

Best, Riccardo

Hi @r_gigante,
Thank you for replying. I'd like to know more about this topic: if I only know the Control Point Index from PointObject.GetPoint(id) of a spline, how could I get the Spline Position of this point?

@r_gigante Thank you for your help! i have another problem that if i only know the spline point index ,how do i get the offset where point locate? just take sample like 1000 and find the nearest sample to the point position？or have faster way in SDK？

@mike What do you mean with "how do i get the offset where point locate?"

You want the offset for a point with a given index? The C++ version of SplineHelp has the function GetSplinePointSegment() which can be used to get the offset for a given index. Unfortunately, this function is not available in the Python API.

best wishes,
Sebastian

@s_bach thank you for your answer! I think I have the same question as mike do, so it has been solved now. Thank you again!

Hi,

Sorry to resurrect this old thread ...
As mentioned above GetSplinePointSegment() is not available in python
Is there no way at all to get to the (real)offset of a spline point with a given index ?

spline --> point index --> offset? (in python)

index

nothing? ...

the only thing i can imagine is to sample the spline using splineHelp.GetPointIndex(), that would get you an initial offset range.
then you could start inserting points at certain offsets and refine the sampling down until a given offset range treshold is reached.
but thats quite complicated and slow as well, of course

maybe somebody with a better idea how to approach this?

Hi nothing changed, in this regards, however I added GetSplinePointSegment on our internal Python parity list
But even when it will be available this will not be backported.

Cheers,
Maxime.

Thanks Maxime,
so GetPointIndex() is the only thing you can work with, I guess.
Already on it to create this sampling function ...

Hi @everyone,

as a little disclaimer, this is my private take on this topic.

Cinema does calculate the length of a spline (at least with SplineHelp) in a rather unpretentious fashion as the sum of the Euclidean norms of the underlying LineObject segments, i.e., there is no fancy arc-length calculation going on. Which makes sense in Cinema, since the discrete LineObject is what does effectively count, and not the smooth SplineObject. Since you can get access to the LineObject of a SplineObject and also convert between SplineObject and LineObject vertex indices with SplineHelp.SplineToLineIndex() , this also means that you can both calculate the length of spline as a total and up to a specific vertex. Which then should mean that you can calculate the relative offset for a spline vertex yourself.

Practically you can cut out the SplineObject altogether when sampling a spline and just use its associated LineObject. The one nice thing that SplineHelp does for you is though, that it implements a parallel transport for the frames of spline points for you. I.e., realizes that the normals of spline points point into a direction humans would consider "correct" (mathematically these normals are not correct, due to them being modified by parallel transport, but they will look smoother, won't flip when then curvature of the spline flips).

Cheers,
Ferdinand