1
$\begingroup$

I want some scalar spline function defined on regular 2D grid $F(x,y)$ with continuous first derivative which is easy to intersect with arbitrary ray/line ${\vec l}(t) = (c_x t,c_y t,c_z t)$.

Finding the intersection means finding roots of equation $f_{c_x,c_y}(t)-c_zt=0$ where $f_{c_x,c_y}(t)=F(c_x t, c_y t )$ are values of the spline along the line ${\vec l}(t)$.

Since it is easy to find roots of quadratic polynominal I want $f_{c_x,c_y}(t)$ to be piecewise quadratic polynominal for any ${c_x,c_y}$.

What is the problem / what is not a solution:

  1. consider bi-quadratic B-spline on rectangular grid created by tensor-product of 1D quadratic B-splines. The result is bi-quadratic which means it is $f_{c_x,c_y}(t)$ is 4-th order polynominal.
  2. function composed of Quadratic Bezier-triangles. While it is certainly just quadratic function along any direction, it is hard ensure continuous derivatives at the boundary between triangles.
|cite|improve this question
$\endgroup$
  • $\begingroup$ I don't think I quite understand. You have an arbitrary ray coming from the origin and you want to the equation of some B-spline surface that intersects it? Or do you want the general form of a B-spline surface-ray intersection? $\endgroup$ – CADJunkie Oct 24 '16 at 18:42
  • $\begingroup$ I just want to compute efficiently (=as fast as possible) intersection between 3D ray and some 2D spline surface. I'm trying to choose best 2D spline for that purpose. While normally ray-surface intersection are computed by quite costly iterative schemes (rayMarching,subdividing...) I was thinking that if any 1D cut would be piecewise-quadratic polynominal than I can easily solve the roots directly. $\endgroup$ – Prokop Hapala Oct 25 '16 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.