From this page Derivatives of a Bézier Curve, I can see that the derivative of a degree $N$ Bezier curve is just a Bezier curve of degree $N-1$ and it explains how to calculate the control points by each control point just being $P_{i+1} - P_i$.

Is it also true that the derivative of a Bezier rectangle of degree $(M,N)$ is also just a Bezier rectangle of degree $(M-1,N-1)$? If so, how would you calculate the control points for the rectangle?

By Bezier rectangle, I mean a tensor product Bezier surface, like the kind described here Wikipediate: Bezier Surface.

I've been trying to work it out on paper but no luck so far. I've been able to make progress on coming up with a derivative, but it has quite a few terms even for a biquadratic patch, and i'd like to do this with higher degrees.

My end goal is that I'm trying to calculate the gradient of a univariate Bezier rectangle that has scalar control points. Specifically, the rectangle takes $X$ and $Z$ values from 0 to 1 and outputs a $Y$ value for the given $(X,Z)$. I want to find the gradient so that I can use it to calculate surface normals, as well as get a distance estimation for sphere tracing (ray marching).

Thanks for any help you guys can provide!

  • $\begingroup$ By a Bezier rectangle, do you mean a rectangle with rounded corners where the curves that make up the corners are Bezier curves? If so, this may be of help. $\endgroup$ – wltrup Jul 31 '15 at 18:17
  • $\begingroup$ No sorry, i mean a tensor product Bezier surface. Question updated, thanks for the excellent question. $\endgroup$ – Alan Wolfe Jul 31 '15 at 18:20

Yes, it's true (almost). If the surface parameters are $u$ and $v$, the partial derivative wrt $u$ is a Bezier patch of degree $(M, N-1)$ and the partial derivative wrt $v$ is a Bezier patch of degree $(M-1, N)$.

If you are using the deCasteljau algorithm to calculate points on your patch, then the partial derivatives will be natural by-products.

Often the best way to handle a surface computation is to reduce it to a curve computation. So, suppose you want to calculate the partial derivative wrt $u$ at parameter values $(\bar u, \bar v)$ on the Bezier patch $(u,v) \mapsto \mathbf{S}(u,v)$. The curve $\mathbf{C}(u) = \mathbf{S}(u,\bar v)$ is a Bezier curve, whose control points are easy to obtain. The desired partial derivative is just the derivative of the Bezier curve $\mathbf{C}$ at the parameter value $u = \bar u$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.