Matching a Cubic Bezier Curve to a Cubic Polynomial

[rumzie]

I'm having a lot of trouble researching this topic to fit my current situation.

As the title suggests. I'm trying to match a cubic Bezier to a cubic polynomial via the Bezier's control points.

The cubic polynomial is parametric, and is intended to be fit together with other cubic polynomials such that one cubic polynomial ends where another begins, and their constraints are identical to preserve continuity. Think of via points along a trajectory.

The parameters of the polynomial are only position and velocity at the beginning and end. 4 terms, so a cubic polynomial.



Is there some way to define the control points of a cubic bezier such that they match a cubic polynomial?
Ideally I'm looking for an algebraic solution.

I first tried to convert the polynomial to a bezier curve using polynumbers. They don't seems to match 1 to 1 though and I'm unsure what is off.




This is my current model for trying to fit the Bezier to the cubic polynomial.

 

[blamocur]

Bezeir curve [imath]\mathbf B(t)[/imath] is defined as a cubic vector polynomial with coefficients derived from the two end points and two control points (https://en.wikipedia.org/wiki/Bézier_curve#Cubic_Bézier_curves). If you compute the derivatives at [imath]t=0[/imath] and [imath]t=1[/imath] you should get nice expressions for your control points (relative to the end points).

 

[rumzie]

Thanks for your reply! This article is fantastic. I think my initial confusion was having the bezier curve in a binomial form. It complicates things a bit when trying to find the derivative of the curve, Wonky graphs.