I have all points where my curve pass through, but I need to get the coordinates of the control points to be able to draw the curve. How can I do to calculate this points?


2 Answers 2


When what you already have is a set of points where your curve must pass through, Bézier is not what you want; you should be using a parametric (cubic) spline.

Assuming that your points $\mathbf{P}_i=(x_i,y_i)$, $i=1\dots n$ represent a general curve (as opposed to a function $f(x)$, for which simply applying the usual cubic spline algorithm on your points suffices), Eugene Lee proposed centripetal parametrization to generate suitable parameter values $t_i$ associated with your $\mathbf{P}_i$. The prescription for generating the $t_i$ (in its most general form) is


where $\left\| \cdot \right\|$ is the (Euclidean) length, $e$ is an adjustable exponent in the interval $[0,1]$, and $t_1=0$. (A usual value for $e$ is 0.5, but $e=1$ is sometimes used as well.)

From this, one applies the usual cubic spline algorithm to the sets $(t_i,x_i)$ and $(t_i,y_i)$, from which you now have your parametric spline. (The periodic spline is recommended for closed curves, and the "not-a-knot" spline for all other cases.)

The MATLAB Spline Toolbox has support for this parametrization scheme, though it shouldn't be too hard to write your own implementation.


It depends how many points you have.

If you have only 4 points, then you will need only one Bezier cubic segment. Suppose your known points (through which you want the curve to pass) are $Q_0, Q_1, Q_2, Q_3$. First you have to choose 4 parameter values $t_0,t_1,t_2,t_3$ to assign to these points. The centripedal approach described in the other answer is good. In simple situations, where your points are fairly equidistant, a much simpler choice is just $(t_0,t_1,t_2,t_3) = (0, 1/3, 2/3, 1)$. Then you have to solve a system of 4 linear equations, as follows. Suppose $P_0,P_1, P_2, P_3$ are the (unknown) control points of the Bezier curve, and denote the curve by $C(t)$. Then we want $C(t_i) = Q_i$ for $i=0,1,2,3$. This means $$ \sum_{j=0}^3{b_j(t_i)P_j} = Q_i$$ for $i=0,1,2,3$, where $b_0, b_1,b_2,b_3$ are the Bernstein polynomials of degree 3. Solve these equations to get $P_0,P_1, P_2, P_3$. If you always use $(t_0,t_1,t_2,t_3) = (0, 1/3, 2/3, 1)$, then the coefficient matrix of the linear system is fixed, and you can just invert it once (exactly), and save the answer. This gives you a simple formula relating the $P$'s and $Q$'s.

If you have more than 4 points, then you can use a Bezier curve with degree higher than 4. The calculations are analogous to those shown above. But a better approach is to use a "spline" that consists of several Bezier cubic segments joined end-to-end. There are many ways to compute splines. One of the easiest (and the most popular amongst graphics folks) is the Catmull-Rom spline. It gives you simple formulae for the control points in terms of the given points $Q_i$.


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