While there is no closed form solution arc length reparameterization for cubic Bezier curves, is there a set of solutions that taken together cover all possible classes of cubic Bezier curves, such that as long as you know which class of curve you're dealing with, the arc length reparameterization is simply a matter of "plugging in the numbers" to transform the parametric Bezier function into a function of arc length?

For instance, cubic Beziers can be split into degenerate (lines) curves, single arches (no inflections), double arches (one inflection), or discontinuous curves (with a cusp), where that last class can be trivially treated as two separate continuous curves with a single shared coordinate without affecting the arc length - and all forms can been rotated/translated without affecting the actual arc length in a way that parts of the parametric functions that define them "disappear".

For example, we can rotate/translate any cubic Bezier such that the first coordinate is $(0,0)$, removing one of the polynomial terms from affecting the reparameterization, with the fourth coordinate on $(x,0)$, removing another term from the reparameterization (either we can do this because both start and end point form an edge of the Bezier hull, or because we can split up the curve into two distinct curves at a known time value such that we end up with two curves for which the start and end points form an edge on their respective hulls).

Is there no reparameterization possible even for those simplified cases, given that we don't want "the" solution, just "one of however many solutions that yields, as absolute results, the correct answer"? E.g. if we rotate/translate the curve such that the entire curve is always in the $+/+$ quadrant, then we don't care if the reparameterization yields solutions that aren't the "true" values, we only care that if we plug in an arc length, we get a coordinate for which $(abs(x),abs(y))$ is the correct coordinate.

(I have not been able to find any material that covers arc length reparameterization in this way, only material that goes "there is no general solution. the end")


1 Answer 1


Arc length reparameterization of a cubic Bezier curve (or Bezier curve of any degree higher) has no close-form solution in general is because we cannot find a close-form representation of the arc length function:


let alone to find its inverse function.

However, there are indeed a few classes of cubic Bezier curves where the $L(t)$ can be represented as a close-form formula. For example, when $P_1=P_0+(P_3-P_0)/3$ and $P_2=P_0+2(P_3-P_0)/3$, the cubic Bezier curve will be a straight line with constant first derivative.

There is another class of curves called Pythagorean Hodograph curves (PH curves) where the $L(t)$ has a close-form representation as well. This class of curves has $x'^2(t)+y'^2(t)+z'^2(t) = \sigma^2(t)$ and therefore their $L(t)$ can be computed exactly.

Even with these two classes of cubic Bezier curves (i.e., straight line with constant first derivative and PH curve) where arc length reparametrization is possible, it remains as "no general solution" for most cubic Bezier curves and has to resort to numeric methods.

  • $\begingroup$ Not sure I understand, quadratic Beziers supposedly have a perfectly fine closed form solution $\endgroup$ Dec 10, 2018 at 22:23
  • $\begingroup$ From the question's title, isn't it about "cubic Bezier"? $\endgroup$
    – fang
    Dec 11, 2018 at 3:23
  • $\begingroup$ I'm remarking on this: "Arc length reparameterization of a cubic Bezier curve (or Bezier curve of any degree) has no close-form solution", which suggests the statement covers quadratic -and linear, I guess- Beziers as well. Going back to the question, though: an arc length function doesn't have to be "one function" so can we just cut up a cubic in a smart way to yield a piecewise arc length reparameterization? $\endgroup$ Dec 11, 2018 at 16:20
  • 1
    $\begingroup$ I see. It should be "...any degree higher". Thanks for catching this. It is corrected now. $\endgroup$
    – fang
    Dec 11, 2018 at 16:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .