Cubic bezier curve segment

Question

If I have the 4 points describing a Bezier curve P1, P2, P3, P4 (where P1 and P4 are the end points of the curve and P2 and P3 are the control points of the curve), how could I find the points that describes only a segment of this bezier curve?

I found this answer which is exactly what I am looking for but the answer seems wrong. If I set t0=0 and t1=1 in the equations which should represent the entire bezier curve, the resulting points are not valid. They are not equal to the original points.

It seems that the solution is related to the De Casteljau's algorithm, but I can't understand how it works.

Answer 1

Yes, De Casteljau's algorithm is the way to go.

Parametrization

If your curve isn't correctly parametrized from t=0 trough t=1, then it seems you're using the wrong equation to descibe your curve. Wikipedia has the correct formula for you:

B(t) = (1−t)³P₁ + 3(1−t)²t P₂ + 3(1−t)t²P₃ + t³P₄

[I adjusted the indices from the zero-based form in Wikipedia to the one-based from your question.]

If you set t=0, you get P₁, your starting point. If you set t=1, you get P₄, your endpoint. In between, the shape of the curve is determined by those points and the two control points P₂ and P₃.

De Casteljau's algorithm

Let t be the parameter where you want to cut your curve. Let's say you want to keep only the initial part. You draw the three lines from P₁ to P₂, from there to P₃ and from there to P₄. Each of these lines you divide at a the fraction t of its length, i.e. the length of the line before the dividing point relates to the entire length as t : 1. You now have three new points P₁₂ through P₃₄. Do the same again to obtain two points P₁₂₃ and P₂₃₄, and again to obtain the single point P₁₂₃₄. This final point is B(t), the endpoint of your truncated curve. The startpoint is P₁ as before. The new control points are P₁₂ and P₁₂₃ the way we just constructed them.

Removing an initial part of the curve works the same way. So in two steps, you can trim both ends of your curve. You obtain a new set of control points which exactly (up to the numeric precision you use) describe a segment of your original curve, with no approximation or similar involved.

You can translate all of the geometric descriptions above into algebraic formulas, and in a perfect world you should come up with the results from this answer to the question you quoted.

Alas, this doesn't appear to be a perfect world. At the time of this writing, those formulas only used polynoms of degree two, so they could not describe endpoints on a third degree curve. The correct formula should be the following:

• P'₁ = u₀u₀u₀P₁ + (t₀u₀u₀ + u₀t₀u₀ + u₀u₀t₀) P₂ + (t₀t₀u₀ + u₀t₀t₀ + t₀u₀t₀) P₃ + t₀t₀t₀P₄
• P'₂ = u₀u₀u₁P₁ + (t₀u₀u₁ + u₀t₀u₁ + u₀u₀t₁) P₂ + (t₀t₀u₁ + u₀t₀t₁ + t₀u₀t₁) P₃ + t₀t₀t₁P₄
• P'₃ = u₀u₁u₁P₁ + (t₀u₁u₁ + u₀t₁u₁ + u₀u₁t₁) P₂ + (t₀t₁u₁ + u₀t₁t₁ + t₀u₁t₁) P₃ + t₀t₁t₁P₄
• P'₄ = u₁u₁u₁P₁ + (t₁u₁u₁ + u₁t₁u₁ + u₁u₁t₁) P₂ + (t₁t₁u₁ + u₁t₁t₁ + t₁u₁t₁) P₃ + t₁t₁t₁P₄

where u₀ = 1 − t₀ and u₁ = 1 − t₁.

Note that in the parenthesized expressions, at least some of the terms are equal and can be combined. I did not do so as the formula as stated here will make the pattern clearer, I believe. You can simply execute those computations independently for the x and y directions to compute your new control points.