Skip to main content

Derive cubic Bezier equation and properties

Short answer

Cubic Bezier: P(u) = (1-u)^3 P0 + 3u(1-u)^2 P1 + 3u^2(1-u) P2 + u^3 P3. It passes through P0 and P3, with end tangents 3(P1-P0) and 3(P3-P2). Properties: lies inside the convex hull, affine invariance, global control, symmetry, variation diminishing, and blending functions are positive and sum to 1.

Derive the equation for cubic Bezier curve. Also write down its properties.

Cubic Bezier equation

A cubic Bezier curve has four control points P0, P1, P2, P3 (n = 3). The general Bezier curve is P(u) = Σ P_i · C(n,i) u^i (1−u)^(n−i), 0 ≤ u ≤ 1. For n = 3:

C(3,0) = 1, C(3,1) = 3, C(3,2) = 3, C(3,3) = 1

P(u) = (1−u)³ P0 + 3u(1−u)² P1 + 3u²(1−u) P2 + u³ P3

Blending functions: B0 = (1−u)³, B1 = 3u(1−u)², B2 = 3u²(1−u), B3 = u³ (they add to 1).

Expanding in powers of u:

P(u) = (−P0 + 3P1 − 3P2 + P3) u³ + (3P0 − 6P1 + 3P2) u² + (−3P0 + 3P1) u + P0

Matrix form:

                  | -1  3 -3  1 | | P0 |
P(u) = [u³ u² u 1]|  3 -6  3  0 | | P1 |
                  | -3  3  0  0 | | P2 |
                  |  1  0  0  0 | | P3 |

Check at u = 0: P = P0. At u = 1: P = P3.

Properties

  1. Starts at P0 and ends at P3; other points are not on the curve.
  2. Tangent at the ends: P'(0) = 3(P1 − P0), P'(1) = 3(P3 − P2).
  3. Lies inside the convex hull of the control points.
  4. Blending functions are positive and sum to 1.
  5. Affine invariance.
  6. Global control: moving a point changes the whole curve.
  7. Symmetric: reversing the point order gives the same curve.
  8. Variation diminishing.
  9. Adjacent sections have C⁰ continuity if P3 of the first = P0 of the second, and C¹ if also the points around the join are collinear with equal spacing.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Computer Graphics chapter 5 questions →