Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 66 / 6 / b

Table of contents
- Solution b

Solution

Use degree elevation of Bernstein coefficients. If

b_{j}

are degree-

p

Bézier curve controls, the identical curve has degree-

(p + 1)

controls

b_{0}^{+} = b_{0}, b_{p + 1}^{+} = b_{p}, b_{j}^{+} = \frac{j}{p + 1} b_{j - 1} + (1 - \frac{j}{p + 1}) b_{j} (1 \leq j \leq p) .

This follows by multiplying each degree-

p

Bernstein basis polynomial by

(1 - t) + t

and collecting the degree-

(p + 1)

terms, so the operation changes the representation, not the curve.

For the quadratic controls in part (a), the cubic controls are

D_{0} = B_{0}, D_{1} = \frac{B _{0} + 2 B _{1}}{3} = \frac{P _{i - 1} + 5 P _{i}}{6}, D_{2} = \frac{2 B _{1} + B _{2}}{3} = \frac{5 P _{i} + P _{i + 1}}{6}, D_{3} = B_{2} .

Raise these cubic Bézier curve controls once more. Then

E_{0} = B_{0}, E_{1} = \frac{1}{4} D_{0} + \frac{3}{4} D_{1} = \frac{1}{2} (B_{0} + B_{1}), E_{2} = \frac{1}{2} (D_{1} + D_{2}) = \frac{1}{6} (B_{0} + 4 B_{1} + B_{2}),

E_{3} = \frac{3}{4} D_{2} + \frac{1}{4} D_{3} = \frac{1}{2} (B_{1} + B_{2}), E_{4} = B_{2} .

Thus the requested quartic Bézier curve controls, expressed in the original polygon, are

E_{0} E_{1} E_{2} E_{3} E_{4} = (P_{i - 1} + P_{i}) /2, = (P_{i - 1} + 3 P_{i}) /4, = (P_{i - 1} + 10 P_{i} + P_{i + 1}) /12, = (3 P_{i} + P_{i + 1}) /4, = (P_{i} + P_{i + 1}) /2.

Each expression is an affine combination whose coefficients sum to one. Endpoint positions and derivatives are preserved at each degree-raising step, as is every point of the span.

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