Building quadratic B-spline

The problems with a single Bezier spline range from the need of a high degree curve to accurately fit a complex shape, which is inefficient to process. To overcome the problems, a piecewise curve is used. Therefore the order of the curve is not dependent on the number of control points.

Building of piecewise B-splines is motivated below. You can skip this and go directly to the explanation of B-splines calculations.

Quadratic Bezier spline subdivision

Any spline point P(t) subdivides Bezier curve in two smaller quadratic splines (with control points P₀, P₀¹, P(t) and P(t), P₁¹, P₂ for quadratic splines to the left). The new curves match the original in position, although they differ in parameterization. E.g. for t = 1/2 you see to the left that triangles (1 P₀¹ 2) and (3 P(t) 2) are equal and the point P(1/4) of the original spline coincides with the point P(t') of the first small spline for t' = 1/2.

Building complex B-spline curves

The first derivatives of these two small splines at P(t) are determined by the P₁¹ P(t) and P(t) P₀¹ control segments. Therefore if we drag only P₀ or P₂ points we get continuous at P(t) curve. Note that both small splines depend only on its own control points.

In a similar way we get piecewise B-spline with continuous (n - 2)-order derivatives from n-order Bezier spline by subdivision and end control points movement. We can repeat subdivisions to get very complex curve.

De Boor points and Cox - be Boor algorithm

As since only 4 control points of the two small quadratic Bezier splines are independent therefore it is convenient to use 4 de Boor control points D_0,1,...,3 (see Fig.2). Then one can evaluate points of k-order B-spline by the Cox - be Boor algorithm similar to the de Casteljau iterations
P_i^j(t) = (1 - τ_i^j)P_i-1^j-1 + τ_i^j P_i^j-1, P_i⁰ = D_i, τ_i^j = (t - t_i)/(t_i+k-j - t_i), 0 < j < k,
where (t₀, ... , t_n+k) are knots and t_k-1 ≤ t ≤ t_n+1.

For the spline in Fig.2 we have n = 3, k = 3, we take the uniform knot vector (-2, -1, 0, 1, 2, 3, 4). For the first spline segment D₀ = P₀ + (P₀ - P₁), 0 ≤ t ≤ 1 and you can check that
τ₁¹ = (t - t₁)/(t₃ - t₁) = (t + 1)/2,
P₁¹ = (1 - τ₁¹)D₀ + τ₁¹D₁ = (1 - t)B₀ + t B₁,
τ₂¹ = (t - t₂)/(t₄ - t₂) = t/2,
P₂¹ = (1 - τ₂¹)D₁ + τ₂¹D₂ = (1 - t)B₁ + t B₂,
τ₂² = (t - t₂)/(t₃ - t₂) = t,
P₂² = (1 - τ₂²)P₁¹ + τ₂²P₂¹ = (1 - t)P₁¹ + tP₂¹.
These are exactly de Casteljau iterations for the first Bezier spline segment (sorry its control points B_i are itroduced too). We can "forget" about this lengthy math and use only de Boor control points and B-spline basis functions in calculations further.

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updated 21 Feb 2013