Parametric Cubic Spline Tutorial

[VBAhack]

Example Problem #2 (Closed Curve)

Let us connect the points in the same sequence as before, but close the loop. In this case, we need a 5^th point in the sequence that is identical to the 1^st:

p₁ = [1,0]
p₂ = [-1, 0]
p₃ = [0, 1]
p₄ = [0, -1]
p₅ = p₁ = [1,0]

For s, arc length approximation could be used as before, but an easier approach will be used so the simplified equations can be employed. We will choose equal spacing: s = [0, 0.25, 0.5, 0.75, 1].

Boundary conditions require special consideration. In the case of a closed loop, the 1^st and last points are joined, which makes them interior points subject to the same 1^st and 2^nd derivative compatibility constraints as the other interior points. Thus, with closed loops, there are no choices to be made for boundary conditions! To derive the 1^st derivative compatibility equation, simply equate 1^st derivatives of p₁ and p₅:

p₁' = p₅'
p₁' = (p₂ – p₁)/h₁ - p₂"h₁/6 - p₁"h₁/3 = p₅' = (p₅ – p₄)/h₄ + p₅"h₄/3 + p₄"h₄/6
6(p₂ – p₁)/h₁ - p₂"h₁ – 2p₁"h₁ = 6(p₅ – p₄)/h₄ + 2p₅"h₄ + p₄"h₄
6[(p₂ – p₁)/h₁ - (p₅ – p₄)/h₄] = p₂"h₁ + 2p₁"h₁ + 2p₅"h₄ + p₄"h₄

The 2^nd derivative boundary conditions are easy: p₁" - p₅" = 0. (Note: using p just means there are 2 sets of equations, one for x and another for y). Also, h₁ = h₂ = h₃ = h₄ = h = s₂ = 0.25, which allows the use of the simplified equations. The resulting matrix equation to be solved is (Note: . means zero):

┌		┐ ┌       ┐          ┌                        ┐      ┌      ┐
│2  1  .  1   2	│ │x1" y1"│          │x2-x1-x5+x4  y2-y1-y5+y4│      │-3  -1│
│1  4  1  .   .	│ │x2" y2"│ = 6/(h*h)│  x3-2x2+x1    y3-2y2+y1│      │ 3   1│
│.  1  4  1   .	│ │x3" y3"│          │  x4-2x3+x2    y4-xy3+y2│ = 96 │-1  -3│
│.  .  1  4   1	│ │x4" y4"│          │  x5-2x4+x3    y5-2y4+y3│      │ 1   3│
│1  .  .  .  -1	│ │x5" y5"│          │          0            0│      │ 0   0│
└   		┘ └       ┘          └                        ┘      └      ┘

The solution is x" = [-120, 120, -72, 72, -120]^T and y" = [-72, 72, -120, 120, -72]^T, from which we can calculate the coefficients of the spline segments and generate the plot.
.