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Parametric Cubic Spline continued...

As a final example, let’s see if a circle can be approximated with the same four points (i.e. 4 spline segments). The connection sequence needed is:

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

┌				┐ ┌       ┐    ┌                                            ┐
│ ?                  		│ │x1" y1"│    │             ?                  ?           │
│h1   2(h1+h2)      h2		│ │x2" y2"│ = 6│(x3-x2)/h2-(x2-x1)/h1  (y3-y2)/h2-(y2-y1)/h1│
│        h2      2(h2+h3)  h3	│ │x3" y2"│    │(x4-x3)/h2-(x3-x2)/h1  (y4-y3)/h2-(y3-y2)/h1│
│                           ?  	│ │x4" y3"│    │             ?                  ?           │
└   				┘ └       ┘    └                                            ┘

┌				┐ ┌       ┐    ┌               ┐
│1                  		│ │x1" y1"│    │ 0        1    │
│0.369   1.261   0.261		│ │x2" y2"│ = 6│ 9.243    3.828│
│        0.261   1.261   0.369	│ │x3" y2"│    │-3.828   -9.243│
│                        1  	│ │x4" y3"│    │ 0        1    │
└   				┘ └       ┘    └               ┘

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

Point p₅, which is identical to p₁, is needed to close the shape. Thus, n = 5. The values of s are easy – since the points are all equidistant, by inspection: s = [0, 0.25, 0.5, 0.75, 1]^T.

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:

Matrix Equation

The solution is x" = [-48, 0, 48, 0, -48]^T and y" = [0, -48, 0, 48, 0]^T, from which we can calculate the coefficients of the spline segments and generate the plot using s = 0 to 1. It is a remarkable fit considering only 4 spline segments were used. Clearly, more points will result in an even better fit.

Conclusion:

We have extended our knowledge of cubic splines to include the powerful and flexible parametric version and have fit various shapes (including approximating a circle) to a set of given points.

 
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* Plot here
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[Squab 14]

...o_O.....

[Jacob Roman]

I was hoping it was XXX literally.

[Squab 14]

lol.

[eSPiYa]

Porn?!

[Jacob Roman]

All your pr0n are belong to us.