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36. Space Curves

Dated: 05-07-2025

\[\vec x(u) = a_x u^3 + b_x u^2 + c_xu + d_x\]

Similarly, \(\vec y(u)\) and \(\vec z(u)\) will be

\[\vec y(u) = a_y u^3 + b_y u^2 + c_yu + d_y\]
\[\vec z(u) = a_z u^3 + b_z u^2 + c_zu + d_z\]

Combining these into one equation, we have

\[\vec p(u) = au^3 + bu^2 + cu + d\]

Again, we will limit \(u\) to \([0, 1]\) and let's take 4 points this time.

  • Starting point (\(u = \frac 0 3\))
  • Intermediate point 1 (\(u = \frac 1 3\))
  • Intermediate point 2 (\(u = \frac 2 3\))
  • Ending point (\(u = \frac 3 3\))

Substituting \(u\) in \(\vec x(u)\)

\[ \begin{aligned} & x_1 = d_x \\ & x_2 = \frac{1}{27}a_x + \frac{1}{9}b_x + \frac{1}{3}c_x + d_x \\ & x_3 = \frac{8}{27}a_x + \frac{4}{9}b_x + \frac{2}{3}c_x + d_x \\ & x_4 = a_x + b_x + c_x + d_x \end{aligned} \]

Finding the variables \(a_x, b_x, c_x\) and \(d_x\), we have

\[ \begin{aligned} & a_x = -\frac{9}{2}x_1 + \frac{27}{2}x_2 - \frac{27}{2}x_3 + \frac{9}{2}x_4 \\ & b_x = 9x_1 - \frac{45}{2}x_2 + 18x_3 - \frac{9}{2}x_4 \\ & c_x = -\frac{11}{2}x_1 + 9x_2 - \frac{9}{2}x_3 + x_4 \\ & d_x = x_1 \end{aligned} \]

Substituting back into \(\vec x(u)\), we have

\[ x(u) = \left(-\frac{9}{2}x_1 + \frac{27}{2}x_2 - \frac{27}{2}x_3 + \frac{9}{2}x_4\right) u^3 + \left(9x_1 - \frac{45}{2}x_2 + 18x_3 - \frac{9}{2}x_4\right) u^2 \\ + \left(-\frac{11}{2}x_1 + 9x_2 - \frac{9}{2}x_3 + x_4\right) u + x_1 \]

Refactoring it, we have

\[ x(u) = \left(-\frac{9}{2}u^3 + 9u^2 - \frac{11}{2}u + 1\right)x_1 + \left(\frac{27}{2}u^3 - \frac{45}{2}u^2 + 9u\right)x_2 \\ + \left(-\frac{27}{2}u^3 + 18u^2 - 9u\right)x_3 + \left(\frac{9}{2}u^3 - \frac{9}{2}u^2 + u\right)x_4 \]

After finding the equivalent expressions for \(\vec y (u)\) and \(\vec z(u)\), we have

\[ \vec P(u) = \left(-\frac{9}{2}u^3 + 9u^2 - \frac{11}{2}u + 1\right)p_1 + \left(\frac{27}{2}u^3 - \frac{45}{2}u^2 + 9u\right)p_2 \\ + \left(-\frac{27}{2}u^3 + 18u^2 - 9u\right)p_3 + \left(\frac{9}{2}u^3 - \frac{9}{2}u^2 + u\right)p_4 \]

Let

\[ \begin{array}{cc} G_1 = \left(-\frac{9}{2}u^3 + 9u^2 - \frac{11}{2}u + 1\right) & G_2 = \left(\frac{27}{2}u^3 - \frac{45}{2}u^2 + 9u\right) \\ G_3 = \left(-\frac{27}{2}u^3 + 18u^2 - 9u\right) & G_4 = \left(\frac{9}{2}u^3 - \frac{9}{2}u^2 + u\right) \end{array} \]

such that

\[\vec P(u) = G_1p_1 + G_2p_2 + G_3 p_3 + G_4 p_4\]
\[\vec P(u) = GP\]

where

\[ G = \begin{bmatrix} G_1 & G_2 & G_3 & G_3 \end{bmatrix} \]
\[ P = \begin{bmatrix} p_1 & p_2 & p_3 & p_3 \end{bmatrix}^T \]

But \(G\) can be further decomposed into

\[G = U N\]

Where

\[ U = \begin{bmatrix} u^3 & u^2 & u & 1 \end{bmatrix}^T \]
\[ N = \begin{bmatrix} -\frac{9}{2} & \frac{27}{2} & -\frac{27}{2} & \frac{9}{2} \\ 9 & -\frac{45}{2} & 18 & -\frac{9}{2} \\ -\frac{11}{2} & 9 & -\frac{9}{2} & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \]
\[ A = \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \\ d_x & d_y & d_z \end{bmatrix} \]
\[\vec p(u) = UA\]
\[\because \vec p(u) = \vec P(u)\]
\[UA = GP\]
\[\because G = UN\]
\[\implies UA = UNP\]
\[\implies A = NP\]