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Uniform Kochanek–Bartels Splines

As a starting point, remember the tangent vectors from Catmull-Rom splines:

\begin{equation*} \boldsymbol{\dot{x}}_i = \frac{ (\boldsymbol{x}_i - \boldsymbol{x}_{i-1}) + (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2} \end{equation*}

Parameters

TCB splines are all about inserting the parameters \(T\), \(C\) and \(B\) into this equation.

Tension

see equation 4 in [KB84]

\begin{equation*} \boldsymbol{\dot{x}}_i = (1 - T_i) \frac{ (\boldsymbol{x}_i - \boldsymbol{x}_{i-1}) + (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2} \end{equation*}

Continuity

Up to now, the goal was to have a continuous first derivative at the control points, i.e. the incoming and outgoing tangent vectors were identical:

\begin{equation*} \boldsymbol{\dot{x}}_i = \boldsymbol{\dot{x}}_i^{(-)} = \boldsymbol{\dot{x}}_i^{(+)} \end{equation*}

This also happens to be the requirement for a spline to be \(C^1\) continuous.

The “continuity” parameter allows us to break this continuity if we so desire, leading to different incoming and outgoing tangent vectors (see equations 5 and 6 in [KB84]):

\begin{align*} \boldsymbol{\dot{x}}_i^{(-)} &= \frac{ (1 - C_i) (\boldsymbol{x}_i - \boldsymbol{x}_{i-1}) + (1 + C_i) (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2}\\ \boldsymbol{\dot{x}}_i^{(+)} &= \frac{ (1 + C_i) (\boldsymbol{x}_i - \boldsymbol{x}_{i-1}) + (1 - C_i) (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2} \end{align*}

Bias

see equation 7 in [KB84]

\begin{equation*} \boldsymbol{\dot{x}}_i = \frac{ (1 + B_i) (\boldsymbol{x}_i - \boldsymbol{x}_{i-1}) + (1 - B_i) (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2} \end{equation*}

All Three Combined

see equations 8 and 9 in [KB84]

\begin{align*} \boldsymbol{\dot{x}}_i^{(+)} &= \frac{ (1 - T_i) (1 + C_i) (1 + B_i) (\boldsymbol{x}_i - \boldsymbol{x}_{i-1}) + (1 - T_i) (1 - C_i) (1 - B_i) (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2}\\ \boldsymbol{\dot{x}}_i^{(-)} &= \frac{ (1 - T_i) (1 - C_i) (1 + B_i) (\boldsymbol{x}_i - \boldsymbol{x}_{i-1}) + (1 - T_i) (1 + C_i) (1 - B_i) (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2} \end{align*}

Note: There is an error in equation (6.11) of [Mil] (all subscripts of \(x\) are wrong, most likely copy-pasted from the preceding equation).

To simplify the result we will get later, we introduce the following shorthands (as suggested in [Mil]):

\begin{align*} a_i &= (1 - T_i) (1 + C_i) (1 + B_i)\\ b_i &= (1 - T_i) (1 - C_i) (1 - B_i)\\ c_i &= (1 - T_i) (1 - C_i) (1 + B_i)\\ d_i &= (1 - T_i) (1 + C_i) (1 - B_i) \end{align*}

This leads to the simplified equations

\begin{align*} \boldsymbol{\dot{x}}_i^{(+)} &= \frac{ a_i (\boldsymbol{x}_i - \boldsymbol{x}_{i-1}) + b_i (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2}\\ \boldsymbol{\dot{x}}_i^{(-)} &= \frac{ c_i (\boldsymbol{x}_i - \boldsymbol{x}_{i-i}) + d_i (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i) }{2} \end{align*}

Calculation

import sympy as sp
sp.init_printing()

from utility import NamedExpression, NamedMatrix

helper.py

from helper import plot_basis

Same control values as Catmull-Rom …

x3, x4, x5, x6 = sp.symbols('xbm3:7')

control_values_KB = sp.Matrix([x3, x4, x5, x6])
control_values_KB

$\displaystyle \left[\begin{matrix}\boldsymbol{x}_{3}\\\boldsymbol{x}_{4}\\\boldsymbol{x}_{5}\\\boldsymbol{x}_{6}\end{matrix}\right]$

… but three additional parameters per vertex. In our calculation, the parameters belonging to \(\boldsymbol{x}_4\) and \(\boldsymbol{x}_5\) are relevant:

T4, T5 = sp.symbols('T4 T5')
C4, C5 = sp.symbols('C4 C5')
B4, B5 = sp.symbols('B4 B5')

a4 = NamedExpression('a4', (1 - T4) * (1 + C4) * (1 + B4))
b4 = NamedExpression('b4', (1 - T4) * (1 - C4) * (1 - B4))
c5 = NamedExpression('c5', (1 - T5) * (1 - C5) * (1 + B5))
d5 = NamedExpression('d5', (1 - T5) * (1 + C5) * (1 - B5))
display(a4, b4, c5, d5)

$\displaystyle a_{4} = \left(1 - T_{4}\right) \left(B_{4} + 1\right) \left(C_{4} + 1\right)$

$\displaystyle b_{4} = \left(1 - B_{4}\right) \left(1 - C_{4}\right) \left(1 - T_{4}\right)$

$\displaystyle c_{5} = \left(1 - C_{5}\right) \left(1 - T_{5}\right) \left(B_{5} + 1\right)$

$\displaystyle d_{5} = \left(1 - B_{5}\right) \left(1 - T_{5}\right) \left(C_{5} + 1\right)$

xd4 = NamedExpression(
    'xdotbm4^(+)',
    sp.S.Half * (a4.name * (x4 - x3) + b4.name * (x5 - x4)))
xd5 = NamedExpression(
    'xdotbm5^(-)',
    sp.S.Half * (c5.name * (x5 - x4)  + d5.name * (x6 - x5)))
display(xd4, xd5)

$\displaystyle \boldsymbol{\dot{x}}^{(+)}_{4} = \frac{a_{4} \left(- \boldsymbol{x}_{3} + \boldsymbol{x}_{4}\right)}{2} + \frac{b_{4} \left(- \boldsymbol{x}_{4} + \boldsymbol{x}_{5}\right)}{2}$

$\displaystyle \boldsymbol{\dot{x}}^{(-)}_{5} = \frac{c_{5} \left(- \boldsymbol{x}_{4} + \boldsymbol{x}_{5}\right)}{2} + \frac{d_{5} \left(- \boldsymbol{x}_{5} + \boldsymbol{x}_{6}\right)}{2}$

display(xd4.subs_symbols(a4, b4))
display(xd5.subs_symbols(c5, d5))

$\displaystyle \boldsymbol{\dot{x}}^{(+)}_{4} = \frac{\left(1 - B_{4}\right) \left(1 - C_{4}\right) \left(1 - T_{4}\right) \left(- \boldsymbol{x}_{4} + \boldsymbol{x}_{5}\right)}{2} + \frac{\left(1 - T_{4}\right) \left(B_{4} + 1\right) \left(C_{4} + 1\right) \left(- \boldsymbol{x}_{3} + \boldsymbol{x}_{4}\right)}{2}$

$\displaystyle \boldsymbol{\dot{x}}^{(-)}_{5} = \frac{\left(1 - B_{5}\right) \left(1 - T_{5}\right) \left(C_{5} + 1\right) \left(- \boldsymbol{x}_{5} + \boldsymbol{x}_{6}\right)}{2} + \frac{\left(1 - C_{5}\right) \left(1 - T_{5}\right) \left(B_{5} + 1\right) \left(- \boldsymbol{x}_{4} + \boldsymbol{x}_{5}\right)}{2}$

Same as with Catmull-Rom, try to find a transformation from cardinal control values to Hermite control values. This can be used to get the full basis matrix.

control_values_H = sp.Matrix([x4, x5, xd4.name, xd5.name])
control_values_H

$\displaystyle \left[\begin{matrix}\boldsymbol{x}_{4}\\\boldsymbol{x}_{5}\\\boldsymbol{\dot{x}}^{(+)}_{4}\\\boldsymbol{\dot{x}}^{(-)}_{5}\end{matrix}\right]$

From the notebook about uniform Hermite splines:

M_H = NamedMatrix(
    r'{M_\text{H}}',
    sp.Matrix([[2, -2, 1, 1],
               [-3, 3, -2, -1],
               [0, 0, 1, 0],
               [1, 0, 0, 0]]))
M_H

$\displaystyle {M_\text{H}} = \left[\begin{matrix}2 & -2 & 1 & 1\\-3 & 3 & -2 & -1\\0 & 0 & 1 & 0\\1 & 0 & 0 & 0\end{matrix}\right]$

M_KBtoH = NamedMatrix(r'{M_{\text{KB$,4\to$H}}}', 4, 4)
M_KB = NamedMatrix(r'{M_{\text{KB},4}}', M_H.name * M_KBtoH.name)
M_KB

$\displaystyle {M_{\text{KB},4}} = {M_\text{H}} {M_{\text{KB$,4\to$H}}}$

NamedMatrix(control_values_H, M_KBtoH.name * control_values_KB)

$\displaystyle \left[\begin{matrix}\boldsymbol{x}_{4}\\\boldsymbol{x}_{5}\\\boldsymbol{\dot{x}}^{(+)}_{4}\\\boldsymbol{\dot{x}}^{(-)}_{5}\end{matrix}\right] = {M_{\text{KB$,4\to$H}}} \left[\begin{matrix}\boldsymbol{x}_{3}\\\boldsymbol{x}_{4}\\\boldsymbol{x}_{5}\\\boldsymbol{x}_{6}\end{matrix}\right]$

If we substitute the above definitions of \(\boldsymbol{\dot{x}}_4\) and \(\boldsymbol{\dot{x}}_5\), we can directly read off the matrix elements:

M_KBtoH.expr = sp.Matrix([
    [expr.coeff(cv) for cv in control_values_KB]
    for expr in control_values_H.subs([xd4.args, xd5.args]).expand()])
M_KBtoH.pull_out(sp.S.Half)

$\displaystyle {M_{\text{KB$,4\to$H}}} = \frac{1}{2} \left[\begin{matrix}0 & 2 & 0 & 0\\0 & 0 & 2 & 0\\- a_{4} & a_{4} - b_{4} & b_{4} & 0\\0 & - c_{5} & c_{5} - d_{5} & d_{5}\end{matrix}\right]$

M_KB = M_KB.subs_symbols(M_H, M_KBtoH).doit()
M_KB.pull_out(sp.S.Half)

$\displaystyle {M_{\text{KB},4}} = \frac{1}{2} \left[\begin{matrix}- a_{4} & a_{4} - b_{4} - c_{5} + 4 & b_{4} + c_{5} - d_{5} - 4 & d_{5}\\2 a_{4} & - 2 a_{4} + 2 b_{4} + c_{5} - 6 & - 2 b_{4} - c_{5} + d_{5} + 6 & - d_{5}\\- a_{4} & a_{4} - b_{4} & b_{4} & 0\\0 & 2 & 0 & 0\end{matrix}\right]$

And for completeness’ sake, its inverse:

M_KB.I

$\displaystyle {M_{\text{KB},4}}^{-1} = \left[\begin{matrix}\frac{b_{4}}{a_{4}} & \frac{b_{4}}{a_{4}} & \frac{b_{4} - 2}{a_{4}} & 1\\0 & 0 & 0 & 1\\1 & 1 & 1 & 1\\\frac{- c_{5} + d_{5} + 6}{d_{5}} & \frac{- c_{5} + d_{5} + 4}{d_{5}} & \frac{- c_{5} + d_{5} + 2}{d_{5}} & 1\end{matrix}\right]$

t = sp.symbols('t')

b_KB = NamedMatrix(r'{b_{\text{KB},4}}', sp.Matrix([t**3, t**2, t, 1]).T * M_KB.expr)
b_KB.T.pull_out(sp.S.Half)

$\displaystyle {b_{\text{KB},4}}^{T} = \frac{1}{2} \left[\begin{matrix}a_{4} t \left(- t^{2} + 2 t - 1\right)\\t^{3} \left(a_{4} - b_{4} - c_{5} + 4\right) + t^{2} \left(- 2 a_{4} + 2 b_{4} + c_{5} - 6\right) + t \left(a_{4} - b_{4}\right) + 2\\t \left(b_{4} + t^{2} \left(b_{4} + c_{5} - d_{5} - 4\right) + t \left(- 2 b_{4} - c_{5} + d_{5} + 6\right)\right)\\d_{5} t^{2} \left(t - 1\right)\end{matrix}\right]$

To be able to plot the basis functions, let’s substitute \(a_4\), \(b_4\), \(c_5\) and \(d_5\) back in (which isn’t pretty):

b_KB = b_KB.subs_symbols(a4, b4, c5, d5).simplify()
b_KB.T.pull_out(sp.S.Half)

$\displaystyle {b_{\text{KB},4}}^{T} = \frac{1}{2} \left[\begin{matrix}t \left(B_{4} + 1\right) \left(C_{4} + 1\right) \left(T_{4} - 1\right) \left(t^{2} - 2 t + 1\right)\\t^{3} \left(\left(B_{4} - 1\right) \left(C_{4} - 1\right) \left(T_{4} - 1\right) - \left(B_{4} + 1\right) \left(C_{4} + 1\right) \left(T_{4} - 1\right) - \left(B_{5} + 1\right) \left(C_{5} - 1\right) \left(T_{5} - 1\right) + 4\right) + t^{2} \left(- 2 \left(B_{4} - 1\right) \left(C_{4} - 1\right) \left(T_{4} - 1\right) + 2 \left(B_{4} + 1\right) \left(C_{4} + 1\right) \left(T_{4} - 1\right) + \left(B_{5} + 1\right) \left(C_{5} - 1\right) \left(T_{5} - 1\right) - 6\right) + t \left(T_{4} - 1\right) \left(\left(B_{4} - 1\right) \left(C_{4} - 1\right) - \left(B_{4} + 1\right) \left(C_{4} + 1\right)\right) + 2\\- t \left(t^{2} \left(\left(B_{4} - 1\right) \left(C_{4} - 1\right) \left(T_{4} - 1\right) + \left(B_{5} - 1\right) \left(C_{5} + 1\right) \left(T_{5} - 1\right) - \left(B_{5} + 1\right) \left(C_{5} - 1\right) \left(T_{5} - 1\right) + 4\right) - t \left(2 \left(B_{4} - 1\right) \left(C_{4} - 1\right) \left(T_{4} - 1\right) + \left(B_{5} - 1\right) \left(C_{5} + 1\right) \left(T_{5} - 1\right) - \left(B_{5} + 1\right) \left(C_{5} - 1\right) \left(T_{5} - 1\right) + 6\right) + \left(B_{4} - 1\right) \left(C_{4} - 1\right) \left(T_{4} - 1\right)\right)\\t^{2} \left(B_{5} - 1\right) \left(C_{5} + 1\right) \left(T_{5} - 1\right) \left(t - 1\right)\end{matrix}\right]$

labels = sp.symbols('xbm_i-1 xbm_i xbm_i+1 xbm_i+2')

plot_basis(
    *b_KB.expr.subs({T4: 0, T5: 0, C4: 0, C5: 1, B4: 0, B5: 0}),
    labels=labels)

plot_basis(
    *b_KB.expr.subs({T4: 0, T5: 0, C4: 0, C5: -0.5, B4: 0, B5: 0}),
    labels=labels)

TODO: plot some example curves