Notes Computer Animation Curves

Curves

8 Sept 2022

§ Motivation

Keyframe

Often in computer animation, an artist creates keyframes which is the state (position, rotation, scale, etc.) at a discrete time step.

Then, an algorithm takes these keyframes and interpolates between them to accomplish a (typically) smooth transition between the keyframes .

Almost anything can be keyframed.

Keyframing: Position and Orientation over Time

An artist may provide positions at timesteps $t_1, t_2,$ and $t_3$ and we generate a curve between them to get the position at any time between $t_1$ and $t_3.$

Keyframing: Joint Angles

We can model an arm and its movement using joint angles $\theta_1, \theta_2,$ and $\theta_3.$

Then, we can define $\theta_i(t)$ to describe how the angles change with respect to time.

This then leaves us with the question of how we would interpolate between keyframes?

§ Representation of Curves

There are a few ways to express a curve:

Type	Form	Example for Circle
Explicit	$y = f(x)$	$y = \pm \sqrt{1 - x^2}$
Implicit	$f(x, y) = 0$	$x^2 + y^2 - 1 = 0$
Parametric	$x = f(t), y = g(t)$	$x = \cos t, y = \sin t$

§ Explicit

Easy to Generate
Only one value of $y$ for each $x$
Heavily dependent on choice of independent variable

§ Implicit

$f(x, y)$ is a measure of the distance of a point to curve (level-set information).
Difficult to generate points on a curve

§ Parametric

Easy to generate points on a curve
Can have multiple $y$ for a given $x$

§ Cubic Bézier Curve

Cubic Bézier Curve

Specify 4 "control" points, $(a_0, b_0), \dots, (a_3, b_3),$ then the cubic Bézier can be given in parametric form by

\begin{align*} x(u) &= a_0 + a_1u + a_2u^2 + a_3u^3\\ y(u) &= b_0 + b_1u + b_2u^2 + b_3u^3 \end{align*}

Alternatively in vector notation:

\begin{align*} \vec{p}(u) &= (1-u)^3\vec{p}_0 + 3u(1-u)^2\vec{p}_1 + 3u^2(1-u)\vec{p}_2 + u^3\vec{p}_3\\ \vec{p}(0) &= \vec{p}_0,\ \vec{p}(1) = \vec{p}_3 \end{align*}

And in matrix form:

\begin{align*} \vec{p}(u) &= \underbrace{\begin{pmatrix} \vec{p}_0 & \vec{p}_1 & \vec{p}_2 & \vec{p}_3 \end{pmatrix}}_{G} \underbrace{\begin{pmatrix} 1&-3&3&-1\\ 0&3&-6&3\\ 0&0&3&-3\\ 0&0&0&1 \end{pmatrix}}_{B} \underbrace{\begin{pmatrix} 1\\ u\\ u^2\\ u^3 \end{pmatrix}}_{\vec{u}}\\ \vec{p}(u) &= G\underbrace{B}_{\mathclap{\textrm{cubic Bezier basis}}}\vec{u} \end{align*}

Why use 4 control points?
The parametric equations have 8 unknowns; each point gives us 2 pieces of information.

§ Continuity in Concatenated Bézier Curves

There different classes of curve continuity

$C^0$ , the curves are connected
$C^1$ , tangents are the same
$G^1$ , tagent directions are the same

They can be visualised with the following illustration

$C^n$ and $G^n$ are families of curves where the $n$ ^th derivative matches, but for animation $C^2$ or $G^2$ is good enough in most cases.

§ Cubic Catmull-Rom Spline

Catmull-Rom Spline

The Catmull-Rom spline is a $C^1$ curve, but it is not $C^2$ . Specifically, not $C^2$ at its control points

The basis for the Catmull-Rom spline can be given by

B = \frac{1}{2} \begin{pmatrix} 0 & -1 & 2 & -1\\ 2 & 0 & -5 & 3\\ 0 & 1 & 4 & -3\\ 0 & 0 & -1 & 1 \end{pmatrix}

Connecting Sequence of Control Points

Connecting a sequence of 3D/2D points can be done using a Catmull-Rom spline.

\begin{align*} \vec{p}(u) &= \langle\vec{p}_0,\ \overbrace{\vec{p}_1,\ \vec{p}_2}^{\mathclap{\text{the curve passes through the middle 2 points}}},\ \vec{p}_3\rangle B\,\vec{u},\ u \in [0, 1]\\ \vec{p}(u) &= \langle\vec{p}_1,\ \vec{p}_2,\ \vec{p}_3,\ \vec{p}_4\rangle B\,\vec{u},\ u \in [0, 1]\\ \vec{p}(u) &= \langle\vec{p}_2,\ \vec{p}_3,\ \vec{p}_4,\ \vec{p}_5\rangle B\,\vec{u},\ u \in [0, 1] \end{align*}

Every set of 4 control points define a segment of the curve
"Concatenated" $u$ goes from $[0, n - 3]$ where $n$ is the number of control points.
To convert between $u$ concatenated to $\hat{u} \in [0, 1]$

\begin{align*} k &= \lfloor u\rfloor\\ \hat{u} &= u_k = u - k\\ G &= \big[\vec{p}_k, \vec{p}_{k+1}, \vec{p}_{k+2}, \vec{p}_{k+3}\big]_{4 \times 4} \end{align*}

Then, the curve can be computed using

\vec{p}(u) = G_kB\vec{u_k}, u \in [0, n - 1]

where $n$ is the number of control points.