Note that the spatial position of the pendulum's mass (endpoint) is $\mathbf{p} = L\begin{bmatrix}\sin(\theta)\\ -\cos(\theta)\end{bmatrix}.$
The pendulum's dynamics are governed Newton's second law (torque form) $$\tau = I\alpha,$$where $\tau$ is the torque about the axis of rotation (origin), $I$ is the moment of inertia, and $\alpha$ is the angular acceleration. Recall that the gravitational force is $mg$, and note the length of the perpendicular lever arm is $L\sin\theta$. Therefore the torque about the axis of rotation is $\tau=mgL\sin\theta$. Recall (via Google) that the moment of inertia of a point mass distance $L$ from its axis of rotation is $I = mL^2$. ✨ Warmup: Solve the above equations for $\alpha$. ✅ $m$ drops out. What does this *mean*?
Choose the state of our system to $\xi = (\theta, \omega)^T$, where $\theta$ is the pendulum's angular position, and $\omega$ is its angular velocity. Recall that $\omega = \dot\theta$ and $\alpha = \dot\omega = \ddot\theta$, where the "dot notation" just denotes the number of time derivatives. I.e., $\dot\theta = \frac{d\theta}{dt}$ and $\ddot\theta = \frac{d^2\theta}{dt^2}$. To simulate, we're going to discretize our physics in time. Call $h$ the simulation time step. The very useful notation $\xi_k$ means "$\xi$ at time $t_k$" ✅ "If the state of the system at time $t_k$ is $\xi_{k}$, then $\xi_{k+1}$ is the state of the system $h$ seconds later, i.e. the state of the system at time $t_{k+1} = t_k + h$."
Here are three of the simplest approaches for integrating our simulation forward in time. ⭐ Forward Euler is $\xi_{k + 1} = \xi_{k} + h \dot\xi_{k}$ I.e., $ \begin{bmatrix}\theta_{k+1}\\\omega_{k+1}\end{bmatrix} = \begin{bmatrix}\theta_{k}\\\omega_{k}\end{bmatrix} + h\begin{bmatrix}\dot\theta_{k}\\\dot\omega_{k}\end{bmatrix}, $ which we can rewrite as $ \begin{bmatrix}\theta_{k+1}\\\omega_{k+1}\end{bmatrix} = \begin{bmatrix}\theta_{k}\\\omega_{k}\end{bmatrix} + h\begin{bmatrix}\omega_{k}\\\alpha_{k}\end{bmatrix}. $ ⭐Semi-implicit Euler can be written as $ \begin{bmatrix}\theta_{k+1}\\\omega_{k+1}\end{bmatrix} = \begin{bmatrix}\theta_{k}\\\omega_{k}\end{bmatrix} + h\begin{bmatrix}\omega_{k + 1}\\\alpha_{k}\end{bmatrix}. $ ⭐Implicit Euler can be written as $ \begin{bmatrix}\theta_{k+1}\\\omega_{k+1}\end{bmatrix} = \begin{bmatrix}\theta_{k}\\\omega_{k}\end{bmatrix} + h\begin{bmatrix}\omega_{k + 1}\\\alpha_{k + 1}\end{bmatrix}. $ Note that for Implicit Euler, $\theta_{k+1}$ will show up on both sides of the equation! To simulate using Implicit Euler we need to solve a nonlinear equation every timestep. We will cover one approach to this (fixed point iteration) in class on Thursday.
Simulate the pendulum. - angular acceleration - potential energy (just used for plotting) -- $\text{PE} = -mgy$ (or, technically speaking, $-mg(y-y_0),$ where $y_0$ is any constant) NOTE $g$ is _negative_ - kinetic energy (just used for plotting) -- $\text{KE} = mv^2/2 = m(L\omega)^2/2$ - explicit euler integration - semi-implicit euler integration - (harder) implicit euler integration HINT: eliminate $\omega_{k+1}$ so you just have an equation of $\theta_{k+1}$. HINT: fixed point iteration is probably the easiest way to solve the resulting equation. ?HINT: if you would like, you can eliminate $\omega_{k+1}$ from the Implicit Euler rule by substitution. ?HINT: you can also get rid of $\alpha_{k+1}$ by recalling that you have $\alpha(\theta)$, a function for $\alpha$ as a function of $\theta$ (true at all time). choose to evaluate it at $t_{k+1}$ and substitute your equation for $\alpha_{k+1}=\alpha(\theta_{k+1})$ into the equation to eliminate $\alpha$ In your glow comment, explain the general behavior (i.e. what happens to the total energy of the system) of the three methods. ✅ Feel free to check your work here. 🚨 Warning: this resource uses a different sign convention for $g$. b. (40 pts) linear blend skinning Let's learn about skinning by implementing it all on the CPU. (In practice, we would probably use a vertex shader.) 👇 I highly recommend drawing a diagram of the setup as you read through this explanation. Consider a skeleton consisting of a chain of $n$ bones. Call the $L_j$ the length of each bone. Call the $\theta_j$ the angle of each bone relative to the previous bone. Let's adopt some convenient conventions. Call the position of a bone the spatial position of its left endpoint. Assume the skeleton's rest pose has it lying along the positive $x$-axis, with the zeroth bone positioned at the origin. I.e., the $j$-th bone has rest position $\mathbf{B}_j = (X_j, 0)^T,$ and the zeroth bone's position $\mathbf{B}_0 = (0, 0)^T$. For simplicity, notate the position of the far end of the last bone in the skeleton $\mathbf{B}_n = \sum_{j=0}^{n-1}L_j.$ ✅ $\mathbf{B}_0 = (0,0)^T$ and $\mathbf{B}_1 = (L_0,0)^T.$
Forward kinematics is the problem of mapping from (relative) joint angles $\theta_0,...,\theta_{n-1}$ to joint positions $\mathbf{b}_0,...,\mathbf{b}_n$.
👩🔬 Optional General-Purpose 3D Derivation
The general solution to forward kinematics is quite elegant, though a bit overkill for this particular 2D problem. For completeness, I include it here, but feel free to skip it. We'll use the horrifying notation from the transforms week :D Call $\mathbf{F}_j$ the coordinate system of the $j$-th bone. Recall ${}_{j-1}\mathbf{F}_{j}$ is the coordinate system of the $j$-th bone written in the coordinate system of the $j-1$-th bone. Say the relative rotation of the $j$-th bone is encoded in $\mathbf{R}_j$. Say the $i$-th bone has length $L_j$. Then ${}_{j-1}\mathbf{F}_j = \mathbf{T}_j\mathbf{R}_j,$ where $\mathbf{T}_j$ is a translation by $(L_j, 0, 0)^T$. This notation had the property that ${}_{\text{World}}\mathbf{F}_{j+1} = {}_{\text{World}}\mathbf{F}_{j}\ {}_{j}\mathbf{F}_{j+1}.$ This is a recurrence, which implies $${}_{\text{World}}\mathbf{F}_{j} = {}_{\text{World}}\mathbf{F}_{0}\ {}_{0}\mathbf{F}_{1} \cdots {}_{j-1}\mathbf{F}_{j}.$$ Broken down into translations and rotations, the $j$-th coordinate system written in world coordinates is $${}_{\text{World}}\mathbf{F}_j = \mathbf{R}_0\mathbf{T}_1\mathbf{R}_1 \cdots \mathbf{T}_j\mathbf{R}_j.$$Skinning involves binding the nodes (vertices of the skin mesh) to the bones through weights. We can think of the weights being stored in a big matrix $\mathbf{W}.$ The entry $W_{ij}$ is the $i$-th node's weighting of the $j$-th bone. Given $\mathbf{W}$ and the joint angles, our job is to compute the node's current position. The notation is going to get gross regardless, so we might as well go all in 🤷♂️ Call ${}_j\mathbf{S}_i$ the rest position of the $i$-th node written in the rest coordinate system of the $j$-th bone. The $j$-th bone has current (absolute) coordinate system $\mathbf{M}_j=\mathbf{T}_{\mathbf{b}_j}\mathbf{R}_{\phi_j}$. ✅ Note that this is just the typical "rotation by absolute orientation followed by translation to absolute position." Linear blend skinning calculates the $i$-th node's current position as $\mathbf{s}_i = \sum_{j=0}^{n-1}W_{ij}\ \mathbf{M}_j\ {{}_j\mathbf{S}_i}.$ ✅ Note that $\mathbf{M}_j\ {{}_j\mathbf{S}_i}$ can be thought of as... - "gluing" the $i$-th node to the $j$-th bone (when both are in their rest positions), - rotating and translating the bone according to $\mathbf{M}_j$, - measuring where the node ends up. ✅ $\sum_{j=0}^{n-1}W_{ij}\ \mathbf{M}_j\ {{}_j\mathbf{S}_i}$ repeats the above operation for all $n$ bones and takes the weighted average of the results.
✨ HINT (${}_{j}\mathbf{S}_i$)
Notate ${}_{\text{World}}\mathbf{S}_i$ the rest position of the $i$-th node in world coordinates. Say the $j$-th bone has rest position (coordinate system origin) $\mathbf{B}_j$. Then ${}_{j}\mathbf{S}_i = {}_{\text{World}}\mathbf{S}_i - \mathbf{B}_j.$Implement skeletal animation. - forward kinematics (skeleton should show up) - skinning (skin should show up) 🌟🌟 HINT To access the weights, use
weights[mode][node_i][bone_i].
(mode is which approach to skinning we're using (i.e. RIGID or SMOOTH))
- (harder) smooth weight selection (press 'k' to switch to smooth mode; should look reasonable -- i.e. skin is smooth, does not self-intersect except at extreme angles, etc.)
HINT the graph of weights (drawn in the top left of the app) should be nice and continuous. 🚨 i did _not_ use sin or cos.
✅ Forward kinematics and skinning require very little code.
c. (20 pts) boids (creative coding)
Using the original 1987 paper (and Google) as inspiration, write an $\mathcal{O}(n^2)$ flocking simulation.
A bare bones solution might simulate ~100 2D boids and draw them as POINTS.
Possible extensions:
- Make the boids follow your mouse.
- Draw the boids as isosceles TRIANLGES, with the tip indicating direction.
- Simulate 3D boids, and drawn them as tets.
- Implement a basic spatial data structure to simulate a *bunch* of boids.
- E.g., repopulate a naive grid at the start of each frame.
Instead of querying every other boid in the world...
...a given boid would just query boids in its grid cell and maybe also the neighboring cells.