How it works

The movement

This applet simulates a phyical pendulum. This means that the swinging body is considered to have a moment of inertia, instead of being a mass concentrated in a single point in space. So in order to simulate such a pendulum for any polygon imaginable we need to find ways to calculate the mass, moment of inertia, and center of mass of any shape.

The differential equation governing the movement of these pendulums is $$ \alpha = \ddot{\theta} = -\frac{mgL}{I}\sin\theta $$ with the following parameters:

The parameters

From the previous paragraph it is apparent that we need to find a few values, namely the mass, center of mass and the moment of inertia. These three properties of the pendulum have fairly straight-forward formulas that allow us to calculate them: $$ m = \iint_{S} \rho(\vec{r}) \mathop{}\!\mathrm{d}{A} $$ $$ \vec{R} = \frac{1}{m} \iint_{S} \rho(\vec{r}) \vec{r} \mathop{}\!\mathrm{d}{A} $$ $$ I = \iint_{S} \rho(\vec{r}) ||\vec{r}||^2 \mathop{}\!\mathrm{d}{A} $$ \(S\) denotes the set of points that lie within our pendulum. So now we have three formulas that we can use to calculate the missing parameters! Except that these equations make use of integrals. We need to find a way to somehow calculate integrals in JavaScript. Thankfully the mathematical field of numerics has already provided us with an algorithm that we can easily implement.

Quadrature

In numerics we don't integrate, we quadrate. Don't ask me why. In this project I settled for the trapezoid method: $$ \int_{x_0}^{x_{N-1}} f(x) \mathop{}\!\mathrm{d}x \approx h\left[ \frac{1}{2} f_0 + f_1 + \cdots + f_{N-2} + \frac{1}{2} f_{N-1} \right] $$ This isn't exactly the most accurate algorithm, but it gets the job done, and we'll later see that it really doesn't matter in this case. The only problem left is that the equations above are double integrals, how can we use the trapezoid method to calculate those? Well let's consider the following integral: $$ \int_{x_1}^{x_2} \int_{y_1(x)}^{y_2(x)} f(x, y) \mathop{}\!\mathrm{d}y \mathop{}\!\mathrm{d}x $$ We notice that the inner integral, once evaluated, only depends on \(x\)! So let's set $$ G(x) = \int_{y_1(x)}^{y_2(x)} f(x, y) \mathop{}\!\mathrm{d}y $$ and we can rewrite that integral as $$ \int_{x_1}^{x_2} G(x) \mathop{}\!\mathrm{d}x $$ Now we can apply the trapezoid rule on this integral. And obviously \(G(x)\) can also be evaluated with the trapezoid rule. But the attentive reader might have noticed another problem. In these numerical examples we always integrated over rectangles, whereas in the formulas above we integrate over any set. We can define a new function \begin{align} \mathbb{I}_S: \mathbb{R}^2 &\longrightarrow \{0, 1\} \\ x &\longmapsto \begin{cases} 1, & \text{if } x \text{ is in } S \\ 0, & \text{else} \end{cases} \end{align} \(\mathbb{I}_S\) is called the characteristic function of \(S\). This allows us to write $$ \iint_{S} f(\vec{r}) \mathop{}\!\mathrm{d}A = \int_{x_1}^{x_2} \int_{y_1(x)}^{y_2(x)} \mathbb{I}_S(x, y) f(x, y) \mathop{}\!\mathrm{d}y \mathop{}\!\mathrm{d}x $$ We'll now see that \(\rho\) is our characteristic function.

Putting it all together

We now have all the tools to calculate our parameters. \(\vec{r}\) is simply the position of the points in 2D-Space, and \(\rho(\vec{r})\) is the density of space at that position. Since the simulated pendulum has the same mass everywhere, it must also have the same density everywhere. So, essentially, the density function looks like this: $$ \rho(\vec{r}) = \begin{cases} \rho, & \text{if } \vec{r} \text{ is in pendulum} \\ 0, & \text{else} \end{cases} $$ Or if we define \(S\) to be the set of all points in the pendulum, it becomes apparent that $$ \rho(\vec{r}) = \rho ~\mathbb{I}_S(x, y) $$ So we need to implement a function that can determine wether a given point is inside our polygon or not. I used a raycasting algorithm I found online for this. Essentially, you cast a ray from the point you're testing in any direction and then count how often you cross an edge of the polygon. If that number is odd, you're inside the polygon. Using that function as our density function, we can now implement the integrals using the trapezoid method, and thus calculate mass, center of mass and the moment of inertia of any polygon in 2D space! With this, simulating the initial differential equation becomes trivial.