I'm trying to figure out how to model projectile motion which can be defined using a simple [quadratic] equation which defines projectile parabola:

$$y = y_0 + v_{0y}t - \dfrac{1}{2} gt^2$$

I came across some slides from Brown University which model the motion as follows:

$$\begin{align} \text{Matrix form: }& \begin{pmatrix} 0 & 100 & -10\end{pmatrix} \\ \\ \text{state: }& \begin{matrix}y & y' & y'' \\ p & v & a\end{matrix} \\ \\ &\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \\ \end{pmatrix} \end{align}$$

But I'm not quite sure how they came up with that transition matrix. Can anyone shed some light how to pick the state variables and how to model the motion?


He started from $$ y = y_0 + v_0 t + \frac 1 2 a t^2$$ but then for some reason best known to himself decided to take $\frac 1 2 a$ as $y''$ rather than a, (and ignored the fact that his $y''(t)$ is no longer equal to $d y'(t) / dt$, but hey, Lewis Carroll was also a mathematician, and this guy, like Humpy Dumpty, can make words mean whatever he feels like they ought to mean).

So he gets the approximate equations $$\begin{matrix} y(t) \\ y'(t) \\ y''(t) \end{matrix} \approx \begin{matrix} y(t-1) + y'(t-1) \\ y'(t-1) + 2y''(y-1) \\ y''(t-1) \end{matrix}$$ or $$\begin{bmatrix} y(t) & y'(t) & y''(t) \end{bmatrix} \approx \begin{bmatrix} y(t-1) & y'(t-1) &y''(t-1) \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1\\ \end{bmatrix} $$ He can't seem to make up his mind whether $y''$ is really a variable, or just the arbitrary constant $-10$, (or was that $+10$, or $\pm 20$ - I've lost the will to care...

It would be more obvious what's going on if you replace the 2's with 1's in his equations, and let $y''(t)$ actually BE the second derivative of $y(t)$, not something close.

Do people actually keep their jobs as university lecturers for teaching like this nowadays? It's 50 years since I was an undergrad, so I don't know....

  • $\begingroup$ Uh, super interesting. I thought I was missing something fundamental there, but your answer makes me think I probably havent - it just seems kinda all over the place in those slides. I dont think I would ever have thought of that. $\endgroup$ Apr 2 '19 at 23:46

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