Mathematical underpinnings of the ball and disk integrator Recently I've been curious about analog computing, and I came across a device called the ball-and-disc integrator. Even though there's lots of information about how they can be used to calculate various functions, there is little to no concrete information about the principles underpinning them, only hand-wavy things. I would appreciate if someone would help me out here.
Thanks for your  time.
 A: I think we can determine the operating principles by looking at the picture on wiki. The disk rotates at a constant angular velocity $\omega$. Let $x$ be the contact point of the sphere on the disk, measured from the disc center to the right. $x$ may be controlled manually and lies in the operating range $-X\leq x\leq X$, where $X$ is determined by the disk radius and the sphere radius. At $x$, the tangential velocity on the disc is $\omega x$. If the sphere has radius $R$ then, at position $x$ it rotates at angular velocity $\Omega=\omega x/R$, which is then also the angular velocity of the output cylinder. Call the angular position of the cylinder $\theta(t)$, then
$$\tag{1}
\theta(t)=\frac{\omega}{R}\int\limits_0^t dt' \ x(t')
$$
$\theta$ is measured by (what I'll call) a protractor attached to the cylinder. We want to relate (1) to the integral $F$ of some given function $f$
$$\tag{2}
F(s)=\int\limits_0^s ds' \ f(s')
$$
Let $g(y)$ be the physical graph (ie. on a sheet of paper) of $f(s)$. The units of $g$ and $y$ are length; the units of $f$ and $s$ are arbitrary. We have
$$\tag{3}
f=Mg \qquad, \qquad s=Ny
$$
Where $M$ and $N$ are the scale factors relating the physical graph $g$ to the function $f$. $M$ must be chosen so that $|g|_\text{max}\leq X$. The integral of $g$ is
$$\tag{4}
G(y)=\int\limits_0^{y(s)} dy' \ g(y')=\frac{1}{MN}\int\limits_0^s ds' f(s')
$$
Or,
$$
F(s)=MN G(s/N)
$$
We imagine$^\dagger$ that $g(y)$ will be arranged vertically (so the $y$ axis increases vertically upwards, and the $+g$ axis in the same direction as $+x$). The graph is to move downwards at a fixed velocity $V$; while it does, the operator moves $x$ so that it matches the graph $g$. Consequently we have
$$\tag{5}
g(Vt)=x(t)
$$
Substituting (5) and (4) into (1)
$$
\theta(t)=\frac{\omega }{MNVR}F(NVt)
$$
Or,
$$\tag{6}
F(s)=\frac{MNVR}{\omega}\theta(s/NV)
$$
On the LHS of (6) is the integral to be computed, on the RHS are constants and the output of the machine after a time $s/NV$.
$\dagger$ It's hard to imagine this with the particular picture on wiki, and it looks like the `real thing' had the operator matching $x$ to $g$ via a lever, but the same principle of operation applies.
Aside: We can modify the procedure to deal with the integral of a product of two functions by letting $\omega=\omega(t)$, then
$$
F(s)=\int\limits_0^s ds' \ f_1(s')f_2(s')
$$
The analog of (1) is then
$$
\theta(t)=\frac{1}{R}\int\limits_0^t dt' \ \omega(t')x(t')
$$
The graph of $f_2$ will follow the same construction as $g$ above. $f_1$ will determine $\omega(t)$ by similar arguments.
