How do I calculate the vertical travel of an oscillating arm with a fulcrum? I'm trying to calculate the vertical travel of the longer arm in a motion like this:

I've had a go at defining the variables involved:

Where the measurables are:

*

*S = Short arm length

*L = Long arm length

*F = Distance from origin of L to the fulcrum

And the unknowns are:

*

*V = Total vertical travel distance of L

I'd like to express V in terms of the measurable variables (perhaps more than defined above, if I've missed any, which is likely).
I don't have much knowledge in this area and no idea how to approach working this out formally, short of brute forcing the measurable variables and working backwards.
 A: Let's call $B\mathord:(b\mathbin;0)$ the apex of the fulcrum, $O\mathord:(0\mathbin;0)$ the fixed point of the short arm, and $M\mathord:(s\cos\vartheta\mathbin;s\sin\vartheta)$ the mobile point of the mechanism; one calculates that the leftmost extremity $E$ of the long arm has these coordinates:
\begin{eqnarray*}
x(\vartheta)\mkern-8mu&=&\mkern-8mu s\cos\vartheta-(s-b)\frac{s\cos\vartheta-b}{\sqrt{b^2+s^2-2bs\cos\vartheta}}\\[5pt]
y(\vartheta)\mkern-8mu&=&\mkern-8mu s\sin\vartheta\left(1-\frac{s-b}{\sqrt{b^2+s^2-2bs\cos\vartheta}}\right)
\end{eqnarray*}
where $s>0$ is the length of the short arm (thus $L=s-b$).
In the following picture, we take as an example $b=-20$ and $s=12$; this corresponds to
$\displaystyle x(\vartheta)=12\cos\vartheta-32\frac{3\cos\vartheta+5}{\sqrt{30\cos\vartheta+34}}$ and
$\displaystyle y(\vartheta)=12\sin\vartheta\left(1-\frac{8}{\sqrt{30\cos\vartheta+34}}\right)\cdot$

Searching for which values of $\vartheta$ is $y$ extremal leads to a most complicated equation (and some pessimism)…
Numerically —according to MacOS Grapher.app— $y$ is maximal for $\vartheta_0\approx3.737$ (about $214^\circ$), and $y(\vartheta_0)\approx11.057)$.

In the special case where $s=-b$ —i.e. when $F=0$— the coordinates of point $E$ become much simpler:
$x(\vartheta)=b(2\cos(\vartheta/2)-\cos\vartheta)$ and $y(\vartheta)=b(2\sin(\vartheta/2)-\sin\vartheta)$ (for $-\pi\leqslant\vartheta\leqslant\pi)$; the corresponding trajectory for $E$ is no longer a closed curve but a piece of cardioid (in red below), limited to $x\leqslant b$.
Then $y(\vartheta)$ is maximal (resp. minimal) at $\vartheta=-\pi$ (resp. $\vartheta=\pi$), and $y(\pm\pi)=\mp2b$.

Once $b$ is fixed are there special values for $s$ (apart from $-b$) that lead to an “explicit” solution?
