This is a home-engineering problem, but there's not much mechanics involved. It's exclusively trig/calculus. I haven't done any of that since I left school more than 40 years ago, so I'm brushing away the cobwebs. I've spent two days trying to puzzle this out, but I just can't crack it.

It's to do with the kind of Roman crossbow called an "inswinger". I'll point it to shoot skywards, so we can talk about "vertical" and "horizontal", to make things easier.

You have two axles 2 feet apart. The bow arms attached to the axles are rigid levers driven by torsion springs. The arms are 1 foot in length, and the bowstring (2 ft long) is strung between the tips. In the rest position, with the bowstring taut, the arms point directly upwards at the 12 o'clock position. To cock the bow you pull the string down, forcing the levers inwards and downwards until they point directly at each other - a 90 deg turn. The tips can be regarded as touching at that point. And obviously then the angles subtended by the two halves of the string is zero. On release, the arms fly upwards and outwards, pulling the bowstring taut while propelling the missile up the arrow shute.

Supposing a constant velocity from the arm tips (to simplify things), I need to calculate the missile velocity at any point during the release. And that is a function of the changing geometrical relationships in the configuration. I can see that the velocity ratio missile/armtip is 1:1 at the start, and something indeterminately large at the finish (infinite in theory). But finding a formula to fix the missile's position at any point is defeating me.

I can simplify a little by ignoring everything to the right of the arrow chute, because it's just the mirror image of the left side. If I draw a line between the axles and call that the "baseline", I can be sure about just one position, apart from the starting and finishing positions. It's where the arm tip is at 60 deg to the baseline. Since the left half of the baseline is the same length as the arm, and the left half of the string is the same length, you have an equilateral triangle. So at that point the missile is positioned exactly on the baseline.

But after that, I can't use the baseline as a reference, because the line between the missile and the axle is no longer horizontal, and no longer the same length. The only line that isn't moving is the chute, so it might be better to deal with a right-angle triangle involving that. The arm-tip is at A, missile at B, and the horizontal from A crosses the chute at C. The angle between the arm and the baseline is $\displaystyle \theta$. The angle of the string to the chute is $\displaystyle \phi$.

The missile is getting its velocity ($\displaystyle \text{v}_m$) from two sources. Call the tip's linear velocity $\displaystyle \text{v}_t$. The vertical component of that, $\displaystyle \text{v}_v$, does not cause the shape of the triangle ABC to change. It just raises it. So there is a 1:1 relationship. One inch of upward movement of the tip causes one inch of movement of the missile.

So I've got the beginnings of an equation.

$\displaystyle \text{v}_m = \text{v}_v+$...something.

...where $\displaystyle \text{v}_v $ must be $\displaystyle \text{v}_t \text{ cos }\theta$, and "something" is the movement of the missile caused by the horizontal movement of the tip ($\displaystyle \text{v}_h$), which changes the angle of the string, thus forcing the missile up the chute. And $\displaystyle \text{v}_h=\text{v}_t\text{ sin } \theta$. But I can't just plug it into the equation, the way I can with $\displaystyle \text{v}_v$. There is no 1:1 relationship. It depends on $\displaystyle \phi$. But how do you calculate that? What is the relationship between $\displaystyle \phi$ and $\displaystyle \theta$? That's where I'm stuck.