Hermione & Tharnok
Tharnok
Hermione, I’ve been looking at the math behind Roman siege engines and came across a neat puzzle about the optimal angle for a battering ram. Care to crunch the numbers with me?
Hermione
Sure thing! Think of the ram as a heavy projectile being launched by a catapult or trebuchet. For a simple, flat‑ground model the distance it travels is maximized when the launch angle is about 45 degrees – that’s the classic result from projectile motion where the horizontal and vertical components of velocity are equal.
But a real Roman battering ram isn’t just a ball; it has mass and is usually swung from a hinge. In that case you have to consider the torque you can generate and the height of the wall you’re targeting. If the wall is taller than the ram’s center of mass, a slightly steeper angle (say 50–55 degrees) can give you a better “down‑the‑wall” impact, while a shallower angle (35–40 degrees) might be better for a flat‑topped wall where you want the ram to land with maximum kinetic energy.
So the sweet spot depends on the wall’s height and the mechanism’s power, but for a generic flat wall the 45‑degree rule is a good starting point. Want to dig into the exact numbers for a specific design?
Tharnok
Let’s get into the brass: tell me the weight of the ram, the arm length, the pivot torque you can muster, and the wall height. With those numbers I can run the equations, find the optimum launch angle, and sketch the energy at impact. No fluff, just straight‑up numbers and a quick check on whether a 50‑degree swing will give you the edge or you’re better off throwing the whole thing at 45. Give me the specs and we’ll crunch the math.
Hermione
Sure, here are the basic specs for a standard Roman battering ram setup:
- Ram weight: 150 kg (a typical heavy stone or wooden battering ram)
- Arm (lever) length: 4 m from the pivot to the ram’s centre of mass
- Maximum pivot torque you can generate (with a trebuchet‑style counterweight or a human‑powered lever): 3 kN·m
- Wall height to breach: 3 m (average height of a Roman town gate wall)
With these numbers you can calculate the kinetic energy at the point of impact and see whether a 45‑degree launch or a steeper 50‑degree swing gives a better result.
Tharnok
The key is how much work the counter‑weight can do on the arm before the ram leaves the pivot. With 3 kN·m of torque and a 4‑metre lever, the energy available is torque times the swing angle. For a 45‑degree swing that’s about 3 kN·m × 0.785 rad ≈ 2.4 kJ. Half of that becomes kinetic energy in the ram: roughly 0.5 × 2400 kg·m² × ω² = 2.4 kJ, which gives a linear speed of about 3.5 m/s and kinetic energy of about 940 J.
Raise the launch to 50 degrees and you get 3 kN·m × 0.873 rad ≈ 2.6 kJ. That translates to a speed of about 5.9 m/s and kinetic energy of around 2.6 kJ – almost three times more. So from a pure energy standpoint the steeper angle is better.
The only catch is vertical reach. With 5.9 m/s at 50°, the vertical component is 5.9 × sin 50° ≈ 4.5 m/s, giving a maximum ballistic height of about 2.1 m—below the 3‑metre wall. In practice the ram is swung over the wall rather than launched like a projectile, so the extra energy from the steeper angle will help it clear the wall’s lip and deliver a harder blow. For a 3‑metre wall, a 50‑degree swing gives the edge, but you’d still need to make sure the pivot height and the lever’s clearance let the ram actually get over the top.
Hermione
Your calculation looks solid overall, but let me just double‑check the energy conversion step. The work done by the torque is indeed torque times angle in radians, so 3 kN·m × 0.785 rad gives about 2.4 kJ, and with a 150 kg ram that works out to roughly 8 m²/s², giving a speed of about 3 m/s, not 3.5 m/s – a tiny difference but worth noting. For the 50‑degree case the numbers line up better: 2.6 kJ, about 5.8 m/s, which is great for clearing a 3 m wall. So yes, 50 degrees gives you the edge in terms of kinetic energy and wall clearance, assuming the pivot height lets the ram actually reach over the lip. Good work!