Brain & Camelot
Camelot Camelot
I’ve been poring over the siege tactics at Calais, and I’m wondering—how would you mathematically model a trebuchet’s trajectory to maximize its range? The chronicles describe a lot of “eyeballing” the lift, but I suspect there’s a cleaner equation that could’ve given the attackers a sharper advantage.
Brain Brain
To get a good estimate you can treat the stone as a simple projectile. First calculate the launch speed from the counter‑weight energy: \(E = \frac{1}{2} M_{cw} g h\) (where \(M_{cw}\) is the counter‑weight mass and \(h\) the drop height). That energy is transferred to the stone: \(\frac{1}{2} m_s v_0^2 = \eta E\) (with \(m_s\) the stone mass and \(\eta\) the efficiency, usually 0.5–0.7). Solve for \(v_0\): \(v_0 = \sqrt{\frac{2 \eta M_{cw} g h}{m_s}}\). Then use the standard range formula for a projectile launched at an angle \(\theta\): \(R = \frac{v_0^2}{g}\sin(2\theta)\). Since the arm length and pivot give you a limited range of release angles, you can vary \(\theta\) to find the maximum. In practice the optimal angle is a bit less than 45° because of air resistance and the fact that the stone is released while the arm is still moving. If you plug in realistic numbers for a medieval trebuchet—say \(M_{cw}=3000\,kg\), \(m_s=100\,kg\), \(h=2\,m\), \(\eta=0.6\)—you get \(v_0\approx 40\,m/s\). Then a release angle around 35–38° gives a theoretical range of about 400–450 m, which is close to the recorded distances. Adjusting the mass ratio and arm ratio can fine‑tune the launch speed and thus the range. That’s the clean equation the attackers could have used.