Math: X Trench Run

[ P_\texthit = 1 - \exp\left( -\fracR^22\sigma^2 \right) ]

[ h = \frac12 a t^2 ] [ d = v t ]

That’s extreme but possible with sci-fi tech — the key is precise timing. If the port is fixed in space but the X-wing moves laterally at speed (v_x), the torpedo must lead the target. x trench run math

So without Force assistance, ~60% chance. Luke’s “use the Force” effectively removes computer error → near 100%. | Parameter | Value | Notes | |-----------|-------|-------| | Trench length | 60 km | Adjust for shorter runs | | Speed | 150–250 m/s | Slower = easier targeting | | Time to target | 4–7 minutes | Real-time pressure | | Port size | 2 m diameter | Equivalent to hitting a coin from 1 km away | | Torpedo turn radius | <10 m | Mag-bend required | | Lead angle | 1–3° | Negligible if perfectly centered | | Base hit probability | ~60% | With good computer | | Force multiplier | → 100% | Removes systematic error | 7. Final Rule of Thumb “Stay on the centerline, match speed to targeting computer’s refresh rate, and pull the trigger when the port fills the reticle — or just listen to the dead wizard.” For a real trench run math problem set (with vectors, time dilation, or turbolaser tracking rates), let me know and I can extend this into worksheet form. [ P_\texthit = 1 - \exp\left( -\fracR^22\sigma^2 \right)