Trench Run Math [top] - X

No lateral motion needed if flying straight down the trench centerline — but if the port is offset, or the X-wing is drifting:

Eliminate (t):

Assume the torpedo enters the port horizontally at speed (v), then experiences a constant vertical acceleration (a) (from the magnetic field) over a short distance (d) (the shaft length before impact). x trench run math

[ P_\texthit \approx 1 - e^-0.913 \approx 1 - 0.401 \approx 0.599 \ (\text59.9%) ]

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. No lateral motion needed if flying straight down

[ h = \frac12 a \left( \fracdv \right)^2 ]

(h) (depth to target):

[ a = \frac2(20)(100^2)10^2 = \frac40 \cdot 10,000100 = 4,000 \ \textm/s^2 \ (\approx 408g) ]