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).
Plug numbers: (\fracR^22\sigma^2 = \frac12(0.74^2) = \frac11.095 \approx 0.913) x trench run math
[ \textLead angle \theta = \arctan\left( \fracv_\textlateralv_\texttorpedo \right) ] Assume the torpedo enters the port horizontally at
[ a = \frac2(20)(100^2)10^2 = \frac40 \cdot 10,000100 = 4,000 \ \textm/s^2 \ (\approx 408g) ] 000100 = 4
If (h \approx 20) m (depth to port), (d \approx 10) m (magnetic bending distance), (v \approx 100) m/s (torpedo speed after launch):