Distance from Secaucus to Aleksandrow Lodzki
The shortest distance (air line) between Secaucus and Aleksandrow Lodzki is 4,209.66 mi (6,774.79 km).
How far is Secaucus from Aleksandrow Lodzki
Secaucus is located in New Jersey, United States within 40° 46' 51.6" N -75° 56' 2.76" W (40.7810, -74.0659) coordinates. The local time in Secaucus is 22:16 (11.11.2025)
Aleksandrow Lodzki is located in Łódzki, Poland within 51° 49' 9.84" N 19° 18' 14.04" E (51.8194, 19.3039) coordinates. The local time in Aleksandrow Lodzki is 04:16 (12.11.2025)
The calculated flying distance from Aleksandrow Lodzki to Aleksandrow Lodzki is 4,209.66 miles which is equal to 6,774.79 km.
Secaucus, New Jersey, United States
Related Distances from Secaucus
Aleksandrow Lodzki, Łódzki, Poland