Distance from Aleksandrow Lodzki to Secaucus
The shortest distance (air line) between Aleksandrow Lodzki and Secaucus is 4,209.66 mi (6,774.79 km).
How far is Aleksandrow Lodzki from Secaucus
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 14:56 (16.12.2025)
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 08:56 (16.12.2025)
The calculated flying distance from Secaucus to Secaucus is 4,209.66 miles which is equal to 6,774.79 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Secaucus, New Jersey, United States