Distance from Sicklerville to Aleksandrow Lodzki
The shortest distance (air line) between Sicklerville and Aleksandrow Lodzki is 4,294.96 mi (6,912.07 km).
How far is Sicklerville from Aleksandrow Lodzki
Sicklerville is located in New Jersey, United States within 39° 44' 42.72" N -75° 0' 23.76" W (39.7452, -74.9934) coordinates. The local time in Sicklerville is 02:26 (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 08:26 (11.11.2025)
The calculated flying distance from Aleksandrow Lodzki to Aleksandrow Lodzki is 4,294.96 miles which is equal to 6,912.07 km.
Sicklerville, New Jersey, United States
Related Distances from Sicklerville
Aleksandrow Lodzki, Łódzki, Poland