Distance from Aleksandrow Lodzki to Sicklerville
The shortest distance (air line) between Aleksandrow Lodzki and Sicklerville is 4,294.96 mi (6,912.07 km).
How far is Aleksandrow Lodzki from Sicklerville
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 00:15 (17.12.2025)
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 18:15 (16.12.2025)
The calculated flying distance from Sicklerville to Sicklerville is 4,294.96 miles which is equal to 6,912.07 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Sicklerville, New Jersey, United States