Distance from Aleksandrow Lodzki to Providence
The shortest distance (air line) between Aleksandrow Lodzki and Providence is 4,061.12 mi (6,535.75 km).
How far is Aleksandrow Lodzki from Providence
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 20:57 (10.12.2025)
Providence is located in Rhode Island, United States within 41° 49' 22.8" N -72° 34' 52.68" W (41.8230, -71.4187) coordinates. The local time in Providence is 14:57 (10.12.2025)
The calculated flying distance from Providence to Providence is 4,061.12 miles which is equal to 6,535.75 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Providence, Rhode Island, United States