Distance from Savage to Aleksandrow Lodzki
The shortest distance (air line) between Savage and Aleksandrow Lodzki is 4,656.95 mi (7,494.64 km).
How far is Savage from Aleksandrow Lodzki
Savage is located in Minnesota, United States within 44° 45' 16.2" N -94° 38' 12.48" W (44.7545, -93.3632) coordinates. The local time in Savage is 08:10 (10.11.2025)
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 15:10 (10.11.2025)
The calculated flying distance from Aleksandrow Lodzki to Aleksandrow Lodzki is 4,656.95 miles which is equal to 7,494.64 km.
Savage, Minnesota, United States
Related Distances from Savage
Aleksandrow Lodzki, Łódzki, Poland