Distance from Aleksandrow Lodzki to Savage
The shortest distance (air line) between Aleksandrow Lodzki and Savage is 4,656.95 mi (7,494.64 km).
How far is Aleksandrow Lodzki from Savage
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 11:18 (16.12.2025)
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 04:18 (16.12.2025)
The calculated flying distance from Savage to Savage is 4,656.95 miles which is equal to 7,494.64 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Savage, Minnesota, United States