Distance from Aleksandrow Lodzki to Sioux Falls
The shortest distance (air line) between Aleksandrow Lodzki and Sioux Falls is 4,824.77 mi (7,764.72 km).
How far is Aleksandrow Lodzki from Sioux Falls
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 21:51 (16.12.2025)
Sioux Falls is located in South Dakota, United States within 43° 32' 22.56" N -97° 16' 8.04" W (43.5396, -96.7311) coordinates. The local time in Sioux Falls is 14:51 (16.12.2025)
The calculated flying distance from Sioux Falls to Sioux Falls is 4,824.77 miles which is equal to 7,764.72 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Sioux Falls, South Dakota, United States