Distance from Waukegan to Aleksandrow Kujawski
The shortest distance (air line) between Waukegan and Aleksandrow Kujawski is 4,545.10 mi (7,314.63 km).
How far is Waukegan from Aleksandrow Kujawski
Waukegan is located in Illinois, United States within 42° 22' 11.28" N -88° 7' 42.24" W (42.3698, -87.8716) coordinates. The local time in Waukegan is 13:19 (17.11.2025)
Aleksandrow Kujawski is located in Włocławski, Poland within 52° 52' 36.12" N 18° 41' 36.96" E (52.8767, 18.6936) coordinates. The local time in Aleksandrow Kujawski is 20:19 (17.11.2025)
The calculated flying distance from Aleksandrow Kujawski to Aleksandrow Kujawski is 4,545.10 miles which is equal to 7,314.63 km.
Waukegan, Illinois, United States
Related Distances from Waukegan
Aleksandrow Kujawski, Włocławski, Poland