Distance from Aleksandrow Kujawski to Lafayette
The shortest distance (air line) between Aleksandrow Kujawski and Lafayette is 5,354.12 mi (8,616.62 km).
How far is Aleksandrow Kujawski from Lafayette
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 00:22 (12.12.2025)
Lafayette is located in Louisiana, United States within 30° 12' 29.52" N -93° 58' 3" W (30.2082, -92.0325) coordinates. The local time in Lafayette is 17:22 (11.12.2025)
The calculated flying distance from Lafayette to Lafayette is 5,354.12 miles which is equal to 8,616.62 km.
Aleksandrow Kujawski, Włocławski, Poland
Related Distances from Aleksandrow Kujawski
Lafayette, Louisiana, United States