Distance from Dunkirk to Aleksandrow Kujawski
The shortest distance (air line) between Dunkirk and Aleksandrow Kujawski is 4,253.01 mi (6,844.55 km).
How far is Dunkirk from Aleksandrow Kujawski
Dunkirk is located in New York, United States within 42° 28' 48.36" N -80° 40' 3.36" W (42.4801, -79.3324) coordinates. The local time in Dunkirk is 14:35 (02.09.2025)
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 20:35 (02.09.2025)
The calculated flying distance from Aleksandrow Kujawski to Aleksandrow Kujawski is 4,253.01 miles which is equal to 6,844.55 km.
Dunkirk, New York, United States
Related Distances from Dunkirk
Aleksandrow Kujawski, Włocławski, Poland