Distance from Aleksandrow Kujawski to Sint-Andries
The shortest distance (air line) between Aleksandrow Kujawski and Sint-Andries is 667.94 mi (1,074.95 km).
How far is Aleksandrow Kujawski from Sint-Andries
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 19:41 (27.04.2025)
Sint-Andries is located in Arr. Brugge, Belgium within 51° 12' 0" N 3° 10' 59.88" E (51.2000, 3.1833) coordinates. The local time in Sint-Andries is 19:41 (27.04.2025)
The calculated flying distance from Sint-Andries to Sint-Andries is 667.94 miles which is equal to 1,074.95 km.
Aleksandrow Kujawski, Włocławski, Poland
Related Distances from Aleksandrow Kujawski
Sint-Andries, Arr. Brugge, Belgium