Distance from Aleksandrow Kujawski to Sainte-Savine
The shortest distance (air line) between Aleksandrow Kujawski and Sainte-Savine is 714.44 mi (1,149.78 km).
How far is Aleksandrow Kujawski from Sainte-Savine
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 08:41 (03.06.2025)
Sainte-Savine is located in Aube, France within 48° 17' 40.92" N 4° 2' 56.04" E (48.2947, 4.0489) coordinates. The local time in Sainte-Savine is 08:41 (03.06.2025)
The calculated flying distance from Sainte-Savine to Sainte-Savine is 714.44 miles which is equal to 1,149.78 km.
Aleksandrow Kujawski, Włocławski, Poland