Distance from Alexander City to Sint-Kruis
The shortest distance (air line) between Alexander City and Sint-Kruis is 4,453.85 mi (7,167.78 km).
How far is Alexander City from Sint-Kruis
Alexander City is located in Alabama, United States within 32° 55' 22.44" N -86° 3' 51.12" W (32.9229, -85.9358) coordinates. The local time in Alexander City is 05:40 (08.01.2025)
Sint-Kruis is located in Arr. Brugge, Belgium within 51° 12' 42.12" N 3° 15' 0" E (51.2117, 3.2500) coordinates. The local time in Sint-Kruis is 12:40 (08.01.2025)
The calculated flying distance from Sint-Kruis to Sint-Kruis is 4,453.85 miles which is equal to 7,167.78 km.
Alexander City, Alabama, United States
Related Distances from Alexander City
Sint-Kruis, Arr. Brugge, Belgium