Distance from Alexander City to Sint-Andries
The shortest distance (air line) between Alexander City and Sint-Andries is 4,451.46 mi (7,163.93 km).
How far is Alexander City from Sint-Andries
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 05:22 (08.01.2025)
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 12:22 (08.01.2025)
The calculated flying distance from Sint-Andries to Sint-Andries is 4,451.46 miles which is equal to 7,163.93 km.
Alexander City, Alabama, United States
Related Distances from Alexander City
Sint-Andries, Arr. Brugge, Belgium