Distance from Sint-Kruis to Alexander City
The shortest distance (air line) between Sint-Kruis and Alexander City is 4,453.85 mi (7,167.78 km).
How far is Sint-Kruis from Alexander City
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 03:35 (04.10.2025)
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 20:35 (03.10.2025)
The calculated flying distance from Alexander City to Alexander City is 4,453.85 miles which is equal to 7,167.78 km.
Sint-Kruis, Arr. Brugge, Belgium
Related Distances from Sint-Kruis
Alexander City, Alabama, United States