Distance from Alexander City to Sassenberg
The shortest distance (air line) between Alexander City and Sassenberg is 4,622.14 mi (7,438.61 km).
How far is Alexander City from Sassenberg
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 20:09 (30.01.2025)
Sassenberg is located in Warendorf, Germany within 51° 59' 22.92" N 8° 2' 26.88" E (51.9897, 8.0408) coordinates. The local time in Sassenberg is 03:09 (31.01.2025)
The calculated flying distance from Sassenberg to Sassenberg is 4,622.14 miles which is equal to 7,438.61 km.
Alexander City, Alabama, United States
Related Distances from Alexander City
Sassenberg, Warendorf, Germany