Distance from Alexander City to Bad Sassendorf
The shortest distance (air line) between Alexander City and Bad Sassendorf is 4,638.71 mi (7,465.28 km).
How far is Alexander City from Bad Sassendorf
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 04:19 (15.01.2025)
Bad Sassendorf is located in Soest, Germany within 51° 34' 59.16" N 8° 10' 0.12" E (51.5831, 8.1667) coordinates. The local time in Bad Sassendorf is 11:19 (15.01.2025)
The calculated flying distance from Bad Sassendorf to Bad Sassendorf is 4,638.71 miles which is equal to 7,465.28 km.
Alexander City, Alabama, United States
Related Distances from Alexander City
Bad Sassendorf, Soest, Germany