Distance from Alexander City to Civitavecchia
The shortest distance (air line) between Alexander City and Civitavecchia is 5,092.66 mi (8,195.85 km).
How far is Alexander City from Civitavecchia
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 17:55 (08.02.2025)
Civitavecchia is located in Roma, Italy within 42° 6' 0" N 11° 48' 0" E (42.1000, 11.8000) coordinates. The local time in Civitavecchia is 00:55 (09.02.2025)
The calculated flying distance from Civitavecchia to Civitavecchia is 5,092.66 miles which is equal to 8,195.85 km.
Alexander City, Alabama, United States