Distance from Jacksonville Beach to Asenovgrad
The shortest distance (air line) between Jacksonville Beach and Asenovgrad is 5,591.41 mi (8,998.51 km).
How far is Jacksonville Beach from Asenovgrad
Jacksonville Beach is located in Florida, United States within 30° 16' 41.52" N -82° 35' 43.8" W (30.2782, -81.4045) coordinates. The local time in Jacksonville Beach is 05:52 (17.11.2024)
Asenovgrad is located in Plovdiv, Bulgaria within 42° 1' 0.12" N 24° 52' 0.12" E (42.0167, 24.8667) coordinates. The local time in Asenovgrad is 12:52 (17.11.2024)
The calculated flying distance from Asenovgrad to Asenovgrad is 5,591.41 miles which is equal to 8,998.51 km.
Jacksonville Beach, Florida, United States