Distance from Jacksonville Beach to Svilengrad
The shortest distance (air line) between Jacksonville Beach and Svilengrad is 5,658.20 mi (9,105.99 km).
How far is Jacksonville Beach from Svilengrad
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 07:20 (17.11.2024)
Svilengrad is located in Haskovo, Bulgaria within 41° 46' 0.12" N 26° 11' 60" E (41.7667, 26.2000) coordinates. The local time in Svilengrad is 14:20 (17.11.2024)
The calculated flying distance from Svilengrad to Svilengrad is 5,658.20 miles which is equal to 9,105.99 km.
Jacksonville Beach, Florida, United States