Distance from Blunsdon Saint Andrew to Jacksonville
The shortest distance (air line) between Blunsdon Saint Andrew and Jacksonville is 4,191.31 mi (6,745.27 km).
How far is Blunsdon Saint Andrew from Jacksonville
Blunsdon Saint Andrew is located in Swindon, United Kingdom within 51° 36' 36" N -2° 12' 36" W (51.6100, -1.7900) coordinates. The local time in Blunsdon Saint Andrew is 00:28 (18.12.2025)
Jacksonville is located in Florida, United States within 30° 19' 55.92" N -82° 19' 30.36" W (30.3322, -81.6749) coordinates. The local time in Jacksonville is 19:28 (17.12.2025)
The calculated flying distance from Jacksonville to Jacksonville is 4,191.31 miles which is equal to 6,745.27 km.
Blunsdon Saint Andrew, Swindon, United Kingdom
Related Distances from Blunsdon Saint Andrew
Jacksonville, Florida, United States