Distance from Blunsdon Saint Andrew to Jacksonville Beach
The shortest distance (air line) between Blunsdon Saint Andrew and Jacksonville Beach is 4,182.66 mi (6,731.34 km).
How far is Blunsdon Saint Andrew from Jacksonville Beach
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 05:39 (18.12.2025)
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 00:39 (18.12.2025)
The calculated flying distance from Jacksonville Beach to Jacksonville Beach is 4,182.66 miles which is equal to 6,731.34 km.
Blunsdon Saint Andrew, Swindon, United Kingdom
Related Distances from Blunsdon Saint Andrew
Jacksonville Beach, Florida, United States