Distance from Aleksandrow Lodzki to Jacksonville
The shortest distance (air line) between Aleksandrow Lodzki and Jacksonville is 5,031.41 mi (8,097.27 km).
How far is Aleksandrow Lodzki from Jacksonville
Aleksandrow Lodzki is located in Łódzki, Poland within 51° 49' 9.84" N 19° 18' 14.04" E (51.8194, 19.3039) coordinates. The local time in Aleksandrow Lodzki is 23:57 (05.12.2025)
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 17:57 (05.12.2025)
The calculated flying distance from Jacksonville to Jacksonville is 5,031.41 miles which is equal to 8,097.27 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Jacksonville, Florida, United States