Distance from Aleksandrow Kujawski to Jacksonville
The shortest distance (air line) between Aleksandrow Kujawski and Jacksonville is 4,975.15 mi (8,006.73 km).
How far is Aleksandrow Kujawski from Jacksonville
Aleksandrow Kujawski is located in Włocławski, Poland within 52° 52' 36.12" N 18° 41' 36.96" E (52.8767, 18.6936) coordinates. The local time in Aleksandrow Kujawski is 23:46 (11.12.2025)
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 17:46 (11.12.2025)
The calculated flying distance from Jacksonville to Jacksonville is 4,975.15 miles which is equal to 8,006.73 km.
Aleksandrow Kujawski, Włocławski, Poland
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Jacksonville, Florida, United States