Distance from Aleksandrow Lodzki to Wilmington Island
The shortest distance (air line) between Aleksandrow Lodzki and Wilmington Island is 4,916.16 mi (7,911.79 km).
How far is Aleksandrow Lodzki from Wilmington Island
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 11:24 (15.01.2026)
Wilmington Island is located in Georgia, United States within 32° 0' 11.88" N -81° 1' 29.28" W (32.0033, -80.9752) coordinates. The local time in Wilmington Island is 05:24 (15.01.2026)
The calculated flying distance from Wilmington Island to Wilmington Island is 4,916.16 miles which is equal to 7,911.79 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Wilmington Island, Georgia, United States